Machine Dynamic Pune University MCQs

 This set of Machine Kinematics Multiple Choice Questions & Answers  focuses on “Precessional Angular Motion”.

1. A disc is spinning with an angular velocity? rad/s about the axis of spin. The couple applied to the disc causing precession will be

a) 1/2Iω 2

b) Iω 2

c) 1/2 Iω ω p

d) Iω ω p

Answer: d

Explanation: Since the rate of change of angular momentum will result by the application of a couple to the disc, therefore the couple applied to the disc causing precession,

Iω ω p

where I = Mass moment of inertia of the disc, and

ω p = Angular velocity of precession of the axis of spin.

2. A disc spinning on its axis at 20 rad/s will undergo precession when a torque 100 N-m is applied about an axis normal to it at an angular speed, if mass moment of inertia of the disc is the 1 kg-m 2

a) 2 rad/s

b) 5 rad/s

c) 10 rad/s

d) 20 rad/s

Answer: b

Explanation: Gyroscopic couple acting on the disc,

C = I. ω. ω p

100 = 1 . 20. ω p

ω p = 100/1.20

= 5 rad/s

3. The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the left. The effect of the gyroscopic couple on the aeroplane will be

a) to raise the nose and dip the tail

b) to dip the nose and raise the tail

c) to raise the nose and tail

d) to dip the nose and tail

Answer: a

Explanation: When the engine or propeller rotates in anticlockwise direction when viewed from the rear or tail end and the aeroplane takes a left turn, then the effect of reactive gyroscopic couple will be to dip the nose and raise the tail of the aeroplane. When the aeroplane takes a right turn, the effect of reactive gyroscopic couple will be to raise the nose and dip the tail of the aeroplane.

4. The air screw of an aeroplane is rotating clockwise when looking from the front. If it makes a left turn, the gyroscopic effect will

a) tend to depress the nose and raise the tail

b) tend to raise the nose and depress the tail

c) tilt the aeroplane

d) none of the mentioned

Answer: b

Explanation: When the engine or propeller rotates in clockwise direction when viewed from the front and the aeroplane takes a left turn, then the effect of reactive gyroscopic couple will be to raise the tail and dip the nose of the aeroplane. When the aeroplane takes a right turn, the effect of reactive gyroscopic couple will be to raise the nose and dip the tail of the aeroplane.

5. The rotor of a ship rotates in clockwise direction when viewed from the stern and the ship takes a left turn. The effect of the gyroscopic couple acting on it will be

a) to raise the bow and stern

b) to lower the bow and stern

c) to raise the bow and lower the stern

d) to lower the bow and raise the stern

Answer: c

Explanation: When the rotor rotates in the clockwise direction when viewed from the bow or fore end and the ship is steering to the left, then the effect of reactive gyroscopic couple will be to raise the stern and lower the bow.

6. When the pitching of a ship is upward, the effect of gyroscopic couple acting on it will be

a) to move the ship towards port side

b) to move the ship towards star-board

c) to raise the bow and lower the stern

d) to raise the stern and lower the bow

Answer: b

Explanation: The effect of the gyroscopic couple is always given on specific position of the axis of spin i.e. whether it is pitching downwards or upwards. The pitching of a ship produces forces on the bearings which act horizontally and perpendicular to the motion of the ship.

7. In an automobile, if the vehicle makes a left turn, the gyroscopic torque

a) increases the forces on the outer wheels

b) decreases the forces on the outer wheels

c) does not affect the forces on the outer wheels

d) none of the mentioned

Answer: a

Explanation: The gyroscopic couple will act over the vehicle outwards i.e. in the anticlockwise direction when seen from the front of the vehicle. The tendency of this couple is to overturn the vehicle in outward direction.

Answer: b

Explanation: When the engine is rotating in the same direction as that of wheels, then the positive sign is used in the above expression and if the engine rotates in opposite direction, then negative sign is used.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Gyroscopic Couple”.

1. The axis of precession is ____________ to the plane in which the axis of spin is going to rotate.

a) parallel

b) perpendicular

c) spiral

d) none of the mentioned

Answer: b

Explanation: The axis of precession is perpendicular to the plane in which the axis of spin is going to rotate. The gyroscopic principle is used in an instrument or toy known as gyroscope.

2. A disc is a spinning with an angular velocity ω rad/s about the axis of spin. The couple applied to the disc causing precession will be

a) 1/2 Iω 2

b) Iω 2

c) 1/2 Iωω p

d) Iωω p

Answer: d

Explanation: None

3. The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the left. The effect of gyroscopic couple on the aeroplane will be

a) to dip the nose and tail

b) to raise the nose and tail

c) to raise the nose and dip of the tail

d) to dip the nose and raise the tail

Answer: c

Explanation: The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the left. The effect of gyroscopic couple on the aeroplane will be to raise the nose and dip of the tail.

The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the right. The effect of gyroscopic couple on the aeroplane will be to dip the nose and raise the tail.

4. The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the right. The effect of gyroscopic couple on the aeroplane will be to dip the nose and raise the tail.

a) True

b) False

Answer: a

Explanation: The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the left. The effect of gyroscopic couple on the aeroplane will be to raise the nose and dip of the tail.

The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the right. The effect of gyroscopic couple on the aeroplane will be to dip the nose and raise the tail.

5. The steering of a ship means

a) movement of a complete ship up and down in vertical plane about transverse axis

b) turning of a complete ship in a curve towards right or left, while it moves forward

c) rolling of a complete ship side-ways

d) none of the mentioned

Answer: b

Explanation: The steering is the turning of a complete ship in a curve towards right or left, while it moves forward.

6. The rolling of a complete ship side-ways is known as pitching of a ship.

a) True

b) False

Answer: b

Explanation: The pitching is the movement of a complete ship up and down in a vertical plane about transverse axis.

7. The rotor of a ship rotates in clockwise direction when viewed from stern and the ship takes a left turn. The effect of gyroscopic couple acting on it will be

a) to raise the bow and stern

b) to lower the bow and stern

c) to raise the bow and lower the stern

d) to raise the stern and lower the bow

Answer: c

Explanation: The rotor of a ship rotates in clockwise direction when viewed from stern and the ship takes a left turn. The effect of gyroscopic couple acting on it will be to raise the bow and lower the stern.

The rotor of a ship rotates in clockwise direction when viewed from stern and the ship takes a right turn. The effect of gyroscopic couple acting on it will be to raise the stern and lower the bow.

8. The rotor of a ship rotates in clockwise direction when viewed from stern and the ship takes a right turn. The effect of gyroscopic couple acting on it will be to raise the stern and lower the bow.

a) True

b) False

Answer: a

Explanation: The rotor of a ship rotates in clockwise direction when viewed from stern and the ship takes a left turn. The effect of gyroscopic couple acting on it will be to raise the bow and lower the stern.

The rotor of a ship rotates in clockwise direction when viewed from stern and the ship takes a right turn. The effect of gyroscopic couple acting on it will be to raise the stern and lower the bow.

9. The pitching of a ship is assumed to take place with simple harmonic motion.

a) True

b) False

Answer: a

Explanation: None.

Answer: a

Explanation: When the pitching of a ship is upward, the effect of gyroscopic couple acting on it will be to move the ship towards star-board.

When the pitching of a ship is downward, the effect of gyroscopic couple, is to turn the ship towards port side.

This set of Machine Dynamics Interview Questions and Answers focuses on “Effect of Gyroscopic Couple on an Aeroplane”.

1. A uniform disc of diameter 300 mm and of mass 5 kg is mounted on one end of an arm of length 600 mm. The other end of the arm is free to rotate in a universal bearing. If the disc rotates about the arm with a speed of 300 r.p.m. clockwise, looking from the front, with what speed will it precess about the vertical axis?

a) 14.7 rad/s

b) 15.7 rad/s

c) 16.7 rad/s

d) 17.7 rad/s

Answer: c

Explanation: Given: d = 300 mm or r = 150 mm = 0.15 m ; m = 5 kg ; l = 600 mm = 0.6 m ;

N = 300 r.p.m. or ω = 2π × 300/60 = 31.42 rad/s

We know that the mass moment of inertia of the disc, about an axis through its centre of gravity and perpendicular to the plane of disc,

I = m.r 2 /2 = 5 2 /2 = 0.056 kg-m 2

and couple due to mass of disc, C = m.g.l = 5 × 9.81 × 0.6 = 29.43 N-m

Let ω P = Speed of precession

We know that couple ,

29.43 = I.ω.ω P = 0.056 × 31.42 × ω P = 1.76 ω P

ω P = 29.43/1.76 = 16.7 rad/s

2. An aeroplane makes a complete half circle of 50 metres radius, towards left, when flying at 200 km per hr. The rotary engine and the propeller of the plane has a mass of 400 kg and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise when viewed from the rear. Find the gyroscopic couple on the aircraft.

a) 10.046 kN-m

b) 11.046 kN-m

c) 12.046 kN-m

d) 13.046 kN-m

Answer: a

Explanation: Given : R = 50 m ; v = 200 km/hr = 55.6 m/s ; m = 400 kg ; k = 0.3 m ;

N = 2400 r.p.m. or ω = 2π × 2400/60 = 251 rad/s

We know that mass moment of inertia of the engine and the propeller,

I = mk 2 = 36 kg-m 2

and angular velocity of precession,

ω P = v/R = 55.6/50 = 1.11 rad/s

We know that gyroscopic couple acting on the aircraft,

C = I. ω. ω P = 36 × 251.4 × 1.11 = 100 46 N-m

= 10.046 kN-m

3. The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration 0.6 m. It rotates at 1800 r.p.m. clockwise, when looking from the stern. Determine the gyroscopic couple, if the ship travels at 100 km/hr and steer to the left in a curve of 75 m radius.

a) 100.866 kN-m

b) 200.866 kN-m

c) 300.866 kN-m

d) 400.866 kN-m

Answer: b

Explanation: Given: m = 8 t = 8000 kg ; k = 0.6 m ; N = 1800 r.p.m. or ω = 2π × 1800/60

= 188.5 rad/s ; v = 100 km/h = 27.8 m/s ; R = 75 m

We know that mass moment of inertia of the rotor,

I = mk 2 = 2880 kg-m 2

and angular velocity of precession,

ω P = v / R = 27.8 / 75 = 0.37 rad/s

We know that gyroscopic couple,

C = I.ω.ω P = 2880 × 188.5 × 0.37 = 200 866 N-m

= 200.866 kN-m

4. The heavy turbine rotor of a sea vessel rotates at 1500 r.p.m. clockwise looking from the stern, its mass being 750 kg. The vessel pitches with an angular velocity of 1 rad/s. Determine the gyroscopic couple transmitted to the hull when bow is rising, if the radius of gyration for the rotor is 250 mm.

a) 4.364 kN-m

b) 5.364 kN-m

c) 6.364 kN-m

d) 7.364 kN-m

Answer: d

Explanation: Given: N = 1500 r.p.m. or ω = 2π × 1500/60 = 157.1 rad/s; m = 750 kg;

ω P = 1 rad/s; k = 250 mm = 0.25 m

We know that mass moment of inertia of the rotor,

I = mk 2 = 46.875 kg-m 2

Gyroscopic couple transmitted to the hull ,

C = I.ω.ω P = 46.875 × 157.1 × 1= 7364 N-m = 7.364 kN-m

5. The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple upon the ship when the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h.

a) 11.27 kN-m

b) 22.27 kN-m

c) 33.27 kN-m

d) 44.27 kN-m

Answer: b

Explanation: Given : m = 3500 kg ; k = 0.45 m; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s

When the ship is steering to the left

Given: R =100 m ; v = km/h = 10 m/s

We know that mass moment of inertia of the rotor,

I = mk 2 = 708.75 kg-m 2

and angular velocity of precession,

ω p = v/R = 10/100 = 0.1 rad/s

Gyroscopic couple,

C = I.ω.ω p = 708.75 × 314.2 × 0.1 = 22 270 N-m

= 22.27 kN-m

6. The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple upon the ship when the ship is pitching in a simple harmonic motion, the bow falling with its maximum velocity. The period of pitching is 40 seconds and the total angular displacement between the two extreme positions of pitching is 12 degrees.

a) 3.675 kN-m

b) 4.675 kN-m

c) 5.675 kN-m

d) 6.675 kN-m

Answer: a

Explanation: Given: t p = 40 s

Since the total angular displacement between the two extreme positions of pitching is 12° , therefore amplitude of swing,

φ = 12 / 2 = 6° = 6 × π/180 = 0.105 rad

and angular velocity of the simple harmonic motion,

ω 1 = 2π / t p = 2π / 40 = 0.157 rad/s

We know that maximum angular velocity of precession,

ω p = φ.ω 1 = 0.105 × 0.157 = 0.0165 rad/s

Gyroscopic couple,

C = I.ω.ω p

= 708.75 × 314.2 × 0.0165 = 3675 N-m

= 3.675 kN-m

7. The mass of the turbine rotor of a ship is 20 tonnes and has a radius of gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the horizontal position. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine Maximum gyroscopic couple.

a) 11.185 kN-m

b) 22.185 kN-m

c) 33.185 kN-m

d) 44.185 kN-m

Answer: c

Explanation: Given : m = 20 t = 20 000 kg ; k = 0.6 m ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s; φ = 6° = 6 × π/180 = 0.105 rad ; t p = 30 s

We know that mass moment of inertia of the rotor,

I = m.k 2 = 20 000  2 = 7200 kg-m 2

and angular velocity of the simple harmonic motion,

ω 1 = 2π / t p = 2π/30 = 0.21 rad/s

Maximum angular velocity of precession,

ω Pmax = φ.ω 1 = 0.105 × 0.21 = 0.022 rad/s

We know that maximum gyroscopic couple,

C max = I.ω.ω Pmax = 7200 × 209.5 × 0.022 = 33 185 N-m

= 33.185 kN-m

8. The mass of the turbine rotor of a ship is 20 tonnes and has a radius of gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the horizontal position. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine Maximum angular acceleration of the ship during pitching.

a) 0.0016 rad/s 2

b) 0.0026 rad/s 2

c) 0.0036 rad/s 2

d) 0.0046 rad/s 2

Answer: d

Explanation: We know that maximum angular acceleration during pitching

= φ(ω 1 ) 2 = 0.105  2 = 0.0046 rad/s 2

9. A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction when viewed from the stern. Find the gyroscopic effect when the ship sails at a speed of 30 km/h and steers to the left in a curve having 60 m radius.

a) 38.5 kN-m

b) 48.5 kN-m

c) 58.5 kN-m

d) 68.5 kN-m

Answer: a

Explanation: Given : m = 5 t = 5000 kg ; N = 2100 r.p.m. or ω = 2π × 2100/60 = 220 rad/s ; k = 0.5 m

v = 30 km / h = 8.33 m / s ; R = 60 m

We know that angular velocity of precession,

ω p = v/R = 8.33/60 = 0.14 rad/s

and mass moment of inertia of the rotor,

I = m.k 2 = 50002 = 1250 kg-m 2

Gyroscopic couple,

C = I.ω.ω p = 1250 × 220 × 0.14 = 38 500 N-m = 38.5 kN-m

Answer: d

Explanation: Given: φ = 6° = 6 × π/180 = 0.105 rad/s ; tp = 20 s

We know that angular velocity of simple harmonic motion,

ω 1 = 2π / t p = 2π / 20 = 0.3142 rad/s

and maximum angular velocity of precession,

ω Pmax = φ.ω 1 = 0.105 × 0.3142 = 0.033 rad/s

Maximum gyroscopic couple,

C max = I.ω.ω Pmax = 1250 × 220 × 0.033 = 9075 N-m

This set of Machine Dynamics Questions and Answers for Freshers focuses on “Effect of Gyroscopic Couple on a Naval Ship during Steering and Pitching”.

1. It is required to lift water at the top of a high building at the rate of 50 litres/min. Assuming the losses due to friction equivalent to 5 m and those due to leakage equal to 5 m head of water, efficiency of pump 90%, decide the kilowatt capacity of motor required to drive the pump.

a) 0.102 kW

b) 0.202 kW

c) 0.302 kW

d) 0.402 kW

Answer: c

Explanation: Work to be done = wQHg

w = 1 Kg/litre

Q = 50 litres per min

H = 20 + 5 + 5 = 30m

Work output/min = 1 x 50 x 30 x 9.81 Nm/min

Input power = 1 x 50 x 30 x 9.81/ 0.9 x 0.9 x1000 x 60

= 0.302 kW.

2. A power screw is rotated at constant angular speed of 1.5 revolutions/sec by applying a steady torque of 1.5Nm. How much work is done per revolution? What is the power required?

a) 141.36 W

b) 241.36 W

c) 341.36 W

d) 441.36 W

Answer: a

Explanation: Work per revolution = Tϴ

= 15 x 2п

= 94.24 Nm per revolution

P = Work/sec

= 94.24 x 1.5 = 141.36 W.

3. Power is transmitted by an electric motor to a machine by using a belt drive. The tensions on the tight and slack side of the belt are 2200 N and 1000 N respectively and diameter of the pulley is 600 mm. If speed of the motor is 1500 r.p.m, find the power transmitted.

a) 46.548 kW

b) 56.548 kW

c) 66.548 kW

d) 76.548 kW

Answer: b

Explanation: Power = Tω

= Frω = Fv

F = T 1 – T 2 = 2200 – 1000 = 1200N

v = пDN/60 = п x 600 x 1500/1000 x 60 = 47.12 m/s

P = 1200 x 47.12/1000 = 56.548 kW.

4. The flywheel of an engine has a mass of 200 kg and radius of gyration equal to 1 m. The average torque on the flywheel is 1200 Nm. Find the angular acceleration of flywheel and the angular speed after 10 seconds starting from rest.

a) 3 rad/s

b) 4 rad/s

c) 5 rad/s

d) 6 rad/s

Answer: d

Explanation: I = mk 2 = 2000 kgm 2

T = Iα

α = T/I = 1200/2000 = 0.6 rad/s 2

ω = ω 0 + αt = 0 + 0.6 x 10 = 6 rad/s.

5. Calculate the moment of inertia and radius of gyration of a solid sphere of mass 10 kg and diameter 6.5m about its centroidal axis.

a) 2.055 m

b) 3.055 m

c) 4.055 m

d) 5.055 m

Answer: a

Explanation: I = 2/5 mr 2 = 49 kgm 2

k = 0.6325 x 3.25 = 2.055 m.

6. Calculate the work done per minute by a punch tool making 20 working strokes per min when a 30 mm diametre hole is punched in 5 mm thick plate with ultimate shear strength og 450 Mpa in each stroke.

a) 10.69 kNm

b) 20.69 kNm

c) 30.69 kNm

d) 40.69 kNm

Answer: a

Explanation: Force required to punch one hole = area shared x ultimate shear strength = пdt x S s

S s = п x 30 x 5 x 450/1000 = 212.05 kN

Work done/min = Average force x Thickness of plate x No. of holes/min

= 212.05/2 x 5/1000 x 20 = 10.69 kNm.

7. In latitude 25.0 S, SA  of a FG  is in position S40E and horizontal. Find the tilt after 6 hours.

a) 61.16 up

b) 61.16 down

c) 51.15 up

d) none of the mentioned

Answer: a

Explanation: PZ= 65, ZX = 90, tilt after 6 hours = 90 – ZY =?

In triangle PZX,

Cos PX = Cos Z Sin PZ

Cos PX = Cos 40 Sin 65

Cos PX = 0.694272

PX = PY = 46.031

Again, in triangle PZX,

Cos P = – Cot PZ Cot PX

Cos P = – Cot 65 Cot 46.031

P or angle ZPX = 116.732

Angle XPY = 6 hours = 90. Thus, angle ZPY = 26.732

In triangle PZY,

Cos ZY = Cos P Sin PZ Sin PY + Cos PZ Cos PY

Cos ZY = Cos 26.732 Sin 65 Sin 46.031 + Cos 65 Cos 46.031

ZY = 28.839 and tilt = 61.16 up.

8. The steering of a ship means

a) movement of a complete ship up and down in vertical plane about transverse axis

b) turning of a complete ship in a curve towards right r left, while it moves forward

c) rolling of a complete ship sideways

d) none of the mentioned

Answer: b

Explanation: Steering means is to turn the vehicle to either left or right.

Answer: a

Explanation: When the rotor rotates in the anticlockwise direction, when viewed from the stern and the ship is steering to the left, then the effect of reactive gyroscopic couple will be to lower the bow and raise the stern. When the ship is steering to the right under similar conditions, then the effect of reactive gyroscopic couple will be to raise the bow and lower the stern.

This set of Advanced Machine Dynamics Questions and Answers focuses on “Effect of Gyroscopic Couple on a Navalship during Pitching & Rolling”.

1. The pitching effect of a naval-ship produces force on which of the following components of the ship?

a) Bearings

b) Lubricant

c) Seals

d) Springs

Answer: a

Explanation: The effect in the form of pitching of a ship produces forces on the bearings which acts in the following direction: horizontal to the motion and perpendicular to the motion in which the ship is travelling. Hence bearings experience the forces due to pitching.

2. Which of the following represent the maximum angular acceleration?

a) α max = Φ(ω 1 ) 2

b) α max = Φ2 (ω 1 ) 2

c) α max = (ω 1 ) 2 / Φ

d) α max = (ω 1 ) 2 / Φ 2

Answer: a

Explanation: The angular acceleration is given by the formula : α = -Φ(ω 1 ) 2 sin (ω 1 t), hence the maximum will occur when sin (ω 1 t)=1.

3. Pitching is the movement of a ship up and down in a vertical plane about transverse axis.

a) True

b) False

Answer: a

Explanation: Pitching occurs about a transverse axis as a longitudinal oscillatory motion which is further assumed to be Simple Harmonic Motion.

4. The pitching of the ship is assumed to take place with which of the following motion?

a) Constant acceleration

b) Simple Harmonic Motion

c) Cycloidal

d) Uniform Velocity

Answer: b

Explanation: Motion of the spin axis about transverse axis undergoes SHM, the motion tends to be longitudinal oscillatory rotation about the transverse axis.

5. Maximum couple during pitching is given by : Misplaced &.

a) Cmax = I. 2

b) Cmax = I. 2

c) Cmax = I. ω. ωpmax

d) Cmax = I 2 . ω. ωpmax

Answer: c

Explanation: Maximum couple depends on both angular velocity of rotor and angular velocity of precession, therefore it is maximum when ωp is maximum.

6. During upward pitching, Gyroscopic couple will tend to move the ship towards ________

a) Star-board

b) Port side

c) Ship movement is unaffected by upward pitching

d) Left side

Answer: a

Explanation: During upward pitching, angular displacement of axis of spin from mean position is positive i.e Anti clockwise, therefore gyroscopic couple moves it towards right side or star board.

7. During downward pitching, Gyroscopic couple will tend to move the ship towards ________

a) Star-board

b) Port side

c) Ship movement is unaffected by upward pitching

d) Right side

Answer: b

Explanation: During downward pitching, angular displacement of axis of spin from mean position is negative i.e Clockwise, therefore gyroscopic couple moves it towards left side or port-side.

8. If the movement of the ship were to occur about a vertical plane about a parallel axis then which of the following factors will be affected while pitching?

a) Upward pitch

b) Downward pitch

c) Pitching will not occur

d) Change in angular velocity of rotor

Answer: c

Explanation: For pitching to occur, motion should be about a vertical plane about a Transverse axis, else the motion will not be considered pitching. If the movement is about the parallel axis then it is called Rolling.

9. What will be the effect of a gyroscopic couple if the axis of precession and axis of spin become parallel?

a) No effect

b) Pitching

c) Steering effect

d) No motion

Answer: a

Explanation: For a gyroscope couple effect to occur the axis of precession should always be perpendicular to the axis of spin, if this condition is not fulfilled then no effect will take place.

10. Which of the following is known as the axis of precession?

a) Latitudinal axis

b) Longitudinal axis

c) Reverse axis

d) Axis of spin

Answer: b

Explanation: The longitudinal axis is known as the axis of precession as it is along the direction of the acting force, about which the precession occurs.

11. The axis of precession is parallel to axis of spin during rolling for only a selected few positions.

a) False

b) True

Answer: a

Explanation: The axis of precession is parallel to the axis of spin during rolling for all positions, if this condition is not met then there can be other effects like pitching which can occur.

12. Which of the following would change during rolling of a navalship?

a) Angular velocity of precession

b) Angular acceleration of precession

c) Angular velocity of spin

d) No change occurs

Answer: d

Explanation: Since the axis of precession is parallel to the axis of spin during rolling, the gyroscopic couple causes no change in motion of the ship, Hence no change occurs.

13. The gyroscope couple does not cause an effect on a naval ship during which of these motions?

a) Pitching

b) Steering

c) Rolling

d) Lifting

Answer: d

Explanation: There is no such motion as “lifting” hence no effect exists, while the effect of a gyroscopic couple is pronounced during pitching steering and rolling.

14. Rolling is the turning of a complete ship in a curve towards left or right, while it moves forward.

a) True

b) False

Answer: b

Explanation: Steering is the turning of a complete ship in a curve towards left or right, while it moves forward, while rolling is rocking from side to side.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Stability of a Four Wheel drive Moving in a Curved Path”.

1. Which of the following relation is correct regarding the net gyroscope couple C acting on a four wheel vehicle?

a) ωw.ωp  2

b) ωw.ωp 

c) ωw.ωp 

d) ωw 2 

Answer: c

Explanation: When a four wheeled vehicle undergoes a motion along a curved path, the gyroscope couple acts on the vehicle. This effect depends on angular velocities of precession & wheels and moment of inertia of wheels and rotating parts of engine.

2. The positive sign is used to determine the gyroscope couple in which of the following cases?

a) When the wheels and rotating parts of the engine rotate in the same direction

b) When the wheels and rotating parts of the engine rotate in the opposite direction

c) When the precession is clockwise

d) When the precession is anticlockwise

Answer: c

Explanation: The positive sign is used when the wheels and rotating parts of the engine rotate in the same direction. If the rotating parts of the engine revolve in opposite direction, then negative sign is used.

3. If the net gyroscope couple is negative then which of the following is true?

a) The reaction will be vertically downwards on the outer wheels and vertically upwards on the inner wheels

b) The reaction will be vertically upwards on the outer wheels and vertically downwards on the inner wheels

c) The reaction will be vertically downwards on the outer wheels and vertically downwards on the inner wheels

d) The reaction will be vertically upwards on the outer wheels and vertically upwards on the inner wheels

Answer: a

Explanation: A negative gyroscope couple would result in opposite direction of the reaction from the ground in the wheels of the vehicle.

4. The relation between overturning couple and height of centre of gravity is given by_______

a) Co = Fcxh

b) Co = Mv 2 xh/R

c) Co = Mv 2 *  2

d) Co = Mv 2 * h/R 2

Answer: a

Explanation: Centrifugal force is given by : Fc = (Mv 2 )/R and the overturning couple is the cross product of h and Fc.

5. In order to maintain contact between inner wheel and ground the sum of vertical reactions at each of the outer and inner wheels should be less than______

a) W

b) W/2

c) W/4

d) W/3

Answer: c

Explanation: The total vertical reaction at each of the inner wheel will be negative if the sum of vertical reactions at each of the outer and inner wheels is less than W/2.

6. Let x be the track width of the vehicle, then which of the following expression is correct for vertical reaction Q at the two outer or inner wheels?

a) Q = (Mv 2 . h)/ 

b) Q = (Mv 2 )/ R

c) Q = Mv 2 . x/ 

d) Q = (Mv 2 )/ 

Answer: a

Explanation: Centrifugal force is given by : Fc = (Mv 2 )/R and the overturning couple is the cross product of h and Fc, and Q=Co/x. Therefore Q = (Mv 2 .h)/ 

7. The overturning couple acting on the four wheeler vehicle is balanced by which of the following forces?

a) Centripetal force

b) Centrifugal force

c) Vertical reactions

d) Horizontal reactions

Answer: c

Explanation: Since the overturning couple tends to overturn the vehicle upside down about a horizontal plane, the vertical reactions provide stability and helps in preventing the overturn.

8. Overturning couple is independent of the track width.

a) False

b) True

Answer: a

Explanation: The overturning is the cross product of height of center of gravity and the centrifugal force. Since the centrifugal force does not depend on the track width, the overturning couple is independent of the track width.

9. Assuming that a vehicle is taking a left turn, The gyroscopic couple acting on the vehicle has a tendency to _____ the inner wheels and ____ the outer wheels.

a) Press, Lift

b) Lift, Press

c) Press, Press

d) Lift, Lift

Answer: b

Explanation: While taking a left turn, a four wheel vehicle undergoes the action of a gyroscopic couple which has an effect on inner and outer wheels. This effect results in lifting of inner wheel and pressing of the outer wheel. This prevents the overturning of vehicle.

10. Which of the following factor is not responsible for the stability of a 4 wheel vehicle while negotiating a turn?

a) Pitching

b) Reaction due to weight of Vehicle

c) Effect of Gyroscopic couple due to Wheel

d) Effect of Gyroscopic Couple due to Engine

Answer: a

Explanation: Pitching effect is not pronounced in four wheel vehicles and thus has a negligible effect on the stability of the vehicle while the other 3 factors play a vital role in preventing the overturning of the vehicle.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Stability of a Two Wheel Vehicle Taking a Turn”.

1. Which of the following quantity represent the gear ratio “G” for a 2 wheeler vehicle? Misplaced &

a) ωw / ωe

b) ωe / ωw

c) rw/ R

d) Ie/ Iw

Answer: b

Explanation: Gear ratio is defined as the ratio of the angular speed of the initial or driving member of a gear train or equivalent mechanism to that of the final or driven member specifically: the number of engine revolutions per revolution of the rear wheels of an automobile.

2. Which of the following is correct regarding angle of heel?

a) It is inclination of the vehicle to the vertical for equilibrium

b) It is inclination of the vehicle to the horizontal for equilibrium

c) It is the angle between the axis of gyroscope couple and horizontal

d) It is the angle between the axis of spin couple and axis of precession

Answer: a

Explanation: The axis of spin is inclined to the horizontal at an angle θ, when a two wheel vehicle is taking a turn this is the angle it makes with the vertical plane and it is known as angle of heel.

3. Which of the following is correct regarding angle of heel?

a) It is the angle between the axis of spin couple and axis of precession

b) It is the angle between the axis of gyroscope couple and horizontal

c) It is inclination of the vehicle to the horizontal for equilibrium

d) It is inclination of the axis of spin to the horizontal for equilibrium

Answer: d

Explanation: While taking a turn, a two wheel vehicle is always at an inclination of an angle θ with the vertical plane. This angle is known as angle of heel and is denoted by theta. In other words, the axis of spin is inclined to the horizontal at an angle θ.

4. When is the positive sign used for the expression of gyroscopic couple?

a) Centre of gravity is lower than the centre of vehicle

b) When the engine is rotating in the opposite direction as that of wheels

c) When the engine is rotating in the same direction as that of wheels

d) Centre of gravity is higher than vehicle centre

Answer: c

Explanation: To decide the direction of rotation between wheels and the engine, conventionally a positive sign represents the same direction and negative sign represents opposite direction.

5. The centrifugal couple’s tendency is to overturn the vehicle.

a) True

b) False

Answer: a

Explanation: When a two wheel vehicle is taking a turn, it experiences a centrifugal force acting on it which has a tendency to turn the vehicle upside down about a horizontal plane, whereas as centripetal force would prevent the overturning.

6. The centrifugal couple affects the stability of a two wheel vehicle taking a turn.

a) True

b) False

Answer: a

Explanation: While taking a turn a two wheel vehicle experiences a centrifugal couple acting on it which has a tendency to turn the vehicle upside down about a horizontal plane. This force has to be balanced by a balancing couple.

7. Balancing couple is given by _______ 

a) m.g.h cos θ

b) m.g.h sin θ

c) m.g.h

d) (m.v 2 ).h/R

Answer: b

Explanation: The balancing couple required for the stability of the two wheel vehicle taking a turn is the cross product of height of centre of gravity and weight of the vehicle.

8. Which of the following is true regarding overturning couple?

a) It is the sum of gyroscope couple and centrifugal couple

b) It is the product of gyroscope couple and centrifugal couple

c) It is the difference of gyroscope couple and centrifugal couple

d) It is the ratio of gyroscope couple and centrifugal couple

Answer: a

Explanation: Since the centrifugal couple has a tendency to overturn the vehicle, therefore Total overturning couple Co is given by: Co = Gyroscopic couple + Centrifugal couple.

9. What is the relation between overturning couple and balancing couple for the stability of vehicle?

a) Independent of each other

b) Overturning couple is greater

c) Balancing couple is greater

d) Equal to each other

Answer: d

Explanation: The balancing couple acts in clockwise direction when seen from the front of the vehicle. Therefore for stability, the overturning couple must be equal to the balancing couple.

10. In the figure below what is known as the angle “x”?

a) Angle of ascent

b) Angle of heel

c) Gyroscope angle

d) Angle of precession

Answer: b

Explanation The vehicle is always inclined at an angle θ with the vertical plane. This angle is known as angle of heel and is denoted by θ.

This set of Machine Dynamics Interview Questions and Answers for freshers focuses on “Effect of Gyroscopic Couple on a Disc Fixed Rigidly at a Certain Angle to a Rotating Shaft”.

1. In the figure given below, the moment of inertia about polar axis OP is _______

machine-dynamics-interview-questions-answers-freshers-q1

a) mr 2 /2

b) mr 2

c) 2mr 2

d) 4mr 2

Answer: a

Explanation: In the figure given above, if m is the mass of the disk and r is the radius, then the moment of inertia about the axis OP is given by mr 2 /2.

2. When seen from the top, the gyroscopic couple acting on the disc will act in the clockwise direction.

a) True

b) False

Answer: b

Explanation: When seen from the top, the gyroscopic couple acting on the disc will act in the anti-clockwise direction.

3. In the figure given below, the moment of inertia about diametral axis is _______

machine-dynamics-interview-questions-answers-freshers-q1

a) m(l 2 /12 + r 2 /4)

b) m(l 2 /6 + r 2 /4)

c) m(l 2 /12 + r 2 /2)

d) m(l 2 /12 + r 2 )

Answer: a

Explanation: If m is the mass, l is the length and r is the radius of the disc, then the moment of inertia about the diametral axis is given by m(l 2 /12 + r 2 /4).

4. If the length of the disc is negligible, then what will be the moment of inertia about the diametral axis?

a) mr 2 /4

b) mr 2

c) mr 2 /2

d) 2mr 2

Answer: a

Explanation: If m is the mass, l is the length and r is the radius of the disc, then the moment of inertia about the diametral axis is given by m (l 2 /12 + r 2 /4), if l is negligible then the expression becomes

mr 2 /4.

5. If the length and radius of the disc is same them moment of inertia is more about which axis?

a) Polar

b) Diametral

c) Shaft

d) Center

Answer: a

Explanation: If the length and radius are same then the moment of inertia will be more about the polar axis. Mr 2 /2 as compared to mr 2 /3.

6. From the following data calculate the gyroscopic couple on the bearing in N-m.

Mass of disc = 30 Kg, diameter = 60 cm

Rotation speed = 1200 rpm

Angle between disc and a plane 90 degree to the axis of shaft = 1°

a) 180

b) 186

c) 190

d) 196

Answer: b

Explanation: ω = 125.7 rad/s

Couple acting on the bearings

= mω 2 r 2 /8 sin2θ

substituting the values we get

C = 186 N-m.

7. If the rotation speed of the disc is increased to 2 times and the mass is reduced to half, overall the Couple acting on bearings will increase.

a) True

b) False

Answer: a

Explanation: If the rotation speed of the disc is increased to 2 times and the mass is reduced to half, overall the Couple acting on bearings will increase as couple on bearing depends on the relation

mω 2 r 2 /8 sin2θ, hence it will increase by two times.

This set of Machine Dynamics written test Questions & Answers focuses on “Resultant Effect of a System of Forces Acting on a Rigid Body”.

1. Which of the following is incorrect regarding inertia force?

a) Imaginary force

b) Acts upon a rigid body

c) Brings the body to equilibrium

d) Same direction as of accelerating force

Answer: d

Explanation: The inertia force is an imaginary force, which on acting upon a rigid body has a tendency to bring it in an equilibrium position. Numerically it is equal to the magnitude of accelerating force, but the direction of this force is opposite.

2. Inertia torque acts in the same direction as the accelerating couple?

a) True

b) False

Answer: b

Explanation: The inertia torque is an imaginary torque, which when applied upon the rigid body, brings it in equilibrium position. It is equal to the accelerating couple in magnitude but opposite in direction.

3. If a force has a line of action at a distance h from the centre of gravity, then the value of h is given by _____

a) I. α/F

b) I. α/m.g

c) I/m.k

d) m.k/I

Answer: a

Explanation: We know that Force, F = Mass × Acceleration = m.a. and F.h = m.K 2 .α = I.α

Therefore, h = I. α/F.

4. D-Alembert’s principle is used for which of the following?

a) Change static problem into a dynamic problem

b) Change dynamic problem to static problem

c) To calculate moment of inertia of rigid bodies

d) To calculate angular momentum of a system of masses

Answer: b

Explanation: D-Alembert’s principle states that the resultant force acting on a body together with the reversed effective force , are in equilibrium. This principle is used to reduce a dynamic problem into an equivalent static problem.

5. In the expression F – m.a = 0, the term – m.a is called _______

a) Reversed effective force

b) Net force

c) Coriolis force

d) Resultant force

Answer: a

Explanation: If the quantity m.a is treated as another force with same line of action as the net force, then the body could be assumed to be in static equilibrium. This force is known as Reversed effective force.

6. Why the inertia torque acts in the opposite direction to the accelerating couple?

a) Bring the body in equilibrium

b) To reduce the accelerating torque

c) Acts as a constraint torque

d) Increase the linear acceleration

Answer: a

Explanation: The inertia torque is an imaginary torque, which when applied upon the rigid body, brings it in equilibrium position. It is equal to the accelerating couple in magnitude but opposite in direction.

7. A body remains in equilibrium if ________

a) Inertia force is applied in the same direction to the resultant force

b) Inertia force is applied in the direction opposite to the resultant force

c) Inertia force is applied in the direction Perpendicular to the resultant force

d) Inertia force is applied in the direction Parallel to the resultant force

Answer: b

Explanation: Inertia force is an imaginary force which tends to act in the direction opposite to the resultant force to bring the body in equilibrium. Numerically it is equal to the magnitude of accelerating force.

8. Inertia force and the reversed effective force are the same.

a) True

b) False

Answer: a

Explanation: The net force m.a is taken as another force acting in the opposite direction to the applied resultant force and is known as inertia force or reversed effective force.

9. In the following picture the G is the center of gravity, the quantity h is known as the “offset”. I is the moment of inertia and k is the radius of gyration. Offset’s value is given by?

a) I.α/F

b) I.α/m.g

c) I/m.k

d) m.k/I

Answer: a

Explanation: We know that Force, F = Mass × Acceleration = m.a. and F.h = m.k 2 .α = I.α

Therefore, h = I.α/F

10. Considering a four bar chain with each link having linear and angular acceleration, applying D-Alembert’s principle will never result in which of the following member?

a) 2- force member

b) 3- force member

c) 4 – force member

d) Non accelerating member

Answer: a

Explanation: Since every link is accelerating it will have inertial torque and inertial force and at least force exerted by a link, thus it can never be a two force member.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Klien’s Construction”.

1. In which of the following cases Klein’s construction can be used?

a) Crank has a uniform angular velocity

b) Crank has a uniform angular acceleration

c) Lever has a uniform angular acceleration

d) When the motion is SHM

Answer: a

Explanation: Klein’s construction can be used in both the cases, i.e when the crack has both uniform and non uniform angular velocity and is independent of this condition.

2. When the acceleration of the piston is 0, then the velocity is _____

a) Maximum

b) Minimum

c) Negative

d) Half the maximum

Answer: a

Explanation: When the crank and the connecting rod are perpendicular to each other, at that instance the acceleration is 0 but the velocity is maximum.

3. If OC is the crank and PC is the connecting rod of a reciprocating steam engine and rotates with uniform angular velocity in clockwise direction in the given figure below: then under which condition the piston will undergo retardation?

machine-dynamics-questions-answers-kliens-construction-q3

a) N lies to the right of O

b) N lies to the left of O

c) N is above O

d) N is below O

Answer: a

Explanation: The point N has to remain in the horizontal line of OP thus both the options of N being above O and N being below are incorrect. When N lies to the left of O the piston undergoes acceleration.

4. In the given figure, the velocity of piston is maximum under which of the following conditions?

machine-dynamics-questions-answers-kliens-construction-q3

a) N lies to the left of O

b) N lies to the right of O

c) N coincides with O

d) N lies above O

Answer: c

Explanation: When the crank and the connecting rod are perpendicular to each other, at that instance the acceleration is 0 but the velocity is maximum.

5. In given figure, Triangle OCM is known as ________

machine-dynamics-questions-answers-kliens-construction-q3

a) Klein’s acceleration diagram

b) Klein’s velocity diagram

c) Klein’s displacement diagram

d) Bennet’s diagram

Answer: b

Explanation: On performing a velocity analysis of the given system, it is easy to figure out that triangle formed by points OCM gives the velocity of each link.

6. With respect to the figure given quadrilateral CQNO is known as _______

machine-dynamics-questions-answers-kliens-construction-q3

a) Klein’s acceleration diagram

b) Klein’s velocity diagram

c) Klein’s displacement diagram

d) Bennet’s diagram

Answer: a

Explanation: Ar is parallel to CO, Ar is parallel to CP, At is parallel to QN and Apo is parallel to PO thus quadrilateral CQNO is the acceleration polygon.

7. A Piston will remain in equilibrium if ________

a) Inertia force is applied in the same direction to the resultant force

b) Inertia force is applied in the direction opposite to the resultant force

c) Inertia force is applied in the direction Perpendicular to the resultant force

d) Inertia force is applied in the direction Parallel to the resultant force

Answer: b

Explanation: Inertia force is an imaginary force which tends to act in the direction opposite to the resultant force to bring the body in equilibrium. The magnitude of this force is equal to that of resultant force.

8. Klein’s constructions can be used to determine the acceleration of various parts at all locations.

a) True

b) False

Answer: a

Explanation: Klein’s constructions provide acceleration diagram of the entire mechanism through which the acceleration of various parts can be calculated.

9. For a slider crank mechanism, the total no. of dead centres are _____

a) 0

b) 1

c) 2

d) 3

Answer: c

Explanation: Dead centres are locations where the velocity of the piston is 0, hence there are only two locations where this is possible in the slider crank mechanism.

10. Klein’s construction gives a graphical construction of a 4 bar chain.

a) True

b) False

Answer: b

Explanation: Klein’s construction is used to draw the graphical construction of acceleration and velocity polygon of a slider crank mechanism.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Ritterhaus Construction”.

1. If OC is the crank and PC is the connecting rod rotating in clockwise direction in the figure given below, then triangle OCM is known as ________

machine-dynamics-questions-answers-ritterhauss-construction-q1

a) Klein’s velocity diagram

b) Klein’s acceleration diagram

c) Ritterhaus’ velocity diagram

d) Ritterhaus’ acceleration diagram

Answer: c

Explanation: Since the crank is rotating in the clockwise direction, then the velocity of C will be perpendicular to OC and it’s value is given by w 2 .OC, hence triangle OCM forms a velocity polygon.

2. Ritterhaus’ construction is used when the motion of the crank is linear shm.

a) True

b) False

Answer: b

Explanation: Ritterhaus’ construction is used when the crank is undergoing a motion which has uniform angular velocity.

3. From figure, acceleration of P with respect to C is given by_________

machine-dynamics-questions-answers-ritterhauss-construction-q1

a) ω 2 .CN

b) ω 2 .QN

c) ω 2 .PC

d) ω 2 .OM

Answer: a

Explanation: Total acceleration is the vector sum of radial and tangential components radial component is given by ω 2 .CQ and tangential component is given by ω 2 .QN.

4. Acceleration of any point D on the connecting rod is given by ________

machine-dynamics-questions-answers-ritterhauss-construction-q1

a) ω 2 .OD1

b) ω 2 .OD2

c) ω 2 .OD

d) ω 2 .PD

Answer: b

Explanation: Acceleration at D is the vector sum of radial acceleration and tangential acceleration components. Hence the net acceleration is ω 2 .OD2.

5. In which of the following cases Ritterhaus’ construction can be used?

a) Crank has a uniform angular velocity

b) Crank has a uniform angular acceleration

c) Lever has a uniform angular acceleration

d) When the motion is SHM

Answer: a

Explanation: Ritterhus’ construction can be used in both the cases, i.e when the crack has both uniform and non uniform angular velocity.

6. From figure, what is the velocity of P with respect to C?

machine-dynamics-questions-answers-ritterhauss-construction-q1

a) ω × OC

b) ω × OM

c) ω × CM

d) ω × QN

Answer: c

Explanation: In the figure, the triangle OCM is known as the velocity diagram, referring to that will provide us the velocity of point P with respect to C. Velocity if C wrt to O is ω × OC and ω × OM is velocity of P wrt to O.

7. From figure, what is the absolute velocity of P, i.e velocity of P with respect to the stationary point O?

machine-dynamics-questions-answers-ritterhauss-construction-q1

a) ω × OC

b) ω × OM

c) ω × CM

d) ω × QN

Answer: b

Explanation: In the figure, the triangle OCM is known as the velocity diagram, referring to that will provide us the absolute velocity of point P. Velocity if C wrt to O is ω × OC and ω × OM is velocity of P wrt to O.

8. Ritterhaus’ construction is used to determine graphically the velocity and acceleration of reciprocating parts of an IC engine.

a) True

b) False

Answer: a

Explanation: The velocity and acceleration of the reciprocating parts of the steam engine or internal combustion engine may be determined by the graphical method or analytical method, Ritterhaus’ construction provides graphical solution.

9. Which of the following construction methods is not used to calculate the velocity and acceleration of reciprocating parts of the internal combustion engine?

a) Klien’s construction

b) Ritterhaus’s construction

c) Bennett’s construction

d) D-Alembert’s constructions

Answer: d

Explanation: D-Alembert’s principle is used to convert a dynamic mechanic problem to a static problem with the help of inertia forces while the other three constructions are used to determine the velocity and acceleration of reciprocating parts of IC engines.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Bennett’s Construction”.

1. If OC is the crank and PC is the connecting rod rotating in clockwise direction in the figure given below, then triangle OCM is known as _________

machine-dynamics-questions-answers-bennetts-construction-q1

a) Klein’s velocity diagram

b) Klein’s acceleration diagram

c) Bennett’ velocity diagram

d) Bennett’ acceleration diagram

Answer: c

Explanation: Since the crank is rotating in the clockwise direction, then the velocity of C will be perpendicular to OC and it’s value is given by w 2 .OC, hence triangle OCM forms a velocity polygon.

2. Bennett’ construction is used when the motion of the crank is linear cycloidal.

a) True

b) False

Answer: b

Explanation: Bennett’ construction is used when the crank is undergoing a motion which has uniform angular velocity.

3. From figure, acceleration of P with respect to C is given by_________

machine-dynamics-questions-answers-bennetts-construction-q1

a) ω 2 .CN

b) ω 2 .QN

c) ω 2 .PC

d) ω 2 .OM

Answer: a

Explanation: Total acceleration is the vector sum of radial and tangential components radial component is given by ω2.CQ and tangential component is given by ω 2 .QN.

4. Acceleration of any point D on the connecting rod is given by ________

a) ω 2 .OD1

b) ω 2 .OD2

c) ω 2 .OD

d) ω 2 .PD

Answer: b

Explanation: Acceleration at D is the vector sum of radial acceleration and tangential acceleration components. Hence the net acceleration is ω 2 .OD2

5. In which of the following cases Bennett’s construction can be used?

a) Crank has a uniform angular velocity

b) Crank has a uniform angular acceleration

c) Lever has a uniform angular acceleration

d) When the motion is SHM

Answer: a

Explanation: Bennett’s construction can be used in both the cases, i.e when the crack has both uniform and non uniform angular velocity.

6. From figure, what is the velocity of P with respect to C?

machine-dynamics-questions-answers-bennetts-construction-q1

a) ω × OC

b) ω × OM

c) ω × CM

d) ω × QN

Answer: c

Explanation: In the figure, the triangle OCM is known as the velocity diagram, referring to that will provide us the velocity of point P with respect to C. Velocity if C wrt to O is ω × OC and ω × OM is velocity of P wrt to O.

7. From figure, what is the absolute velocity of P, i.e velocity of P with respect to the stationary point O?

machine-dynamics-questions-answers-bennetts-construction-q1

a) ω × OC

b) ω × OM

c) ω × CM

d) ω × QN

Answer: b

Explanation: In the figure, the triangle OCM is known as the velocity diagram, referring to that will provide us the absolute velocity of point P. Velocity if C wrt to O is ω × OC and ω × OM is velocity of P wrt to O.

8. Bennett’ construction is used to determine graphically the velocity and acceleration of reciprocating parts of an IC engine.

a) True

b) False

Answer: a

Explanation: The velocity and acceleration of the reciprocating parts of the steam engine or internal combustion engine are determined by either graphical method or analytical method, Bennett’s construction provides graphical solution for the same.

9. Velocity of any point D on the connecting rod is given by ________

machine-dynamics-questions-answers-bennetts-construction-q1

a) ω.OD1

b) ω.OD2

c) ω.OD

d) ω.PD

Answer: a

Explanation: Linear velocity is obtained by a product of angular velocity of the crank and the perpendicular distance of the respective point. In this case ω is the angular velocity and OD1 is the perpendicular distance.

This set of Machine Dynamics Questions and Answers for Experienced people focuses on “Angular Velocity and Acceleration of the Connecting Rod”.

1. If the crank and the connecting rod are 300 mm and 1 m long respectively and the crank rotates at a constant speed of 250 r.p.m., determine the crank angle at which the maximum velocity occurs is ____

a) 45

b) 75

c) 90

d) 60

Answer: b

Explanation: For maximum velocity of the piston : cosθ + 2cos/2n = 0

n = l/r : 1/.3 = 3.33

therefore, θ = 75°

2. If the crank and the connecting rod are 600 mm and 2 m long respectively and the crank rotates at a constant speed of 250 r.p.m, determine maximum velocity of the piston in m/s is _____

a) 15

b) 17.5

c) 20

d) 10.5

Answer: b

Explanation: maximum velocity is given by V=w.r

Maximum velocity occurs at 75, n = 2/.6 = 3.33

substituting values gives Vmax = 17.5 m/s.

3. The crank and connecting rod of a steam engine are 0.3 m and 1.5 m in length. The crank rotates at 150 r.p.m. clockwise, determine velocity of the piston when the crank is at an angle 40 degrees from IDC.

a) 4.19

b) 5

c) 3.49

d) 3.36

Answer: c

Explanation: We know that ratio of lengths of the connecting rod and crank, n = l/r = 1.5/0.3 = 5,

substituting the values into the formula of velocity V=w.r gives V = 3.49 m/s.

4. From the data given:

Crank and connecting rod of a steam engine are 0.3 m and 1.5 m in length; The crank rotates at 150 r.p.m. clockwise.

Determine the acceleration in m/s 2 of the piston for the same position.

a) 59.27

b) 55.25

c) 65.3

d) 50.41

Answer: a

Explanation: We know that ratio of lengths of the connecting rod and crank, n = l/r = 1.5/0.3 = 5,

substituting the values into the formula of acceleration a= ω 2 .r  = 59.27 m/s 2 .

5. Which of the following expression represent the angular acceleration α of the connecting rod?

a) −ω 2 . sin θ/n

b) −ω 2 . cos θ.n

c) ω 2 . cos θ/n

d) ω 2 . sin θ.n

Answer: a

Explanation: Angular acceleration is the rate of change of angular velocity, for the connecting rod it depends on the length of the rod and crank’s radius. The negative signs indicate that the sense of rotation tends to reduce the angle phi.

6. In a slider crank mechanism, the length of the crank and connecting rod are 150 mm and 600 mm respectively. The crank position is 60° from inner dead centre. The crank shaft speed is 400 r.p.m. . Velocity of the slider is ________

a) 6.9 m/s

b) 6.12 m/s

c) 7.32 m/s

d) 6.66 m/s

Answer: b

Explanation: We know that ratio of the length of connecting rod and crank, n = l / r = 0.6 / 0.15 = 4, substituting the values into the formula of velocity V=w.r gives V = 6.12 m/s.

7. From the data given:

The length of the crank and connecting rod are 150 mm and 600 mm

The crank position is 60° from inner dead centre. The crank shaft speed is 400 r.p.m.

Find the acceleration in m/s 2 of the slider.

a) 101.5

b) 100.6

c) 98.6

d) 97.6

Answer: c

Explanation: We know that ratio of lengths of the connecting rod and crank, n = l/r = 600/150 = 4,

substituting the values into the formula of acceleration a= ω 2 .r  = 98.6 m/s 2 .

8. From the data given:

The length of the crank and connecting rod are 150 mm and 600 mm

The crank position is 60° from inner dead centre. The crank shaft speed is 400 r.p.m.

Find the angular acceleration in rad/s 2 of the connecting rod.

a) 421

b) 400

c) 379

d)388

Answer: c

Explanation: Angular acceleration of the connecting rod is given by (ω 2 .sinθ)/n hence substituting the values will give the result as 379 rad/s 2 .

9. While calculating angular acceleration of the connecting rod, sin 2  term is neglected.

a) True

b) False

Answer: a

Explanation: sin 2  term is neglected while calculating angular acceleration of the connecting rod as it is negligible and has a very low impact in the final calculations.

10. In a slider crank mechanism, the length of the crank and connecting rod are 180 mm and 540 mm respectively. The crank position is 45° from inner dead centre. The crank shaft speed is 450 r.p.m. , calculate angular velocity of the connecting rod in rad/s.

a) 10.3

b) 11.1

c) 12.2

d) 11.8

Answer: b

Explanation: n = l/r : 540/180 = 3

angular velocity is given by ω.cos θ/n = .cos45/3 = 11.1 rad/s.

This set of Tricky Machine Dynamics Questions and Answers focuses on “Forces on the Reciprocating Parts of an Engine Neglecting Weight of the Connecting Rod”.

1. The net force acting on the crosshead pin is known as __________

a) Crank pin effort

b) Crank effort

c) Piston effort

d) Shaft effort

Answer: c

Explanation: Crank pin effort is the force acting along the direction perpendicular to the crank. Crank effort is the torque acting on the crank, piston effort is the force acting on the piston or the crosshead pin.

2. Piston effort acts along the line of stroke.

a) True

b) False

Answer: a

Explanation: Piston effort is the force acting along the line of action of the stroke on the crosshead pin or the piston.

3. In a horizontal engine, reciprocating parts are accelerated when the piston moves from _______

a) TDC to BDC

b) BDC to TDC

c) Midway to TDC

d) BDC to midway

Answer: a

Explanation: The reciprocating parts of a horizontal engine are accelerated during the period when the piston moves from inner dead center to outer dead center.

4. In a horizontal engine, reciprocating parts are retarded when the piston moves from _________

a) TDC to BDC

b) BDC to TDC

c) Midway to TDC

d) BDC to midway

Answer: b

Explanation: The reciprocating parts of a horizontal engine are accelerated during the latter half of the stroke, i.e when the piston moves from inner dead center to outer dead center and retards when the piston moves from outer dead center to inner dead center.

5. When the piston is accelerated, the piston effort is given by which of the following the equation?

a) F – F

b) F + F

c) F ± F

d) F – F + R

Answer: b

Explanation: During acceleration, piston effort is the sum total of the net load on piston and inertia forcer. The presence of resistance force also effects the expression.

6. In the presence of frictional resistance, the expression for piston effort is _________

a) F – F

b) F + F

c) F ± F – R

d) F – F + R

Answer: c

Explanation: Piston effort depends on the net piston load and inertia force, in presence of frictional resistance and additional term of R is subtracted from the overall expression.

7. Crank effort is the product of crank pin radius and _______

a) Thrust on sides

b) Crankpin effort

c) Force acting along connecting rod

d) Piston effort

Answer: b

Explanation: Crank effort also known as turning moment or torque on the crank shaft is the product of the crankpin effort a.k.a FT and the crank pin radius a.k.a r. It is applied at crank pin perpendicular to the crank.

8. In a horizontal engine, the weight of the reciprocating parts also add/subtract to the piston effort.

a) True

b) False

Answer: b

Explanation: The weight of the reciprocating parts of the engine add or subtract to the piston effort only in a vertical engine and it depends on the motion of the piston.

9. For the given data of an Internal combustion engine : Mass of parts = 180 kg bore = 175 mm, length of stroke = 200 mm, engine speed = 500 r.p.m., length of connecting rod = 400 mm and crank angle = 60° from T.D.C, find the inertia force.

a) 17.56 N

b) 19.2 N

c) 18.53 N

d) 18.00 N

Answer: c

Explanation: Ratio of length of connecting rod and crank n = l/r = 2*200/ = 4,

We know that inertia force is m.ω 2 .r

inserting values will give F = 18.53 N.

10. The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the reciprocating parts is 250 kg. When the crank has travelled 30° from T.D.C., the difference between the driving and the back pressures is 0.45 N/mm 2 . The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter is neglected, calculate the net load on piston.

a) 88000 N

b) 90560 N

c) 78036 N

d) 88357 N

Answer: d

Explanation: Net piston load is given by F = (p 1 -p 2 )πD 2 ÷4

p 1 -p 2 = 0.45 N/mm 2

D = 500 mm

Therefore F = 88357 N.

11. From the data given:

crank-pin circle radius = 300mm

mass of the reciprocating parts = 250kg

difference between the driving and the back pressures is 0.45 N/mm 2

The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m.

engine runs at 250 r.p.m & 30° from T.D.C.

Find the piston effort.

a) 32.4 kN

b) 35.2 kN

c) 37.3 kN

d) 40.2 N

Answer: c

Explanation: Net piston load is given by F = (p 1 -p 2 )πD 2 ÷4

Fp = Fl – Fi

Fi = m.ω 2 .r

= 51020 N

Therefore, Fp = 37.3 kN.

This set of Machine Dynamics Question Paper focuses on “Correction Couple to be Applied to Make the Two Mass Systems Dynamically Equivalent”.

1. Correction couple is applied when masses are placed arbitrarily and to maintain _________

a) Static equilibrium

b) Dynamic equilibrium

c) Stable equilibrium

d) Unstable equilibrium

Answer: b

Explanation: Difference between the torques required to accelerate the two-mass system and the torque required to accelerate the rigid body is called correction couple and this couple must be applied, when the masses are placed arbitrarily to make the system dynamical equivalent.

2. The sum of torques required to accelerate a 2 mass system and to accelerate a rigid body is called correction couple.

a) True

b) False

Answer: b

Explanation: Difference between the torques required to accelerate the two-mass system and the torque required to accelerate the rigid body is called correction couple.

3. The correction couple does not depend on _________

a) Distance between arbitrary masses

b) Distance between the two masses for a true dynamically equivalent system

c) Radius of gyration of equivalent system

d) Distance between fixed masses

Answer: d

Explanation: Distance between arbitrary masses, Distance between the two masses for a true dynamically equivalent system and Radius of gyration of equivalent system all effect the correction couple.

4. In the figure given below, A and D are arbitrary masses placed Quantity L is known as _______

machine-dynamics-question-papers-q4

a) Distance between the two masses for a true dynamically equivalent system

b) Distance between fixed masses

c) Distance between arbitrary masses

d) Equivalent radius

Answer: a

Explanation: The distance between the two masses for a true dynamically equivalent system is given by L and has the expression of (Kg 2 + l1 2 )/l1.

5. An Internal combustion engine has a connecting rod of mass 2 kg and the distance between the centre of crank and centre of gudgeon pin is 25 cm. The Center of Gravity lies at a point 10 cm from the gudgeon pin along the line of centres. The radius of gyration of this system about an axis through the Center of Gravity perpendicular to the plane of rotation is 11 cm. Find the mass located at gudgeon pin in Kg.

a) 0.9

b) 0.8

c) 1.1

d) 1.2

Answer: c

Explanation: From the given data l = 25 cm, l 1 = 10cm kg=11cm

l 1 .l 2 =kg 2 , l 2 = 12.1 cm

M= l 2 m/( l 1 + l 2 )

= 12.1x 2/

= 1.1 Kg.

6. From the data given:

54tr xc

Connecting rod of mass 2 kg and the distance between the centre of crank and centre of gudgeon pin is 25 cm. The Center of Gravity lies at a point 10 cm from the gudgeon pin along the line of centres.

The radius of gyration of this system about an axis through the Center of Gravity perpendicular to the plane of rotation is 11 cm.

Find the mass placed at distance l2 from center of gravity.

a) 0.9

b) 0.8

c) 1.1

d) 1.2

Answer: a

Explanation: From the given data l = 25 cm, l1 = 10cm kg=11cm

l 1 .l 2 =kg 2 , l 2 = 12.1 cm

M = l 1 .m/( l 1 + l 2 )

= 10×2

= 0.9 kg.

7. An Internal combustion engine has a connecting rod of mass 2 kg and the distance between the centre of crank and centre of gudgeon pin is 25 cm. The Center of Gravity lies at a point 10 cm from the gudgeon pin along the line of centres. The radius of gyration of this system about an axis through the Center of Gravity perpendicular to the plane of rotation is 11 cm. If two masses are used instead of the connecting rod, one at the gudgeon pin and other at the crank pin. What will be the new radius of gyration?

a) 0.212m

b) 0.122m

c) 0.1m

d) 0.145m

Answer: b

Explanation: Let l 3 be the distance between the masses, then l 3 = l- l 1 = 15cm

If K 1 is the new radius of gyration then, K 1 2 = l 1 . l 3 = 0.1×0.15 = 0.015 m 2

Therefore, K 1 = 0.122 m.

8. Mass moment of inertia of two arbitrary masses placed will be same as the mass moment inertia of the rigid body.

a) True

b) False

Answer: b

Explanation: Mass moment of inertia of two arbitrary masses placed will be different as the mass moment inertia of the rigid body, this is why there is a need to calculate the correction couple.

9. An Internal combustion engine has a connecting rod of mass 2 kg and the distance between the centre of crank and centre of gudgeon pin is 25 cm. The Center of Gravity lies at a point 10 cm from the gudgeon pin along the line of centres. The radius of gyration of this system about an axis through the Center of Gravity perpendicular to the plane of rotation is 11 cm. If two masses are used instead of the connecting rod, one at the gudgeon pin and other at the crank pin, if the angular acceleration of the rod is 23 000 rad/s 2 , then find the correction couple in N-m.

a) 140.2

b) 133.4

c) 136.8

d) 135.6

Answer: b

Explanation: We know that correction couple is given by T’ = m.(k 1 2 – kg 2 ).α

T’ = 2. (0.015 – 0.11 2 ). 23000

= 133.4 N-m.

This set of Machine Dynamics Interview Questions and Answers for Experienced people focuses on “Inertia Forces in a Reciprocating Engine Considering the Weight of Connecting Rod”.

1. Force which does not act on the connecting rod is ______

a) Weight of connecting rod

b) Inertia force of connecting rod

c) Radial force

d) Coriolis force

Answer: d

Explanation: Since there is no accelerated frame of reference having different velocities, therefore coriolis force does not act on the connecting rod.

2. Inertia forces on the reciprocating parts acts along the line of stroke.

a) True

b) False

Answer: a

Explanation: The reciprocating parts of the engine experience an Inertia force of magnitude m.ω 2 .r, this inertia force acts along the line of stroke.

3. When mass of the reciprocating parts is neglected then the inertia force is _____

a) Maximum

b) Minimum

c) 0

d) Not defined

Answer: c

Explanation: Inertia force for neglected mass of reciprocating parts is 0 as it depends on the mass, since it has a value, it is defined.

4. For a steam engine, the following data is given:

Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m

calculate inertia force at θ=30 degrees from IDC.

a) 19000 N

b) 19064 N

c) 19032 N

d) 20064 N

Answer: b

Explanation: l 1 = l – G.C = 1.5 – 0.5 = 1m

Fi = .ω 2 .r

Mr=300 Kg

Mc=250 kg

ω=13.1 rad/s r=0.3m

substituting these values will give Fi = 19064 N.

5. Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m

Find the equivalent length L of a simple pendulum swung about an axis.

a) 1.35 m

b) 1.42 m

c) 1.48 m

d) 1.50 m

Answer: b

Explanation: We know that equivalent length L is given by the expression

L = (Kg 2 + l 1 2 )/l 1

Kg = 0.65m l 1 = 1m

therefore L = 1.42 m.

6. From the data given:

Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m

Find the correcting couple in N-m?

a) 52.7

b) 49.5

c) 59.5

d)56.5

Answer: c

Explanation: The correction couple depends on equivalent length, l 1 mass of connecting rod and angular position and velocity

Tc = Mc.l 1 ..(ω 2 .sin2θ/2n 2 )

substituting the values into the equation results in

Tc = 59.5 N-m.

7. Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m

Find angular acceleration of connecting rod in rad/s 2 .

a) 16.782

b) 17.824

c) 15.142

d) 17.161

Answer: b

Explanation: α = -ω 2 .sinθ/n

= 13.1 2 .sin3θ/5

= 17.161 rad/s 2 .

8. Torque due to weight of the connecting rod affects the torque due to connecting rod.

a) True

b) False

Answer: b

Explanation: The torque due to connecting rod remains same irrespective of the torque caused by the weight of the connecting rod.

This set of Machine Dynamics Objective Questions & Answers focuses on “Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine”.

1. In the figure given below, the quantity represented by the arrow is known as ___________

machine-dynamics-objective-questions-answers-q1

a) Maximum Torque

b) Minimum Torque

c) Maximum Force

d) Mean resisting torque

Answer: d

Explanation: In a turning moment diagram given above, the line pointed out by the arrow is the value of mean torque and hence it is called mean resisting torque.

2. Turning moment is maximum when the crank angle is 90 degrees.

a) True

b) False

Answer: a

Explanation: Referring the turning moment diagram for a single cylinder double acting steam engine, the minimum values occur at 0, 180 and 360 degrees and maximum values occur at 90 and 270 degrees.

3. The curve abc in figure below is known as _________

machine-dynamics-objective-questions-answers-q1

a) Instroke

b) Outstroke

c) Positive couple

d) Negative couple

Answer: b

Explanation: In the given figure, the curve during the curve abc there is an ascent in the motion. The ascent of motion is known as outstroke thus curve abc is called outstroke.

4. The curve cde in figure below is known as _________

machine-dynamics-objective-questions-answers-q1

a) Instroke

b) Outstroke

c) Positive couple

d) Negative couple

Answer: a

Explanation: In the given figure, the curve during the curve cde there is a descent in the motion. The descent of motion is known as intstroke, thus curve cde is called instroke.

5. When engine torque is more than mean resisting torque, then the flywheel _______

a) Has uniform velocity

b) Has 0 velocity

c) Has acceleration

d) Has retardation

Answer: c

Explanation: When the difference between T-Tmean is positive the work is done by the steam and the crankshaft accelerated which results in acceleration of flywheel.

6. When engine torque is less than mean resisting torque, then the flywheel _______

a) Has uniform velocity

b) Has 0 velocity

c) Has acceleration

d) Has retardation

Answer: d

Explanation: When the difference between T-Tmean is negative the work is done on the steam and the crankshaft decelerates which results in retardation of flywheel.

7. Area of the turning moment diagram represents _______

a) Work done

b) Work done per revolution

c) Power generated

d) Power generated per revolution

Answer: b

Explanation: Since the work done is the product of the turning moment and the angle turned, therefore the area of the turning moment diagram represents the work done per revolution.

8. In actual practice, engine is assumed to work against the mean resisting torque.

a) True

b) False

Answer: a

Explanation: In actual practice, the engine is assumed to work against the mean resisting torque, therefore the area under the curve gives the work done against the mean resisting torque.

This set of Machine Dynamics test focuses on “Turning Moment Diagram for a Four Stroke Cycle Internal Combustion Engine”.

1. Which of the following process is not the part of working of a 4 stroke I.C Engine?

a) Compression

b) Suction

c) Exhaust

d) Turbine flow

Answer: d

Explanation: 4 stroke IC engine works through suction compression working or power stroke and then exhaust. It does not use turbine hence no turbine flow is used.

2. 4-stroke engine has one power stroke per 2 revolutions hence turning moment diagram is uniform.

a) True

b) False

Answer: b

Explanation: One power stroke per 2 revolutions of crank shaft hence its turning moment is not uniform so heavier flywheel is needed.

3. Work produced by the 4-stroke engine is less because _______

a) One power stroke per 2 revolutions

b) One power stroke per revolution

c) One power stroke per 4 revolution

d) Four power stroke per revolution

Answer: a

Explanation: One power stroke per 2 revolutions of crank shaft hence its turning moment is not uniform so heavier flywheel is needed and hence results in less power generation.

4. A negative loop is formed in the turning moment diagram because the pressure inside the cylinder is _______ than the atmospheric pressure.

a) Equal

b) More

c) Less

d) No pressure inside

Answer: c

Explanation: A negative loop is formed in the turning moment diagram of a 4 stroke IC engine because during suction the pressure inside the engine is reduced due to intake and has a lower value than the atmospheric pressure.

5. When the work is done on the gases, which of the following effect is observed on the turning moment diagram?

a) Formation of negative loop

b) Formation of positive loop

c) Formation of an infinite loop

d) Formation of no loop

Answer: a

Explanation: When the work is done on the gases, the pressure inside the cylinder decreases and has a value lower than the atmospheric pressure, as a result a negative loop is formed.

6. When the work is by on the gases, which of the following effect is observed on the turning moment diagram?

a) Formation of negative loop

b) Formation of positive loop

c) Formation of an infinite loop

d) Formation of no loop

Answer: b

Explanation: When the work is done by the gases, the pressure inside the cylinder increases and has a value higher than the atmospheric pressure, as a result a positive loop is formed.

7. Compression stroke results in ________

a) Formation of negative loop

b) Formation of positive loop

c) Formation of an infinite loop

d) Formation of no loop

Answer: a

Explanation: During compression stroke, the work is done on the gases, the pressure inside the cylinder decreases and has a value lower than the atmospheric pressure, as a result a negative loop is formed.

8. 4 strokes engine have a lower thermal efficiency than 2 stroke engine.

a) True

b) False

Answer: b

Explanation: 4-stroke IC engines despite having low power output per cycle have a higher thermal efficiency than a 2 stroke cycle because in 2 stroke engines, the fuel escapes with the exhaust.

9. While drawing the turning moment diagram, which of the forces in not taken into account?

a) Inertia force

b) Force on connecting rod

c) Force on crank

d) Coriolis force

Answer: d

Explanation: When a turning moment diagram is drawn, it takes into account all the forces inertia, connecting rod and crank however there is no coriolis force acting in the engine thus it is not taken into account.

10. In the given turning moment diagram, the process X is known as ______

machine-dynamics-questions-answers-test-q10

a) Suction

b) Compression

c) Power

d) Exhaust

Answer: c

Explanation: Since the area of the loop is maximum during the process X hence it is the power stroke since most the power is generated in this process which makes the pressure rise and create a positive loop.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Coefficient of Fluctuation of Energy”.

1. In the figure given below, the areas BbC, CcD represent _______

machine-dynamics-questions-answers-coefficient-fluctuation-energy-q1

a) Power generated

b) Power lost

c) Fluctuation of energy

d) Change in momentum

Answer: c

Explanation: The variations of energy above and below the mean resisting torque line are called fluctuations of energy. The areas under the following curves: BbC, CcD, DdE and so on represent fluctuations of energy.

2. The fluctuation of energy is the variation of energy above the line of minimum torque.

a) True

b) False

Answer: b

Explanation: The given statement is false because the variations of energy above and below the mean resisting torque line are called fluctuations of energy. It is represented by the enclosing curves above and below the mean line.

3. From the turning moment diagram shown below, the maximum speed will occur at point _____

machine-dynamics-questions-answers-coefficient-fluctuation-energy-q1

a) p

b) b

c) q

d) c

Answer: c

Explanation: The maximum engine speed occurs at point q because the flywheel absorbs energy when the crank moves from point p to q. This results in a gain of speed.

4. From the turning moment diagram shown below, the minimum speed will occur at point _____

machine-dynamics-questions-answers-coefficient-fluctuation-energy-q1

a) p

b) b

c) q

d) c

Answer: a

Explanation: The minimum engine speed occurs at point p because the flywheel gives out some of its energy when the crank moves from point a to p. This results in loss of speed.

5. Coefficient of fluctuation of energy is ratio of _______

a) Mean fluctuation energy to work done per cycle

b) Maximum fluctuation energy to work done per cycle

c) Minimum fluctuation energy to work done per cycle

d) Mean fluctuation energy to power generated per cycle

Answer: b

Explanation: The ratio of the maximum fluctuation of energy to the work done per cycle is known as the coefficient of fluctuation of energy. Mathematically, coefficient of fluctuation of energy  = Maximum fluctuation of energy/Work done per cycle.

6. Work done per cycle is calculated as _______

a) Tmean × θ

b) Tmin × θ

c) Tmax × θ

d) Tmax × θ/2

Answer: a

Explanation: The work done per cycle  may be obtained by using the following relations: Work done per cycle = Tmean × θ, where Tmean = Mean torque, Mean torque is same as the mean resisting torque.

7. For a 4 stroke IC engine, the angle θ assumes a value equal to _______

a) π

b) 2π

c) 4π

d) π/2

Answer: c

Explanation: For a 4 stroke internal combustion engine, a power stroke takes place every 2 revolution therefore the total angle turned is equal to 4π.

8. To calculate power generated, product of maximum torque and angular speed is taken.

a) True

b) False

Answer: b

Explanation: To calculate power generated, product of mean effective torque and angular speed is taken.

P = Tmean × ω

9. For a 4 stroke IC engine, Tmean = 1875 N-m, then find the work done per cycle.

a) 23.56 kJ

b) 11.78 kJ

c) 5.89 kJ

d) 2.94 kJ

Answer: a

Explanation: For a 4 stroke internal combustion engine, a power stroke takes place every 2 revolution therefore the total angle turned is equal to 4π. Hence θ=4π

therefore work done per cycle = 23.56 kJ.

10. A shaft fitted with a flywheel rotates at 250 r.p.m. and drives a machine, the torque fluctuated about 18750 N-m, find the power required to drive the machine.

a) 49.125 kW

b) 24.56 kW

c) 12.28 kW

d) 6.14 kW

Answer: a

Explanation: ω = 2π × 250/60 = 26.2 rad/s,

P = Tmean ×ω

= 1875 x 26.2

= 49.125 kW.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Coefficient of Fluctuation of Speed”.

1. Coefficient of fluctuation is the ratio of ______

a) The maximum fluctuation of speed to the mean speed

b) The minimum fluctuation of speed to the mean speed

c) The maximum speed to the mean speed

d) The minimum speed to the maximum speed

Answer: a

Explanation: Maximum fluctuation of speed is defined as the difference between the maximum and minimum speeds during a cycle. The ratio of the maximum fluctuation of speed to the mean speed is known as the coefficient of fluctuation of speed.

2. Coefficient of steadiness is reciprocal of coefficient of fluctuation of speed.

a) True

b) False

Answer: a

Explanation: Coefficient of steadiness is defined as the ratio of mean speed to the ratio of maximum fluctuation of speed, hence it is the reciprocal of coefficient of fluctuation of speed.

3. If the minimum speed of the engine is half the maximum speed, then coefficient of fluctuation is _____

a) 0.5

b) 1.5

c) 2

d) 0.66

Answer: d

Explanation: Let N be the maximum speed, then minimum speed is N/2

mean speed = /2 = 3N/4

Maximum fluctuation = N – N/2 = N/2

therefore coefficient of fluctuation = 2/3 = 0.66.

4. If the mean speed is 3/4 th of the maximum speed, the coefficient of steadiness is ______

a) 4/3

b) 3/2

c) 3/4

d) 1

Answer: b

Explanation: Let N be the mean speed and N 1 and N 2 be the maximum and minimum speed respectively,

given N=3N 1 /4 , N 1 = 4N/3

also N = (N 1 +N 2 )/2

N = 2N/3 + N 2 /2 : N 2 = 2N/3

maximum fluctuation = N 1 – N 2 = 4N/3 – 2N/3 = 2N/3

mean speed = N

therefore coefficient of steadiness = N/

= 3/2.

5. The moment of inertia of a flywheel is 6500 kg-m 2 , from the turning moment diagram, it is found that fluctuation of energy is 56000 N-m, if mean speed is 120 rpm, then find maximum speed.

a) 121

b) 119

c) 122

d)118

Answer: a

Explanation: We know that ΔE= I ω 2 . Cs

= I ω. (ω 1 -ω 2 )

= 6500. 120x 2 ( N 1 – N 2 )

N 1 – N 2 = 2

N=120

Therefore, N 1 =121.

6. The moment of inertia of a flywheel is 6500 kg-m 2 , from the turning moment diagram, it is found that fluctuation of energy is 56000 N-m, if mean speed is 120 rpm, find the minimum speed.

a) 121

b) 119

c) 122

d) 118

Answer: b

Explanation: We know that ΔE = I ω 2 . Cs

= I ω. (ω 1 -ω 2 )

= 6500. 120x 2 (N 1 – N 2 )

N 1 – N 2 = 2

N=120

Therefore, N 2 =119.

7. In order to reduce the heavy weight of the flywheel which of the following mechanical component is used?

a) Bearings

b) Shaft

c) Gears

d) Engine

Answer: c

Explanation: The mass of the flywheel depends on the velocity at which it rotates, the velocity can be moderated by installing a set of gears. This results in reduction of mass of flywheel.

8. The main function of a flywheel is to cause variations in the speed of a shaft caused by torque fluctuations.

a) True

b) False

Answer: b

Explanation: The primary function of a fly wheel is to smoothen out variations in the speed of a shaft caused by torque fluctuations, this is basically to lessen the fluctuations of the torque. If the source of the driving torque or load torque is fluctuating in nature, then a flywheel is used.

9. When the flywheel absorbs energy, its speed ________

a) Remains unaffected

b) Increases

c) Decreases

d) Goes down to 0

Answer: b

Explanation: When a flywheel absorbs energy, its speed increases and when it gives up energy, its speed decreases. This effect is reflected in the turning moment diagram of the flywheel.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Turning Moment Diagram for a Multicylinder Engine”.

1. For a multicylinder engine, the moment of inertia of a flywheel is 6500 kg-m 2 , from the turning moment diagram, it is found that fluctuation of energy is 56000 N-m, if mean speed is 120 rpm, then find maximum speed.

a) 121

b) 119

c) 122

d) 118

Answer: a

Explanation: We know that ΔE= I ω 2 . Cs

= I ω. (ω 1 -ω 2 )

= 6500. 120x 2 (N 1 -N 2 )

N 1 -N 2 = 2

N=120

Therefore, N 1 =121.

2. Multi cylinder engines are usually placed at an equal inclination to each other.

a) True

b) False

Answer: a

Explanation: In case of multi cylinder engines, the cylinders are placed in such a way that the performance of the engine is optimum hence placing them at equal inclination improves the performance.

3. Consider a three cylinder engine, the resultant turning moment diagram is the _____ of three cylinders.

a) Sum

b) Difference

c) Product

d) Independent

Answer: a

Explanation: For a multicylinder engine, the turning moment diagram is formed by combining the turning moment diagrams of individual cylinders.

4. For a multicylinder engine, the coefficient of fluctuation of speed would vary with _________

a) Number of cylinders

b) Remains unaffected

c) Length of connecting rod

d) Input temperature

Answer: b

Explanation: The coefficient of fluctuation depends only on the speed of the engine and is unaffected by the no. of cylinders.

5. For a 4 cylinder engine, if the minimum speed of the engine is half the maximum speed, then coefficient of fluctuation is _________

a) 0.5

b) 1.5

c) 2

d) 0.66

Answer: d

Explanation: Let N be the maximum speed, then minimum speed is N/2

mean speed = /2 = 3N/4

Maximum fluctuation = N – N/2 = N/2

therefore coefficient of fluctuation = 2/3.

6. In the turning moment diagram of a multicylinder engine, the work done during exhaust stroke is by ______

a) The gases

b) On the gases

c) Piston wall

d) Valve

Answer: b

Explanation: During exhaust stroke, the area under the loop is negative. This indicates that the work is done on the gases and power has not been generated during the process.

7. For a 4 cylinder engine, when the pressure inside the cylinders exceeds the atmospheric pressure then.

a) Work is done by the gases

b) Work is done on the gases

c) Work is done on the piston wall

d) Work is done by the piston wall

Answer: a

Explanation: When the pressure inside the cylinders exceeds the atmospheric pressure then the area under the loop is positive. This indicates that the work is done by the gases.

8. The flywheel of a steam engine has a mass moment of inertia of 2500 Kg-m 2 . The starting torque of the steam engine is 1500 N-m and may be assumed constant, using this data find the angular acceleration of the flywheel in rad/s 2 .

a) 0.4

b) 0.6

c) 0.3

d) 1.2

Answer: a

Explanation: We know that T = I.α

T = 1500 Nm 

I = 2500 kg-m 2

therefore, &aplpha; = 15/25 = 0.6 rad/s 2 .

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Fluctuation of Energy”.

1. The maximum fluctuation of speed is the

a) difference of minimum fluctuation of speed and the mean speed

b) difference of the maximum and minimum speeds

c) sum of the maximum and minimum speeds

d) variations of speed above and below the mean resisting torque line

Answer: b

Explanation: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called coefficient of fluctuation of speed.

2. The coefficient of fluctuation of speed is the _____________ of maximum fluctuation of speed and the mean speed.

a) product

b) ratio

c) sum

d) difference

Answer: b

Explanation: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called coefficient of fluctuation of speed.

3. In a turning moment diagram, the variations of energy above and below the mean resisting torque line is called

a) fluctuation of energy

b) maximum fluctuation of energy

c) coefficient of fluctuation of energy

d) none of the mentioned

Answer: a

Explanation: The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation.The variations of energy above and below the mean resisting torque line are called fluctuation of energy.

4. If E = Mean kinetic energy of the flywheel, C S = Coefficient of fluctuation of speed and Δ E = Maximum fluctuation of energy, then

a) ΔE = E / C S

b) ΔE = E 2 × C S

c) ΔE = E × C S

d) ΔE = 2 E × C S

Answer: d

Explanation: ΔE = Maximum K.E. — Minimum K.E. = 2 E × C S

5. The ratio of the maximum fluctuation of energy to the ___________ is called coefficient of fluctuation of energy.

a) minimum fluctuation of energy

b) workdone per cycle

c) coefficient of fluctuation of energy

d) none of the mentioned

Answer: b

Explanation: Coefficient of fluctuation of energy is defined as the ratio of the maximum fluctuation of energy to the work done per cycle. It is usually denoted by C E .

6. Due to the centrifugal force acting on the rim, the flywheel arms will be subjected to

a) tensile stress

b) compressive stress

c) shear stress

d) none of the mentioned

Answer: a

Explanation: The following types of stresses are induced in the rim of a flywheel:

1. Tensile stress due to centrifugal force,

2. Tensile bending stress caused by the restraint of the arms, and

3. The shrinkage stresses due to unequal rate of cooling of casting. These stresses may be very high but there is no easy method of determining. This stress is taken care of by a factor of safety.

7. The tensile stress in the flywheel rim due to the centrifugal force acting on the rim is given by

a) ρ v 2 /4

b) ρ v 2 /2

c) 3ρ v 2 /4

d) ρ v 2

Answer: d

Explanation: Tensile stress in the flywheel rim due to the centrifugal force = ρ v 2

where ρ = Density of the flywheel material, and

v = Linear velocity of the flywheel.

8. The cross-section of the flywheel arms is usually

a) elliptical

b) rectangular

c) I-section

d) L-section

Answer: a

Explanation: The cross-section of the arms is usually elliptical with major axis as twice the minor axis, and it is designed for the maximum bending stress.

9. In order to find the maximum bending moment on the arms, it is assumed as a

a) simply supported beam carrying a uniformly distributed load over the arm

b) fixed at both ends  and carrying a uniformly distributed load over the arm.

c) cantilever beam fixed at the hub and carrying a concentrated load at the free end of the rim

d) none of the mentioned

Answer: c

Explanation: Due to the torque transmitted from the rim to the shaft or from the shaft to the rim, the arms will be subjected to bending, because they are required to carry the full torque load. In order to find out the maximum bending moment on the arms, it may be assumed as a centilever beam fixed at the hub and carrying a concentrated load at the free end of the rim.

Answer: b

Explanation: The diameter of hub is usually taken as twice the diameter of shaft and length from 2 to 2.5 times the shaft diameter. It is generally taken equal to width of the rim.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Flywheel”.

1. In a four stroke I.C. engine, the turning moment during the compression stroke is

a) positive throughout

b) negative throughout

c) positive during major portion of the stroke

d) negative during major portion of the stroke

Answer: b

Explanation: Since the pressure inside the engine cylinder is less than the atmospheric pressure during the suction stroke, therefore a negative loop is formed. During the compression stroke, the work is done on the gases, therefore a higher negative loop is obtained.

2. The maximum fluctuation of energy is the

a) difference between the maximum and minimum energies

b) sum of the maximum and minimum energies

c) variations of energy above and below the mean resisting torque to the

d) ratio of the mean resisting torque to the workdone per cycle

Answer: a

Explanation: The difference between the maximum and the minimum energies is known as maximum fluctuation of energy.

3. The co-efficient of fluctuation of energy is the ratio of maximum energy to the minimum energy.

a) True

b) False

Answer: b

Explanation: The co-efficient of fluctuation of energy is the ratio of the maximum fluctuation of energy to the work done per cycle.

4. Which of the following statement is wrong?

a) The difference between the maximum and minimum energies is called maximum fluctuation of energy.

b) The co-efficient of fluctuation of speed is the ratio of maximum fluctuation of speed to the mean speed.

c) The variations of energy above and below the mean resisting torque line is known as fluctuation of energy.

d) None of the mentioned

Answer: d

Explanation: All the statements are correct.

The difference between the maximum and minimum energies is called maximum fluctuation of energy.

The co-efficient of fluctuation of speed is the ratio of maximum fluctuation of speed to the mean speed.

The variations of energy above and below the mean resisting torque line is known as fluctuation of energy.

5. The ratio of maximum fluctuation of energy to the workdone per cycle is called

a) fluctuation of energy

b) maximum fluctuation of energy

c) coefficient of fluctuation of speed

d) none of the mentioned

Answer: c

Explanation: The co-efficient of fluctuation of speed is the ratio of maximum fluctuation of speed to the mean speed.

6. Maximum fluctuation of energy in a flywheel is equal to

a) Iω(ω 1 – ω 2 )

b) Iω 2 C S

c) 2EC S

d) all of the mentioned

Answer: d

Explanation: None

7. A flywheel is fitted to the crankshaft of an engine having W as the amount of indicated work per revolution and permissible limits of coefficient of fluctuation of energy and speed as C E and C S respectively. The kinetic energy of the flywheel is given by

a) 2WC E /C S

b) WC E /2C S

c) WC E /C S

d) WC S /2C E

Answer: b

Explanation: None

8. If the rotating mass of a rim type flywheel is distributed on another rim type flywheel whose mean radius is half the mean radius of the former, then energy stored in the latter at the same speed will be

a) four times the first one

b) same as the first one

c) one fourth of the first one

d) one and a half times the first one

Answer: c

Explanation: None

9. The ratio of maximum fluctuation of speed to the mean speed is called

a) fluctuation of energy

b) maximum fluctuation of energy

c) coefficient of fluctuation of speed

d) none of the mentioned

Answer: c

Explanation: The co-efficient of fluctuation of speed is the ratio of maximum fluctuation of speed to the mean speed.

10. The flywheel of a machine having weight of 4500 N and radius of gyration of 2 m has cyclic fluctuation of speed from 125 r.p.m to 120 r.p.m. Assuming g = 10m/s 2 , the maximum fluctuation of energy is

a) 12822 N-m

b) 24200 N-m

c) 14822 N-m

d) 12100 N-m

Answer: d

Explanation: Mass of flywheel = weight of flywheel/Acceleration due to gravity = 4500/10kg

Moment of Inertia = mk 2

= 1800 kgm 2

ω 1 = 2π/60 x 125rad/sec

ω 2 = 2π/60 x 120rad/sec

E max = 1/2 I 2

= 12087.2 N-m

= 12100 Nm

11. A circular solid disc of uniform thickness 20 mm, radius 200 mm and mass 20 kg, is used as a flywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in Joules is

a) 395

b) 790

c) 1580

d) 3160

Answer: b

Explanation: For flywheel K.E = 1/2Iω 2

ω = 2πN/60 = 62.83 rad/s

I  = 1/2mR 2 = 0.4 kg m 2

Hence, K.E = 790 Joules.

12. The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kg/m2 is

a) 0.10

b) 0.20

c) 0.30

d) 0.40

Answer: a

Explanation: Given ω 1 = 210 rad/ sec, ω 2 = 190 rad/ sec, ΔE= 400 Nm

As the speed of flywheel changes from ω 1 to ω 2 , the maximum fluctuation of energy,

ΔE = 1/2I [(ω 1 ) 2 (ω 2 ) 2 ] I = 0.10 kgm 2

13. For a certain engine having an average speed of 1200 rpm, a flywheel approximated as a solid disc, is required for keeping the fluctuation of speed within 2% about the average speed. The fluctuation of kinetic energy per cycle is found to be 2 kJ. What is the least possible mass of the flywheel if its diameter is not to exceed 1 m ?

a) 40 kg

b) 51 kg

c) 62 kg

d) 73 kg

Answer: b

Explanation: Given N = 1200 rpm, ΔE = 2kJ = 2000 J, D = 1m, C s = 0.02

Mean angular speed of engine,

ω = 2πN/60 = 125.66 rad/ sec

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Energy Stored in a Flywheel”.

1. The maximum fluctuation of speed is the

a) difference of minimum fluctuation of speed and the mean speed

b) difference of the maximum and minimum speeds

c) sum of the maximum and minimum speeds

d) variations of speed above and below the mean resisting torque line

Answer: b

Explanation: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed.

The ratio of the maximum fluctuation of speed to the mean speed is called coefficient of fluctuation of speed.

2. The coefficient of fluctuation of speed is the _________ of maximum fluctuation of speed and the mean speed.

a) product

b) ratio

c) sum

d) difference

Answer: b

Explanation: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed.

The ratio of the maximum fluctuation of speed to the mean speed is called coefficient of fluctuation of speed.

3. In a turning moment diagram, the variations of energy above and below the mean resisting torque line is called

a) fluctuation of energy

b) maximum fluctuation of energy

c) coefficient of fluctuation of energy

d) none of the mentioned

Answer: a

Explanation: The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation.The variations of energy above and below the mean resisting torque line are called fluctuation of energy.

4. If E = Mean kinetic energy of the flywheel, C S = Coefficient of fluctuation of speed and Δ E = Maximum fluctuation of energy, then

a) ΔE = E / C S

b) ΔE = E 2 × C S

c) ΔE = E × C S

d) ΔE = 2 E × C S

Answer: d

Explanation: ΔE = Maximum K.E. — Minimum K.E.

ΔE = 2 E × C S

5. The ratio of the maximum fluctuation of energy to the ……. is called coefficient of fluctuation of energy.

a) minimum fluctuation of energy

b) workdone per cycle

c) maximum fluctuation of energy

d) none of the mentioned

Answer: b

Explanation: The difference between the maximum and the minimum energies is known as maximum fluctuation of energy.

6. Due to the centrifugal force acting on the rim, the flywheel arms will be subjected to

a) tensile stress

b) compressive stress

c) shear stress

d) none of the mentioned

Answer: a

Explanation: The tensile stress in the rim due to the centrifugal force, assuming that the rim is unstrained by the arms, is determined in a similar way as a thin cylinder subjected to internal pressure.

7. The tensile stress in the flywheel rim due to the centrifugal force acting on the rim is given by

a) ρ v 2 /4

b) ρ v 2 /2

c) 3ρ v 2 /4

d) ρ v 2

Answer: d

Explanation: Tensile stress = ρ v 2

where ρ = Density of the flywheel material, and

v = Linear velocity of the flywheel.

8. The cross-section of the flywheel arms is usually

a) elliptical

b) rectangular

c) I-section

d) L-section

Answer: a

Explanation: The cross-section of the arms is usually elliptical with major axis as twice the minor axis.

9. In order to find the maximum bending moment on the arms, it is assumed as a

a) simply supported beam carrying a uniformly distributed load over the arm

b) fixed at both ends

c) cantilever beam fixed at the hub and carrying a concentrated load at the free end of the rim

d) none of the mentioned

Answer: c

Explanation: Due to the torque transmitted from the rim to the shaft or from the shaft to the rim, the arms will be subjected to bending, because they are required to carry the full torque load. In order to find out the maximum bending moment on the arms, it may be assumed as a cantilever beam fixed at the hub and carrying a concentrated load at the free end of the rim.

Answer: b

Explanation: The diameter of hub is usually taken as twice the diameter of shaft and length from 2 to 2.5 times the shaft diameter. It is generally taken equal to width of the rim.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Flywheel in a Punching Press”.

1. In a punching press, which of the following quantity is constant?

a) Load

b) Torque

c) Angular velocity

d) Angle of rotation

Answer: b

Explanation: The punching press works on the principle that the input torque acting remains constant when the action is performed whereas the load is varied.

2. In a punching press, load is 0 at the time of punching.

a) True

b) False

Answer: b

Explanation: In a punching press, the load acts only during the time of punching and remains 0 for the rest of the time.

3. The maximum shear force required for punching depends on ________

a) Sheared area

b) Length of the plate

c) Speed of the flywheel

d) Total load

Answer: a

Explanation: Maximum shear force is given by the equation

Fs =  x ultimate shear stress.

4. A machine punching 38 mm holes in 32 mm thick plate requires 7 N-m of energy per sq. mm of sheared area, find the maximum shear force required.

a) 26.7 kN

b) 53.4 kN

c) 13.35 kN

d) 106.8 kN

Answer: a

Explanation: Given: d = 38 mm; t = 32 mm; 2 E1 = 7 N-m/mm of sheared area

A = π.38.32 mm 2

F = 7.A N = 26.7 kN.

5. The relation between stroke punch s and radius of crank r is ______

a) s=r

b) s=2r

c) s=4r

d) s=r/2

Answer: b

Explanation: Stroke punch is the distance travelled from top to bottom at the time of punching. Hence is equal to the diameter of the crank.

6. If the stroke punch is 100mm, find the radius of the crank in mm.

a) 200

b) 100

c) 50

d) 400

Answer: c

Explanation: Stroke punch is the distance travelled from top to bottom at the time of punching. Hence is equal to the diameter of the crank. Now s=100mm therefore r=s/2mm.

7. In a punching press, the requirement is to punch 40 mm diameter holes in a plate having a thickness of 15 mm. The rate at which the holes should be punched is 30 holes/min. Energy requirement is of 6 N-m per mm2 of sheared area. If the punching takes time 1/10 of a second and the speed of the flywheel varies from 160 to 140 rpm, determine the mass of the flywheel having radius of gyration of 1 m.

a) 327 Kg

b) 654 Kg

c) 163.5 Kg

d) 200 Kg

Answer: a

Explanation: N = /2 = 150 rpm

E1 = 11 310 N-m 

E2 = 565.5 N-m 

ΔE = 10744.5 N-m

therefore m = 327 Kg.

8. Energy during actual punching operation is same as the energy supplied by the motor.

a) True

b) False

Answer: b

Explanation: Energy during actual punching  operation is different as the energy supplied by the motor, E2 = E1x.

9. The balance energy required for punching is supplied by the flywheel by ________

a) Increase in its kinetic energy

b) Decrease in its kinetic energy

c) Decrease in its potential energy

d) By variation of mass

Answer: b

Explanation: The balance energy is to be supplied by the flywheel by the decrease in its kinetic energy when its speed falls from maximum to minimum.

10. When the length of the connecting rod is unknown then the value (θ 2 –θ 1 )/2π is equal to ________

a) t/s

b) t/2s

c) t/2r

d) t/r

Answer: b

Explanation: The values of θ 1 and θ 2 may be determined only if the crank radius , length of connecting rod  and the relative position of the job with respect to the crankshaft axis are known, in their absence we use the relation (θ 2 –θ 1 )/2π = t/2s.

This set of Machine Dynamics Quiz focuses on “Steam Engine Valves and Reversing Gears”.

1. In a steam engine, the distance by which the outer edge of the D-slide valve overlaps the steam port is called

a) lead

b) steam lap

c) exhaust lap

d) none of the mentioned

Answer: b

Explanation: Steam lap is the distance by which the outer edge of the D-slide valve overlaps the steam port.

2. In a steam engine, the distance by which the inner edge of the D-slide valve overlaps the steam port is called lead.

a) True

b) False

Answer: b

Explanation: In actual practise, the displacement of the D-slide valve is greater than the steam lap by a distance known as lead of the valve.

3. The displacement of the D-slide valve is ______________ the steam lap by a distance known as lead of the valve.

a) greater than

b) less than

c) equal to

d) none of the mentioned

Answer: a

Explanation: In actual practise, the displacement of the D-slide valve is greater than the steam lap by a distance known as lead of the valve.

4. The D-slide valve is also known as

a) inside admission valve

b) outside admission valve

c) piston slide valve

d) none of the mentioned

Answer: b

Explanation: Since the steam is admitted from outside the steam chest, therefore, D-slide valve is also known as outside admission valve.

5. The piston slide valve is an inside admission valve.

a) True

b) False

Answer: a

Explanation: Since the steam enters from the inside of the two pistons, therefore, the piston valve is also known as inside admission valve.

6. The distance by which the _____________ of the D-slide valve overlaps the steam port is called exhaust lap.

a) inner edge

b) outer edge

c) centre edge

d) none of the mentioned

Answer: a

Explanation: Steam lap or exhaust lap is the distance by which the outer edge of the D-slide valve overlaps the steam port.

7. Which of the following statement is correct?

a) The power absorbed in operating the piston valve is less than D-slide valve.

b) The wear of the piston valve is less than the wear of the D-slide valve.

c) The D-slide valve is also called outside admission valve.

d) all of the mentioned

Answer: d

Explanation: All the statements are correct.

8. The distance moved by the valve from one end to the other end is called valve travel.

a) True

b) False

Answer: a

Explanation: Valve travel is the distance moved by the valve from one end to the other end.

9. The throw of the eccentric is equal of the valve travel.

a) True

b) False

Answer: b

Explanation: The throw of the eccentric is equal to half of the valve travel.

Answer: b

Explanation: In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve is by increasing the angle of advance of the eccentric while the throw of the eccentric, steam lap and exhaust lap are kept constant. The method will reduce length of effective stroke of piston.

In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve may be obtained by increasing the angle of advance of the eccentric but reducing the throw of the eccentric and keeping the steam lap and exhaust lap constant. The method will cause withdrawing or throttling of steam.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Reversing Gears”.

1. In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve may be obtained by increasing the angle of advance of the eccentric but reducing the throw of the eccentric and keeping the steam lap and exhaust lap constant. The method will cause withdrawing or throttling of steam.

a) True

b) False

Answer: a

Explanation: In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve is by increasing the angle of advance of the eccentric while the throw of the eccentric, steam lap and exhaust lap are kept constant. The method will reduce length of effective stroke of piston.

In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve may be obtained by increasing the angle of advance of the eccentric but reducing the throw of the eccentric and keeping the steam lap and exhaust lap constant. The method will cause withdrawing or throttling of steam.

2. In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve is by increasing the steam lap and the angle of the eccentric but keeping constant the travel and lead of the valve. This method will

a) cause withdrawing or throttling of steam

b) reduce length of effective stroke of piston

c) reduce maximum opening opening of port to steam

d) all of the mentioned

Answer: c

Explanation: In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve is by increasing the steam lap and the angle of the eccentric but keeping constant the travel and lead of the valve. This method will reduce maximum opening opening of port to steam.

In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve is by increasing the angle of advance of the eccentric while the throw of the eccentric, steam lap and exhaust lap are kept constant. The method will reduce length of effective stroke of piston.

In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve may be obtained by increasing the angle of advance of the eccentric but reducing the throw of the eccentric and keeping the steam lap and exhaust lap constant. The method will cause withdrawing or throttling of steam.

3. In Meyer’s expansion valve, the main valve is driven by an eccentric having and angle of advance from

a) 10 0 -15 0

b) 15 0 -25 0

c) 25 0 -30 0

d) 30 0 -40 0

Answer: c

Explanation: In Meyer’s expansion valve, the main valve is driven by an eccentric having and angle of advance from 25 0 -30 0 and the expansion valve is driven by an eccentric having angle of advance 80 0 -90 0 .

4. The virtue or equivalent eccentric for the Meyer’s expansion valve is defined as an eccentric having such a length and angle of advance that will cause cut-off to take place at the same position, as is caused by the combined effect of main eccentric and expansion eccentric.

a) True

b) False

Answer: a

Explanation: None

5. In Meyer’s expansion valve, the expansion valve is driven by an eccentric having an angle of advance from

a) 50 0 -60 0

b) 60 0 -70 0

c) 70 0 -8 0

d) 80 0 -90 0

Answer: d

Explanation: In Meyer’s expansion valve, the main valve is driven by an eccentric having and angle of advance from 25 0 -30 0 and the expansion valve is driven by an eccentric having angle of advance 80 0 -90 0 .

6. The function of a reversing gear in a steam engine is

a) to control the supply of steam

b) to alter the point of cut-off while the engine is running

c) to reverse the direction of motion of the crank shaft

d) both b and c

Answer: d

Explanation: The function of a reversing gear in a steam engine is to alter the point of cut-off while the engine is running and to reverse the direction of motion of the crank shaft.

7. When the load on the engine increases, it becomes necessary to increase the supply of working fluid and when the load decreases, less working fluid is required. The supply of the working fluid to the engine is controlled by a

a) D-slide valve

b) governor

c) Meyer’s expansion valve

d) flywheel

Answer: b

Explanation: When the load on the engine increases, it becomes necessary to increase the supply of working fluid and when the load decreases, less working fluid is required. The supply of the working fluid to the engine is controlled by a governor.

Flywheel controls the speed variation caused by the fluctuations of the engine turning moment during each cycle of operation.

8. A flywheel is used to control the mean speed of an engine caused by the variations in load.

a) True

b) False

Answer: b

Explanation: The function of a governor is to regulate the mean speed of an engine caused by the variations in load.

Answer: c

Explanation: The function of a governor is to regulate the mean speed of an engine caused by the variations in load.

Flywheel controls the speed variation caused by the fluctuations of the engine turning moment during each cycle of operation.

This set of Machine Kinematics Multiple Choice Questions & Answers  focuses on “Governors”.

1. The height of a Watt’s governor  in equal to

a) 8.95/N 2

b) 89.5/N 2

c) 895/N 2

d) 8950/N 2

Answer: c

Explanation: Height of a Watt governor = g/ω 2

= 9.81/ 2

= 895/N 2

where N = Speed of the arm and ball about the spindle axis.

2. The ratio of the height of a Porter governor  to the height of a Watt’s governor is

a) m/m+M

b) M/m+M

c) m + M/m

d) m + M/M

Answer: c

Explanation: Mass of the central load  increases the height of governor in the ratio m + M/m

where m = Mass of the ball, and

M = Mass of the load on the sleeve.

3. When the sleeve of a Porter governor moves upwards, the governor speed

a) increases

b) decreases

c) remains unaffected

Answer: a

Explanation: When the loaded sleeve moves up and down the spindle, the frictional force acts on it in a direction opposite to that of the motion of sleeve. The + sign is used when the sleeve moves upwards or the governor speed increases and negative sign is used when the sleeve moves downwards or the governor speed decreases.

4. A Hartnell governor is a

a) pendulum type governor

b) spring loaded governor

c) dead weight governor

d) inertia governor

Answer: b

Explanation: Since in hartnell governor spring is used, therefore it is the spring loaded governor.

5. Which of the following governor is used to drive a gramophone ?

a) Watt governor

b) Porter governor

c) Pickering governor

d) Hartnell governor

Answer: c

Explanation: A Pickering governor is mostly used for driving gramophone. It consists of three straight leaf springs arranged at equal angular intervals round the spindle. Each spring carries a weight at the centre. The weights move outwards and the springs bend as they rotate about the spindle axis with increasing speed.

6. Which of the following is a spring controlled governor?

a) Hartnell

b) Hartung

c) Pickering

d) all of the mentioned

Answer: d

Explanation: None

7. For two governors A and B, the lift of sleeve of governor A is more than that of governor B, for a given fractional change in speed. It indicates that

a) governor A is more sensitive than governor B

b) governor B is more sensitive than governor A

c) both governors A and B are equally sensitive

d) none of the mentioned

Answer: a

Explanation: The greater the lift of the sleeve corresponding to a given fractional change in speed, the greater is the sensitiveness of the governor. It may also be stated in another way that for a given lift of the sleeve, the sensitiveness of the governor increases as the speed range decreases.

8. The sensitiveness of a governor is given by

a) ω mean /ω 2 – ω 1

b) ω 2 – ω 1 / ω mean

c) ω 2 – ω 1 / 2ω mean

d) none of the mentioned

Answer: b

Explanation: Let N 1 = Minimum equilibrium speed,

N 2 = Maximum equilibrium speed, and

N = Mean equilibrium speed = N 1 + N 2 / 2

Sensitiveness of the governor = N 1 – N 2 / 2

= 2(N 1 + N 2 ) / N 1 + N 2

= ω 2 – ω 1 / ω mean

where ω 1 and ω 2 = Minimum and maximum angular speed, and

ω mean = Mean angular speed.

9. In a Hartnell governor, if a spring of greater stiffness is used, then the governor will be

a) more sensitive

b) less sensitive

c) isochronous

d) none of the mentioned

Answer: b

Explanation: Stiffness is directly proportional to sensitiveness. Therefore, it stifness is greater, the governor will be less sensitive.

10. A governor is said to be hunting, if the speed of the engine

a) remains constant at the mean speed

b) is above the mean speed

c) is below the mean speed

d) fluctuates continuously above and below the mean speed

Answer: d

Explanation: A governor is said to be hunt if the speed of the engine fluctuates continuously above and below the mean speed. This is caused by a too sensitive governor which changes the fuel supply by a large amount when a small change in the speed of rotation takes place.

11. A hunting governor is

a) more stable

b) less sensitive

c) more sensitive

d) none of the mentioned

Answer: c

Explanation: None

12. Isochronism in a governor is desirable when

a) the engine operates at low speeds

b) the engine operates at high speeds

c) the engine operates at variable speeds

d) one speed is desired under one load

Answer: d

Explanation: The isochronous governor is not of practical use because the sleeve will move to one of its extreme positions immediately the speed deviates from the isochronous speed.

13. The power of a governor is equal to

a) (c 2 /1 + 2c ) h

b) (2c 2 /1 + 2c ) h

c) (3c 2 /1 + 2c ) h

d) (4c 2 /1 + 2c ) h

Answer: d

Explanation: None

14. When the relation between the controlling force (F C ) and radius of rotation  for a spring controlled governor is F C = a.r + b, then the governor will be

a) stable

b) unstable

c) isochronous

d) none of the mentioned

Answer: b

Explanation: For the governor to be stable, the controlling force (F C ) must increase as the radius of rotation  increases, i.e. F C / r must increase as r increases.

Answer: a

Explanation: None

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Types of Governors”.

1. The height of a Watt’s governor is equal to

a) 8.95/N 2

b) 89.5/N 2

c) 895/N 2

d) 8950/N 2

Answer: c

Explanation: If N is the speed of the arm and ball about the spindle axis, then the height of the governor  is given by

h = 895/N 2 metres

2. The height of a Watt’s governor is

a) directly proportional to speed

b) directly proportional to 

c) inversely proportional to speed

d) inversely proportional to 

Answer: d

Explanation: If N is the speed of the arm and ball about the spindle axis, then the height of the governor  is given by

h = 895/N 2 metres

From this expression, we see that the height height of a Watt’s governor is inversely proportional to N 2

3. A Watt’s governor can work satisfactorily at speeds from

a) 60 to 80 r.p.m

b) 80 to 100 r.p.m

c) 100 to 200 r.p.m

d) 200 to 300 r.p.m

Answer: a

Explanation: A watt’s governor may only work satisfactorily at low speeds i.e. from 60 to 80 r.p.m.

4. The ratio of height of Porter governor to the height of Watt’s governor is

a) m/m + M

b) M/ m + M

c) m + M/m

d) m + M/M

Answer: c

Explanation: The ratio of height of a Porter governor  to the height of Watt’s governor is m + M/n, where m and M re the masses of the ball and sleeve respectively.

5. When the sleeve of a porter governor moves upwards, the governor speed

a) increases

b) decreases

c) remains unaffected

d) first increases and then decreases

Answer: a

Explanation: When the sleeve of a porter governor moves upwards, the governor speed increases and when the sleeve moves downwards, the governor speed decreases.

6. When the sleeve of a Porter governor moves downwards, the governor speed

a) increases

b) decreases

c) remains unaffected

d) first increases and then decreases

Answer: b

Explanation: When the sleeve of a porter governor moves upwards, the governor speed increases and when the sleeve moves downwards, the governor speed decreases.

7. In a Porter governor, the balls are attached to the extension of lower links.

a) True

b) False

Answer: a

Explanation: Porter governor is a modification of Watt’s governor, with a central load attached to the sleeve.

8. A Hartnell governor is a

a) dead weight governor

b) pendulum type governor

c) spring loaded governor

d) inertia governor

Answer: c

Explanation: A Hartnell governor is a spring loaded governor.

Watt’s governor is a pendulum type governor.

9. A Watt’s governor is a spring loaded governor.

a) True

b) False

Answer: b

Explanation: A Hartnell governor is a spring loaded governor.

Watt’s governor is a pendulum type governor.

Answer: a

Explanation: Watt’s governor is a pendulum type governor.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Watt Governor”.

1. In a Hartnell governor, the lift of the sleeve is given by

a) (r 1 + r 2 y/x

b) (r 1 + r 2 x/y

c) (r 1 – r 2 y/x

d) (r 1 – r 2 x/y

Answer: c

Explanation: h = (r 1 – r 2 y/x

and stiffness of the spring, s = S 2 – S 1 /h

2. In a Hartnell governor, the compression of the spring is ______________ the lift of the sleeve.

a) equal to

b) less than

c) greater than

d) none of the mentioned

Answer: a

Explanation: None

3. In a Hartnell governor, the stiffness of the spring is given by

a) S 1 + S 2 /h

b) S 2 – S 1 /h

c) S 1 + S 2 /2h

d) S 1 – S 2 /2h

Answer: b

Explanation: h = (r 1 – r 2 y/x

and stiffness of the spring, s = S 2 – S 1 /h

4. Which of the following is a spring controlled governor?

a) Hartnell governor

b) Hartung governor

c) Wilson-Hartnell governor

d) All of the mentioned

Answer: d

Explanation: None

5. A pendulum type governor is a Watt governor.

a) True

b) False

Answer: a

Explanation: A Hartnell governor is a spring loaded governor.

Watt’s governor is a pendulum type governor.

6. A Porter governor is a

a) dead weight governor

b) pendulum type governor

c) spring loaded governor

d) inertia governor

Answer: b

Explanation: A Hartnell governor is a spring loaded governor.

Watt’s governor is a pendulum type governor.

As Porter governor is the modification of Watt’s governor, so it is also a pendulum type governor.

7. Which of the following governor is used to drive a gramophone?

a) Watt’s governor

b) Porter governor

c) Pickering governor

d) Hartnell governor

Answer: c

Explanation: A Pickering governor is mostly used for driving gramophone. It consists of three straight leaf springs arranged at equal angular intervals round the spindle. Each spring carries a weight at the centre.

8. The sensitiveness of a governor depends upon the lift of the sleeve.

a) True

b) False

Answer: a

Explanation: In general, the greater the lift of the sleeve corresponding to a given fractional change in speed, the greater is the sensitiveness of the governor.

9. For two governors A and B, the lift of sleeve of governor A is more than that of governor B, for a given fractional change in speed. It indicates that

a) governor A is more sensitive than governor B

b) governor B is more sensitive than governor A

c) both governors A and B are equally sensitive

d) none of the mentioned

Answer: a

Explanation: In general, the greater the lift of the sleeve corresponding to a given fractional change in speed, the greater is the sensitiveness of the governor.

Answer: c

Explanation: For a given lift of the sleeve, the sensitiveness of the governor increases as the speed range decreases.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Porter Governor”.

1. Sensitiveness of the governor is defined as the ratio of the

a) mean speed to the maximum equilibrium speed

b) mean speed to the minimum equilibrium speed

c) difference of the maximum and minimum equilibrium speeds to the mean speed

d) sum of the maximum and minimum equilibrium speeds to the mean speed

Answer: c

Explanation: The sensitiveness is defined as the ratio of the difference between the maximum and minimum equilibrium speeds to the mean equilibrium speed.

2. A governor is said to be stable, if the

a) radius of rotation of balls increases as the equilibrium speed decreases

b) radius of rotation of balls decreases as the equilibrium speed decreases

c) radius of rotation of balls increases as the equilibrium speed increases

d) radius of rotation of balls decreases as the equilibrium speed increases

Answer: c

Explanation: A governor is said to be stable when for every speed within the working range there is a definite configuration i.e. there is only one radius of rotation of the governor balls at which the governor is in equilibrium. For a stable governor, if the equilibrium speed increases, the radius of governor balls must also increase.

3. When the radius of rotation of balls ______________ as the equilibrium speed increases, the governor is said to be unstable.

a) remains constant

b) decreases

c) increases

d) none of the mentioned

Answer: b

Explanation: A governor is said to be unstable, if the radius of rotation decreases as the speed increases.

4. A governor is said to be isochronous when range of speed is zero for all radii of rotation of the balls within the working range, neglecting friction.

a) True

b) False

Answer: a

Explanation: A governor is said to be isochronous when the equilibrium speed is constant  for all radii of rotation of the balls within the working range, neglecting friction. The isochronism is the stage of infinite sensitivity.

5. A Porter governor can not be isochronous.

a) True

b) False

Answer: a

Explanation: None

6. In a Hartnell governor, if a spring of greater stiffness is used, then the governor will be

a) less sensitive

b) more sensitive

c) unaffected of sensitivity

d) isochronous

Answer: a

Explanation: In general, the greater the lift of the sleeve corresponding to a given fractional change in speed, the greater is the sensitiveness of the governor.

7. When the speed of the engine fluctuates continuously above and below the mean speed, the governor is said to be

a) stable

b) unstable

c) isochronous

d) hunt

Answer: d

Explanation: The isochronous governor is not of practical use because the sleeve will move to one of its extreme positions immediately the speed deviates from the isochronous speed.

8. A very sensitive governor will cause hunting.

a) True

b) False

Answer: a

Explanation: A governor is said to be hunt if the speed of the engine fluctuates continuously above and below the mean speed. This is caused by a too sensitive governor which changes the fuel supply by a large amount when a small change in the speed of rotation takes place.

9. For isochronous Hartnell governor

a) mg + S 1 / mg + S 2 = r 1 /r 2

b) mg – S 1 / mg – S 2 = r 2 /r 1

c) S 1 /S 2 = r 1 /r 2

d) S 2 /S 1 = r 1 /r 2

Answer: a

Explanation: None

Answer: a

Explanation: The effort of a governor is the mean force exerted at the sleeve for a given percentage change of speed .

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Hartnell Governor”.

1. Power of a governor is the

a) mean force exerted at the sleeve for a given percentage change of speed

b) workdone at the sleeve for maximum equilibrium speed

c) mean force exerted at the sleeve for maximum equilibrium speed

d) none of the mentioned

Answer: d

Explanation: The power of a governor is the work done at the sleeve for a given percentage change of speed. It is the product of the mean value of the effort and the distance through which the sleeve moves. Mathematically,

Power = Mean effort × lift of sleeve

2. The effort of a Porter governor is equal to

a) cg

b) cg

c) C/g

d c/g

Answer: b

Explanation: The effort of a Porter governor is equal to cg.

The power of a Porter governor is equal to 4c 2 /1 + 2c gh.

3. The power of a Porter governor is equal to

a) c 2 /1 + 2c gh

b) 2c 2 /1 + 2c gh

c) 3c 2 /1 + 2c gh

d) 4c 2 /1 + 2c gh

Answer: d

Explanation: The effort of a Porter governor is equal to cg.

The power of a Porter governor is equal to 4c 2 /1 + 2c gh.

4. For the isochronous Porter governor, the controlling force curve is a straight line passing through the origin.

a) True

b) False

Answer: a

Explanation: For the isochronous governor, the controlling force curve is a straight line passing through the origin. The angle φ will be constant for all values of the radius of rotation of the governor.

5. The controlling force diagram for a spring controlled governor is a curve passing through the origin.

a) True

b) False

Answer: b

Explanation: The controlling force diagram for the spring controlled governors is a straight line.

6. A spring controlled governor is said to be unstable when the controlling force

a) increases as the radius of rotation decreases

b) increases as the radius of rotation increases

c) decreases as the radius of rotation decreases

d) remains constant for all radii of rotation

Answer: c

Explanation: If the equilibrium speed of the governor decreases with an increase of the radius of rotation of balls, then the governor is said to be unstable.

7. In a spring controlled governor, when the controlling force _____________ as the radius of rotation increases, it is said to be a stable governor.

a) remains constant

b) decreases

c) increases

d) none of the mentioned

Answer: c

Explanation: For the governor to be stable, in spring controlled governor, the controlling force must increase as the radius of rotation increases.

A spring controlled governor is said to be isochronous when the controlling force remains constant for all radii of rotation.

8. A spring controlled governor is said to be isochronous when the controlling force

a) increases as the radius of rotation decreases

b) increases as the radius of rotation increases

c) decreases as the radius of rotation decreases

d) remains constant for all radii of rotation

Answer: d

Explanation: For the governor to be stable, in spring controlled governor, the controlling force must increase as the radius of rotation increases.

A spring controlled governor is said to be isochronous when the controlling force remains constant for all radii of rotation.

9. A spring controlled governor is found unstable. It can be made stable by

a) increasing the spring stiffness

b) decreasing the spring stiffness

c) increasing the ball mass

d) decreasing the ball mass

Answer: b

Explanation: For the governor to be stable, in spring controlled governor, the controlling force must increase as the radius of rotation increases.

Answer: b

Explanation: A spring controlled governor is said to be stable if the controlling force line when produced intersects the Y-axis below the origin.

If the controlling force line for a spring controlled governor when produced intersects the Y-axis at the origin, then the governor is said to be isochronous.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Hartung Governor”.

1. If the controlling force line for a spring controlled governor when produced intersects the Y-axis at the origin, then the governor is said to be

a) stable

b) unstable

c) isochronous

d) none of the mentioned

Answer: c

Explanation: If the controlling force line for a spring controlled governor when produced intersects the Y-axis at the origin, then the governor is said to be isochronous.

A spring controlled governor is said to be stable if the controlling force line when produced intersects the Y-axis below the origin.

If the controlling force line for a spring controlled governor when produced intersects the Y-axis at the origin, then the governor is said to be unstable.

2. If the controlling force line for a spring controlled governor when produced intersects the Y-axis at the origin, then the governor is said to be unstable.

a) True

b) False

Answer: a

Explanation: A spring controlled governor is said to be stable if the controlling force line when produced intersects the Y-axis below the origin.

If the controlling force line for a spring controlled governor when produced intersects the Y-axis at the origin, then the governor is said to be unstable.

3. The relation between the controlling force (F c ) and radius of rotation  for a stable spring controlled governor is

a) F c = ar + b

b) F c = ar – b

c) F c = ar

d) F c = a/r + b

Answer: b

Explanation: The relation between the controlling force (F C ) and the radius of rotation  for the stability of spring controlled governors is given by the following equation

F C = a.r – b

4. A spring controlled governor is said to be unstable, if the relation between the controlling force (F c ) and radius of rotation is F c = ar

a) True

b) False

Answer: b

Explanation: A governor is said to be unstable and the relation between the controlling force and the radius of

rotation is, therefore

F C = a.r + b

5. When the relation between the controlling force (F c ) and radius of rotation  for a spring controlled governor is F c = ar + b, then the governor will be

a) stable

b) unstable

c) isochronous

d) none of the mentioned

Answer: b

Explanation: A governor is said to be unstable and the relation between the controlling force and the radius of

rotation is, therefore

F C = a.r + b

6. A Hartnell governor has its controlling force (F c ) given by F c = ar + b, where r is the radius of rotation and a and b are constants. The governor becomes isochronous when

a) a is + ve and b = 0

b) a = 0 and b is +ve

c) a is +ve and b is -ve

d) a is +ve and b is also +ve

Answer: a

Explanation: None

7. Calculate the vertical height of a Watt governor when it rotates at 60 r.p.m.

a) 0.248 m

b) 0.248 m

c) 0.448 m

d) 0.548 m

Answer: a

Explanation: Given : N 1 = 60 r.p.m. ; N 2 = 61 r.p.m.

Initial height

We know that initial height,

h 1 = 895/(N 1 ) 2

= 895/60 2 = 0.248 m

8. The power of a governor is the work done at

a) the governor balls for change of speed

b) the sleeve for zero change of speed

c) the sleeve for a given rate of change of change

d) each governor ball for given percentage change of speed

Answer: c

Explanation: Power of Governor: The work done by the governor on the sleeve to its equilibrium position for the fractional change in speed of governor is known as power of governor.

It is actually a work done. Power = Main force × Sleeve movement.

9. In a governor, if the equilibrium speed is constant for all radii of rotation of balls, the governor is said to be

a) Stable

b) unstable

c) inertial

d) isochronous

Answer: d

Explanation: The governor is said to be Isochronous if the equilibrium speed is constant for all radii of rotation of balls.

Answer: b

Explanation: Isochronism in governor means constant equilibrium speed for all the radii of rotation.

This set of Machine Dynamics MCQs focuses on “Wilson-Hartnell and Pickering Governor”.

1. The power of a governor is the work done at

a) the governor balls for change of speed

b) the sleeve for zero change of speed

c) the sleeve for a given rate of change of change

d) each governor ball for given percentage change of speed

Answer: c

Explanation: Power of Governor: The work done by the governor on the sleeve to its equilibrium position for the fractional change in speed of governor is known as power of governor.

It is actually a work done. Power = Main force × Sleeve movement.

2. In a governor, if the equilibrium speed is constant for all radii of rotation of balls, the governor is said to be

a) Stable

b) unstable

c) inertial

d) isochronous

Answer: d

Explanation: The governor is said to be Isochronous if the equilibrium speed is constant for all radii of rotation of balls.

3. A governor is said to be isochronous when the equilibrium speed is

a) variable for different radii of rotation of governor balls

b) constant for all radii of rotation of the balls within the working range

c) constant for particular radii of rotation of governor balls

d) constant for only one radius of rotation of governor balls

Answer: b

Explanation: Isochronism in governor means constant equilibrium speed for all the radii of rotation.

4. A centrifugal Watt governor is fitted with two balls, each of mass 2.5 kg. Find the height of the governor, when it is running at 75 r.p.m.

a) 139 mm

b) 149 mm

c) 159 mm

d) 169 mm

Answer: c

Explanation: Mass of flyballs = 2.5 kg and N speed of governor = 75 r.p.m.

So the angular velocity of the governor ω = 2пN/60 = 2п75/60 = 2.5п rad/s

Height of the governor, h = g/ω 2 = 9.81/2.5п 2 = 0.159m = 159 mm.

5. A centrifugal Watt governor is fitted with two balls, each of mass 2.5 kg. Find the sped of the governor, when the balls rise by 20 mm.

a) 80.2 r.p.m.

b) 90.2 r.p.m.

c) 100.2 r.p.m.

d) 110.2 r.p.m.

Answer: a

Explanation: Mass of flyballs = 2.5 kg and N speed of governor = 75 r.p.m.

So the angular velocity of the governor ω = 2пN/60 = 2п75/60 = 2.5п rad/s

Height of the governor, h = g/ω 2 = 9.81/2.5п 2 = 0.159m = 159 mm

Speed of the governor when balls rise by 20 mm

The height of the governor h reduces to h 1 = 159 – 20 = 139 mm = 0.139 m

Hence, the angular velocity ω 1 corresponding to height h 1 ,

ω 1 2 = g/h 1 = 9.81/0.139 = 70.5

ω 1 = 8.4 rad/s

Speed in r.p.m., N 1 = 60ω 1 /2п = 60 x 8.4/2п = 80.2 r.p.m.

6. A centrifugal Watt governor is fitted with two balls, each of mass 2.5 kg. Find the sped of the governor, when the balls fall by 20 mm.

a) 60 r.p.m.

b) 70 r.p.m.

c) 70.2 r.p.m.

d) 80 r.p.m.

Answer: c

Explanation: Mass of flyballs = 2.5 kg and N speed of governor = 75 r.p.m.

So the angular velocity of the governor ω = 2пN/60 = 2п75/60 = 2.5п rad/s

Height of the governor, h = g/ω 2 = 9.81/2.5п 2 = 0.159m = 159 mm

Speed of the governor when balls fall by 20 mm

The height of the governor h increases to h 2 = 159 + 20 = 179 mm = 0.179 m

Hence, the angular velocity ω 2 corresponding to height h 2 ,

ω 2 2 = g/h 2 = 9.81/0.179 = 54.7

ω 2 = 7.4 rad/s

Speed in r.p.m., N 2 = 60ω 2 /2п = 60 x 7.4/2п = 70.7 r.p.m.

7. It is required to lift water at the top of a high building at the rate of 50 litres/min. Assuming the losses due to friction equivalent to 5 m and those due to leakage equal to 5 m head of water, efficiency of pump 90%, decide the kilowatt capacity of motor required todrive the pump.

a) 0.102 kW

b) 0.202 kW

c) 0.302 kW

d) 0.402 kW

Answer: c

Explanation: Work to be done = wQHg

w = 1 Kg/litre

Q = 50 litres per min

H = 20 + 5 + 5 = 30m

Work output/min = 1 x 50 x 30 x 9.81 Nm/min

Input power = 1 x 50 x 30 x 9.81/ 0.9 x 0.9 x1000 x 60

= 0.302 kW.

8. A power screw is rotated at constant angular speed of 1.5 revolutions/sec by applying a steady torque of 1.5Nm. How much work is done per revolution? What is the power required?

a) 141.36 W

b) 241.36 W

c) 341.36 W

d) 441.36 W

Answer: a

Explanation: Work per revolution = Tϴ

= 15 x 2п

= 94.24 Nm per revolution

P = Work/sec

= 94.24 x 1.5 = 141.36 W.

9. Power is transmitted by an electric motor to a machine by using a belt drive. The tensions on the tight and slack side of the belt are 2200 N and 1000 N respectively and diameter of the pulley is 600 mm. If speed of the motor is 1500 r.p.m, find the power transmitted.

a) 46.548 kW

b) 56.548 kW

c) 66.548 kW

d) 76.548 kW

Answer: b

Explanation: Power = Tω

= Frω = Fv

F = T 1 – T 2 = 2200 – 1000 = 1200N

v = пDN/60 = п x 600 x 1500/1000 x 60 = 47.12 m/s

P = 1200 x 47.12/1000 = 56.548 kW.

10. The flywheel of an engine has a mass of 200 kg and radius of gyration equal to 1 m. The average torque on the flywheel is 1200 Nm. Find the angular acceleration of flywheel and the angular speed after 10 seconds starting from rest.

a) 3 rad/s

b) 4 rad/s

c) 5 rad/s

d) 6 rad/s

Answer: d

Explanation: I = mk 2 = 2000 kgm 2

T = Iα

α = T/I = 1200/2000 = 0.6 rad/s 2

ω = ω 0 + αt = 0 + 0.6 x 10 = 6 rad/s.

11. Calculate the moment of inertia and radius of gyration of a solid sphere of mass 10 kg and diameter 6.5m about its centroidal axis.

a) 2.055 m

b) 3.055 m

c) 4.055 m

d) 5.055 m

Answer: a

Explanation: I = 2/5 mr 2 = 49 kgm 2

k = 0.6325 x 3.25 = 2.055 m.

Answer: a

Explanation: Force required to punch one hole = area shared x ultimate shear strength = пdt x S s

S s = п x 30 x 5 x 450/1000 = 212.05 kN

Work done/min = Average force x Thickness of plate x No. of holes/min

= 212.05/2 x 5/1000 x 20 = 10.69 kNm.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Pickering Governor”.

1. For driving a gramophone, which of the following governor is used?

a) Watt governor

b) Hartnell governor

c) Porter governor

d) Pickering governor

Answer: d

Explanation: For driving a gramophone a Pickering governor is used. It consists of three straight leaf springs arranged at equal angular intervals round the spindle. Its speed is controlled by dissipating excess kinetic energy.

2. The weights move inwards and the springs bend as they rotate about the spindle axis with increasing speed.

a) True

b) False

Answer: b

Explanation: For a Pickering governor, the weights move outwards and the springs bend as they rotate about the spindle axis with increasing speed. Hence, the given statement is false.

3. A Pickering governor is driving a gramophone. Each disc attached to the centre of a leaf spring has a mass of 20 g. The width of each spring is 5 mm and thickness of 0.125 mm. The effective length of each spring is 40 mm. The distance from the spindle axis to the centre of gravity of the mass when the governor is at rest, is 10 mm. Find the speed of the governor in rpm when the sleeve has risen to a height of 0.8 mm. (E = 210kN/mm 2 )

a) 25.5

b) 2.43

c) 51.0

d) 12.25

Answer: a

Explanation: I = bt 3 /12

=0.0008mm 2

Length between fixed ends 

Calculating delta from central deflection will give

δ = 3.65 mm

3.65=0.51ω 2

ω=2.675 rad/s.

4. A Pickering governor is driving a gramophone. Each disc attached to the centre of a leaf spring has a mass of 20 g. The width of each spring is 5 mm and thickness of 0.125 mm. The effective length of each spring is 40 mm. The distance from the spindle axis to the centre of gravity of the mass when the governor is at rest, is 10 mm, find the speed of the turntable in rpm if ratio of the governor speed to the turntable speed is 10.5.

a) 25.5

b) 2.43

c) 51.0

d) 12.25

Answer: b

Explanation: I = bt 3 /12

= 0.0008 (mm 2 ) 2

Length between fixed ends 

Calculating delta from central deflection will give

δ = 3.65 mm

3.65=0.51ω 2

ω=2.675 rad/s

N = 25.5 rpm

N1 = 2.43 rpm.

5. From the following data calculate the speed of governor in rpm.

Dimensions of spring = 6mm wide, 0.12 mm thick and length = 48 mm

mass attached to each leaf at centre = 25g

Distance between spindle axis and C.O.M when the governor is at rest = 8mm

Sleeve lift = 0.6mm

E = 200 GN/mm 2

a) 575

b) 625

c) 600

d) 550

Answer: a

Explanation: I = bt 3 /12

= 0.864×10 -15 (m 2 ) 2

δ = 0.003464 m

= Fl 3 /192EI

ω=60.22 rad/s

N=575 rpm.

6. From the data given:

Dimensions of spring = 6mm wide, 0.12 mm thick and length = 48 mm

mass attached to each leaf at centre = 25g

Distance between spindle axis and C.O.M when the governor is at rest = 8mm

Sleeve lift = 0.6mm

E= 200 GN/mm 2

Find the speed of the turntable in rpm if ratio of the governor speed to the turntable speed is 10.

a) 57.5

b) 62.5

c) 60

d) 55

Answer: a

Explanation: I = bt 3 /12

= 0.864×10 -15 (m 2 ) 2

δ = 0.003464 m

= Fl 3 /192EI

ω=60.22 rad/s

N=575 rpm

N 1 = N/10

= 57.5 rpm.

7. In case of pickering governor, the central load is_______

a) Centripetal force

b) Centrifugal force

c) Torque

d) Normal reaction

Answer: b

Explanation: W = FC = m.ω 2 , this is the expression of the central load, hence the central load is the centrifugal force.

8. Porter governor is used to vary the speed of the turntable in a gramophone.

a) True

b) False

Answer: b

Explanation: Pickering governor is used to vary the speed of the turntable in a gramophone. Porter governor is used to regulate the speed of the engine.

9. Which of the following part of the spring is attached to the sleeve?

a) Upper end

b) Lower end

c) Middle part

d) At a height 3 ⁄ 4 of total length

Answer: b

Explanation: The working of a Pickering governor includes: attachment of the lower end of the spring to the sleeve which is free to slide on the spindle. The spindle runs in a bearing at each end and is driven through gearing by the motor.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Sensitiveness and Stability of Governors”.

1. For a given fraction of change in speed, a more sensitive governor will have a ______

a) Higher lift

b) Lower lift

c) More effective length

d) Less effective length

Answer: a

Explanation: For an independent governor, the greater the lift of the sleeve corresponding to a given fractional change in speed, the greater is the sensitiveness of the governor. Hence the sleeve lift is directly proportional to the fractional change in speed.

2. For a given lift of the sleeve, the sensitiveness of the governor increases as the speed range decreases

a) True

b) False

Answer: a

Explanation: The greater the lift of the sleeve corresponding to a given fractional change in speed, the greater is the sensitiveness of the governor, this is for an independent governor. Hence the given statement is true.

3. The equilibrium speed of a governor varies from 30 rpm to 20 rpm, find the sensitiveness of the governor.

a) 0.4

b) 1.5

c) 0.66

d)1.25

Answer: a

Explanation: Mean speed = /2 = 25 rpm

Difference in max and min speed = 30-20 = 10 rpm

sensitiveness = 10/25 = 0.4.

4. When a governor is fitted into an engine, then sensitiveness is defined as the ratio of ________ of maximum and minimum speed to the mean speed.

a) Sum

b) Difference

c) Product

d) Root mean square

Answer: b

Explanation: For an independent governor, steadiness can be defined in terms of sleeve lift and change in speed, but however when the governor is fitted into an engine then the sensitiveness is defined as the ratio of the difference between the maximum and minimum equilibrium speeds to the mean equilibrium speed.

5. Consider two governors A and B running at the same speed. When this speed increases or decreases by a certain amount, the lift of the sleeve of governor A is greater than the lift of the sleeve of governor B. Which of the following statement is true for the above situation?

a) Governor A is more sensitive than B

b) Governor B is more sensitive than A

c) Governor A is equally sensitive as B

d) Lift does not affect sensitivity

Answer: a

Explanation: For an independent governor, the greater the lift of the sleeve for a given fractional change in speed, the greater will be the sensitiveness of the governor. As a consequence, we can establish a relation in which the sensitiveness of the governor is related to sleeve lift and fractional change in speed.

6. For a governor, the sensitivity is 0.5 and the mean speed is 20 rpm, find the maximum speed.

a) 15 rpm

b) 25 rpm

c) 30 rpm

d) 40 rpm

Answer: b

Explanation: 20 = (N 1 +N 2 )/2 :: 40 = N 1 + N 2

0.5×20 = N 1 -N 2

therefore N 1 = 25 rpm.

7. For a governor, the sensitivity is 0.5 and the mean speed is 20 rpm, find the minimum speed.

a) 15 rpm

b) 25 rpm

c) 30 rpm

d) 40 rpm

Answer: a

Explanation: 20 = (N 1 +N 2 )/2 :: 40 = N 1 + N 2

0.5×20 = N 1 -N 2

10 = N 1 -N 2

therefore N 1 = 25 rpm

N 2 = 15 rpm.

8. For a stable governor, if the equilibrium speed increases, the radius of governor balls must also increase.

a) True

b) False

Answer: a

Explanation: For a stable governor: the radius of governor balls increases with increase in equilibrium speed, a governor is said to be unstable if the radius of rotation decreases with increase in speed.

9. On which of the following factors, the stability of governor does not depend?

a) Speed

b) Radius of rotation

c) Governor equilibrium

d) Mass of the engine

Answer: d

Explanation: For the stability of the governor, with the increase in the equilibrium speed, the radius of governor balls should also increase, a governor is said to be unstable if the radius of rotation decreases with increase in speed.

10. When the governor is too sensitive, then which of the following process occurs?

a) Hunting

b) More stability

c) Less variation in speed

d) Increased steadiness

Answer: a

Explanation: Hunting is a process in which the speed of the engine is continuously fluctuating above and below the mean speed. This is something which happens when a governor is too sensitive.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Isochronous Governor”.

1. When equilibrium speed is constant, then the governor is called ______

a) Pickering

b) Hartung

c) Porter

d) Isochronous

Answer: d

Explanation: Pickering, Porter and Hartung are different categories of governor whereas Isochronous describes the property of the governor by the virtue of which the range in equilibrium speed is zero.

2. For isochronism, range of speed should be equal to mean speed.

a) True

b) False

Answer: b

Explanation: Isochronism describes the property of the governor by the virtue of which the range in equilibrium speed is zero.

3. The sensitiveness of an isochronous governor is ______

a) 0

b) 2

c) 1/2

d) 1

Answer: a

Explanation: For isochronisms N 1 – N 2 = 0, since this range is 0, the sensitiveness will also be equal to as

sensitiveness = (N 1 – N 2 )÷N.

4. Which of the following type of governor cannot be isochronous?

a) Pickering

b) Porter

c) Hartung

d) Hartnell

Answer: b

Explanation: For a porter governor it is impossible to have the two balls moving at the same speed, thus the range cannot be 0 which makes it impossible to be an isochronous governor.

5. An isochronous governor is not practical because of the following reason.

a) Friction at the sleeve

b) Weight of the spring

c) Impossible to achieve 0 range

d) High use of porter governors

Answer: a

Explanation: The isochronous governor is not of practical use because the sleeve will move to one of its extreme positions immediately the speed deviates from the isochronous speed.

6. The sensitivity of a governor is 0, and the mean speed is 25 rpm, then it is not possible that the governor will be ________

a) Pickering

b) Porter

c) Hartung

d) Hartnell

Answer: b

Explanation: For a porter governor it is impossible to have the two balls moving at the same speed, thus the range cannot be 0 which makes it impossible to be an isochronous governor. Since sensitiveness is 0, therefore range is 0 hence the governor cannot be porter.

7. For a Hartnell governor which of the following relation regarding r₁/r₂ is true for isochronisms?

a) (M.g + S 1 ) ÷ (M.g + S 2 )

b) (M.g + S 2 ) ÷ (M.g + S 1 )

c) S 2 / S 1

d) S 1 / S 2

Answer: a

Explanation: For isochronisms the equilibrium speed should remain constant throughout, i.e Range should be 0.

Applying that condition we have r 1 /r 2 = (M.g + S 1 ) ÷ (M.g + S 2 )

8. Pendulum type governor cannot be isochronous

a) True

b) False

Answer: a

Explanation: For isochronisms, the equilibrium speed should remain constant throughout, i.e Range should be 0 and for pendulum type governors it is impossible to achieve same speed for different balls.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Effort and Power of a Governor”.

1. The effort of a governor is the _________ force exerted at the sleeve for a given percentage change of speed.

a) Maximum

b) Minimum

c) Mean

d) Reciprocal of

Answer: c

Explanation: The effort of a governor is the mean force exerted at the sleeve for a given percentage change of speed or lift of the sleeve.

2. When there is no force on the sleeve, then the governor must be running steadily.

a) True

b) False

Answer: a

Explanation: When a governor is running steadily then it is under the condition that it experiences no force on its sleeve. Forces acting on the sleeve results in change of speed of the governor.

3. Resistance in a sleeve which results in opposing the motion is caused by _______

a) Change of speed

b) Change of weight

c) Isochronism

d) Change in type of governor

Answer: a

Explanation: When the speed at which the governor is undergoing motion changes, there is a resistance at the sleeve which opposes its motion.

4. Power of a governor depends on which of the following factor?

a) Mean effort

b) Maximum effort

c) Minimum effort

d) Weight of the engine

Answer: a

Explanation: The power of a governor is the work done at the sleeve for a given percentage change of speed. It is the product of the mean value of the effort and the distance through which the sleeve moves.

5. Find the power of the governor if the Mean effort is 50 N and sleeve lift is 100 mm.

a) 5N-m

b) 25N-m

c) 250N-mm

d) 50N-mm

Answer: a

Explanation: Power is given by

 x 

= 50×0.1 N-m

= 5N-m.

6. If the sleeve lift is 80 mm and Power generated by the governor is 6 N-m, then find the mean effort.

a) 7.5 N

b) 15 N

c) 20 N

d) 30 N

Answer: a

Explanation: Power is given by

mean effort x sleeve lift

6 = 0.8xmean effort

6/0.8 = 7.5 N.

7. If the mean effort is doubled and the sleeve lift is halved, If P was the original power then what will be the power post the changes?

a) 2P

b) P

c) 4P

d) P/2

Answer: b

Explanation: Power is given by:

mean effort x sleeve lift

since power is directly proportional to the sleeve lift and mean effort, hence doubling one factor and reducing the other to half has no effect on the power.

8. Power of a governor is the work done by the sleeve for a given percentage change in speed.

a) True

b) False

Answer: b

Explanation: Power of the governor is the work done at the sleeve for a given percentage change of speed. It is important to note that the work should be done at the sleeve and not at any other part of the governor.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Effort and Power of a Porter Governor”.

1. For a porter governor, Each arm has a length of 250mm and pivoted on the axis of rotation. Sleeves carry a mass of 25kg and each ball’s mass is 5Kg. Radius of rotation: 150mm at the beginning of lift and 200mm at the maximum speed of governor. Find range in speed neglecting friction in rpm.

a) 25

b) 35

c) 45

d) 15

Answer: a

Explanation: For minimum position, h1 = 200 mm

N 1 2 = /m x 895/h1 = 26850 rpm

For maximum position,

h 2 =150 mm

N 2 2 = /m x 895/h 2 = 35800 rpm

Range = N 2 – N 1 = 25 rpm

2. A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Neglecting friction at the sleeve, find the sleeve lift.

a) 0.2m

b) 0.1m

c) 0.5m

d) 200 mm

Answer: b

Explanation: For minimum position, h 1 = 200 mm

For maximum position,

h 2 =150 mm

sleeve lift = 2(h 1 – h 2 ) = 0.1m.

3. A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Neglecting friction at the sleeve, find the governor effort.

a) 44.7 N

b) 22.35 N

c) 89.4 N

d) 50 N

Answer: a

Explanation: If c = Percentage increase in speed, then

cN 1 = N 2 – N 1 = 25 rpm

c = 25/164 = 0.152

governor effort = cg = 44.7 N.

4. A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Neglecting friction at the sleeve, find the power of the governor in N-m.

a) 447

b) 44.7

c) 4.47

d) 5.0

Answer: c

Explanation: We know that power of a governor is given by

effort of governor x sleeve lift

= 44.7 x 0.1 = 4.47 N-m.

5. A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Considering friction at the sleeve of 10N, find the range of speed in rpm.

a) 31.4

b) 35.4

c) 45.2

d) 15.6

Answer: a

Explanation: When friction is taken into account

N 1 2 = /mg x 895/h 1

h 1 = 0.2m

therefore N 1 = 161 rpm

N 2 2 = /mg x 895/h 2

N 2 = 192.4 rpm

Therefore range = 31.4 rpm.

6. A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Considering friction at the sleeve of 10N, find the governor effort.

a) 44.7 N

b) 57.4 N

c) 88.4 N

d) 53.8 N

Answer: b

Explanation: If c is the percentage change in speed

cN 1 = N 2 – N 1 = 31.4

c = 31.4/161 = 0.195

P = c

= 0.195  = 57.4 N.

7. A Porter governor has equal arms each of length 250 mm and pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central load has a mass on the sleeve of 25 kg. The radius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Considering friction at the sleeve of 10N, find the power of the governor in N-m

a) 447

b) 44.7

c) 4.47

d) 5.0

Answer: c

Explanation: We know that the power of a governor is given by the relation

Power = governor effort x sleeve lift

= 57.4 x 0.1 N-m

= 5.74 N-m.

8. When friction acts at the sleeve the maximum angular velocity increases?

a) True

b) False

Answer: a

Explanation: If F is the friction force acting on the sleeve, then the value of maximum speed is given

N = /mg x 895/h 2

hence maximum angular velocity increases.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Controlling Force”.

1. The controlling force acting on the governor is also known as ________

a) Centripetal force

b) Centrifugal force

c) Governor effort

d) Piston effort

Answer: a

Explanation: When a governor rotates at a steady speed, an inward force acts on the body which is centripetal in nature. This force is called a controlling force.

2. Controlling force is in the same direction as centrifugal force.

a) True

b) False

Answer: b

Explanation: When a governor rotates at a steady speed, an inward force acts on the body which is centripetal in nature. This force is called controlling force and acts in the opposite direction to that of centrifugal force.

3. If F=m.ω 2 .r represents the centrifugal force then which of the following expressions represents controlling force.

a) F

b) 2F

c) -2F

d) -F

Answer: d

Explanation: Inward force acts on the body of the governor which is centripetal in nature. This force is called controlling force and acts in the opposite direction to that of centrifugal force.

4. Controlling force in the porter governors is due to which of the following components?

a) Spring

b) Weight of sleeve

c) Weight of governor

d) Piston effort

Answer: b

Explanation: The controlling force is provided by the weight of the sleeve and balls as in Porter governor and by the spring and weight as in Hartnell governor.

5. Controlling force in the Hartnell governors is due to which of the following components?

a) Weight of balls

b) Weight of sleeve

c) Spring

d) Piston effort

Answer: b

Explanation: The controlling force is provided by the weight of the sleeve and balls in case of Porter governor and by the spring and weight in case of Hartnell governor.

6. If the mass of the ball is 7.5 kg and the controlling force is 200N with an angular velocity of 100 rpm, find the radius.

a) 24 cm

b) 28 cm

c) 30 cm

d) 20 cm

Answer: a

Explanation: F=m.ω 2 .r

200 = 7.5  2 .r

Therefore r = 0.243m.

7. If the mass of the ball is 25 kg and the controlling force is 450 N with an angular velocity of 150 rpm, find the radius.

a) 7.2 cm

b) 7.9 cm

c) 8.3 cm

d) 12.3 cm

Answer: b

Explanation: F = m.ω 2 .r

450 = 25  2 .r

18 = 246.7.r

Therefore r = 0.072m.

8. Control force diagram enables to examine stability and sensitivity of the governor.

a) True

b) False

Answer: a

Explanation: When the graph is drawn between the controlling force  as y-axis and radius of rotation of the balls  as x-axis, then the graph obtained is known as controlling force diagram of the governor. This diagram enables the stability and sensitiveness of the governor to be examined and also shows clearly the effect of friction.

9. Which of the following quantities cannot be examined from controlling force diagram?

a) Sensitivity

b) Stability

c) Effect of friction

d) Power of governor

Answer: d

Explanation: The graph between the controlling force  as y-axis and radius of rotation of the balls  as x-axis is known as controlling force diagram. This diagram enables us to examine the stability and sensitiveness of the governor along with the effect of friction.

This set of Machine Dynamics Multiple Choice Questions & Answers focuses on “Controlling Force Diagram for a Porter Governor”.

1. The graph plotted between the controlling force of the governor and Radius of rotation is known as _______

a) Controlling force diagram

b) Radius of rotation diagram

c) Governor diagram

d) Sensitivity diagram

Answer: a

Explanation: When the graph between the controlling force  as ordinate and radius of rotation of the balls  as abscissa is drawn, then the graph obtained is known as controlling force diagram of the governor.

2. Controlling force is provided by spring and weight in the porter governor.

a) True

b) False

Answer: b

Explanation: Controlling force is provided by the weight of the sleeve and ball in the porter governor and by the spring and the weight in Hartnell governor.

3. In the given figure, the x – axis represents radius of rotation and the y axis represents the controlling force. This graph is known as _________

machine-dynamics-multiple-choice-questions-answers-q3

a) Controlling force diagram of Porter governor

b) Controlling force diagram of Hartnell governor

c) Controlling force diagram of isochronous governor

d) Controlling force diagram of a general governor

Answer: a

Explanation: The given figure is a plot between the controlling force and the radius of rotation, from the curve it is factually known that it represents the controlling force diagram of a porter governor.

4. What should be the stability condition for the angle made by OA and the x axis? 

machine-dynamics-multiple-choice-questions-answers-q3

a) Angle must increase

b) Angle must decrease

c) Angle must remain constant

d) Stability is independent of this angle

Answer: a

Explanation: In case the governor satisfies the condition for stability, the angle between OA and the x-axis must increase with the radius of rotation of the governor balls.

5. How should the equilibrium speed vary with the increase of radius of rotation of the governor balls for stability?

a) Must increase

b) Must decrease

c) Must remain constant

d) Equilibrium speed independent of radius of rotation

Answer: a

Explanation: In order to achieve stability in a porter governor, the equilibrium speed must increase with the increase of radius of rotation of the governor balls.

6. For a more sensitive governor, the change in angle between OA and the x axis with change in radius of rotation should be? 

machine-dynamics-multiple-choice-questions-answers-q3

a) Small

b) Large

c) Remain same

d) Is independent of this angle

Answer: a

Explanation: For the governor to be more sensitive, the change in the value of angle between OA and the x-axis over the change of radius of rotation should be as small as possible.

7. In the given figure, the straight line OA represents _________ 

machine-dynamics-multiple-choice-questions-answers-q3

a) Isochronous governor

b) Hartnell governor

c) Non – Isochronous governor

d) Stable governor

Answer: a

Explanation: For the isochronous governor, the controlling force curve is a straight line passing through the origin, thus in the above figure, AO is passing through the origin.

8. In the given figure if the angle between OA and x-axis is made 90 degrees, the rotation speed will become 0. 

machine-dynamics-multiple-choice-questions-answers-q3

a) True

b) False

Answer: b

Explanation: In the given figure if the angle between OA and x-axis is made 90 degrees, the rotation speed will become infinite, as the speed is directly proportional to the root of tangent of that angle.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Controlling Force Diagram for a Spring-controlled Governor”.

1. Let r represent the radius of rotation, then for stability of a spring controlled governor, which of the following relation regarding controlling force is true? 

a) Fc = ar-b

b) Fc = ar +b

c) Fc = ar

d) Fc = a

Answer: a

Explanation: The relation between the controlling force  and the radius of rotation  for the stability of spring controlled governors is given by the following equation:

Fc = ar-b.

2. If the line intersects the Controlling force axis in positive direction in the controlling force diagram of the spring controlled governor, then it is stable.

a) True

b) False

Answer: b

Explanation: If the line intersects the Controlling force axis in negative direction in the controlling force diagram of the spring controlled governor, then it is stable.

Fc = ar-b

3. Let r represent the radius of rotation, then for an unstable spring controlled governor, which of the following relation regarding controlling force is true? 

a) Fc = ar-b

b) Fc = ar +b

c) Fc = ar

d) Fc = a

Answer: b

Explanation: The relation between the controlling force  and the radius of rotation  for the unstability of spring controlled governors is given by the following equation:

Fc = ar+b.

4. Let r represent the radius of rotation, then for an isochronous spring controlled governor, which of the following relation regarding controlling force is true? 

a) Fc = ar-b

b) Fc = ar +b

c) Fc = ar

d) Fc = a

Answer: c

Explanation: The relation between the controlling force  and the radius of rotation  for the instability of spring controlled governors is given by the following equation:

Fc = ar.

5. In the given figure x axis is the radius of rotation and y axis is the controlling force, identify the correct option.

machine-dynamics-questions-answers-controlling-force-diagram-spring-controlled-governor-q5

a) Curve represents isochronous governor

b) Curve represents stable governor

c) Curve represents unstable governor

d) Curve represents sensetive governor

Answer: a

Explanation: In the given figure, the straight line is passing through the origin thus following a relation

Fc = ar. Since there are no constants it represents an isochronous governor.

6. In the given figure x axis is the radius of rotation and y axis is the controlling force, identify the correct option. 

machine-dynamics-questions-answers-controlling-force-diagram-spring-controlled-governor-q6

a) Curve represents isochronous governor

b) Curve represents stable governor

c) Curve represents unstable governor

d) Curve represents sensetive governor

Answer: b

Explanation: In the given figure, the straight line is below the axis representing the controlling force thus following a relation

Fc = ar – b. Since there is a negative constant it represents a stable governor.

7. In the given figure x axis is the radius of rotation and y axis is the controlling force, identify the correct option. 

machine-dynamics-questions-answers-controlling-force-diagram-spring-controlled-governor-q7

a) Curve represents isochronous governor

b) Curve represents stable governor

c) Curve represents unstable governor

d) Curve represents sensetive governor

Answer: c

Explanation: In the given figure, the straight line is above the axis representing the controlling force thus following a relation

Fc = ar + b. Since there is a negative constant it represents an unstable governor.

8. It is possible that the equilibrium speed of the governor decreases with an increase of the radius of rotation of balls.

a) True

b) False

Answer: b

Explanation: If the equilibrium speed of the governor decreases with an increase of the radius of rotation of balls then Fc/r decreases as r increases. This is impracticable.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Coefficient of Insensitiveness”.

1. When the speed of the governor decreases, which of the following effect does friction causes?

a) Prevents downward movement of sleeve

b) Prevents downward movement of sleeve

c) Prevents radial outward movement of balls

d) Increases downward movement of sleeve

Answer: a

Explanation: In actual conditions there is always friction acting at joints and operating mechanisms of the governor, because of this while slowing down friction prevents the downward movement of sleeve.

2. Because of friction acting at the joints, while speeding up it prevents the downward movement of sleeve.

a) True

b) False

Answer: b

Explanation: In actual conditions there is always friction acting at joints and operating mechanisms of the governor, because of this while speeding up friction prevents the upward movement of sleeve.

3. When the speed of the governor increases, which of the following effect does friction causes?

a) Prevents downward movement of sleeve

b) Prevents downward movement of sleeve

c) Prevents radial inward movement of balls

d) Increases upward movement of sleeve

Answer: b

Explanation: In actual conditions there is always friction acting at joints and operating mechanisms of the governor, as a consequence when the speed of rotation increases.

4. For a porter governor, the following data is given:

Mass of each ball = 3kg

Mass on the sleeve = 15Kg

Length of each arm = 20cm

Ball path at life = 12 cm

Path at maximum speed = 16 cm

Calculate range of speed.

a) 14.15 rpm

b) 28.3 rpm

c) 56.9 rpm

d) 32.4 rpm

Answer: b

Explanation: h 1 = 0.16m

h 2 = 0.12m

N 1 2 = /m) x895/h 1 = x895/0.16 = 33563

N 1 = 183.2 rpm

N 2 2 = /m) x895/h 2 = x895/0.12 = 44750

N 2 = 211.5 rpm

therefore range = 28.3 rpm.

5. For a porter governor, the following data is given:

Mass of each ball = 3kg

Mass on the sleeve = 15Kg

Length of each arm = 20cm

Ball path at life = 12 cm

Path at maximum speed = 16 cm, If there is a friction at the sleeve of 10 N, find the coefficient of insensitiveness.

a) 11.3%

b) 5.00%

c) 8.66%

d) 5.66%

Answer: d

Explanation: We know that coefficient of insensitiveness is given by

F/g

= 10/9.81 = 5.66%.

6. For a porter governor, the following data is given:

Mass of each ball = 5kg

Mass on the sleeve = 20Kg

Length of each arm = 20cm

Ball path at life = 12 cm

Path at maximum speed = 16 cm

Calculate range of speed in rpm.

a) 25.9

b) 51.8

c) 12.9

d) 31.2

Answer: a

Explanation: h 1 = 0.16m

h 2 = 0.12m

N 1 2 = /m) x895/h 1 = x895/0.16 = 27968

N 1 = 167.2 rpm

N 2 2 = /m) x895/h 2 = x895/0.12 = 37291

N 2 = 193.1 rpm

therefore range = 25.9 rpm.

7. For a porter governor, the following data is given:

Mass of each ball = 5kg

Mass on the sleeve = 20Kg

Length of each arm = 20cm

Ball path at life = 12 cm

Path at maximum speed = 16 cm, If there is a friction at the sleeve of 10 N, find the coefficient of insensitiveness.

a) 4%

b) 6%

c) 5%

d) 7%

Answer: a

Explanation: We know that coefficient of insensitiveness is given by

F/g

= 10/9.81 = 4%.

8. With increase in friction, coefficient of insensitiveness increases.

a) True

b) False

Answer: a

Explanation: Coefficient of insensitiveness is directly proportional to the friction at the sleeve and inversely proportional to the sum of masses of ball and the weight at sleeve, hence increase in friction results in increase in coefficient of insensitiveness.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Brakes and Dynamometers”.

1. The brakes commonly used in railway trains is

a) shoe brake

b) band brake

c) band and block brake

d) internal expanding brake

Answer: a

Explanation: A single block or shoe brake consists of a block or shoe which is pressed against the rim of a revolving brake wheel drum. The block is made of a softer material than the rim of the wheel. This type of a brake is commonly used on railway trains and tram cars.

2. The brake commonly used in motor cars is

a) shoe brake

b) band brake

c) band and block brake

d) internal expanding brake

Answer: d

Explanation: The internal expanding type of brake is commonly used in motor cars and light trucks.

3. When brakes are applied to all the four wheels of a moving car, the distance travelled by the car before it is brought to rest, will be

a) maximum

b) minimum

c) equal

d) none of the mentioned

Answer: b

Explanation: When the brakes are applied to all the four wheels, the braking distance  will be the least. It is due to this reason that the brakes are applied to all the four wheels.

4. Which of the following is an absorption type dynamometer ?

a) prony brake dynamometer

b) epicyclic-train dynamometer

c) torsion dynamometer

d) none of the mentioned

Answer: a

Explanation: Absorption type dynamometers are

1. Prony brake dynamometer,

2. Rope brake dynamometer.

5. The transmission type of dynamometer is a

a) prony brake dynamometer

b) epicyclic-train dynamometer

c) torsion dynamometer

d) none of the mentioned

Answer: c

Explanation: Transmission type dynamometers are

1. Epicyclic-train dynamometer,

2. Belt transmission dynamometer,

3. Torsion dynamometer.

6. The material used for the brake lining should have

a) It should have low wear rate.

b) It should have high heat resistance.

c) It should have high heat dissipation capacity.

d) All of the mentioned

Answer: d

Explanation: The material used for the brake lining should have the following characteristics:

1. It should have high coefficient of friction with minimum fading.

2. It should have low wear rate.

3. It should have high heat resistance.

4. It should have high heat dissipation capacity.

5. It should have adequate mechanical strength.

6. It should not be affected by moisture and oil.

7. The capacity of a brake depends upon

a) The unit pressure between the braking surfaces,

b) The coefficient of friction between the braking surfaces,

c) The peripheral velocity of the brake drum

d) All of the mentioned

Answer: d

Explanation: The capacity of a brake depends upon the following factors :

1. The unit pressure between the braking surfaces,

2. The coefficient of friction between the braking surfaces,

3. The peripheral velocity of the brake drum,

4. The projected area of the friction surfaces, and

5. The ability of the brake to dissipate heat equivalent to the energy being absorbed.

Answer: d

Explanation: For high speed engines, the cam follower should move with cyclodial motion. For S.H.M. cam, the acceleration of the follower at the ends of the stroke and aimed stroke respectively, is

maximum and zero.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Brakes”.

1. Which of the following type of brakes can bring the member to an absolute rest?

a) Hydraulic

b) Electric

c) Mechanical

d) Fluid agitator

Answer: c

Explanation: The electric and hydraulic brakes cannot bring the member to rest and are mostly used where large amounts of energy are to be transformed.

2. Electric brakes are used to control the speed of vehicle during downhill travel.

a) True

b) False

Answer: a

Explanation: The electric brake cannot bring the member to rest and are mostly used where large amounts of energy are to be transformed, hence retards heavy vehicles during downhill travel.

3. In which of the following brakes the force acting on the brake drum is in axial direction?

a) Block brake

b) Shoe brake

c) Band brake

d) Cone brake

Answer: d

Explanation: Block, shoe and band brakes are radial brakes, in these brakes the force acts in the radial direction whereas cone brake is an example of axial brake.

4. The combined mass of bicycle and rider is 100 kg, they are travelling at the speed of 16 km/h on a level road. Diameter of rear wheel is 0.9m. How far will the bicycle travel after the application of brake? The pressure applied on the brake is 100 N and μ = 0.05.

a) 200.1 m

b) 197.2 m

c) 98.6 m

d) 124.3 m

Answer: b

Explanation: Ft = μ.Rn = 0.05×100 = 5N

Let x be the distance travelled by the bicycle before coming to rest.

then

Ft.x = mv 2 /2

x = 100×4.44 2 /

= 197.2m.

5. The combined mass of bicycle and rider is 100 kg, they are travelling at the speed of 16 km/h on a level road. Diameter of rear wheel is 0.9m. The pressure applied on the brake is 100 N and μ = 0.05., find the no. of revolutions made by the bicycle before it comes to rest.

a) 35

b) 140

c) 70

d) 84

Answer: c

Explanation: Ft = μ.Rn = 0.05×100 = 5N

Let x be the distance travelled by the bicycle before coming to rest.

then

Ft.x = mv 2 /2

x = 100×4.44 2 /

= 197.2m

The distance travelled by the bicycle before coming to rest is 197.2m

now, If N is the no. of revolutions then:

197.2 = πDN

197.2/πx0.9 = N

N = 70.

6. If the angle of contact is greater than 60 degrees, then unit pressure normal to the surface of contact is __________

a) Less at the ends than at the centre

b) More at the ends then centre

c) Remains equal at the ends and the centre

d) Remains unaffected

Answer: a

Explanation: If the angle of contact is greater than 60°, then the unit pressure normal to the surface of contact is less at the ends than at the centre. For less than 60° the pressure is uniform.

7. Drawback of a single block brake is __________

a) Bending of shaft

b) Twisting of shaft

c) Tension in shaft

d) Tension in bearings

Answer: a

Explanation: In case of a single block brake, when it is applied to a rolling wheel, an additional load is thrown on the shaft bearings due to the normal force. This produces bending of the shaft.

8. For a double block shoe brake, the braking action is halved.

a) True

b) False

Answer: b

Explanation: Because of the presence of two blocks instead of one, the braking action is doubled, and practically these blocks can be applied with the same force as required for a single brake.

9. The brake shown in figure has a coefficient of friction 0.25 and is to operate using a maximum force F of 500 N. Assume width of the band to be 50 mm. Find the maximum band tension. All dimensions are in mm.

machine-dynamics-questions-answers-brakes-q9

a) 4449.6 N

b) 4440.8 N

c) 5432.4 N

d) 4882.3 N

Answer: a

Explanation:

machine-dynamics-questions-answers-brakes-q9a

ΣMo = 0

= 115P 2 – 365F = 0

P 2 = 1586.96N

α = 33.59°

Φ = 270 – 33.59 = 4.126 radians

P 1 = 4449.6N.

10. The brake shown in figure has a coefficient of friction 0.25 and is to operate using a maximum force F of 500 N. Assume width of the band to be 50 mm, find the braking torque.

machine-dynamics-questions-answers-brakes-q9

a) 380.2 N

b) 357.8 N

c) 444.3 N

d) 320.3 N

Answer: b

Explanation:

machine-dynamics-questions-answers-brakes-q9a

ΣMo=0

= 115P 2 – 365F =0

P 2 = 1586.96N

α = 33.59°

Φ = 270 – 33.59 = 4.126 radians

P 1 = 4449.6N

Braking torque = (P 1 – P 2 )D/2

= 357.8 N.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Dynamometer”.

1. Which of the following is a characteristic feature of a dynamometer?

a) It can measure torque

b) It can measure frictional resistance

c) It can measure the balancing force

d) It can act as a speedometer

Answer: b

Explanation: Dynamometer is a device that acts as a brake but in addition to this, it can also measure frictional resistance and can measure the power of the engine.

2. Prony brake dynamometer is an absorption dynamometer.

a) True

b) False

Answer: b

Explanation: Pronybrake dynamometer absorbs all the energy produced by the engine through frictional resistances hence it is an absorption dynamometer.

3. In which of the following dynamometers does the entire energy or power produced by the engine is absorbed by the friction resistances of the brake?

a) Prony brake dynamometer

b) Torsional dynamometer

c) Epicyclic train dynamometer

d) Belt transmission dynamometer

Answer: a

Explanation: Prony brake dynamometer is an absorption dynamometer, in absorption dynamometers the entire energy or power produced by the engine is absorbed by the friction resistances of the brake and is transformed into heat.

4. In which of the following dynamometers does the energy produced by the engine is used for doing work?

a) Prony brake dynamometer

b) Rope brake dynamometer

c) Absorption dynamometer

d) Epicyclic train dynamometer

Answer: d

Explanation: An epicyclic train dynamometer is an example of transmission dynamometer. In these types of dynamometers the energy is not wasted in friction but is used for doing work.

5. Flywheel diameter 1.2 m; diameter of the rope 12.5 mm; engine speed 200 r.p.m.; dead load on the brake 600 N; spring balance reading 150 N. Using this data, calculate the brake power of the engine in kW.

a) 5.715

b) 5.525

c) 6.125

d)5.625

Answer: a

Explanation: The brake power of an engine is given by the relation:

B.P = [π N]/60

= 5715 W.

6. Flywheel diameter 1.2 m; diameter of the rope 12.5 mm; engine speed 400 r.p.m.; dead load on the brake 600 N; spring balance reading 150 N. Using this data, calculate the brake power of the engine in kW.

a) 12.82

b) 11.43

c) 13.64

d) 10.22

Answer: b

Explanation: The brake power of an engine is given by the relation:

B.P = [π N]/60

= 11430 W.

7. For a rope brake dynamometer, the flywheel is cooled with soapy water because________

a) Energy is absorbed by the dynamometer

b) Energy is used to do work

c) Energy is provided by the motor

d) Energy supplied is more

Answer: a

Explanation: A rope brake dynamometer is an example of absorption dynamometer, in absorption dynamometers the entire energy or power produced by the engine is absorbed by the friction resistances of the brake and is transformed into heat. Hence to overcome this soapy water is used.

8. Belt transmission dynamometer is an absorption dynamometer.

a) True

b) False

Answer: b

Explanation: Belt transmission dynamometer is a transmission dynamometer as a result the energy is not wasted in friction but is used for doing work.

9. For which of the following dynamometers the driving torque on the shaft is not uniform and it is subjected to severe oscillations?

a) Pony brake

b) Band brake

c) Belt transmission

d) Epicyclic train

Answer: a

Explanation: For a pony brake dynamometer, When the driving torque on the shaft is not uniform, it is subjected to severe oscillations.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Differential Band Brake”.

1. When the fulcrum is between the two ends of a lever, the type of brake is _____

a) Simple band brake

b) Differential band brake

c) Shoe brake

d) Double block shoe brake

Answer: b

Explanation: When the fulcrum is at one end of the lever, the brake is called a simple band brake, whereas when the fulcrum is between the two ends it is known as differential band brake.

2. For a simple band brake, the fulcrum is at the end of the lever.

a) True

b) False

Answer: a

Explanation: For a simple band brake, the fulcrum is at the end of the lever and for a differential band brake, the fulcrum is between the ends of the lever.

3. For the given figure, the brake can sustain a torque of 350 N-m and coefficient of friction between the band and drum is 0.3, find necessary force required for clockwise rotation of the drum in N.

machine-dynamics-questions-answers-differential-band-brake-q3

a) 32

b) 64

c) 96

d) 128

Answer: b

Explanation: 2.3 log(T 1 /T 2 ) = μ.θ = 0.3×3.93 = 1.179

log(T 1 /T 2 ) = 1.129/2.3

= 0.51

T 1 /T 2 = 3.255

Braking torque = (T 1 -T 2 )r

T 1 -T 2 = 2000

therefore T 1 = 2887N T 2 = 887N

Px500 = T 2 x150 – T 1 x35

P = 64N.

4. For the given figure, the brake can sustain a torque of 350 N-m and coefficient of friction between the band and drum is 0.3, If the drum rotates in anticlockwise direction, find force in N.

machine-dynamics-questions-answers-differential-band-brake-q3

a) 804

b) 756

c) 850

d) 900

Answer: a

Explanation: 2.3 log(T 1 /T 2 ) = μ.θ = 0.3×3.93 = 1.179

log(T 1 /T 2 ) = 1.129/2.3

= 0.51

T 1 /T 2 = 3.255

Braking torque = (T 1 -T 2 )r

T 1 -T 2 = 2000

therefore T 1 = 2887N T 2 = 887N

Px500 = T 1 x150 – T 2 x35

P = 804N.

5. For the given figure, the brake can sustain a torque of 350 N-m and coefficient of friction between the band and drum is 0.3, find OA for self locking condition during clockwise rotation.

machine-dynamics-questions-answers-differential-band-brake-q3

a) 114mm

b) 130mm

c) 57mm

d) 88mm

Answer: a

Explanation: For self locking condition, P=0

therefore T 1 x35 = T 2 xOA

OA = 114mm.

6. Which of the following brake does not has self locking?

a) Disk

b) Differential band

c) Simple band

d) Depends on the acting force

Answer: a

Explanation: Disk brakes do not have self locking However, it is possible to achieve self locking in differential band brakes.

7. Braking torque is independent of the radius of drum.

a) True

b) False

Answer: b

Explanation: Braking torque is given by the equation

T = (T 1 -T 2 )r

hence it is directly proportional to the radius of the drum.

8. With the help of following data, calculate the braking torque in N-m.

T 1 = 2500 N

T 2 = 200 N

diameter of drum = 350mm

a) 402.5

b) 805

c) 201.5

d) 560

Answer: a

Explanation: T = T = (T 1 -T 2 )r

r = 350/2000

T = 350/2000

= 2300×350/2000 N-m

= 402.5 N-m.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Types of Dynamometers”.

1. The entire energy or power produced by the engine is absorbed by the friction resistances of the brake in which of the following dynamometers does?

a) Torsional brake dynamometer

b) Prony dynamometer

c) Belt Transmission dynamometer

d) Epicyclic train dynamometer

Answer: b

Explanation: In absorption dynamometers the entire energy or power produced by the engine is absorbed by the friction resistances of the brake and is transformed into heat. Prony brake is an absorption dynamometer.

2. Belt transmission dynamometer is a transmission dynamometer.

a) True

b) False

Answer: a

Explanation: Belt transmission dynamometer is a transmission dynamometer as a result the energy is not wasted in friction but is used for doing work.

3. Which of the following dynamometer belongs to a different type.

a) Prony brake dynamometer

b) Torsional dynamometer

c) Epicyclic train dynamometer

d) Belt transmission dynamometer

Answer: a

Explanation: Prony brake dynamometer is an absorption dynamometer whereas the remaining three are transmission dynamometer.

4. Which of the following is not a characteristic of the rope brake dynamometer?

a) Energy loss in friction

b) Energy absorbed by frictional resistances

c) Energy utilization in work

d) Energy can be either utilized to do work or can be reused.

Answer: b

Explanation: Rope brake dynamometer is an absorption dynamometer hence all the energy is absorbed by frictional resistances and gets converted into heat.

5. Which of the following is not a characteristic of the Torsional dynamometer.

a) Conversion of energy into heat

b) Conversion of energy into work

c) Utilization of engine power

d) Energy can be either utilized to do work or can be reused.

Answer: a

Explanation: Torsional dynamometer is a transmission dynamometer as a result the energy is not wasted in friction but is used for doing work. In case of absorption dynamometers, the energy is wasted into heat.

6. Flywheel diameter 1.2 m; diameter of the rope 12.5 mm; engine speed 100 r.p.m.; dead load on the brake 600 N; spring balance reading 150 N. Using this data, calculate the brake power of the engine in kW.

a) 2.85

b) 5.71

c) 11.4

d) 4.88

Answer: a

Explanation: The brake power of an engine is given by the relation:

B.P = [π N]/60

= π 100]/60 W

B.P = 2.85 kW.

7. On which of the following factor, the power of a torsion dynamometer does not depend?

a) Speed

b) Torque

c) Angle of twist

d) Friction

Answer: d

Explanation: Power is directly proportional to Torque and Speed,

Torque is directly proportional to the angle of twist. Thus power also depends on the angle of twist.

8. Transmission dynamometer uses the energy is used to do work.

a) True

b) False

Answer: a

Explanation: Absorption dynamometers lose their energy through friction but transmission dynamometer utilizes the same energy for doing work.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Epicyclic-train Dynamometers”.

1. Which of the following gears is not a part of an epicyclic train dynamometer?

a) Spur gear

b) Pinion gear

c) Annular gear

d) Helical gear

Answer: d

Explanation: An epicyclic train dynamometer consists of simple epicyclic gear trains including spur gear, annular gear with internal toothing and pinion gear.

2. An epicyclic-train dynamometer consists of a simple epicyclic train of gears.

a) True

b) False

Answer: a

Explanation: An epicyclic train dynamometer consists of simple epicyclic gear trains including pinion gear, annular gear with internal teeth and spur gear.

3. Which of the following statement is true regarding epicyclic train dynamometer.

a) Spur gear rotates in clockwise direction

b) Annular gear rotates in clockwise direction

c) Spur gear rotates is meshes with both annular and pinion

d) Annular gear rotates in anticlockwise direction

Answer: b

Explanation: The spur gear is keyed to the driving shaft and rotates in anticlockwise direction. The annular gear rotates in clockwise direction and is keyed to the engine shaft.

4. Which of the following dynamometers have a pinion meshing with two gears?

a) Prony brake dynamometer

b) Epicyclic train dynamometer

c) Torsional dynamometer

d) Belt transmission dynamometer

Answer: b

Explanation: In an epicyclic train dynamometer, three gears are used namely Spur gear annular gear and pinion gear. The pinion gear is in mesh with annular and spur gear.

5. Which of the following dynamometer type differs from an Epicyclic train dynamometer?

a) Belt transmission

b) Torsional dynamometer

c) Transmission dynamometer

d) Prony brake dynamometer

Answer: d

Explanation: Prony brake dynamometer is an absorption dynamometer and uses all the energy obtained by the engine to convert it into heat, whereas the remaining three are transmission dynamometers.

6. If the friction of the pin on which the pinion rotates is neglected, which of the following force is equal to the tangential effort exerted by the spur gear on the pinion

a) Radial effort exerted by the spur gear on the pinion

b) Tangential reaction of the annular gear on the pinion

c) Radial reaction of the annular gear on the pinion

d) Piston effort

Answer: b

Explanation: If the friction of the pin on which the pinion rotates is neglected, then the tangential effort P exerted by the spur gear on the pinion and the tangential reaction of the annular gear on the pinion are equal.

7. The upward force is balanced by ________

a) Radial effort exerted by the spur gear on the pinion

b) Tangential reaction of the annular gear on the pinion

c) Radial reaction of the annular gear on the pinion

d) Dead weight

Answer: d

Explanation: The reaction force by the gears acts in the upward direction tends to rotate the lever about its fulcrum. These upward forces are balanced by the dead weight at the end of the lever.

8. The dead weight at the end of the fulcrum causes rotation of lever about fulcrum.

a) True

b) False

Answer: b

Explanation: The reaction effort by the gears acts in the upward direction which has a tendency to rotate the lever about its fulcrum. These upward forces are balanced by the dead weight at the end of the lever.

This set of Machine Dynamics online test focuses on “Belt Transmission Dynamometer-Froude or Throneycraft Transmission Dynamometer”.

1. Which of the following is a function of a belt transmission dynamometer?

a) Measure the difference in tension of the belts

b) Measuring speed of an engine

c) Measuring power of an engine

d) Measuring friction

Answer: a

Explanation: The tangential force on the pulley which is driven is equal to the difference between the tensions in the tight and slack sides of the belt. A belt dynamometer is inserted to measure directly the difference between the tensions of the belt.

2. Belt transmission dynamometer is an absorption dynamometer.

a) True

b) False

Answer: b

Explanation: Belt transmission dynamometer uses the energy from the engine to do useful work, unlike absorption dynamometer where this energy gets converted into heat. Hence it is a transmission dynamometer.

3. Which of the following dynamometers work in a similar way to belt transmission dynamometer?

a) Epicyclic train dynamometer

b) Rope brake dynamometer

c) Prony brake dynamometer

d) Froude or Throneycroft Transmission Dynamometer Dynamometer

Answer: d

Explanation: Rope brake and Prony brake dynamometers are absorption dynamometers hence they function differently. Froude or Throneycroft Transmission Dynamometer Dynamometer is another name for belt transmission dynamometer.

4. Brake power of an engine is independent of the following quantity _______

a) Difference in band torques

b) Diameter of puller

c) Speed of engine

d) Friction

Answer: d

Explanation: Braking power of the engine is given by the equation:

B.P = (T 1 -T 2 )πDN/60

hence it is independent of friction.

5. For a belt transmission dynamometer, if the diameter of pully is doubled and speed of rotation is halved, then the new braking power will be ________

a) Halved

b) Doubled

c) Quadrupled

d) Remain same

Answer: d

Explanation: Braking power of the engine is given by the equation

B.P = (T 1 -T 2 )πDN/60

if Diameter is doubled and speed is halved it will have no effect in the existing brake power.

6. Calculate brake power from the following data:

T 1 = 1500N

T 2 = 800N

Diameter of pulley = 30cm

Speed of engine shaft = 600rpm

a) 6.5 kW

b) 13 kW

c) 3.2 kW

d) 7 kW

Answer: a

Explanation: B.P = (T 1 -T 2 )πDN/60

T 1 = 1500N

T 2 = 800N

D = 0.3 m

N = 600 rpm

therefore BP = 6.5 kW.

7. To prevent the rotation caused by frame, which of the following quantity is added?

a) Weight at a distance

b) Weight at the centre

c) Ball bearings

d) Welded joints

Answer: a

Explanation: The frame causes a rotation in an anticlockwise direction, to prevent this rotation a weight is added at a certain distance from the shaft.

8. Brake power of an engine is directly proportional to the speed of engine.

a) True

b) False

Answer: a

Explanation: Brake power of an engine depends on the following factors:

Difference in tension of the bands, Speed of engine and diameter of pulley.

It is given by the equation : B.P = (T 1 -T 2 )πDN/60.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Bevis Gibson Flash Light Torsion Dynamometer”.

1. Which of the following dynamometers works on the assumption of uniform density of air?

a) Belt transmission

b) Epicyclic train dynamometer

c) Torsional dynamometer

d) Bevis-Gibson Flash Light Torsion Dynamometer

Answer: d

Explanation: Bevis-Gibson Flash Light Torsion Dynamometer’s working depends on the fact that the density of air is uniform and light has an infinite velocity during its propagation.

2. Bevis-Gibson Flash Light Torsion Dynamometer is a transmission dynamometer.

a) True

b) False

Answer: a

Explanation: Bevis-Gibson Flash Light Torsion Dynamometer is not absorption and is a transmission dynamometer. It uses the propagation of light to rotate the shaft.

3. The lamp and the eye piece must be moved _______ so as to bring them into line.

a) Radially

b) Tangentially

c) Parallel

d) In opposite plane

Answer: a

Explanation: In a Bevis-Gibson Flash Light Torsion Dynamometer, the setup consisting of lamp and the eyepiece must be moved radially in order to bring them into one line.

4. When the shaft is not transmitting torque, which of the following phenomenon occurs?

a) Visible Flashing of light

b) Flashing of light cannot be seen

c) Angle of torsion cannot be measured

d) Torsion meter stops working

Answer: a

Explanation: When the shaft is not transmitting torque, the position of slit does not change relative to one another, as a result flashing of light can be seen. This phenomenon occurs only because the relative position of slits does not change.

5. When the shaft is transmitting torque, which of the following phenomenon occurs?

a) Visible Flashing of light

b) Flashing of light cannot be seen

c) Angle of torsion cannot be measured

d) Torsion meter stops working

Answer: b

Explanation: When the shaft is transmitting torque, the position of slit does change relative to one another, as a result flashing of light cannot be seen. This phenomenon occurs only because the relative position of slits is changing.

6. When the shaft is transmitting torque, which of the following phenomenon can cause visibility of light in the eyepiece?

a) Moving eyepiece equal to the lag amount

b) Moving eyepiece lesser than the lag amount

c) Moving eyepiece greater than the lag amount

d) Independent of eyepiece position

Answer: a

Explanation: During transmission of torque by the shaft, the light is not visible as is does not reach the eyepiece however if the eyepiece is moved to the same amount as the lag of the disc, then light can be seen on the eyepiece.

7. To measure the angle of twist, which of the following device is used in the Bevis-Gibson Flash Light Torsion Dynamometer?

a) Torsion meter

b) Prony brake dynamometer

c) Rope brake dynamometer

d) Belt transmission dynamometer

Answer: b

Explanation: Torsion meter is the device used in Bevis-Gibson Flash Light Torsion Dynamometer to measure the angle of twist in the shaft. This helps in calculating the lag of the disc which in turn is essential for making the light visible at the eyepiece.

8. Light is visible when the torque is being transmitted by the shaft.

a) True

b) False

Answer: b

Explanation: When the torque being transmitted by the shaft, the relative position of disc changes and therefore the light cannot be obtained at the eyepiece.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Cams”.

1. The size of a cam depends upon

a) base circle

b) pitch circle

c) prime circle

d) pitch curve

Answer: a

Explanation: Base circle is the smallest circle that can be drawn to the cam profile.

Pitch circle is a circle drawn from the centre of the cam through the pitch points.

Prime circle is the smallest circle that can be drawn from the centre of the cam and

tangent to the pitch curve.

2. The angle between the direction of the follower motion and a normal to the pitch curve is called

a) pitch angle

b) prime angle

c) base angle

d) pressure angle

Answer: d

Explanation: It is the angle between the direction of the follower motion and a normal to the pitch curve. This angle is very important in designing a cam profile. If the pressure angle is too large, a reciprocating follower will jam in its bearings.

3. A circle drawn with centre as the cam centre and radius equal to the distance between the cam centre and the point on the pitch curve at which the pressure angle is maximum, is called

a) base circle

b) pitch circle

c) prime circle

d) none of the mentioned

Answer: b

Explanation: Base circle is the smallest circle that can be drawn to the cam profile.

Pitch circle is a circle drawn from the centre of the cam through the pitch points.

Prime circle is the smallest circle that can be drawn from the centre of the cam and

tangent to the pitch curve.

4. The cam follower generally used in automobile engines is

a) knife edge follower

b) flat faced follower

c) spherical faced follower

d) roller follower

Answer: c

Explanation: When the contacting end of the follower is of spherical shape, it is called a spherical faced follower. It may be noted that when a flat-faced follower is used in automobile engines, high surface stresses are produced. In order to minimise these stresses, the flat end of the follower is machined to a spherical shape.

5. The cam follower extensively used in air-craft engines is

a) knife edge follower

b) flat faced follower

c) spherical faced follower

d) roller follower

Answer: d

Explanation: When the contacting end of the follower is a roller, it is called a roller follower. Since the rolling motion takes place between the contacting surfaces , therefore the rate of wear is greatly reduced. In roller followers also the side thrust exists between the follower and the guide. The roller followers are extensively used where more space is available such as in stationary gas and oil engines and aircraft engines.

6. In a radial cam, the follower moves

a) in a direction perpendicular to the cam axis

b) in a direction parallel to the cam axis

c) in any direction irrespective of the cam axis

d) along the cam axis

Answer: a

Explanation: In radial cams, the follower reciprocates or oscillates in a direction perpendicular to the cam axis.

7. A radial follower is one

a) that reciprocates in the guides

b) that oscillates

c) in which the follower translates along an axis passing through the cam centre of rotation.

d) none of the mentioned

Answer: a

Explanation: When the motion of the follower is along an axis passing through the centre of the cam, it is known as radial follower.

8. Ofset is provided to a cam follower mechanism to

a) minimise the side thrust

b) accelerate

c) avoid jerk

d) none of the mentioned

Answer: a

Explanation: When the motion of the follower is along an axis away from the axis of the cam centre, it is called off-set follower.

9. For low and moderate speed engines, the cam follower should move with

a) uniform velocity

b) simple harmonic motion

c) uniform acceleration and retardation

d) cycloidal motion

Answer: b

Explanation: None

10. For high speed engines, the cam follower should move with

a) uniform velocity

b) simple harmonic motion

c) uniform acceleration and retardation

d) cycloidal motion

Answer: d

Explanation: Since with high speed engines, maximum acceleration is required and that is possible only through cycloidal motion.

11. Which of the following displacement diagrams should be chosen for better dynamic performance of a cam-follower mechanism ?

a) simple hormonic motion

b) parabolic motion

c) cycloidal motion

d) none of the mentioned

Answer: c

Explanation: Only cycloidal motion gives maximum acceletation. Hence, it is considered the most dynamic cam- follower mechanism.

12. The linear velocity of the reciprocating roller follower when it has contact with the straight flanks of the tangent cam, is given by

a) ω(r 1 -r 2 )sinθ

b) ω(r 1 -r 2 )cosθ

c) ωr 1 +r 2 )sinθsec 2 θ

d) ω(r 1 +r 2 )cosθcosec 2 θ

Answer: c

Explanation: Velocity = ωr 1 +r 2 )sinθsec 2 θ

where ω = Angular velocity of the cam shaft,

r 1 = Minimum radius of the cam,

r 2 = Radius of the roller, and

θ = Angle turned by the cam from the beginning of the displacement for contact of roller with the straight flanks.

13. The displacement of a flat faced follower when it has contact with the flank of a circular arc cam, is given by

a) R

b) R

c) (R-r 1 )

d) (R-r 1 )

Answer: c

Explanation: Displacement = (R-r 1 )

where R = Radius of the flank,

r 1 = Minimum radius of the cam, and

θ = Angle turned by the cam for contact with the circular flank.

Answer: a

Explanation: Retardation is maximum when α−θ = 0 or θ = α ,

Maximum retardation = ω 2 ×OQ

where OQ = Distance between the centre of circular flank and centre of nose.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Classification of Cams”.

1. Cam size depends upon

a) base circle

b) pitch circle

c) prime circle

d) outer circle

Answer: a

Explanation: Cam size depends upon base circle. The pressure angle of a cam depends upon offset between centre lines of cam and follower,lift of follower and angle of ascent.

2. Angle of ascent of cam is defined as the angle

a) during which the follower returns to its initial position

b) of rotation of the cam for a definite displacement of the follower

c) through which the cam rotates during the period in which the follower remains in highest position

d) moved by the cam from the instant the follower begins to rise, till it reaches its highest position

Answer: d

Explanation: Angle of ascent of cam is defined as the angle moved by the cam from the instant the follower begins to rise, till it reaches its highest position. Angle of action of cam is defined as the angle moved by the cam from beginning of ascent to the termination of descent.

3. The pressure angle of a cam depends upon

a) offset between centre lines of cam and follower

b) lift of follower

c) angle of ascent

d) all of the mentioned

Answer: d

Explanation: Cam size depends upon base circle. The pressure angle of a cam depends upon offset between centre lines of cam and follower,lift of follower and angle of ascent.

4. Angle of action of cam is defined as the angle

a) of rotation of the cam for a definite displacement of the follower

b) through which the cam rotates during the period in which the follower remains in the highest position

c) moved by the cam from the instant the follower begins to rise, till it reaches its highest position

d) moved by the cam from beginning of ascent to the termination of descent

Answer: d

Explanation: Angle of ascent of cam is defined as the angle moved by the cam from the instant the follower begins to rise, till it reaches its highest position. Angle of action of cam is defined as the angle moved by the cam from beginning of ascent to the termination of descent.

5. Throw of a cam is the maximum distance of the follower from

a) base circle

b) pitch circle

c) root circle

d) prime circle

Answer: a

Explanation: Cam size depends upon base circle. Throw of a cam is the maximum distance of the follower from base circle.

6. Cylindrical cams can be classified as

a) circular

b) tangent

c) reciprocating

d) none of the mentioned

Answer: d

Explanation: A cylindrical cam or barrel cam is a cam in which the follower rides on the surface of a cylinder. In the most common type, the follower rides in a groove cut into the surface of a cylinder. These cams are principally used to convert rotational motion to linear motion parallel to the rotational axis of the cylinder. A cylinder may have several grooves cut into the surface and drive several followers.

7. The cam follower generally used in aircraft engines is

a) knife edge follower

b) flat faced follower

c) spherical faced follower

d) roller follower

Answer: d

Explanation: The roller followers are extensively used where more space is available such as in stationary gas and oil engines and aircraft engines.

8. For S.H.M. cam, the acceleration of the follower at the ends of the stroke and aimed stroke respectively, is

a) maximum and zero

b) zero and maximum

c) minimum and maximum

d) zero and minimum

Answer: a

Explanation: For high speed engines, the cam follower should move with cyclodial motion. For S.H.M. cam, the acceleration of the follower at the ends of the stroke and aimed stroke respectively, is

maximum and zero.

Answer: b

Explanation: Cam angle is defined as the angle of rotation of the cam for a definite displacement of the follower. Angle of ascent of cam is defined as the angle moved by the cam from the instant the follower begins to rise, till it reaches its highest position. Angle of action of cam is defined as the angle moved by the cam from beginning of ascent to the termination of descent.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Tangent Cam with Reciprocating Roller Follower”.

1. When the flanks of the cam are straight and tangential to the base circle and nose circle, then the cam is known as a ____________

a) Tangent cam

b) Reciprocating cam

c) Roller cam

d) Radial cam

Answer: a

Explanation: When the flanks of the cam are straight and tangential to the base circle and nose circle, then the cam is known as a Tangent cam.

2. Tangent cam with reciprocating follower is an example of cams with specified contours.

a) True

b) False

Answer: a

Explanation: Tangent cams are symmetric about the center line of the cam shaft and hence they fall into a category of cams with specified contours.

3. Tangent cams are symmetric about __________

a) Centre line of cam shaft

b) Tangent line of cam shaft

c) Radial line of cam shaft

d) Along the roller

Answer: a

Explanation: Tangent cams are symmetrical about the center line of the cam shaft. Not in all cases the roller is radial to the cam and has an offset, in these cases the cam is not symmetric about the roller.

4. What is the use of tangent cams?

a) To operate inlet and exhaust valves of I C engines.

b) To operate the inlet valve of I C engines

c) To operate the exhaust valve of I C engines

d) To operate the crankshaft of an IC engine

Answer: a

Explanation: Cams are used to change or transfer one type of motion to another type, in internal combustion engines, Tangent cams are used to operate inlet and exhaust valves.

5. In the given figure, the cam has contact with _______

machine-dynamics-questions-answers-tangent-cam-reciprocating-roller-follower-q5

a) Nose

b) Straight flank

c) Arm

d) Shaft

Answer: a

Explanation: The line connecting the smaller circle to the bigger circle is known as the Straight flank, in the figure given above, the follower circle has contact with the straight flank.

6. If θ is the angle turned by the cam from the beginning of the roller displacement, then minimum acceleration pf the follower occurs at what value of θ in degrees?

a) 0

b) 30

c) 45

d) 60

Answer: a

Explanation: Acceleration of the follower is given by the equation:

ω 2 (r 1 +r 2 )(2-cos 2 θ)/cos 3 θ

This will occur at θ=0 degrees.

7. From the given data, calculate the acceleration of follower in m/s 2 at the beginning of the lift for a symmetrical tangent cam operating a roller follower.

Least radius of the cam is 30 mm; Roller radius is 17.5 mm. The angle of ascent is 75° and the total lift is 17.5 mm. The speed of the cam shaft is 600 r.p.m.

a) 187.6

b) 185.5

c) 183.2

d) 190.1

Answer: a

Explanation: At the beginning of the lift the angle turned by the cam from the beginning of the roller displacement is 0, hence acceleration is minimum which is given by the equation

ω 2 (r 1 +r 2 ).

8. Maximum acceleration of the follower is independent of the angle theta.

a) True

b) False

Answer: b

Explanation: The maximum acceleration of the follower is given by the equation

ω 2 (r 1 +r 2 )(2-cos 2 θ)/cos 3 θ

hence it is not independent of theta.

9. In the given figure, the cam has contact with _______

machine-dynamics-questions-answers-tangent-cam-reciprocating-roller-follower-q9

a) Base circle

b) Nose

c) Straight flank

d) Shaft

Answer: b

Explanation: In the given figure the cam has 2 circular parts. The circular part with a bigger radius is called the base circle and the one with the smaller radius is called the nose. Hence the roller has contact with the nose.

10. If θ is the angle turned by the cam from the beginning of the roller displacement, how the velocity changes with the change in theta?

a) Increases with increase in theta

b) Decreases with increase in theta

c) Increases with negative change in theta

d) Independent of theta

Answer: a

Explanation: The velocity of the follower is given by the equation

V = ω(r 1 +r 2 )(sinθ/cos 2 θ)

Hence with increasing theta the velocity increases.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Velocity and Acceleration Diagrams”.

1. The given figure is a velocity time diagram for which of the follower motion?

machine-dynamics-questions-answers-velocity-acceleration-diagrams-q1

a) Simple harmonic

b) Uniform acceleration

c) Uniform velocity

d) Uniform retardation

Answer: a

Explanation: The given figure is the velocity time diagram of the simple harmonic motion of the follower.

This is similar to the sine curve as the velocity relation is a sine function.

2. For a follower Simple Harmonic Motion having a small period of dwell at the beginning of motion, the jerk is 0.

a) True

b) False

Answer: b

Explanation: For a follower undergoing simple harmonic motion, starting off with a small dwell period, when it begins to ascend the acceleration is suddenly increased to a great amount from 0, hence the jerk is also infinite.

3. The given figure is a velocity time diagram for which of the follower motion?

machine-dynamics-questions-answers-velocity-acceleration-diagrams-q3

a) Simple harmonic

b) Uniform acceleration

c) Uniform velocity

d) Uniform retardation

Answer: c

Explanation: The given figure is the velocity time diagram of the uniform velocity motion of the follower with initial dwell. This is a constant curve as the velocity relation does not change with time.

4. Which of the following motion is not suitable from a practical point of view?

a) Simple harmonic

b) Uniform acceleration

c) Uniform velocity

d) Uniform retardation

Answer: c

Explanation: Due to infinite jerks, consequently infinite inertia forces arise as result, this makes the uniform velocity method of follower non-practical.

5. If a follower is undergoing simple harmonic motion, then at what value of angle of ascent the acceleration is maximum?

a) 0

b) 30

c) 45

d) 60

Answer: a

Explanation: At 0 degrees, the change in velocity in the motion of follower is maximum under the shm as a result a maximum value of acceleration is obtained at 0 degrees.

6. The given figure is a velocity time diagram for which of the follower motion?

machine-dynamics-questions-answers-velocity-acceleration-diagrams-q6

a) Simple harmonic

b) Uniform acceleration

c) Uniform velocity

d) Uniform acceleration and retardation

Answer: d

Explanation: The given figure is the velocity time diagram of the uniform acceleration and retardation motion of the follower with initial dwell. This is a linear curve as the velocity relation changes linearly with time.

7. Which of the following motion is used only up to moderate speeds?

a) Simple harmonic

b) Uniform acceleration

c) Uniform velocity

d) Uniform acceleration and retardation

Answer: d

Explanation: It is observed from the plots of uniform acceleration and retardation that there are abrupt changes in the acceleration at the beginning, midway and the end of the follower motion. At midway, an infinite jerk is produced.

8. While constructing cam profile, Kinematic inversion is used.

a) True

b) False

Answer: a

Explanation: While cam profile is being constructed, it is assumed that the cam is stationary and the follower moves in the opposite direction to that of cam in reality.

9. From the given figure below, at what value of theta the acceleration will be maximum?

machine-dynamics-questions-answers-velocity-acceleration-diagrams-q9

a) 0

b) 30

c) 60

d) Constant acceleration

Answer: d

Explanation: The given figure is the displacement diagram of the follower undergoing a constant acceleration and retardation motion, thus as a result of that there is a constant value of acceleration and has no maxima and minima.

10. Which of the following motion has the maximum number of infinite jerks in one rotation of the cam shaft?

a) Simple harmonic

b) Uniform acceleration

c) Uniform velocity

d) Uniform acceleration and retardation

Answer: b

Explanation: Uniform velocity motion is proved to be useless for practical purposes as there are infinite inertia force results due to many number of infinite jerks between the motion, so from the given options it is observed from the graph that uniform velocity motion has maximum number of jerks in one rotation of the cam shaft.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Circular Arc Cam with Flatfaced Follower”.

1. When the flanks of the cam connecting the base circle and nose are of convex circular arcs, then the cam is known as _______

a) Circular frame cam

b) Tanget arc cam

c) Tangent frame

d) Circular arc cam

Answer: d

Explanation: When the flanks of the cam connecting the base circle and nose are of convex circular arcs, then the cam is known as Circular arc cam, there are no cams such as tangent frame, circular frame cams.

2. The size of cam is independent of the base circle.

a) True

b) False

Answer: b

Explanation: The size of cam is defined by the radius of the base circle which is also known as the smallest circle and is independent of the pitch circle.

3. On which of the following factors does the pressure angle of the cam does not depend?

a) Offset between centre lines of follower and cam

b) Angle of ascent

c) Lift of the follower

d) Shape of cam

Answer: d

Explanation: The pressure angle of the cam depends on the following factors”

offset between center lines of follower and cam, angle of ascent and lift of follower, it is independent of the shape of cam.

4. If φ is the angle of action of cam on the circular flank, and r 1 is the base circle of the cam, then the maximum velocity of the follower is given by the equation.

a) ω(R-r 1 )sin φ

b) ω(R+r 1 )sin φ

c) ω(R-r 1 )cos φ

d) ω(R+r 1 )cos φ

Answer: a

Explanation: The velocity of the follower is dependent of the following factors:

base circle radius, rotation speed of the cam, angle of action of cam on the circular flank and Radius of circular flank.

5. If φ is the angle of action of cam on the circular flank, and r 1 is the base circle of the cam, then the maximum acceleration of the follower is given by the equation.

a) ω 2 (R-r 1 )

b) ω 2 (R+r 1 )sin φ

c) ω 2 (R-r 1 )cos φ

d) ω 2 (R+r 1 )cos φ

Answer: a

Explanation: The maximum acceleration of the follower is dependent of the following factors: base circle radius, rotation speed of the cam and radius of circular flank.

6. If φ is the angle of action of cam on the circular flank, and r 1 is the base circle of the cam, then the minimum acceleration of the follower is given by the equation.

a) ω 2 (R-r 1 )

b) ω 2 (R+r 1 )sin φ

c) ω 2 (R-r 1 )cos φ

d) ω 2 (R+r 1 )cos φ

Answer: c

Explanation: The minimum acceleration of the follower is dependent of the following factors:

base circle radius, rotation speed of the cam, cosine of angle of action of cam on the circular flank and radius of circular flank.

7. Flat faced follower’s acceleration when in contact with a circular arc cam is given by _________

a) ω 2 (R-r 1 )

b) ω 2 (R+r 1 )sin φ

c) ω 2 (R-r 1 )cos φ

d) ω 2 (R+r 1 )cos φ

Answer: c

Explanation: The acceleration of the follower is dependent of the following factors:

base circle radius, rotation speed of the cam, cosine of angle of action of cam on the circular flank and radius of circular flank.

8. The maximum acceleration of the follower occurs when the angle of action of cam on the circular flank is 90 degrees

a) True

b) False

Answer: b

Explanation: The acceleration of the follower is dependent of the following factors:

base circle radius, rotation speed of the cam, cosine of angle of action of cam on the circular flank and radius of circular flank. Since at φ = 90 degrees the cosine will become 0 and hence it will not be maximum.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Balancing of Rotating Masses”.

1. Often an unbalance of forces is produced in rotary or reciprocating machinery due to the ______

a) Centripetal forces

b) Centrifugal forces

c) Friction forces

d) Inertia forces

Answer: d

Explanation: Often an unbalance of forces is produced in rotary or reciprocating machinery due to the inertia forces associated with the moving masses.

2. Balancing of single rotating mass by balancing masses in same plane and in different planes cannot take place.

a) True

b) False

Answer: b

Explanation: Balancing of single rotating mass by balancing masses in same plane and in different planes can be done with the help of both static balancing and dynamic balancing.

3. Which of the following is true for centrifugal force causing unbalance?

a) Direction changes with rotation

b) Magnitude changes with rotation

c) Direction and magnitude both change with rotation

d) Direction and magnitude both remain unchanged with rotation

Answer: a

Explanation: Force acting radially outwards acts on the axis of rotation and is known as centrifugal force. This is a disturbing force on the axis of rotation, the magnitude of which is constant but the direction changes with the rotation of the mass.

4. In a revolving rotor, the centrifugal force remains balanced as long as the centre of the mass of the rotor lies ___________

a) Below the axis of shaft

b) On the axis of the shaft

c) Above the axis of shaft

d) Away from the axis of shaft

Answer: b

Explanation: In a revolving rotor, the centrifugal force remains balanced as long as the centre of the mass of the rotor lies on the axis of shaft. In other cases, the system will be unbalanced.

5. If the unbalanced system is not set right then.

a) Static forces develop

b) Dynamic forces develop

c) Tangential forces develop

d) Radial forces develop

Answer: a

Explanation: It is very essential that all the rotating and reciprocating parts should be completely balanced as far as possible, otherwise the dynamic forces will be set up.

6. What is not the effect of unbalanced forces?

a) Load on bearings

b) Dangerous vibrations

c) Stresses in various members

d) Violation of conservation of mass principle

Answer: d

Explanation: The unbalanced forces increase the loads on bearings and stresses in the various members, as well as it produces unpleasant and even dangerous vibrations.

7. What is the effect of a rotating mass of a shaft on a system?

a) Bend the shaft

b) Twist the shaft

c) Extend the shaft

d) Compress the shaft

Answer: a

Explanation: Whenever an object containing mass is attached to a rotating shaft, it exerts some centrifugal force. This centrifugal force’s effect is to bend the shaft and to produce vibrations in it.

8. Balancing of reciprocating masses is the process of providing the second mass in order to counteract the effect of the centrifugal force of the first mass.

a) True

b) False

Answer: b

Explanation: The process of providing the second mass in order to counteract the effect of the centrifugal force of the first mass, is called balancing of rotating masses. The main objective of balancing is to minimize the unbalanced forces.

This set of Machine Dynamics online quiz focuses on “Balancing of a Single Rotating Mass By a Single Mass Rotating in the Same Plane”.

1. Let the disturbing mass be 100 kg and the radius of rotation be 20 cm and the rotation speed be 50 rad/s, then calculate the centrifugal force in kN.

a) 50

b) 25

c) 50000

d) 25000

Answer: a

Explanation: Centrifugal force is given by the equation

Fc = mω 2 r

using the given data

Fc = 50000 N.

2. Centrifugal force acts radially inwards and bends the shaft.

a) True

b) False

Answer: b

Explanation: We know that the centrifugal force exerted by the mass on the shaft, this centrifugal force acts radially outwards and thus produces bending moment on the shaft.

3. The mass used to balance the mass defect is known as ______

a) Balancing mass

b) Defect mass

c) Replacement mass

d) Fixing mass

Answer: a

Explanation: Disturbing mass causes centrifugal force and bending of the shaft, this mass is balanced by counteracting the shaft with another mass known as balancing mass.

4. Let the disturbing mass be 100 kg and the radius of rotation be 10 cm and the rotation speed be 50 rad/s, then calculate the centrifugal force in kN.

a) 50

b) 25

c) 50000

d) 25000

Answer: b

Explanation: Centrifugal force is given by the equation

Fc = mω 2 r

using the given data

Fc = 25000 N.

5. Let the disturbing mass be 200 kg and the radius of rotation be 20 cm and the rotation speed be 50 rad/s, then calculate the centripetal force in kN.

a) -50

b) -25

c) -100

d) -2500

Answer: c

Explanation: Centripetal force is given by the equation

Fc = – mω 2 r

This is the negative of the centrifugal force.

using the given data

Fc = -100000 N.

6. Let the centrifugal force in kN be 25 and the radius of rotation be 20 cm and the rotation speed be 50 rad/s, then calculate the mass defect in Kg.

a) 50

b) 25

c) 50000

d) 25000

Answer: a

Explanation: Centrifugal force is given by the equation

Fc = mω 2 r

using the given data

Fc = 25000 N

r = 0.2 m

ω = 50 rad/s

m = 50 kg.

7. From the given data, calculate the unbalanced centrifugal force in N s 2

Distance from shaft = 0.2 m

Mass = 100 kg

Rotating speed = 1000 rpm.

a) 20

b) 2×10 7

c) 200

d) 20000

Answer: a

Explanation: Centrifugal force is given by

Mω 2 r

In N-s 2 = m.r

= 100×0.2 = 20 N-s 2 .

8. In balancing of single-cylinder engine, the rotating unbalance is ____________

a) completely made zero and so also the reciprocating unbalance

b) completely made zero and the reciprocating unbalance is partially reduced

c) partially reduced and the reciprocating unbalance is completely made zero

d) partially reduced and so also the reciprocating unbalance

Answer: b

Explanation: In balancing of single-cylinder engine, the rotating unbalance is completely made zero and the reciprocating unbalance is partially reduced.

This set of Tough Machine Dynamics Questions and Answers focuses on “Balancing of a Single Rotating Mass By Two Masses Rotating in Different Planes”.

1. Let the disturbing mass be 50 Kg, with radius of rotation = 0.1m, if one of the balancing mass is 30 Kg at a radius of rotation 0.1m then find the other balancing mass situated at a distance of 0.2m.

a) 80

b) 40

c) 20

d) 10

Answer: d

Explanation: mr = m 1 mr + m 2 r 2

50×0.1 = 30×0.1 + m 2 x0.2

5=3 + 0.2m

m 2 =10 Kg.

2. A single mass defect needs to be balanced by a single balancing mass

a) True

b) False

Answer: b

Explanation: A single mass defect lying on a single plane can be countered with the help of different masses lying on different planes with the defective mass either being at center or to the either side.

3. In a system two masses are used to balance the unbalanced forces. Find the mass of the balancing mass which has to be situated at a distance of 20cm, if the disturbing mass is of 100 Kg having radius of rotation of 0.1m. One of the balancing mass is 30 Kg with RoR of 10cm.

a) 70

b) 35

c) 20

d) 10

Answer: b

Explanation: For a balanced system

m.r = m 1 r + m 2 r 2

100×0.1 = 30×0.1 + m 2 x0.2

10=3 + 0.2m

m 2 =35 Kg.

4. In the given figure, m 1 =20 Kg, m 2 =30Kg and m=50 Kg, if r 1 =0.2m and r=0.3m, l=1m, find l 2 .

tough-machine-dynamics-questions-answers-q4

a) 0.26m

b) 0.52m

c) 1.04m

d)0.13m

Answer: a

Explanation: M 1 r 1 l = mrl 2

20×0.2×1 = 50 x 0.3 xl 2

4 = 15xl 2

l 2 = 0.26m.

5. In the given figure, m 1 =10, Kg m 2 =30Kg and m=50 Kg, if r 1 =0.2m and r=0.3m, l=1m, find l 2 .

tough-machine-dynamics-questions-answers-q4

a) 0.26m

b) 0.52m

c) 1.04m

d) 0.13m

Answer: d

Explanation: M 1 r 1 l = mrl 2

10×0.2×1 = 50 x 0.3 xl 2

2 = 15xl 2

l 2 = 0.13m.

6. In the given figure, m 1 =10 Kg, m 2 =30Kg and m=50 Kg, if r=0.3m, l=1m, find l 2 = 0.5m, find r1 in m.

tough-machine-dynamics-questions-answers-q4

a) 1.5

b) 0.75

c) 3

d) 6

Answer: b

Explanation: M 1 r 1 l = mrl 2

10xr 1 x1 = 50 x 0.3 x0.5

10r 1 = 7.5

l 2 = 0.75m.

7. Unbalanced mass leads to vibrations.

a) True

b) False

Answer: a

Explanation: The unbalanced forces not only increase the loads on bearings and stresses in the various members, but also produce unpleasant and even dangerous vibrations.

8. Which of the following statement is correct?

a) In any engine, 100% of the reciprocating masses can be balanced dynamically

b) In the case of balancing of multicylinder engine, the value of secondary force is higher than the value of the primary force

c) In the case of balancing of multimass rotating systems, dynamic balancing can be directly started without static balancing done to the system

d) none of the mentioned

Answer: c

Explanation: None.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Balancing of Several Masses Rotating in the Same Plane”.

1. From the given data, find the balancing mass in Kg if r=0.2m required in the same plane.

Masses = 200kg, 300kg, 240 kg, 260Kg, corresponding radii = 0.2m, 0.15m, 0.25m and 0.3m.

Angles between consecutive masses = 45, 75 and 135 degrees.

a) 116

b) 58

c) 232

d) 140

Answer: a

Explanation: To make the calculation simple, we will calculate centrifugal force as m.r

hence

horizontal forces = mrcosθ = 21.6 Kg-m

vertical forces = mrsinθ = 8.5 Kg-m

Resultant = 23.2 Kg-m

R = 0.2 m

therefore, M = 116Kg

2. Graphical method gives the best results.

a) True

b) False

Answer: b

Explanation: In order to find the balancing mass, analytical and graphical approach can be used, analytical method gives the accurate results as there is less scope for error.

3. From the given data, find the balancing mass’s inclination in degrees if r=0.2m required in the same plane.

Masses = 200kg, 300kg, 240 kg, 260Kg, corresponding radii = 0.2m, 0.15m, 0.25m and 0.3m.

Angles between consecutive masses = 45, 75 and 135 degrees.

a) 201.48

b) 200.32

c) 210.34

d) 202.88

Answer: a

Explanation: horizontal forces = mrcosθ = 21.6 Kg-m

vertical forces = mrsinθ = 8.5 Kg-m

Resultant = 23.2 Kg-m

R = 0.2 m

therefore, M =116Kg

theta = vertical / horizontal

= 180 + 21.48 = 201.48.

4. If all the masses are in one plane, then what is the maximum no. of masses which can be placed in the same plane?

a) 3

b) 4

c) 6

d) No limitation

Answer: d

Explanation: while balancing in one plane, any number of masses can be placed, the net result will only depend on the sum of vertical and horizontal components and the resultant should be equal to the unbalance.

5. If the rotation speed of the shaft increases then the balancing mass must also increase.

a) True

b) False

Answer: b

Explanation: The balancing mass to be placed in the same plane as the other unbalanced masses, is independent of the speed at which the shaft rotates.

6. If all the masses are in one plane, then in which of the following condition is possible?

a) Resultant only in horizontal

b) Resultant only in vertical

c) System remains unbalanced

d) No limitation

Answer: d

Explanation: While balancing in one plane, there is no restriction of placing masses, the net result will only depend on the vector sum of vertical and horizontal components and the resultant should be equal to the unbalance.

7. The secondary unbalanced force produced by the reciprocating parts of a certain cylinder of a given engine with crank radius r and connecting rod length l can be considered as equal to primary unbalanced force produced by the same weight having

a) an equivalent crank radius r 2 /4l and rotating at twice the speed of the engine

b) r 2 /4l as equivalent crank radius and rotating at engine speed

c) equivalent crank length of r 2 /4l and rotating at engine speed

d) none of the mentioned

Answer: a

Explanation: We know that secondary force = primary force

F S = F P

Therefore, to balance the force the primary force should contain an equivalent crank radius r 2 /4l and rotating at twice the speed of the engine.

8. The effect of hammer blow in a locomotive can be reduced by ______________

a) decreasing the speed

b) using two or three pairs of wheels coupled together

c) balancing whole of the reciprocating parts

d) decreasing the speed and using two or three pairs of wheels coupled together

Answer: d

Explanation: None.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Balancing of Several Masses Rotating in Different Planes”.

1. Which of the following statements are associated with complete dynamic balancing of rotating systems?

1. Resultant couple due to all inertia forces is zero.

2. Support reactions due to forces are zero but not due to couples.

3. The system is automatically statically balanced.

4. Centre of masses of the system lies on the axis of rotation.

a) 1, 2, 3 and 4

b) 1, 2, and 3 only

c) 2, 3 and 4 only

d) 1, 3 and 4 only

Answer: d

Explanation: Support reactions due to forces & couple both need to be zero.

2. Which of the following statements is correct about the balancing of a mechanical system?

a) If it is under static balance, then there will be dynamic balance also

b) If it is under dynamic balance, then there will be static balance also

c) Both static as well as dynamic balance have to be achieved separately

d) None of the mentioned

Answer: c

Explanation: Complete balancing means both static & dynamic should be balanced.

3. The magnitude of swaying couple due to partial balance of the primary unbalancing force in locomotive is

a) inversely proportional to the reciprocating mass

b) directly proportional to the square of the distance between the centre lines of the two cylinders

c) inversely proportional to the distance between the centerlines of the two cylinders

d) directly proportional to the distance between the centerlines of the two cylinders

Answer: d

Explanation: Torque about “C”

Swaying couple a b

4. In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to

a) minimize the effect of primary forces

b) minimize the effect of secondary forces

c) have perfect balancing

d) start the locomotive conveniently

Answer: b

Explanation: As l increases FS decreases

Hence the ratio is kept large to avoid secondary forces for each balancing.

5. In case of partial balancing of locomotives, the maximum magnitude of the unbalanced force perpendicular to the line of stroke is called hammer blow and this has to be limited by proper choice of the balancing mass and its radial position.

a) True

b) False

Answer: a

Explanation: The effect of hammer blow is to cause variation in pressure between the wheel and the rail, and it may sometimes cause the lifting of wheels from the rails.

6. Multi-cylinder engines are desirable because

a) only balancing problems are reduced

b) only flywheel size is reduced

c) both  and 

d) none of the mentioned

Answer: c

Explanation: None.

7. When the primary direct crank of a reciprocating engine makes an angle θ with the line of stroke, then the secondary direct crank will make an angle of . . . . . with the line of stroke.

a) θ /2

b) θ

c) 2 θ

d) 4 θ

Answer: c

Explanation: The secondary direct crank angle is always twice of primary direct crank angle.

8. Secondary forces in reciprocating mass on engine frame are

a) of same frequency as of primary forces

b) twice the frequency as of primary forces

c) four times the frequency as of primary forces

d) none of the mentioned

Answer: b

Explanation: None.

This set of Machine Dynamics Question Bank focuses on “Primary and Secondary Unbalanced Forces of Reciprocating Masses”.

1. The primary unbalanced force is maximum when the angle of inclination of the crank with the line of stroke is

a) 0°

b) 90°

c) 180°

d) 360°

Answer: c

Explanation: The primary unbalanced force is maximum, when θ = 0° or 180°. Thus, the primary force is maximum twice in one revolution of the crank. The maximum primary unbalanced force is given by

F P = m⋅ω 2 ⋅r

2. The partial balancing means

a) balancing partially the revolving masses

b) balancing partially the reciprocating masses

c) best balancing of engines

d) all of the mentioned

Answer: b

Explanation: To balance the reciprocating masses partially is known as partial balancing.

3. In order to facilitate the starting of locomotive in any position, the cranks of a locomotive, with two cylinders, are placed at ______________ to each other.

a) 45°

b) 90°

c) 120°

d) 180°

Answer: b

Explanation: Due to the partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke and also an unbalanced primary force perpendicular to the line of stroke.

4. In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to

a) minimise the effect of primary forces

b) minimise the effect of secondary forces

c) have perfect balancing

d) start the locomotive quickly

Answer: b

Explanation: The secondary unbalanced force is maximum, when θ = 0°, 90°,180° and 360°. Thus, the secondary force is maximum four times in one revolution of the crank. Keeping large distance between connecting rod and crank, minimises the effect of secondary forces.

5. If c be the fraction of the reciprocating parts of mass m to be balanced per cyclinder of a steam locomotive with crank radius r, angular speed ω, distance between centre lines of two cylinders a, then the magnitude of the maximum swaying couple is given by

a) 1 – c / 2 mrω 2 a

b) 1 – c / √2mrω 2 a

c) √2mrω 2 a

d) none of the mentioned

Answer: b

Explanation: The swaying couple is maximum or minimum when  is maximum or minimum. For  to be maximum or minimum

∴ tanθ =1 or θ = 45° or 225°

Maximum value of the swaying couple = 1 – c / √2mrω 2 a

6. The swaying couple is maximum or minimum when the angle of inclination of the crank to the line of stroke  is equal to

a) 45° and 135°

b) 90° and 135°

c) 135° and 225°

d) 45° and 225°

Answer: d

Explanation: The swaying couple is maximum or minimum when  is maximum or minimum. For  to be maximum or minimum

∴ tanθ =1 or θ = 45° or 225°

7. The tractive force is maximum or minimum when the angle of inclination of the crank to the line of stroke  is equal to

a) 90° and 225°

b) 135° and 180°

c) 180° and 225°

d) 135° and 315°

Answer: d

Explanation: The tractive force is maximum or minimum when  is maximum or minimum. For  to be maximum or minimum,

∴ tanθ = −1 or θ =135° or 315°

8. The swaying couple is due to the

a) primary unbalanced force

b) secondary unbalanced force

c) two cylinders of locomotive

d) partial balancing

Answer: a

Explanation: The unbalanced forces along the line of stroke for the two cylinders constitute a couple. This couple has swaying effect about a vertical axis, and tends to sway the engine alternately in clockwise and anticlockwise directions. Hence the couple is known as swaying couple.

9. In a locomotive, the maximum magnitude of the unbalanced force along the perpendicular to the line of stroke, is known as

a) tractive force

b) swaying couple

c) hammer blow

d) none of the mentioned

Answer: c

Explanation: The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as hammer blow. The effect of hammer blow is to cause the variation in pressure between the wheel and the rail.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Partial Balancing of Locomotives”.

1. What is the total no. of cylinders in a locomotive having crank at right angles?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: The locomotives, usually, have two cylinders with cranks placed at right angles to each other, this is done in order to have uniformity in turning moment diagram.

2. For a two cylinder locomotives, the crank are placed at right angles to each other.

a) True

b) False

Answer: a

Explanation: The locatives in general have two cylinders having cranks at an angle of 90 degrees to each other in order to have uniform turning moment diagram.

3. In an inside cylinder locomotive, the position of the driving wheels is _________

a) Outside cylinder

b) Inside cylinder

c) Down to cylinder

d) Above the cylinder

Answer: a

Explanation: In an inside cylinder locomotive, the two cylinders are placed in between the two driving wheels at an angle of 90 degrees to each other.

4. In an outside cylinder locomotive, the position of the driving wheels is _________

a) Outside cylinder

b) Inside cylinder

c) Down to cylinder

d) Above the cylinder

Answer: b

Explanation: In an outside cylinder locomotive, the two cylinders are placed outside the two driving wheels at an angle of 90 degrees to each other.

5. A single or uncoupled locomotive is one, in which the effort is transmitted to _____

a) Both the pair of wheels

b) Alternatively between the wheels

c) One pair of wheels

d) Neither of the wheels

Answer: c

Explanation: A single or uncoupled locomotive is one, in which the effort is transmitted to one pair of the wheels only.

6. In coupled locomotives, the driving wheels are connected to the leading and trailing wheel by an _______ coupling rod.

a) Inside

b) Outside

c) Sideway

d) Bottom

Answer: b

Explanation: In coupled locomotives, the driving wheels are connected to the leading and trailing wheel by an outside coupling rod.

7. Identify the type of locomotive from the given figure.

machine-dynamics-questions-answers-partial-balancing-locomotives-q7

a) Inside cylinder

b) Outside cylinder

c) Top cylinder

d) Bottom cylinder

Answer: a

Explanation: The given figure is of an inside cylinder locomotive, In an inside cylinder locomotive, the two cylinders are placed in between the two driving wheels at an angle of 90 degrees to each other.

8. In the given figure, the locomotive is a 4 cylinder locomotive.

machine-dynamics-questions-answers-partial-balancing-locomotives-q7

a) True

b) False

Answer: b

Explanation: The given figure is of an inside cylinder locomotive, since there are only two cranks at right angles to each other, it is two cylinder locomotive.

9. Identify the type of locomotive from the given figure.

machine-dynamics-questions-answers-partial-balancing-locomotives-q9

a) Inside cylinder

b) Outside cylinder

c) Top cylinder

d) Bottom cylinder

Answer: b

Explanation: The given figure is of an outside cylinder locomotive, In an outside cylinder locomotive, the two cylinders are on the either side of the two driving wheels at an angle of 90 degrees to each other.

This set of Machine Dynamics Puzzles focuses on “Effect of Partial Balancing of Reciprocating Parts of Two Cylinder Locomotives”.

1. What is the effect of partial balancing of the reciprocating parts?

a) Unbalanced primary force

b) Unbalanced secondary force

c) Balanced primary force

d) Balanced secondary force

Answer: a

Explanation: The reciprocating parts are only partially balanced. Due to this partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke.

2. Due to partial balancing there is unbalanced force only along the line of stroke.

a) True

b) False

Answer: b

Explanation: Partial balancing results in generation of unbalanced force acting along the line of stroke and in a direction perpendicular to the line of stroke.

3. The effect of an unbalanced primary force along the line of stroke is to produce ________

a) Swaying couple

b) Constant tractive force

c) Piston effort

d) Crank effort

Answer: a

Explanation: The major effect of an unbalanced primary force along the line of stroke is to produce Swaying couple and change in Tractive force along the line of stroke.

4. The effect of an unbalanced primary force perpendicular to the line of stroke is to produce _____

a) Variation in pressure on the rails

b) Variation in tractive force

c) Swaying couple

d) Piston effort

Answer: a

Explanation: The effect of an unbalanced primary force along the line of stroke is to produce Swaying couple and variation in tractive force along the line of stroke and The effect of an unbalanced primary force perpendicular to the line of stroke is to produce variation in pressure on the rails.

5. What is the effect of variation of pressure on the rails?

a) Hammering action

b) Piston effort

c) Variation in tractive force

d) Swaying couple

Answer: a

Explanation: The effect of an unbalanced primary force perpendicular to the line of stroke is to produce variation in pressure on the rails, which results in hammering action on the rails.

6. Hammering action on the rails is caused by _______

a) Variation of pressure on the rails

b) Variation of temperature on the rails

c) Variation of thermal diffusibility of the rails

d) Variation of crank effort

Answer: a

Explanation: The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke of rails is the hammering action caused by the variation in pressure on the rails due to unbalanced forces.

7. Hammer blow is along the line of stroke of the unbalanced force.

a) True

b) False

Answer: b

Explanation: Hammer blow is defined as the maximum magnitude of the unbalanced force along the perpendicular to the line of action of the stroke, it’s effect can be seen as the hammering effect on the rails.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Variation of Tractive Force”.

1. The resultant unbalanced force due to the two cylinders, along the line of stroke, is known as ______

a) Tractive force

b) Swaying couple

c) Hammer blow

d) Hammer couple

Answer: a

Explanation: The resultant unbalanced force due to the two cylinders, along the line of stroke, is known as tractive force. The value of tractive force depends on the angle of the crank.

2. Tractive force is produced due to unbalanced forces perpendicular to the line of stroke of action.

a) True

b) False

Answer: b

Explanation: Tractive force is produced due to unbalanced forces along the line of stroke of action. The value of tractive force depends on the angle of the crank.

3. At which of the following angles in degrees does the tractive force attains a maximum value?

a) 135

b) 45

c) 90

d) 60

Answer: a

Explanation: The tractive force is maximum or minimum when  is maximum or minimum, differentiating it with respect to theta gives tan θ = -1

hence maximum/minimum value occurs at 135 or 315 degrees.

4. At which of the following angles in degrees does the tractive force attains a minimum value?

a) 315

b) 45

c) 90

d) 60

Answer: a

Explanation: The tractive force is minimum when  is minimum, differentiating it with respect to theta gives tan θ = -1

hence maximum/minimum value occurs at 135 or 315 degrees.

5. Find the maximum value of tractive force in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a) 6978

b) 7574

c) 6568

d) 7374

Answer: a

Explanation: The maximum value of tractive force is given by

+√2 mω 2 r

substituting the values we get

F = 6978 N.

6. Find the variation in tractive force in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a) 13956

b) 17574

c) 16568

d) 17374

Answer: a

Explanation: The variation in tractive force is given by

±√2 mω 2 r

substituting the values we get

F = 13956 N.

7. Find the minimum value of tractive force in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a) -6978

b) -7574

c) -6568

d) -7374

Answer: a

Explanation: The minimum value of tractive force is given by

-√2 mω 2 r

substituting the values we get

F = 6978 N.

8. Tractive force attains a maxima or minima when the crank angle is 90 degrees.

a) True

b) False

Answer: b

Explanation: The tractive force is maximum or minimum when the expression  is maximum or minimum, differentiating it with respect to theta gives tan θ = -1

hence maximum/minimum value occurs at 135 or 315 degrees.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Swaying Couple”.

1. The unbalanced forces along the line of stroke for the two cylinders constitute a ______ about the centre line between the cylinders.

a) Tension

b) Compression

c) Torsion

d) Couple

Answer: d

Explanation: The unbalanced forces along the line of stroke for the two cylinders constitute a couple about the centre line between the cylinders. This couple is known as swaying couple as it has a swaying effect.

2. This couple has swaying effect about a horizontal axis.

a) True

b) False

Answer: b

Explanation: This couple has swaying effect about a vertical axis, and tends to sway the engine alternately in clockwise and anticlockwise directions. About horizontal axis no swaying effect is observed.

3. Which of the following has a tendency to sway engine in alternately in clockwise and anticlockwise directions?

a) Swaying couple

b) Tractive force

c) Hammer blow

d) Hammer couple

Answer: a

Explanation: A couple having a swaying effect about a vertical axis, and a tendency to sway the engine alternately in clockwise and anticlockwise directions is known as swaying couple.

4. At which of the following angles in degrees does the swaying couple attains a maximum value?

a) 135

b) 45

c) 90

d) 60

Answer: b

Explanation: The swaying couple is maximum or minimum when  is maximum or minimum. This results in a relation tanθ=1

hence theta = 45 degrees or 225 degrees.

5. If the maximum value occurs at 45 degrees then at which of the following angles in degrees will the swaying couple attains its maximum value again?

a) 135

b) 45

c) 225

d) 60

Answer: c

Explanation: The swaying couple is maximum or minimum when the relation  is maximum or minimum. This results in a relation tan θ=1

hence theta = 45 degrees or 225 degrees.

6. Find the maximum value of swaying couple in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a = 2\ 6978

b) 7574

c) 6568

d) 7374

Answer: a

Explanation: The maximum value of Swaying couple is given by

+a/\mω 2 r

substituting the values we get

F = 6978 N.

7. Find the minimum value of swaying couple in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a = 2\ -6978

b) -7574

c) -6568

d)- 7374

Answer: a

Explanation: The minimum value of Swaying couple is given by

-mω 2 r

substituting the values we get

F = -6978 N.

8. Swaying couple attains a maxima or minima when the crank angle is 45 degrees.

a) True

b) False

Answer: a

Explanation: The swaying couple is maximum or minimum when the expression  is maximum or minimum, differentiating it with respect to theta gives tan θ = 1

hence maximum/minimum value occurs at 45 or 225 degrees.

9. Find the variation in swaying couple in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a = 2\ 13956

b) 17574

c) 16568

d) 17374

Answer: a

Explanation: The variation in tractive force is given by

±a/\mω 2 r

substituting the values we get

F = 13956 N.

10. Find the maximum value of swaying couple in newtons from the following data:

m = 200Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a = 2\ 13956

b) 17574

c) 16568

d) 17374

Answer: a

Explanation: The maximum value of Swaying couple is given by

+a/\mω 2 r

substituting the values we get

F = 13956 N.

This set of Machine Dynamics Questions and Answers for Entrance exams focuses on “Balancing of Coupled Locomotives & Secondary Forces of Multi-cylinder Inline Engines”.

1. In a coupled locomotive, the driving wheels are connected to the leading and trailing wheels by a/an __________

a) Outside coupling rod

b) Inside coupling rod

c) Backside coupling rod

d) Frontend coupling rod

Answer: a

Explanation: In a coupled locomotive, the driving wheels are connected to the leading and trailing wheels by an outside coupling rod.

2. For a coupled locomotive, the driving wheels and the leading and trailing wheels are connected by an inside coupling rod.

a) True

b) False

Answer: b

Explanation: In a coupled locomotive, the driving wheels are connected to the leading and trailing wheels by an outside coupling rod.

3. On connecting a wheel through outside coupling rod, which utility comes of use?

a) Tractive

b) Swaying

c) Hammer

d) Piston

Answer: a

Explanation: For a coupled locomotive, the driving wheels and the leading and trailing wheels are connected by an outside coupling rod. A significantly greater portion of the engine mass is utilized by tractive purposes.

4. In coupled locomotives, the coupling rod cranks are placed diametrically opposite to the adjacent ________

a) Driving cranks

b) Driving rods

c) Driving piston

d) Driving valves

Answer: a

Explanation: In coupled locomotives, the coupling rod cranks are placed at a position which is diametrically opposite to the adjacent driving cranks which are also known as the main cranks.

5. Coupling rods with connecting rods and pins can be treated as ________

a) Rotating masses

b) Masses in linear motion

c) Reciprocating masses

d) Massless

Answer: a

Explanation: The coupling rods together along with cranks and pins can be treated as rotating masses, and completely balanced by masses in the respective wheels i.e trailing, driving and leading.

6. The variation of pressure between the wheel and the rail  may be reduced by _______

a) Equal distribution of balanced masses

b) More mass towards driving wheel

c) More mass towards trailing wheel

d) More mass towards leading wheel

Answer: a

Explanation: The variation of pressure between the wheel and the rail also known as hammer blow, it can be reduced by equal distribution of the balanced mass between the three wheels i.e driving, leading and trailing wheels respectively.

7. In coupled locomotives, where are the coupling rod cranks placed with respect to the main crank?

a) Diametrically opposite

b) At 90 degrees

c) At 60 degrees

d) At 120 degrees

Answer: b

Explanation: In coupled locomotives, the coupling rod cranks are placed at a position which is diametrically opposite to the adjacent main cranks .

8. If the connecting rod is not too long, then which of the following phenomenon occurs?

a) Secondary disturbing forces

b) Primary disturbing forces

c) Secondary stabilizing forces

d) Primary stabilizing forces

Answer: a

Explanation: When the connecting rod isn’t too long or in other words when the obliquity of the connecting rod is taken into consideration, then the secondary disturbing force due to the reciprocating mass arises.

9. When the connecting rod is too long, secondary forces arise due to reciprocating masses.

a) True

b) False

Answer: b

Explanation: When the obliquity of the connecting rod is taken into consideration which means that the long is not too long, then the secondary disturbing force due to the reciprocating mass arises.

10. Which of the following is the correct expression for secondary force?

a) mω 2 r.cos2θ/n

b) mω 2 r.sin2θ/n

c) mω 2 r.tan2θ/n

d) mω 2 r.cosθ/n

Answer: a

Explanation: The secondary force depends on the following factors:

mass, rotating speed, radius of crank and the cosine of the crank angle.

11. In a multicylinder inline engine, each imaginary secondary crank with a mass attached to the crankpin is inclined to the line of stroke at which angle?

a) Twice the angle of crank

b) Half the angle of crank

c) Thrice the angle of crank

d) Four times the angle of crank

Answer: a

Explanation: In multi-cylinder in-line engines, each of the imaginary secondary crank with a mass attached to the crankpin is inclined to the line of stroke at twice the angle of the actual crank.

12. The numerical values of the secondary forces and secondary couples couples may be obtained by considering the ___________

a) Revolving mass

b) Reciprocating mass

c) Translating mass

d) Rotating mass

Answer: a

Explanation: Each imaginary secondary crank with a mass attached to the crankpin is inclined to the line of stroke at two times the angle of the actual crank. The values of the secondary forces and couples can be obtained by considering the revolving mass.

13. For the secondary balancing of the engine, which of the condition is necessary?

a) Secondary force polygon must be close

b) Secondary force polygon must be open

c) Primary force polygon must be close

d) Primary force polygon must be open

Answer: a

Explanation: In order to achieve the secondary balancing of the engine. The sum  of the secondary forces must be equal to zero. In other words, the secondary force polygon must close.

14. For the secondary balancing of the engine, which of the condition is necessary?

a) Secondary couple polygon must be close

b) Secondary couple polygon must be open

c) Primary force polygon must be close

d) Primary force polygon must be open

Answer: a

Explanation: In order to achieve the secondary balancing of the engine. The sum  of the secondary couples must be equal to zero. In other words, the secondary couple polygon must close.

15. If the secondary couple polygon is open and the force polygon is closed then stability is achieved.

a) True

b) False

Answer: b

Explanation: In order to achieve stability, the sum of secondary forces and secondary couples must be zero and their respective polygons must be closed.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Balancing of V-engines”.

1. A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 11.5 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons?

a) 736

b) 836

c) 936

d) 636

Answer: b

Explanation: Maximum resultant secondary force is given by the equation:

\xω 2 .r

substituting the values we get

Fsmax = 836 N.

2. If the mass of the reciprocating parts is doubled, then the primary force is halved.

a) True

b) False

Answer: b

Explanation: The primary force has a direct relation with the mass of the reciprocating parts thus doubling the mass will result in an increment of two times in the primary force.

3. A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 23 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons?

a) 7172

b) 1672

c) 1122

d) 1272

Answer: b

Explanation: Maximum resultant secondary force is given by the equation:

\xω 2 .r

substituting the values we get

Fsmax = 1672 N.

4. A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 34.5 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in kN?

a) 2.238

b) 2.508

c) 2.754

d) 2.908

Answer: b

Explanation: Maximum resultant secondary force is given by the equation:

\xω 2 .r

substituting the values we get

Fsmax = 2508 N.

5. From the following data of a 60 degree V-twin engine, determine the maximum value for secondary forces in newtons:

Reciprocating mass per cylinder = 1.5 Kg

Stroke length = 10 cm

Length of connecting rod = 25 cm

Engine speed = 2500 rpm

a) 890.3

b) 760.4

c) 580.3

d) 970.6

Answer: a

Explanation: The maximum value for secondary force is given by the equation:

\.ω 2 .r

substituting the values we get

Fpmax = 890.3 N.

6. From the following data of a 60 degree V-twin engine, determine the maximum value for primary forces in newtons:

Reciprocating mass per cylinder = 1.5 Kg

Stroke length = 10 cm

Length of connecting rod = 25 cm

Engine speed = 2500 rpm

a) 7711

b) 4546

c) 2508

d) 8764

Answer: a

Explanation: The primary force is maximum, when θ = θ°

therefore the equation for maximum primary force becomes

ω 2 rx3

substituting the values we get

Fpmax = 7711 N.

7. From the following data of a 60 degree V-twin engine, determine the minimum value for primary forces in newtons:

Reciprocating mass per cylinder = 1.5 Kg

Stroke length = 10 cm

Length of connecting rod = 25 cm

Engine speed = 2500 rpm

a) 7711

b) 4546

c) 2570

d) 8764

Answer: c

Explanation: The primary force is maximum, when θ = 90°

therefore the equation for maximum primary force becomes

ω 2 r

substituting the values we get

Fpmax = 2570 N.

8. For a V-twin engine, which of the following means can be used to balance the primary forces?

a) Revolving balance mass

b) Rotating balance mass

c) Reciprocating balance mass

d) By the means of secondary forces

Answer: a

Explanation: We know that primary resultant force depends on speed of revolution, mass and radius of rotation. Hence revolving mass can be used to balance the primary forces.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Vibratory Motion”.

1. When there is a reduction in amplitude over every cycle of vibration, then the body is said to have

a) free vibration

b) forced vibration

c) damped vibration

d) none of the mentioned

Answer: c

Explanation: When no external force acts on the body, after giving it an initial displacement, then the body is said to be under free or natural vibrations. The frequency of the free vibrations is called free or natural frequency.

When there is a reduction in amplitude over every cycle of vibration, the motion is said to be damped vibration.

2. Longitudinal vibrations are said to occur when the particles of a body moves

a) perpendicular to its axis

b) parallel to its axis

c) in a circle about its axis

d) none of the mentioned

Answer: b

Explanation: When the particles of the shaft or disc moves parallel to the axis of the shaft, then the vibrations are known as longitudinal vibrations.

When the particles of the shaft or disc move approximately perpendicular to the axis of the shaft, then the vibrations are known as transverse vibrations.

3. When a body is subjected to transverse vibrations, the stress induced in a body will be

a) shear stress

b) tensile stress

c) compressive stress

d) none of the mentioned

Answer: b

Explanation: In transverse vibrations,the shaft is straight and bent alternately and bending stresses are induced in the shaft.

4. The natural frequency  of free longitudinal vibrations is equal to

a) 1/2π√s/m

b) 1/2π√g/δ

c) 0.4985/δ

d) all of the mentioned

Answer: d

Explanation: Natural Frequency, f n = 0.4985/δ

where m = Mass of the body in kg,

s = Stiffness of the body in N/m, and

δ = Static deflection of the body in metres.

5. The factor which affects the critical speed of a shaft is

a) diameter of the disc

b) span of the shaft

c) eccentricity

d) all of the mentioned

Answer: d

Explanation: To determine the critical speed of a shaft which may be subjected to point loads, uniformly distributed load or combination of both, we find the frequency of transverse vibration which is equal to critical speed of a shaft in r.p.s. The Dunkerley’s method may be used for calculating the frequency.

6. The equation of motion for a vibrating system with viscous damping is

d 2 x/dt 2 + c/m X dx/dt + s/m X x = 0

If the roots of this equation are real, then the system will be

a) over damped

b) under damped

c) critically damped

d) none of the mentioned

Answer: a

Explanation: When the roots are real, overdamping takes place.

When the roots are complex conjugate underdamping takes place.

7. In under damped vibrating system, if x 1 and x 2 are the successive values of the amplitude on the same side of the mean position, then the logarithmic decrement is equal to

a) x 1 /x 2

b) log (x 1 /x 2 )

c) loge (x 1 /x 2 )

d) log (x 1 .x 2 )

Answer: b

Explanation: None

8. The ratio of the maximum displacement of the forced vibration to the deflection due to the static force, is known as

a) damping factor

b) damping coefficient

c) logarithmic decrement

d) magnification factor

Answer: d

Explanation: Magnificiant Factor is the ratio of maximum displacement of the forced vibration (x max ) to the deflection due to the static force F(x o ).

Damping Factor is the ratio of amping coefficient for the actual system, and damping coefficient for the critical damped system.

9. In vibration isolation system, if ω/ω n is less than √2 , then for all values of the damping factor, the transmissibility will be

a) less than unity

b) equal to unity

c) greater than unity

d) zero

where ω = Circular frequency of the system in rad/s, and

ω n = Natural circular frequency of vibration of the system in rad/s.

Answer: c

Explanation: The value of ω/ω n must be greater than √2 if ε is to be less than 1 and it is the numerical value of ε , independent of any phase difference between the forces that may exist which is important.

Answer: c

Explanation: There is a phase difference of 180° between the transmitted force and the disturbing force.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Types of Vibratory Motion and Natural Frequency of Free Longitudinal Vibrations”.

1. In vibration isolation system, if ω/ω n , then the phase difference between the transmitted force and the disturbing force is

a) 0°

b) 90°

c) 180°

d) 270°

Answer: c

Explanation: In Vibration-isolation system, If the phase difference between transmitted force and the disturbing force is 180°C, then ω/ω n = 1.

2. When a body is subjected to transverse vibrations, the stress induced in a body will be

a) shear stress

b) bending stress

c) tensile stress

d) compressive stress

Answer: b

Explanation: The critical speed of a shaft with a disc supported in between is equal to the natural frequency of the system in transverse vibrations and the stress induced is bending stress.

3. The critical speed of a shaft with a disc supported in between is equal to the natural frequency of the system in

a) transverse vibrations

b) torsional vibrations

c) longitudinal vibrations

d) none of the mentioned

Answer: a

Explanation: The critical speed of a shaft with a disc supported in between is equal to the natural frequency of the system in transverse vibrations and the stress induced is bending stress.

4. In steady state forced vibrations, the amplitude of vibrations at resonance is _____________ damping coefficient.

a) equal to

b) directly proportional to

c) inversely proportional to

d) independent of

Answer: c

Explanation: Forced vibration is when a time-varying disturbance  is applied to a mechanical system. The disturbance can be a periodic and steady-state input, a transient input, or a random input. The periodic input can be a harmonic or a non-harmonic disturbance.

5. When there is a reduction in amplitude over every cycle of vibration, then the body is said to have

a) free vibration

b) forced vibration

c) damped vibration

d) under damped vibration

Answer: c

Explanation: The vibrations of a body whose amplitude goes on reducing over every cycle of vibrations are known as damped vibrations. This is due to the fact that a certain amount of energy possessed by the vibrating body is always dissipated in overcoming frictional resistance to the motion.In these vibrations, the amplitude of the vibrations decreases exponentially due to damping forces like frictional force, viscous force, hysteresis etc.

6. In vibration isolation system, if ω/ω n < 2, then for all values of damping factor, the transmissibility will be

a) less than unity

b) equal to unity

c) greater than unity

d) zero

Answer: c

Explanation: For underdamped systems the maximum amplitude of excitation has a definite value and it occurs at a frequency ω/ω n 1. For damping factor, the transmissibility will be greater than unity.

7. The accelerometer is used as a transducer to measure earthquake in Richter scale. Its design is based on the principle that

a) its natural frequency is very low in comparison to the frequency of vibration

b) its natural frequency is very high in comparison to the frequency of vibration

c) its natural frequency is equal to the frequency of vibration

d) measurement of vibratory motion is without any reference point

Answer: c

Explanation: Natural frequency need to be equal to frequency of vibration so that resonance exists and it should show the indication of earthquake.

8. While calculating the natural frequency of a spring-mass system, the effect of the mass of the spring is accounted for by adding X times its value to the mass, where X is

a) 1/2

b) 1/3

c) 1/4

d) 3/4

Answer: b

Explanation: Velocity at a distance “y” from fixed End = Velocity at free end /length of spring x y

∆k = 1/2 x M/3 x v 2 .

9. Critical speed is expressed as

a) rotation of shaft in degrees

b) rotation of shaft in radians

c) rotation of shaft in minutes

d) natural frequency of the shaft

Answer: d

Explanation: Critical speed is expressed as natural frequency of the shaft.

10. The first critical speed of an automobile running on a sinusoidal road is calculated by 

a) Resonance

b) Approximation

c) Superposition

d) Rayleigh quotient

Answer: a

Explanation: Frequency of automobile and road are same.

11. The natural frequency of a spring-mass system on earth is ω n . The natural frequency of this system on the moon (g moon = g earth /6) is

a) ω n

b) 0.408ω n

c) 0.204ω n

d) 0.167ω n

Answer: a

Explanation: We know natural frequency of a spring mass system is,

ω n = √k/m ………..

This equation  does not depend on the g and weight 

So, the natural frequency of a spring mass system is unchanged on the moon.

Hence, it will remain ω n , i.e. ω moon = ω n .

12. A vehicle suspension system consists of a spring and a damper. The stiffness of the spring is 3.6 kN/m and the damping constant of the damper is 400 Ns/m. If the mass is 50 kg, then the damping factor  and damped natural frequency (f n ), respectively, are

a) 0.471 and 1.19 Hz

b) 0.471 and 7.48 Hz

c) 0.666 and 1.35 Hz

d) 0.666 and 8.50 Hz

Answer: a

Explanation: Given k = 3.6 kN/m, c = 400 Ns/m, m = 50 kg

We know that, Natural Frequency

ω n = √k/m = 8.485 rad/ sec

And damping factor is given by,

d = c/c c = c/2√km = 0.471

Damping Natural frequency,

ω d = √1 – d 2 ω n

2пf d = √1 – d 2 ω n

f d = 1.19 Hz.

Answer: c

Explanation: For an under damped harmonic oscillator resonance occurs when excitation

frequency is equal to the undamped natural frequency

ω d = ω n .

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Natural Frequency of Free Longitudinal Vibrations”.

1. Find the natural frequency in Hz of the free longitudinal vibrations if the displacement is 2mm.

a) 11.14

b) 12.38

c) 11.43

d) 11.34

Answer: a

Explanation: We know that the natural Frequency of Free Longitudinal Vibration is given by the equation

f = 0.4985/\(\sqrt{s}\)

where s is the displacement of the spring.

2. If the spring displacement is high then the frequency of the spring increases.

a) True

b) False

Answer: b

Explanation: The natural frequency of the free longitudinal vibration depends on the displacement and gravitational acceleration  by the relation:

f= 0.4985/\(\sqrt{s}\)

where s is the displacement of the spring.

since it is not directly proportional, the given statement is false.

3. Find the displacement in mm of the free longitudinal vibrations if the Natural frequency is 15 Hz.

a) 1.1

b) 1.2

c) 1.5

d) 1.6

Answer: a

Explanation: We know that the natural Frequency of Free Longitudinal Vibration is given by the equation

f = 0.4985/\(\sqrt{s}\)

where s is the displacement of the spring.

substituting the given values we get

s=1.1 mm.

4. Find the displacement in mm of the free longitudinal vibrations if the Natural frequency is 20 Hz.

a) 0.1

b) 0.2

c) 0.5

d) 0.6

Answer: d

Explanation: We know that the natural Frequency of Free Longitudinal Vibration is given by the equation

f= 0.4985/\(\sqrt{s}\)

where s is the displacement of the spring.

Substituting the given values we get:

s=0.6 mm.

5. Which of the following methods will give an incorrect relation of the frequency for free vibration?

a) Equilibrium method

b) Energy method

c) Reyleigh’s method

d) Klein’s method

Answer: d

Explanation: Equilibrium, Energy and Reyleigh’s method give the same relation between the natural frequency and displacement of free vibration whereas klein’s method is used to calculate velocity and acceleration of the parts of the mechanisms.

6. A cantilever shaft has a diameter of 6 cm and the length is 40cm, it has a disc of mass 125 kg at its free end. The Young’s modulus for the shaft material is 250 GN/m². Calculate the static deflection in nm.

a) 0.001

b) 0.083

c) 1.022

d) 0.065

Answer: a

Explanation: Area = πd 2 /4 = 0.00282 m 2

s = W.l/A.E

= 0.001 nm.

7. Static deflection and frequency are independent of each other.

a) True

b) False

Answer: b

Explanation: The natural frequency of the free longitudinal vibration depends on the static displacement and gravitational acceleration  by the relation:

f = 0.4985/\(\sqrt{s}\).

8. A cantilever shaft having 50 mm diameter and length of 300 mm has a disc of mass 100 kg at its free end. The Young’s modulus for the shaft material is 200 GN/m 2 . Calculate the natural longitudinal frequency in Hz.

a) 575

b) 625

c) 525

d) 550

Answer: a

Explanation: Area = πd 2 /4 = 0.00196 m 2

s = W.l/A.E = = 0.751 µm

I = 0.3×10 -6 m 4

f = 0.4985/\(\sqrt{s}\)

= 575 Hz.

9. If the mass is of 10 Kg, find the natural frequency in Hz of the free longitudinal vibrations. The displacement is 0.01mm.

a) 44.14

b) 49.85

c) 43.43

d) 46.34

Answer: b

Explanation: We know that the natural Frequency of Free Longitudinal Vibration is given by the equation

f = 0.4985/\(\sqrt{s}\)

where s is the displacement of the spring.

substituting the given values we get

f=49.85 Hz

It is to be noted that mass has no effect on the natural frequency as it only depends on the displacement.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Natural Frequency of Free Transverse Vibrations”.

1. A cantilever shaft having 50 mm diameter and a length of 300 mm has a disc of mass 100 kg at its free end. The Young’s modulus for the shaft material is 200 GN/m 2 . Determine the frequency of transverse vibrations of the shaft.

a) 31

b) 35

c) 37

d) 41

Answer: d

Explanation: We know that deflection is given by the relation:

Wl 3 /3E.I

I = 0.3×10 -6 m 4

d = 0.147×10 -3

f = 0.4985/\(\sqrt{d}\)

Thus f = 41 Hz.

2. For the same dimensions of a beam, transverse vibrations have a lower frequency than longitudinal frequency.

a) True

b) False

Answer: a

Explanation: Keeping the dimensions of the testing beam same, it is noted that the natural frequency of vibrations obtained in longitudinal waves is larger than the one obtained in transverse waves.

3. A cantilever shaft having 50 mm diameter and a length of 300 mm has a disc of mass 100 kg at its free end. The Young’s modulus for the shaft material is 200 GN/m 3 . Determine the static deflection of the shaft in mm.

a) 0.147

b) 0.213

c) 0.132

d) 0.112

Answer: a

Explanation: We know that deflection is given by the relation:

Wl 3 /3E.I

W = 100xg, l=0.3m, I = I = 0.3×10 -6 m 4

substituting values we get d = 0.147mm.

4. For the same dimensions of the shaft which of the following has the greater natural frequency?

a) Transverse

b) Longitudinal

c) Depends on thickness

d) Depends upon length

Answer: b

Explanation: If the dimensions of the testing beam are kept same, it is observed that the natural frequency of vibrations obtained in longitudinal waves is larger than the one obtained in transverse waves.

5. Calculate the natural frequency of transverse vibrations if the static deflection is 0.01mm.

a) 157.6

b) 144.8

c) 173.2

d) 154.1

Answer: a

Explanation: We know that the natural Frequency of Free Transverse Vibration is given by the equation

f = 0.4985/\(\sqrt{s}\)

where s is the displacement of the spring.

substituting the given values we get

f=157.6 Hz.

6. Increasing mass will result in lower frequency.

a) True

b) False

Answer: a

Explanation: We know that deflection is given by the relation :

Wl 3 /3E.I

Increasing the mass will result in an increased deflection which, frequency decreases as the deflection increases hence increasing mass will reduce the frequency of vibration.

7. Calculate the static deflection in µm of transverse vibrations if the frequency is 200Hz.

a) 6

b) 0.6

c) 60

d) 0.006

Answer: a

Explanation: We know that the natural Frequency of Free Transverse Vibration is given by the equation

f = 0.4985/\(\sqrt{s}\)

where s is the displacement of the spring.

substituting the given values we get

f = 0.00000621255625‬ m.

This set of Machine Dynamics Questions and Answers for Campus interviews focuses on “Critical or Whirling Speed of a Shaft”.

1. Assuming the shaft to be freely supported. Calculate the whirling speed of the shaft : 2 cm diameter and 60 cm long carrying a mass of 1 kg at its mid-point. The density of the shaft material is 40 Mg/m 3 , and Young’s modulus is 200 GN/m 2 .

a) 2598

b) 2434

c) 2756

d) 2634

Answer: a

Explanation: I = πd 4 /64 = 7.855×10 -9 m 4

d due to 1 kg mass = 28×10 -6 m

d due to mass of the shaft = 0.133×10 -3 m

substituting these values into the frequency relation of transverse vibrations

we get

f = 43.3 Hz

Therefore Nc = 43.3×60 = 2598 rpm.

2. When the centre of gravity of the rotor lies between the centre line of the shaft and the centre line of the bearing, e is taken positive.

a) True

b) False

Answer: b

Explanation: If the centre of gravity of the rotor does not lie between the centre line of the shaft and the centre line of the bearing, then the value of e is taken positive.

3. The speed at which the shaft runs so that the additional deflection from the axis of rotation of the shaft becomes infinite, is known as _________

a) Whirling speed

b) Rotational speed

c) Stabilizing speed

d) Reciprocating speed

Answer: a

Explanation: The rotational speed at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes infinite, is known as critical or whirling speed.

4. From the following data, calculate the critical speed of the shaft in rps.

Shaft diameter = 5mm

length = 200mm

Mass of disc = 50Kg at centre of shaft

E = 200GN/m 2

Centre of disc at 0.25m away from centre of axis of shaft.

a) 8.64

b) 9.64

c) 10.64

d) 11.64

Answer: a

Explanation: I = πd 4 /64 = 30.7×10 -12 m 4

d = 3.33×10 -3 m

substituting these values into the frequency relation of transverse vibrations

we get

Nc = 8.64 rps.

5. From the following data, calculate the static deflection in mm.

Critical speed = 8.64 rps

Mass of disc = 100Kg at centre of shaft

E = 100GN/m 2

Centre of disc at 0.25m away from centre of axis of shaft.

a) 3.32

b) 9.64

c) 10.64

d) 11.64

Answer: a

Explanation: I = πd 4 /64 = 30.7×10 -12 m 4

Nc = 8.64 rps

substituting these values into the frequency relation of transverse vibrations

we get

d = 3.32mm.

6. If the static deflection is 1.665×10 -3 m, calculate the critical speed of the shaft in rps.

Centre of disc at 0.25m away from centre of axis of shaft.

a) 8.64

b) 9.64

c) 10.64

d) 12.2

Answer: d

Explanation: d = 1.665×10 -3 m

substituting these values into the frequency relation of transverse vibrations

we get

Nc = 12.2 rps.

7. A shaft supported in ball bearings is assumed to be a simply supported.

a) True

b) False

Answer: a

Explanation: A shaft supported in short bearings is assumed to be a simply supported shaft while the shaft supported in journal bearings is assumed to have both ends fixed.

8. From the following data, calculate the static deflection in mm.

Shaft diameter = 5mm

length = 200mm

Mass of disc = 100Kg at centre of shaft

E = 100GN/m 2

Centre of disc at 0.25m away from centre of axis of shaft.

a) 4.32

b) 9.64

c) 10.64

d) 11.64

Answer: a

Explanation: I = πd 4 /64 = 30.7×10 -12 m 4

d = 13.32×10 -3 m.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Frequency of Free Damped Vibrations”.

1. Fluid resistance causes damping which is known as ______

a) Resistance damping

b) Fluid damping

c) Viscous damping

d) Liquid damping

Answer: c

Explanation: Damping due to the resistance offered by the fluid is known as viscous damping. This is because of the reduction in the amplitude caused by the viscous forces of the fuild.

2. In damped vibrations, the amplitude of the resulting vibration gradually diminishes.

a) True

b) False

Answer: a

Explanation: Since a certain amount of energy is always dissipated while overcoming the resistance forces, In damped vibrations, the amplitude of the resulting vibration gradually diminishes.

3. In damped vibrations, the amplitude of the resulting vibration gradually reduces. This is due to the reason that an amount of energy is always dissipated to overcome the ________

a) Frictional resistance

b) Work done

c) Fluid pressure

d) Air pressure

Answer: a

Explanation: In case of damped vibrations, the amplitude of the resulting vibration gradually diminishes. This is due to the reason that a certain amount of energy is always dissipated to overcome the frictional resistance.

4. The resistance to the motion of the body is provided by ______

a) Medium of vibration

b) Speed of vibration

c) Length of the material

d) External friction

Answer: a

Explanation: The resistance to the motion of the vibrating body is provided partly by the medium in which the vibration takes place and partly by the internal friction.

5. In which direction does the damping force acts?

a) Opposite to the motion

b) Along the motion

c) Perpendicular to motion

d) Variable

Answer: a

Explanation: It is assumed that the frictional resistance to the motion of the body is directly proportional to the speed of, therefore Damping force or frictional force on the mass acts in opposite direction to the motion of the mass.

6. In which direction does the accelerating force acts?

a) Opposite to the motion

b) Along the motion

c) Perpendicular to motion

d) Variable

Answer: b

Explanation: It is assumed that the frictional resistance to the motion of the body is directly proportional to the speed, therefore accelerating force on the mass acts in same direction to the motion of the mass.

7. In which of the following cases, overdamping occurs?

a) Roots are real

b) Roots are complex conjugate

c) Roots are equal

d) Independent of the equation

Answer: a

Explanation: If the roots k1 and k2 are real but negative the case is of overdamping or large damping and the mass moves slowly to the equilibrium position. This motion is known as aperiodic.

8. In which of the following cases, underdamping occurs?

a) Roots are real

b) Roots are complex conjugate

c) Roots are equal

d) Independent of the equation

Answer: b

Explanation: If the roots k1 and k2 are complex, the case is of underdamping or small damping and the mass moves slowly to the equilibrium position.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Damping Factor & Magnification Factor”.

1. The ratio of the actual damping coefficient  to the critical damping coefficient  is known as _________

a) Damping factor

b) Damping coefficient

c) Resistive factor

d) Resistive coefficient

Answer: a

Explanation: The ratio of the actual damping coefficient  to the critical damping coefficient  is known as damping factor or damping ratio given by c/Cc.

2. Calculate critical damping coefficient in Ns/m from the following data.

mass = 200Kg

ω = 20rad/s

a) 25,132

b) 26,132

c) 27,132

d) Not possible

Answer: d

Explanation: We know that critical damping coefficient is given by the relation

Cc = 2πxmxω

inserting the values we get

Cc = 25132 N/m/s

The above value is in N/m/s and damping factor cannot be calculated in Ns/m.

3. Calculate critical damping coefficient in N/m/s from the following data:

mass = 100Kg

ω = 40rad/s

a) 25,132

b) 26,132

c) 27,132

d) 28,132

Answer: a

Explanation: We know that critical damping coefficient is given by the relation

Cc = 2πxmxω

inserting the values we get

Cc = 25132 N/m/s.

4. Calculate critical damping coefficient in N/m/s from the following data:

mass = 100Kg

ω = 10rad/s

a) 5,132

b) 6,283

c) 7,132

d) 8,132

Answer: a

Explanation: We know that critical damping coefficient is given by the relation

Cc = 2πxmxω

inserting the values we get

Cc = 6283 N/m/s.

5. Calculate damping ratio from the following data:

mass = 200Kg

ω = 20rad/s

damping coefficient = 800 N/m/s

a) 0.03

b) 0.04

c) 0.05

d) 0.06

Answer: a

Explanation: We know that critical damping coefficient is given by the relation

Cc = 2πxmxω

inserting the values we get

Cc = 25132 N/m/s

ratio = 800/25132.

6. Calculate damping ratio from the following data:

mass = 200Kg

ω = 20rad/s

damping coefficient = 1000 N/m/s

a) 0.03

b) 0.04

c) 0.05

d) 0.06

Answer: b

Explanation: We know that critical damping coefficient is given by the relation:

Cc = 2πxmxω

inserting the values we get

Cc = 25132 N/m/s

ratio = 1000/25132 = 0.04.

7. Unit of damping factor is N/m/s.

a) True

b) False

Answer: b

Explanation: Damping factor is a ratio of damping coefficient to the critical damping coefficient hence, it has no unit and is a dimensionless quantity.

8. Magnification factor is the ratio of the maximum displacement due to forced vibrations to the deflection due to _______

a) Static force

b) Dynamic force

c) Torsion

d) Compression

Answer: a

Explanation: Magnification factor or the Dynamic magnifier is the ratio of maximum displacement of the forced vibration to the deflection due to the static force.

9. Maximum displacement due to forced vibration is dependent on deflection due to static force.

a) True

b) False

Answer: a

Explanation: Maximum displacement due to forced vibration is directly proportional to the displacement due to static force, hence it is dependent on deflection due to static force.

10. In which of the cases the factor c = 0?

a) When there is damping

b) No damping

c) Resonance

d) c is never 0

Answer: b

Explanation: When there is no damping, the factor c becomes 0 and magnification factor becomes independent of the damping coefficient.

11. A body of mass 20 kg is suspended from a spring which deflects 20mm under this load. Calculate the frequency of free vibrations in Hz.

a) 3.5

b) 5

c) 6

d) 7

Answer: a

Explanation: Frequency of free vibrations is given by √

substituting the value, we get

f = 3.5 Hz.

12. If the mass increases, then the frequency of the free vibrations increases.

a) True

b) False

Answer: b

Explanation: Frequency of free vibrations is given by √, from the given formula it is evident that it is independent of mass, Hence the given statement is false.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Vibration Isolation and Transmissibility”.

1. Which of the following systems produce a vibration in the foundation?

a) Unbalanced machine

b) Balanced machine

c) Coupled machine

d) Uncoupled machine

Answer: a

Explanation: It has been observed that when an unbalanced machine is installed on the foundation, it produces vibration in the foundation which needs to be countered to create smooth working of the machine.

2. When a periodic disturbing force is applied to a machine, the force is transmitted to the foundation by the means of spring.

a) True

b) False

Answer: a

Explanation: Springs or dampers are mounted on the machine in order to reduce the vibrations produce but however they also transmit the disturbing periodic force to the machine.

3. Which of the following is correct regarding isolation factor?

a) Dimensionless quantity

b) Has Newton as its unit

c) Has joule as its Unit

d) Has Hz as its unit

Answer: a

Explanation: Isolation factor is a ratio of transmitted force to the applied force, since it is a ratio of two quantities having same dimension, it is a dimensionless quantity.

4. Which of the following is a type of transmitted force to the foundation?

a) Damping force

b) Undamping force

c) Tensile force

d) Torsional force

Answer: a

Explanation: The transmitted force consists of the following two forces which are perpendicular to each other namely: Spring or elastic force and damping force. This happens as both spring and damper is used in order to minimize vibrations.

5. If the damper is not provided and the system is in resonance, which of the following is the correct isolation factor?

a) 0

b) 1/2

c) 1/4

d) Infinity

Answer: d

Explanation: When the damper is not involved, the damping coefficient c becomes zero and isolation factor is given by:

1/)

Since during resonance, ω=ωn

isolation factor becomes infinite.

6. If isolation factor is negative, then what is the phase difference between transmitted and disturbing force?

a) 180°

b) 90°

c) 450°

d) 360°

Answer: a

Explanation: If the isolation factor is negative then the ratio ω/ωn is greater than 1, in this case the phase difference between the disturbing force and the transmitted force is 180°.

7. Which of the following is true regarding Ɛ>1?

a) Transmitted force is greater than applied force

b) Transmitted force is less than applied force

c) Spring force is less than applied force

d) Damping force is less than applied force

Answer: a

Explanation: if the isolation factor is greater than 1, then the ratio ω/ωn < √2, this means that there is a phase difference between the transmitted force and the disturbing force, where the transmitted force is greater than the applied force.

8. Isolation factor is twice the transmissibility ratio.

a) True

b) False

Answer: b

Explanation: The given statement is false as the isolation factor and the transmissibility ratio have the same value which is given by ratio of Transmitted force to the Applied force.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Effect of Inertia of the Constraint on Torsional Vibrations”.

1. A flywheel is mounted on a vertical shaft. The both ends of a shaft are fixed and its diameter is 50 mm. The flywheel has a mass of 500 kg, the modulus of rigidity for the shaft material is 80 GN/m 2 and its radius of gyration is 0.5 m. Find the natural frequency of torsional vibrations.

a) 5.32

b) 5.86

c) 5.65

d) 6.66

Answer: a

Explanation: We know that J = πd4/32 = 0.6×10 -6 m 4

q 1 = 56×10 3

q 2 = 84×10 3

q = 140×10 3

therefore using the formula for fn

we get fn = 5.32 Hz.

2. If a mass whose moment of inertia is Ic/3 is placed at the free end and the constraint is assumed to be of negligible mass, then the kinetic energy is ______

a) 1/6 Icω 2

b) 1/2Icω 2

c) 1/3Icω 2

d) 1/12Icω 2

Answer: a

Explanation: Kinetic energy is given by the equation

0.5Iω 2

inserting Ic/3 in place of I will give kinetic energy as

1/6 Icω 2 .

3. If Ic = 125 Kg-m 2 and ω= 20 rad/s, calculate the kinetic of the constraint.

a) 8333 J

b) 7333 J

c) 6333 J

d) 9333 J

Answer: a

Explanation: The kinetic energy of the constraint is given by the equation:

K.E = 1/6 Icω 2

substituting the value we get

K.E = 8333J.

4. If Ic = 500 Kg-m 2 and ω = 10 rad/s, calculate the kinetic of the constraint.

a) 8333 J

b) 7363 J

c) 4578 J

d) 9333 J

Answer: a

Explanation: The kinetic energy of the constraint is given by the equation:

K.E = 1/6 Icω 2

substituting the value we get

K.E = 8333J.

5. If the mass of the constraint is negligible then what is the kinetic energy of the system?

a) 0

b) Half the value

c) Double the value

d) Infinite

Answer: a

Explanation: If the mass of the constraint is negligible, then the mass moment of inertia will be 0. As a result the expression of the kinetic energy will become 0.

6. If the mass of the constraint is negligible and a mass is attached at the end of the rod having following values: Ic = 500 Kg-m 2 and ω= 10 rad/s, then calculate the kinetic of the constraint system.

a) 8333 J

b) 7363 J

c) 4578 J

d) 9333 J

Answer: a

Explanation: The kinetic energy of the constraint system in the absence of mass of constrant is given by the equation:

K.E = 1/6 Icω 2

substituting the value we get

K.E = 8333J.

7. Free torsional vibrations will occur in a two rotor system only if both rotors have same frequency.

a) True

b) False

Answer: a

Explanation: In a two rotor system the free torsional vibration will occur only if both the rotors have the same frequency of vibration, it follows the following relation between length and mass moment of inertia: LI = LI.

8. Increasing which of the following factor would result in increase of free torsional vibration?

a) Radius of gyration

b) Mass moment of inertia

c) Polar moment of inertia

d) Length

Answer: c

Explanation: The free torsional vibration frequency of a single rotor system depends by the following relation

f = \(\frac{1}{2}\sqrt{

 

}\)

Hence increase in polar moment of inertia results in increase in vibration frequency.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Free Torsional Vibrations of a Single Rotor System”.

1. Which of the following relation is correct regarding free torsional vibrations of a single motor system?

a) Independent of modulus of rigidity

b) Independent of polar moment of inertia

c) Dependent on mass moment of inertia

d) Independent of length of shaft

Answer: c

Explanation: The free torsional vibrations of a single motor system depends on the following factors:

Modulus of rigidity, Polar moment of inertia, mass moment of inertia and length of the shaft.

2. Free torsional vibrations of a single motor system increases with increase in polar moment of inertia.

a) True

b) False

Answer: b

Explanation: The free torsional vibrations of a single motor system depends on the following factors:

√Polar moment of inertia as directly proportional to the square root of polar moment of inertia.

3. If the polar moment of inertia is increased to four times, then what will be the effect on free torsional vibrations of a single motor system?

a) Increases 4 times

b) Increases 2 times

c) Decreases 4 times

d) Decreases 2 times

Answer: b

Explanation: Since the free torsional vibrations of a single motor system depend on the square root of the polar moment of inertia of the system, increasing 4 times will lead to increase in two times the initial vibration.

4. If the mass moment of inertia is increased to four times, then what will be the effect on free torsional vibrations of a single motor system?

a) Increases 4 times

b) Increases 2 times

c) Decreases 4 times

d) Decreases 2 times

Answer: d

Explanation: Since the free torsional vibrations of a single motor system depend on the inverse of square root of the mass moment of inertia of the system, increasing 4 times will lead to decrease in two times the initial vibration.

5. Calculate the free torsional vibrations of a single motor system from the following data:

C = 8 GN/m 2 , L=9m, I = 600 Kg-m 2 , J = 8×10 4 m 4

a) 162,132

b) 172,132

c) 182,132

d) 192,132

Answer: b

Explanation: The free torsional vibrations of a single motor system is given by

\(\frac{1}{2}\sqrt{}\)

Substituting the given values gives f = 172132 Hz.

6. If the length inertia is decreased to nine times, then what will be the effect on free torsional vibrations of a single motor system?

a) Increases 3 times

b) Increases 9 times

c) Decreases 9 times

d) Decreases 3 times

Answer: a

Explanation: Since the free torsional vibrations of a single motor system depends on the inverse of square root of the length, decreasing 9 times will lead to increase in 3 times the initial vibration.

7. Calculate the Polar moment of inertia in m 4 of a single motor system from the following data:

C = 8 GN/m 2 , L=9m, I = 600 Kg-m 2 , f=10 Hz

a) 0.00027

b) 0.00032

c) 0.00045

d) 0.00078

Answer: a

Explanation: The free torsional vibrations of a single motor system is given by

\(\frac{1}{2}\sqrt{}\)

Substituting the given values gives J = 0.00027 m 4 .

8. If radius of gyration increases then the torsional free vibration increases.

a) True

b) False

Answer: b

Explanation: Increase in the radius of gyration will result in increase in mass moment of inertia since mass moment of inertia is not directly dependent, the given statement is false.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Free Torsional Vibrations of a Two Rotor System”.

1. Consider P and Q as the shaft having two rotors at the end of it, what is the point N known as in the given figure?

machine-dynamics-questions-answers-free-torsional-vibrations-two-rotor-system-q1

a) Node

b) Elastic point

c) Inelastic point

d) Breaking point

Answer: a

Explanation: The torsional vibrations occur in a two-rotor system only if the rotors move in the opposite direction, at point N there is no vibration hence it is known as node.

2. In a two rotor system, torsional vibration occurs only if the rotors are moving in the same direction.

a) True

b) False

Answer: b

Explanation: For a two rotor system, the torsional vibration will occur only if the rotors are moving in the direction opposite to each other.

3. In the given figure representing a two rotor system, what is the line LNM called?

machine-dynamics-questions-answers-free-torsional-vibrations-two-rotor-system-q1

a) Node line

b) Elastic line

c) Inelastic line

d) Breaking line

Answer: b

Explanation: The given figure represents the shaft PQ with rotors attached to the end of it making it a two rotor system, in the figure, the line LNM is known as the elastic line of the shaft.

4. For occurrence of free torsional vibration which of the condition is necessary?

a) Rotors moving in same direction

b) Rotors having same frequency

c) Rotors having different frequency

d) Rotors rotate in the same sense

Answer: b

Explanation: In a two rotor torsional vibration system, the system will rotate if the rotors are rotating in opposite direction with the same frequency.

5. In the given figure, considering PN as one shaft, If the mass moment of inertia is increased to four times, then what will be the effect on free torsional vibrations of a rotor at P?

machine-dynamics-questions-answers-free-torsional-vibrations-two-rotor-system-q1

a) Increases 4 times

b) Increases 2 times

c) Decreases 4 times

d) Decreases 2 times

Answer: d

Explanation: If N is the node then PN can be considered as a shaft of single rotor system, since the free torsional vibrations of a single motor system depends on the inverse of square root of the mass moment of inertia of the system, increasing 4 times will lead to decrease in two times the initial vibration.

6. In the given figure if N is the node then NQ acts as which of the following system?

machine-dynamics-questions-answers-free-torsional-vibrations-two-rotor-system-q1

a) Single rotor system

b) Two rotor system

c) Three rotor system

d) Four rotor system

Answer: a

Explanation: If N is the node in the given figure, then N can act as a fixed end of the shaft NQ where the rotor is connected at Q, hence it acts like a single rotor system.

7. Keeping the mass moment of inertia of both the shafts in a two rotor system same, if the length of one shaft is doubled what should be the effect on the length of other shaft?

a) Doubled

b) Halved

c) Constant

d) Increased to 4 times

Answer: a

Explanation: The vibration will occur in a two rotor system only if the frequencies of both the rotors are same, hence LI = LI. Therefore the relation is directly proportional.

8. For a two rotor system, the mass moment of inertia of one shaft is twice the other, then what is the relation between the length of the shafts.

a) 2L = L

b) L = 2L

c) L = L

d) 2L = 3L

Answer: a

Explanation: For a two rotor system, we have LI = LI

Now I=2I

Therefore, 2L = L.

9. For a two rotor system, the length of one shaft is twice the other, then what is the relation between the Mass moment of inertia of the shafts.

a) 2I = I

b) I = 2I

c) I = I

d) 2I = 3I

Answer: a

Explanation: For a two rotor system, we have LI = LI

Now L=2L

Therefore, 2I = I.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Free Torsional Vibrations of a Three Rotor System”.

1. In the figure given below, the points N1 and N2 are known as_______

machine-dynamics-questions-answers-free-torsional-vibrations-three-rotor-system-q1

a) Nodes

b) Elastic points

c) Inelastic points

d) Breaking points

Answer: a

Explanation: The torsional vibrations occur in a three rotor system only if there are one node or two nodes, at points N1 and N2 there is no vibration hence they are known as nodes.

2. In a three rotor system, free torsional vibration cannot occur if there is only one node.

a) True

b) False

Answer: b

Explanation: In a three rotor system there may occur one or two nodes, free torsional vibration may occur in presence of either one or two nodes.

3. In which of the following condition torsional vibration will not take place, considering 3 rotors A, B and C. A is rotating in clockwise direction.

a) B in clockwise C in anticlockwise

b) C in clockwise B in anticlockwise

c) B and C in clockwise

d) B and C in anticlockwise

Answer: c

Explanation: For a three rotor system torsional vibration will occur only if two of the three rotors are rotating in the same direction and the third one is rotating in the opposite direction.

4. For occurrence of free torsional vibration in a three rotor system which of the condition is necessary?

a) Rotors moving in same direction

b) Rotors having same frequency

c) Rotors having different frequency

d) Rotors rotate in the same sense

Answer: b

Explanation: In a three rotor torsional vibration system, the system will rotate if the two rotors are rotating in same direction and the third one with opposite direction all with the same frequency.

5. the given figure, considering left end has one rotor, If the mass moment of inertia of the shaft till node N1 is increased to four times, then what will be the effect on free torsional vibrations of a rotor at left end of N1?

machine-dynamics-questions-answers-free-torsional-vibrations-three-rotor-system-q1

a) Increases 4 times

b) Increases 2 times

c) Decreases 4 times

d) Decreases 2 times

Answer: d

Explanation: If N1 is the node then left end and N1 can be considered as a shaft of single rotor system, since the free torsional vibrations of a single motor system depends on the inverse of square root of the mass moment of inertia of the system, increasing 4 times will lead to decrease in two times the initial vibration.

6. In the given figure if N1 is the node then N1Q acts as which of the following system?

machine-dynamics-questions-answers-free-torsional-vibrations-three-rotor-system-q1

a) Single rotor system

b) Two rotor system

c) Three rotor system

d) Four rotor system

Answer: b

Explanation: If N1 is the node in the given figure, then N1 can act as a fixed end of the shaft N1N2 where the rotor is connected at Q and N1, hence it acts like a two rotor system.

7. Keeping the mass moment of inertia of left end and the right end shafts in a three rotor system same, if the length of one shaft is doubled what should be the effect on the length of other shaft?

a) Doubled

b) Halved

c) Constant

d) Increased to 4 times

Answer: a

Explanation: The vibration will occur in a two rotor system only if the frequencies of both the rotors are same, hence LI = LI. Therefore the relation is directly proportional.

8. Free torsional vibrations will occur in a three rotor system only if all rotors have same frequency.

a) True

b) False

Answer: a

Explanation: In a three rotor system the free torsional vibration will occur only if all the rotors have the same frequency of vibration and two have common rotation sense.

9. What is the total number of nodes formed in a three rotor system if the rotors at one of the ends and the one in the middle rotate in the same direction?

a) 0

b) 1

c) 2

d) 3

Answer: b

Explanation: In a three rotor system, if the rotor at one of the ends and the one in the middle are rotating in the same direction, then there is formation of only one node.

10. For a three rotor system in the figure given below, the length of one shaft is twice the other, then what is the relation between the Mass moment of inertia of the shafts.

machine-dynamics-questions-answers-free-torsional-vibrations-three-rotor-system-q1

a) 2I = I

b) I = 2I

c) I = I

d) 2I = 3I

Answer: a

Explanation: For a two rotor system, we have LI = LI

Now L=2L

Therefore, 2I = I.

11. In a three rotor system, for the middle rotor, if the stiffness of both the length either side of the rotor is increased to two times what will be the effect on total stiffness of the middle rotor?

a) Remains constant

b) Decreases by two times

c) Increases by two times

d) Increases by 4 times

Answer: c

Explanation: In a three rotor system, each length to either side of the middle rotor is twisted through the same angle therefore the total stiffness is the sum of individual stiffness.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Torsionally Equivalent Shaft”.

1. From the following data, calculate the equivalent length of shaft in m.

l 1 =0.6m, l 2 =0.5m, l 3 =0.4m

d 1 =0.095m, d 2 =0.06m, d 3 =0.05m

a) 8.95

b) 7.95

c) 6.95

d) 5.95

Answer: a

Explanation: We know that torsional equivalent shaft length is given by

l 1 /(d 1 ) 4 + l 2 /(d 2 ) 4 + l 3 /(d 3 ) 4

substituting the values we get

L = 8.95 m.

2. In a system with different shaft parameters, the longest shaft is taken for calculations.

a) True

b) False

Answer: b

Explanation: In a system with different shaft parameters the equivalent shaft length is taken which depends on the length and diameters of each shaft.

3. From the following data, calculate natural frequency of free torsional vibrations in Hz.

l 1 =0.6m, l 2 =0.5m, l 3 =0.4m

d 1 =0.095m, d 2 =0.06m, d 3 =0.05m

Ma = 900 Kg, Mb = 700 Kg

ka = 0.85m, kb = 0.55m

C = 80 GN/m 2

a) 3.37

b) 7.95

c) 6.95

d) 5.95

Answer: a

Explanation: We know that torsional equivalent shaft length is given by

l 1 /(d 1 ) 4 + l 2 /(d 2 ) 4 + l 3 /(d 3 ) 4

substituting the values we get

L = 8.95 m

now calculating J

we get polar moment of inertia J = 8×10 6 m 4

Now we know that natural frequency is given by the formula

f = \(\frac{1}{2}\sqrt{

 

}\)

f = 3.37 Hz.

4. From the following data, calculate the location of node from the left end of shaft (l 1 ).

l 1 =0.6m, l 2 =0.5m, l 3 =0.4m

d 1 =0.095m, d 2 =0.06m, d 3 =0.05m

Ma = 900 Kg, Mb = 700 Kg

ka = 0.85m, kb = 0.55m

a) 0.855m

b) 0.795m

c) 0.695m

d) 0.595m

Answer: a

Explanation: We know that for the system to be in vibration, the frequency should be same for all rotors

therefore,

LaIa = LbIb

Where La is the length of the left end shaft.

Substituting the values we get,

l = 0.855m.

5. For a vibration system having different shaft parameters, calculate which of the following cannot be the diameter of the equivalent shaft if the diameters of shafts in m are: 0.05, 0.06, 0.07.

a) 0.05

b) 0.06

c) 0.07

d) 0.08

Answer: d

Explanation: In a torsionally equivalent shaft, it is assumed that the equivalent shaft has the diameter which is equal to one of the diameters of the shaft and equivalent length is calculated.

6. A single cylinder oil engine works with a three rotor system, the shaft length is 2.5m and 70mm in diameter, the middle rotor is at a distance 1.5m from one end. Calculate the free torsional vibration frequency for a single node system in Hz if the mass moment of inertia of rotors in Kg-m 2 are: 0.15, 0.3 and 0.09. C=84 kN/mm 2

a) 171

b) 181

c) 191

d) 201

Answer: a

Explanation: For a single node system, the node occurs at a distance 1.146m from left end.

The polar moment of inertia = 2.36×10 6 m 4

La = 0.4356m 

Substituting these values into the frequency relation we get

f = 171 Hz.

7. A single cylinder oil engine works with a three rotor system, the shaft length is 2.5m and 70mm in diameter, the middle rotor is at a distance 1.5m from one end. Calculate the free torsional vibration frequency for a two node system in Hz if the mass moment of inertia of rotors in Kg-m 2 are: 0.15, 0.3 and 0.09. C=84 kN/mm 2

a) 257

b) 281

c) 197

d) 277

Answer: d

Explanation: For a two node system, the first node occurs at a distance 0.4356 m from left end.

The polar moment of inertia = 2.36×10 6 m 4

La = 0.4356m 

Lc = 0.726 m

Substituting these values into the frequency relation we get

f = 277 Hz.

8. Frequency is independent of the no. of nodes.

a) True

b) False

Answer: b

Explanation: The frequency of the free torsional vibration depends on the number of nodes it forms, it also depends on the no. of rotors as the no. of rotors have an effect on the number of nodes.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Free Torsional Vibrations of a Geared System”.

1. For the free torsional vibration of a geared system which of the following statement is true?

a) Backlash in geared system

b) High inertia of shaft

c) Gearing teeth always in contact

d) High inertia of gear

Answer: c

Explanation: For the free torsional vibration of a geared system there should be no backlash in geared system and the inertia of the shaft, gears should be considered negligible and the gearing system should always be in contact.

2. In a geared system free torsional vibration will take place when the Inertia of gear is taken into consideration.

a) True

b) False

Answer: b

Explanation: In a geared system free torsional vibration will take place when the Inertia of gear and shaft is not taken into consideration.

3. If the pinion has 20 teeth and the gear has 32 teeth, then what is the gear ratio?

a) 5/8

b) 8/5

c) 25/64

d) 64/25

Answer: a

Explanation: Gear ratio is defined as the ratio of no. of teeth on the pinion to the no. of teeth on the gear, there in this question the gear ratio is 5/8.

4. A geared system possessed a kinetic energy if 200kJ, later it was converted into a two-rotor system connected by a shaft. The kinetic energy of the equivalent system will be_____

a) 200 kJ

b) less than 200 kJ

c) more than 200 kJ

d) Dependent on the length of the equivalent shaft

Answer: a

Explanation: If the geared system is converted into an equivalent system of rotors connected by a shaft then the kinetic energy of the original and the equivalent system must be same.

5. A geared system possessed a strain energy if 150kJ, later it was converted into a two-rotor system connected by a shaft. The Strain energy of the equivalent system will be _____

a) 150 kJ

b) less than 150 kJ

c) more than 150 kJ

d) Dependent on the length of the equivalent shaft

Answer: a

Explanation: If the geared system is converted into an equivalent system of rotors connected by a shaft then the Strain energy of the original and the equivalent system must be same.

6. When inertia of gearing is taken into consideration, then which of the following should be taken into account.

a) Addition of rotor

b) Addition of gear

c) Addition of shaft

d) Addition of pump

Answer: a

Explanation: When inertia of gearing is taken into consideration, then an additional rotor must be introduced in the equivalent system of the rotors connected by the shaft.

7. In a gearing system, pump speed is one third of the motor. Shaft from motor is 6 cm in diameter and 30cm long, the impellar shaft is 10cm diameter and 60cm long. Mass moment of inertia is 1500 Kg-m 2 , C = 80Gn/m 2 . Neglecting the inertia of shaft and gears calculate the frequency of free torsional vibrations in Hz.

a) 4.7

b) 5.7

c) 4.5

d) 5.5

Answer: a

Explanation: Moment of inertia of rotor B : Ib = 166.7 Kg-m 2

total length of equivalent shaft = 1000mm = 1m

Position of node = 294mm from left end 

J = 1.27×10 6 m 4

substituting the values into the free torsional vibration relation we get

f = 4.7 Hz.

8. When inertia of gearing is taken into consideration, then an additional shaft must be introduced in the equivalent system of the rotors connected by the shaft.

a) True

b) False

Answer: b

Explanation: For gearing system, if the inertia of the gear is taken into consideration then an additional rotor must be placed in the equivalent system of vibration.

9. At a nodal point in the shaft, the frequency of vibration is _________

a) Zero

b) Double than at the ends

c) Minimum

d) Maximum

Answer: a

Explanation: A point of zero vibration is known as a node, hence at a nodal point the amplitude of frequency of free torsional vibration is zero.

10. For a gearing system, the equivalent system is made consisting of two rotors. How many nodes will this new equivalent system will have?

a) 0

b) 1

c) 2

d) 3

Answer: b

Explanation: When a gearing system is converted into an equivalent system consisting of two rotors, it will have only a single node on the shaft as the two rotor systems only produce one node.

11. For a gearing system, the equivalent system is made consisting of three rotors. How many nodes will this new equivalent system will have?

a) 0

b) 1

c) 2

d) 3

Answer: c

Explanation: When a gearing system is converted into an equivalent system consisting of three rotors, it will have two nodes on the shaft as the three-rotor system produces two nodes.

This set of Basic Machine Dynamics Questions and Answers focuses on “Computer Aided Analysis for Four Bar Mechanism”.

1. In a CAD package, mirror image of a 2D point P  is to be obtained about a line which passes through the origin and makes an angle of 45° counterclockwise with the X-axis. The coordinates of the transformed point will be

a) 

b) 

c) 

d) 

Answer: b

Explanation: Coordinates of D in x-axis is  and in y-axis is 

So coordinates are 

2. In a CNC program block, N002 G02 G91 X40 Z40…, G02 AND G91 refer to

a) Circular interpolation in counterclockwise direction and incremental dimension

b) Circular interpolation in counterclockwise direction and absolute dimension

c) Circular interpolation in clockwise direction and incremental dimension

d) Circular interpolation in clockwise direction and absolute dimension

Answer: c

Explanation: G02 – Circular interpolation in clockwise direction

G91 – incremental dimension

3. For generating Coons patch we require

a) A set of grid points on surface

b) A set of control points

c) Four bounding curves defining surface

d) Two bounding curves and a set of grid control points

Answer: c

Explanation: Coons patch or surface is generated by the interpolation of 4 edge curves. A linear interpolation between four bounded curves is used to generate a Coons surface, also called as Coons patch. The method is credited to S. Coons who developed this concept for generating a surface.

4. NC contouring is an example of

a) Continuous path positioning

b) Point-to-point positioning

c) Absolute positioning

d) Incremental positioning

Answer: a

Explanation: Contouring is the most complex flexible and the most expensive type of machine tool control. It is capable of performing both PTP and straight cut operations. In addition to the distinguishing feature of the contouring NC system is their capacity for simultaneous control of more than one axis movement of machine tool illustrates the versatility of continuous path NC. Milling and Turning are the common examples of the use of conturing control.

5. The tool of an NC machine has to move along a circular arc from  to  while performing an operation. The center of the arc is at . Which one of the following NC tool path commands performs the above mentioned operation?

a) N010 G02 X10 Y10 X5 Y5 R5

b) N010 G03 X10 Y10 X5 Y5 R5

c) N010 G01 X5 Y5 X10 Y10 R5

d) N010 G02 X5 Y5 X10 Y10 R5

Answer: d

Explanation: Tool-Radius Compensation

Left hand G41

Right hand G42

Cancel tool-radius compensation G40

Tool-Height Compensation

Positive G43

Negative G44

Cancel tool-height compensation G49

Tool-radius compensations make it possible to program directly from the drawing, and thus eliminate the tool-offset calculation

G41  Dxx

Dxx: the radius of tool to compensate is saved in a memory unit that is named Dxx

G41/G42 is directly related with direction of tool movement and which side of part is cut.

6. During the execution of a CNC part program block NO20 GO2 X45.0 Y25.0 R5.0 the type of tool motion will be

a) Circular Interpolation – clockwise

b) Circular Interpolation – counterclockwise

c) Linear Interpolation

d) Rapid feed

Answer: a

Explanation: G02 – Circular interpolation in clockwise direction

G91 – incremental dimension

7. In a 2-D CAD package, clockwise circular arc of radius, 5, specified from P1  to P2 will have its center at

a) 

b) 

c) 

d) 

Answer: a

Explanation: Coordinates of P in x-axis is  and in y-axis is 

So coordinates are 

8. In an NC machining operation, the tool has to be moved from point  to point  along a circular path with center at . Before starting the operation, the tool is at . The correct G and M code for this motion is

a) N010 G03 X7.0 Y2.0 I5.0 J2.0

b) N010 G02 X7.0 Y2.0 I5.0 J2.0

c) N010 G01 X7.0 Y2.0 I5.0 J2.0

d) N010 G00 X7.0 Y2.0 I5.0 J2.0

Answer: b

Explanation: G02 – Circular interpolation in clockwise direction

G91 – incremental dimension

9. In computer aided drafting practice, an arc is defined by

a) Two end points only

b) Center and radius

c) Radius and one end point

d) Two end points and center

Answer: d

Explanation: Two end points and center defines an arc.

Answer: a

Explanation: Radius of trajectory = 5cm

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Programme for Four Bar Mechanism”.

1. With reference to NC machines, which of the following statements is wrong

a) Both closed-loop and open-loop control systems are used

b) Paper tapes, floppy tapes and cassettes are used for data storage

c) Digitizers may be used as interactive input devices

d) Post processor is an item of hardware

Answer: c

Explanation: None.

2. In a point-to-point type of NC system

a) Control of position and velocity of the tool is essential

b) Control of only position of the tool is sufficient

c) Control of only velocity of the tool is sufficient

d) Neither position nor velocity need to be controlled

Answer: b

Explanation: Point to point is also sometimes called a positioning system. In PTP the objective of the machine tool control system is to move the cutting tool to predefined location. The speeder path by which this movement is accomplished is not important in point to point NC. Once the tool reaches the desired location, the machining operation is performed at that position.

3. GUI is the acronym for Graphical User Interface.

a) True

b) False

Answer: a

Explanation: None.

4. The heart of a computer is:

a) CPU

b) ALU

c) Monitor

d) Keyboard

Answer: a

Explanation: Keyboard is input device, monitor is output device.

5. The widely employed computer architecture for CAD/CAM applications is:

a) Mainframe-based system

b) Minicomputer-based system

c) Microcomputer-based system

d) Workstation-based system

Answer: d

Explanation: Workstation based system is widely used for CAD/cAM architecture.

6. Keyboard is a ___________ input device.

a) Graphical

b) Text

c) Numericals

d) All of the mentioned

Answer: d

Explanation: None

7. Locating devices are classified as:

a) Text input devices

b) Graphics input devices

c) Both a and b

d) None of the mentioned

Answer: b

Explanation: None.

8. Mouse is a __________ type of input device.

a) Text

b) Graphics

c) Locating

d) All of the mentioned

Answer: c

Explanation: Through mouse we locate the required data or information.

9. Light pen is a:

a) Writing device

b) Drawing device

c) Locating device

d) Lighting device

Answer: b

Explanation: Light pen is used to draw on screen.

Answer: c

Explanation: The Digitizer is a completely separate component than the LCD, although the LCD and Digitizer are usually fused together during manufacturing with an optical adhesive.

This set of Machine Dynamics Questions and Answers for Aptitude test focuses on “Computer Aided Analysis for Slider Crank Mechanism”.

1. Thumbwheels are usually mounted on:

a) Keyboard

b) Monitor

c) CPU

d) Mouse

Answer: b

Explanation: None

2. The screen is scanned from left to right, top to bottom all the time to generate graphics by:

a) Raster scans

b) Random scan

c) Vector scan

d) Stoke writing

Answer: a

Explanation: In a raster- scan system, the electron beam is swept across the screen, one row at a time from top to bottom. As the electron beam moves across each row, the beam intensity is turned on and off to create a pattern of illuminated spots.

3. Color raster display uses three electron guns, namely:

a) Red, green and blue

b) Red, green and yellow

c) White, blue and black

d) Red, black and white

Answer: a

Explanation: Usual color image  is a raster consisting of red, blue and green band. Single band layers typically represent either continuous variables.

4. The software that is used for file manipulations, managing directories and subdirectories, programming and accounts setups is known as:

a) Graphics software

b) Operating system

c) Application software

d) Programming language

Answer: c

Explanation: Application software can be divided into two general classes: systems software and applications software. Applications software  include such things as database programs, word processors, Web browsers and spreadsheets.

5. The software that provides users with various functions to perform geometric modelling and construction. Editing and manipulation of existing geometry. drafting and documentation is known as:

a) Operating system

b) Application software

c) Graphics software

d) Programming language

Answer: c

Explanation: In computer graphics, graphics software or image editing software is a program or collection of programs that enable a person to manipulate visual images on a computer. These are the application software which lets the user to create and manipulate any type of computer graphics with the use of an operating system.

6. The software used for the purpose of mass property calculations, assembly analysis, tolerance analysis, finite element analysis, mechanisms analysis, sheet metal design, analysis of plastic injection molding and animation techniques is:

a) Graphics software

b) Operating system

c) Application software

d) Programming language

Answer: c

Explanation: Application software can be divided into two general classes: systems software and applications software. Applications software  include such things as database programs, word processors, Web browsers and spreadsheets.

7. The software that enables the user to implement custom applications or modify the system for specialized needs is known as:

a) Programming language

b) Operating system

c) Application software

d) Graphics software

Answer: a

Explanation: A programming language is a formal constructed language designed to communicate instructions to a machine, particularly a computer. Programming languages can be used to create programs to control the behavior of a machine or to express algorithms.

8. The following is not a graphics standard:

a) GKS

b) IGBS

c) UNIX

d) PHIGS

Answer: c

Explanation: UNIX is a computer Operating System which is capable of handling activities from multiple users at the same time.

9. In the following geometric modelling techniques which are not three-dimensional modelling?

a) Wireframe modelling

b) Drafting

c) Surface modelling

d) Solid modelling

Answer: b

Explanation: Technical drawing, also known as drafting or draughting, is the act and discipline of composing drawings that visually communicate how something functions or is to be constructed.

Answer: c

Explanation: Compared to surface and solid modeling, wireframe modeling is the least complex method for representing 3D images.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Classifications of Synthesis Problem”.

1. In the following geometric modelling techniques, which cannot be used for finite element analysis:

a) Wireframe modelling

b) Surface modelling

c) Solid modelling

d) None of the mentioned

Answer: d

Explanation: None

2. In the following geometric primitives, which is not a solid entity of CSG modelling:

a) Box

b) Cone

c) Cylinder

d) Circle

Answer: d

Explanation: In 3D computer graphics and CAD CSG is often used in procedural modeling. CSG can also be performed on polygonal meshes, and may or may not be procedural and/or parametric.Typically they are the objects of simple shape: cuboids, cylinders, prisms, pyramids, spheres, cones.

3. The number of lines required to represent a cube, in a wireframe model is:

a) 8

b) 6

c) l2

d) I6

Answer: c

Explanation: None

4. Which of the following is not an analytical entity?

a) Line

b) Circle

c) Spline

d) Parabola

Answer: c

Explanation: In mathematics, a spline is a numeric function that is piecewise-defined by polynomial functions, and which possesses a high degree of smoothness at the places where the polynomial pieces connect.

5. Which of the following is not a synthetic entity?

a) Hyperbola

b) Bezier curve

c) B-spline curve

d) Cubic spline curve

Answer: a

Explanation: he hyperbola is one of the three kinds of conic section, formed by the intersection of a plane and a double cone.

6. Which one of the following does not belong to the family of conics?

a) Parabola

b) Ellipse

c) Hyperbola

d) Line

Answer: d

Explanation: A line is a straight one-dimensional figure having no thickness and extending infinitely in both directions. A line is sometimes called a straight line or, more archaically, a right line.

7. The number of tangents required to describe cubic splines is:

a) 2

b) 1

c) 3

d) 4

Answer: b

Explanation: Since the order of cubic spline is one, therefore only only tangent is required to describe it.

8. The order of the cubic spline is the

a) 2″ order

b) 3″ order

c) 1″ order

d) 4″ order

Answer: c

Explanation: As only one tangent is required to describe cubic spline, therefore its order is 1.

9. The shape of the Bezier curve is controlled by:

a) Control points

b) Knots

c) End points

d) All of the mentioned

Answer: a

Explanation: A control point  is a marked way point used in orienteering and related sports such as rogaining and adventure racing.

Answer: b

Explanation: A B-spline is a generalization of the Bezier curve.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Precision Points for Function Generation”.

1. The degree of the Bezier curve with n control points is:

a) n + 1

b) n – I

c) n

d) 2n

Answer: a

Explanation: The degree of a Bézier curve defined by n+1 control points is n:

In each basis function, the exponent of u is i +  = n. Therefore, the degree of the curve is n.

2. The degree of the B-spline with varying knot vectors:

a) Increases with knot vectors

b) Decreases with knot vectors

c) Remains constant

d) none of the mentioned

Answer: a

Explanation: Changing the degree of the curve due to the increase of knots will change the shape of the curve globally and will not be considered. Therefore, inserting a new knot causes a new control point to be added. In fact, some existing control points are removed and replaced with new ones by corner cutting.

3. C” continuity refers to:

a) Common tangent

b) Common point

c) Common curvature

d) Common normal

Answer: a

Explanation: C‘ continuity refers to Common curvature .

C 0 continuity refers to Common point.

C” continuity refers to Common tangent.

4. C‘ continuity refers to:

a) Common tangent

b) Common point

c) Common curvature

d) Common normal

Answer: c

Explanation: C‘ continuity refers to Common curvature .

C 0 continuity refers to Common point.

C” continuity refers to Common tangent.

5. C 0 continuity refers to:

a) Common tangent

b) Common point

c) Common curvature

d) Common normal

Answer: b

Explanation: C‘ continuity refers to Common curvature .

C 0 continuity refers to Common point.

C” continuity refers to Common tangent.

6. The number of non-coincidental points required to define the simplest surface are:

a) 4

b) 3

c) 2

d) 5

Answer: b

Explanation: None.

7. Convex hull property is satisfied by the following surface:

a) Bezier

b) B-spline

c) NURBS

d) All of the mentioned

Answer: b

Explanation: The curve that follows a convex hull property is B-spline.

8. The tensor product technique constraints surfaces by two curves.

a) Adding

b) Subtraction

c) Multiplying

d) Dividing

Answer: c

Explanation: None.

9. The degrees of freedom of a two-node bar element are:

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: None.

Answer: a

Explanation: None.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Programme using Least Square Technique”.

1. The sum of the shape functions over the element is always equal to:

a) Zero

b) Infinity

c) Unity

d) None of the mentioned

Answer: c

Explanation: A shape function is always equal to one at its designated node and zero at the other nodes of the element.

2. Stiffness is ______________ to the length of the element.

a) Inversely proportional

b) Directly proportional

c) Exponential

d) Independent

Answer: d

Explanation: None.

3. Computer will perform the data processing functions in

a) NC

b) CNC

c) DNC

d) ACS

Answer: c

Explanation: In NC it performs numerical problems.

In DNC it will perform data processing functions.

4. The first commercial CNC machine was developed in the year:

a) 1970

b) 1972

c) 1976

d) 1980

Answer: a

Explanation: The first NC machines were built in the 1940s and 1950s.

CNC was developed in 1970.

5. CNC drilling machine is considered to be a:

a) Point-to-point controlled machine

b) Straight line controlled machine

c) Continuous path controlled machine

d) Servo-controlled machine

Answer: a

Explanation: None.

6. The lost motion in CNC machine tool is on account of:

a) Backlash in gearing

b) Wind-up of drive shafts

c) Deflection of machine tool members

d) All of the mentioned

Answer: a

Explanation: The system must provide sufficient feedback and programming capacity to allow for backlash and lost motion between motor and load. Computer numerical control  of machine tools has embodied position control for a number of years.

7. The axes of turning machine arc:

a) Z and X-axes

b) X and Y-axes

c) Z and Y-axes

d) X, Y and Z-axes

Answer: a

Explanation: None.

8. On turning lathes the machine zero point is generally at the:

a) Head stock of lathe spindle nose face

b) Dead center of tail stock

c) Tool point mounted on tool post

d) none of the mentioned

Answer: b

Explanation: On turning lathes, the machine zero point is generally at the centre of the spindle nose face.

9. Dwell is defined by:

a) G04

b) G03

c) G02

d) G01

Answer: a

Explanation: G02 – Circular interpolation in clockwise direction

G91 – incremental dimension.

Answer: a

Explanation: M30 – Program Stopping and Ending.

This set of Machine Dynamics Assessment Questions and Answers focuses on “Computer Aided Synthesis of Four Bar Mechanism With Coupler Point”.

1. A micro programmed control unit

a) is faster than hardwired control unit

b) facilitates easy implementation of new instructions

c) is useful when very small programs are to be run

d) usually refers to the control unit of microprocessor

Answer: b

Explanation: A micro programmed control unit facilitates easy implementation of new instructions.

2. Preparing a magnetic disk for data storage is called

a) booting

b) formatting

c) debuffing

d) buffing

Answer: b

Explanation: Preparing a magnetic disk for data storage is called formatting.

3. The addressing mode used in the instruction PUSH B is

a) direct

b) register

c) register indirect

d) immediate

Answer: b

Explanation: There are four types of instruction: –

PUSH A, PUSH B, add α popc

Wherein addressing mode used are as follows: –

PUSH A – Direct

PUSH B – Register

Add – Register indirect

Pop.C – Immediate

4. Index register in a microprocessor is used for

a) direct addressing

b) address modification

c) pointing to the stack

d) loop execution

Answer: c

Explanation: Index Register is also called special purpose Register, which is used as stack pointer to programme stack i.e., is used to hold the address of the top of stack.

5. In the Fortran program

M = 0

DO 100I = 1,2

DO 200J = 1,2

M = M + I + J

200 CONTINUE

100 CONTINUE

STOP

END

the value of M in the end will be

a) 10

b) 11

c) 12

d) 14

Answer: c

Explanation: m = 0,

I = 1,

J = 1

m = m + I + J = 0 + 1 + 1 = 2

m = 2, I = 1, J = 2

m = 2 + 1 + 2 = 5

m = 5, I = 2, I = 1

m = 5 + 2 + 1 = 8

m = 8, I = 2, J = 2

m = 8 + 2 + 2 = 12

6. In a FORTRAN program

a) all statements must be numbered

b) the numbered statements must be referred

c) the statements referred must be numbered

d) all statements must be referred

Answer: c

Explanation: In Fortran programming the statements referred must be numbered.

7. Program status word  contains various status of

a) Program

b) CPU

c) ALU

d) Register

Answer: d

Explanation: PSW is a collection of data 8 bytes or  long maintained by the o.s and it keeps track of current status of system registers. It describes: –

1. Interrupt masks

2. Privillage states

3. condition code

4. Instruction address

8. In NC machining, coordinated movement of separately driven axes motion is required to achieve the desired path of tool relative to workpiece. The generation of these reference signals is accomplished through a device called

a) approximator

b) interpolator

c) coordinator

d) director

Answer: b

Explanation: Interpolator is a device which manages the axis coordination in multi-axes machines like more than 3-axis machines.

9. MRP input requires:

a) MPS

b) BOM

c) Inventory file

d) All of the mentioned

Answer: d

Explanation: None

Answer: b

Explanation: The BOM is stored in a database and it is used in Materials Requirement Planning.

This set of Machine Dynamics Questions & Answers for Exams focuses on “Synthesis of Four Bar Mechanism for Body Guidance”.

1. Which of the following mathematical tool is used to synthesize the mechanism graphically?

a) Tangent

b) Perpendicular bisector

c) Chord

d) Arc of contact

Answer: b

Explanation: When the spatial position of the link or the mechanism is given, then the mathematical tool used to locate the position of joints is the perpendicular bisector.

2. Graphical errors occur during the function generation solely due to choice of wrong scale.

a) True

b) False

Answer: b

Explanation: Graphical errors occur during function generation due to the following two reasons:

1. choice of wrong scale

2. wrong graphical procedures.

3. Which of the following given below is not a type of error which could be identified during function regeneration?

a) Structural

b) Mechanical

c) Graphical

d) Theoretical

Answer: d

Explanation: When a function is being regenerated, then it is possible that we may encounter structural, Mechanical or simply graphical error but it is not possible to have a theoretical error.

4. Which of the following is not classified under synthesis problem?

a) Body guidance

b) Function generation

c) Path generation

d) Graphical methodology

Answer: d

Explanation: Graphical methodology can result in errors in the synthesis of a mechanism hence, it is classified as a synthesis problem. However, the remaining options do not classify as synthesis problems.

5. Order defect can be defined as _________

a) Improper configuration of precision points

b) Non covering of all points by the mechanism

c) Too many configuration points

d) Too many coordinate systems

Answer: a

Explanation: Order defect occurs when the precision points are not configured properly. It is to be noted that too many configuration points does not lead to order defect.

6. For errors in function generation, which of the following statement is true?

a) Structural errors are 0 at precision points

b) Structural and mechanical errors are coexistent

c) Increase in precision points will increase structural error

d) Decrease in precision points will decrease structural error

Answer: a

Explanation: Precision points are defined as the points where structural errors are 0. Structural errors can be reduced by increasing the number of precision points.

7. If the given axes are Y and X axis and the numbers represent the spatial points, then what will be the position of one of the pivots?

machine-dynamics-questions-answers-synthesis-four-bar-mechanism-body-guidance-q7

a) Below X axis

b) Above X axis

c) Left of Y axis

d) Between Y axis and 1

Answer: a

Explanation: If the points are joined and then a perpendicular bisector of the line formed by the joined points is made to intersect, then the intersection point gives the position of the pivot. In the above figure, it lies below the X axis and in the IVth quadrant.

8. Structural errors are 0 at precision points.

a) True

b) False

Answer: a

Explanation: Precision points are known as points having zero structural error, thus increasing the number of structural points will result in decrease in structural error.

9. Which of the following linkage is observed to have branch defect?

a) Grashof type 4 bar linkage

b) Klein constructions

c) Ritterhaus’ constructions

d) Non- Grashof type 4 bar linkage

Answer: a

Explanation: Grashof type 4 bar linkage has been oserved to have branch defect. Klein and ritterhau’s construction methods are used to calculate velocity and acceleration of the links in a mechanism.

10. Which method is used in the synthesis of function generation?

a) Overlay method

b) Klein’s method

c) Chebychev’s inequality

d) Freudenstein’s equation

Answer: a

Explanation: Overlay method is used in the synthesis of function generation, and it is used to generate precision points in a 4 bar linkage.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Analytical Synthesis for Slider Crank Mechanism”.

1. If the crank rotation angle varies from 45 degrees to 135 degrees, Using three precision points with chebyshev’s spacing, synthesize a slider crank mechanism where where displacement of slider is square of the crank rotation. From the above information calculate ‘a’ in mm.

a) 48.2

b) 58.2

c) 45.6

d) 42.2

Answer: a

Explanation: From the above data θs=45° and θf=135°, assuming starting displacement and final displacement of slider as 100mm and 30mm respectively,

We calculate k 1 = 96.4mm

k 1 = 2a

Therefore a = 48.2mm.

2. If the crank rotation angle varies from 45 degrees to 135 degrees, Using three precision points with chebyshev’s spacing, synthesize a slider crank mechanism where where displacement of slider is square of the crank rotation. From the above information calculate ‘c’ in mm.

a) 148.2

b) 158.2

c) 135.7

d) 120.2

Answer: c

Explanation: From the above data θs=45° and θf=135°, assuming starting displacement and final displacement of slider as 100mm and 30mm respectively,

We calculate k 1 = 96.4mm

k 1 = 2a

Therefore a = 48.2mm

Now k 2 = 2.a.c = 13084

Therefore c = 135.7mm.

3. If the crank rotation angle varies from 45 degrees to 135 degrees, Using three precision points with chebyshev’s spacing, synthesize a slider crank mechanism where where displacement of slider is square of the crank rotation. From the above information calculate ‘b’ in mm.

a) 148.2

b) 158.2

c) 135.7

d) 120.2

Answer: d

Explanation: From the above data θs=45° and θf=135°, assuming starting displacement and final displacement of slider as 100mm and 30mm respectively,

We calculate k 1 = 96.4mm

k 1 = 2a

Therefore a = 48.2mm

Now k 2 = 2.a.c = 13084

Therefore c = 135.7mm

Now b 2 = a 2 + c 2 – k 3

Therefore b = 120.2mm.

4. What is the relation between displacement of slider and crank angle over an interval?

a) Quadratic

b) Directly proportional

c) Linear

d) Inversely proportional

Answer: c

Explanation: The displacement of the slider depends on the crank angle in a linear mathematical relation. It also depends on the starting crank angle and starting point.

5. Which of the following is true regarding synthesis of mechanisms?

a) It determines the dimensions of the link

b) It determines the input and output angles of the links

c) It determines velocity of link

d) It determines acceleration of link

Answer: a

Explanation: For computer aided synthesis of mechanism, determination of the dimensions of the given link is the main objective.

6. What is the unit of constant of proportionality C used in the relation between displacement of slider and crank angle?

a) m/rad

b) m/s-rad

c) m-s/rad

d) m/s 2

Answer: a

Explanation: The displacement of the slider is proportional to the crank angle hence a proportionality constant C is introduced; this has the units metre/radians. m/rad.

7. Synthesis of mechanism is used to determine the acceleration of the link.

a) True

b) False

Answer: b

Explanation: Synthesis of mechanism is used to determine the dimensions of the link, acceleration of the link can be calculated by using methods of klein’s constructions.

8. Constant of proportionality is always positive.

a) True

b) False

Answer: b

Explanation: The constant of proportionality used in the relation between slider displacement and the crank angle can take both positive and negative values.

9. From the given data calculate the constant of proportionality.

s = 10mm, Ss = 5mm θ=50 degrees and θs = 45 degrees.

a) 1

b) -1

c) 1/2

d) -1/2

Answer: a

Explanation: We know that constant of proportionality is related by:

s – Ss = C

substituting the values we get

C = 1.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Automatic Control”.

1. The result of the act of adjustment is called

a) response

b) command

c) process control

d) process controller

Answer: b

Explanation: The result of the act of adjustment is called command. The subsequent result of the system to the command is known as response. The automatic control of variables is known as process control. The device which controls a process is called a process controller.

2. The subsequent result of the system to the command is known as

a) response

b) command

c) process control

d) process controller

Answer: a

Explanation: The result of the act of adjustment is called command. The subsequent result of the system to the command is known as response. The automatic control of variables is known as process control. The device which controls a process is called a process controller.

3. The automatic control of variables is known as

a) response

b) command

c) process control

d) process controller

Answer: c

Explanation: The result of the act of adjustment is called command. The subsequent result of the system to the command is known as response. The automatic control of variables is known as process control. The device which controls a process is called a process controller.

4. The device which controls a process is called a

a) response

b) command

c) process control

d) process controller

Answer: d

Explanation: The result of the act of adjustment is called command. The subsequent result of the system to the command is known as response. The automatic control of variables is known as process control. The device which controls a process is called a process controller.

5. The device used to keep the variables at a constant desired value is called as

a) regulator

b) kinetic control

c) feed back

d) error detector

Answer: a

Explanation: The device used to keep the variables at a constant desired value is called as regulator. The instrument measuring the output of the machine for comparison with the input to the machine is known as feedback. The automatic control of the displacement or velocity or acceleration of a member of a machine is called as kinetic control. A differential device used to measure the actual controlled quantity and to compare it continuously with the desired value is called error detector.

6. The instrument measuring the output of the machine for comparison with the input to the machine is known as

a) regulator

b) kinetic control

c) feed back

d) error detector

Answer: c

Explanation: The device used to keep the variables at a constant desired value is called as regulator. The instrument measuring the output of the machine for comparison with the input to the machine is known as feedback. The automatic control of the displacement or velocity or acceleration of a member of a machine is called as kinetic control. A differential device used to measure the actual controlled quantity and to compare it continuously with the desired value is called error detector.

7. The automatic control of the displacement or velocity or acceleration of a member of a machine is called as

a) regulator

b) kinetic control

c) feed back

d) error detector

Answer: b

Explanation: The device used to keep the variables at a constant desired value is called as regulator. The instrument measuring the output of the machine for comparison with the input to the machine is known as feedback. The automatic control of the displacement or velocity or acceleration of a member of a machine is called as kinetic control. A differential device used to measure the actual controlled quantity and to compare it continuously with the desired value is called error detector.

8. A differential device used to measure the actual controlled quantity and to compare it continuously with the desired value is called

a) regulator

b) kinetic control

c) feed back

d) error detector

Answer: d

Explanation: The device used to keep the variables at a constant desired value is called as regulator. The instrument measuring the output of the machine for comparison with the input to the machine is known as feedback. The automatic control of the displacement or velocity or acceleration of a member of a machine is called as kinetic control. A differential device used to measure the actual controlled quantity and to compare it continuously with the desired value is called error detector.

9. It is a device to change a signal which is in one physical form to a corresponding signal in another physical form.

a) amplification

b) transducer

c) feed back

d) none of the mentioned

Answer: b

Explanation: Increasing the amplitude of the signal without affecting its waveform is known as amplification. The device to change a signal which is in one physical form to a corresponding signal in another physical form is known as transducer.

10. Increasing the amplitude of the signal without affecting its waveform is known as

a) amplification

b) transducer

c) feed back

d) none of the mentioned

Answer: a

Explanation: Increasing the amplitude of the signal without affecting its waveform is known as amplification. The device to change a signal which is in one physical form to a corresponding signal in another physical form is known as transducer.

This set of Machine Dynamics Multiple Choice Questions & Answers  focuses on “Open-Loop Transfer Function”.

1. In an open loop control system

a) Output is independent of control input

b) Output is dependent on control input

c) Only system parameters have effect on the control output

d) None of the mentioned

Answer: a

Explanation: When the input to a system is independent of the output from the system, then the system is called an open-loop or unmonitored system.

2. For open control system which of the following statements is incorrect ?

a) Less expensive

b) Recalibration is not required for maintaining the required quality of the output

c) Construction is simple and maintenance easy

d) Errors are caused by disturbances

Answer: b

Explanation: Most measuring instruments are open-loop control systems, as for the same input signal, the readings will depend upon things like ambient temperature and pressure.

3. A control system in which the control action is somehow dependent on the output is known as

a) Closed loop system

b) Semiclosed loop system

c) Open system

d) None of the mentioned

Answer: a

Explanation: When output of a system is measured and is continuously compared with the required value, then it is known as closed-loop or monitored system.

4. In closed loop control system, with positive value of feedback gain the overall gain of the system will

a) decrease

b) increase

c) be unaffected

d) none of the mentioned

Answer: a

Explanation: In closed loop control system, the output is measured and through a feedback transducer, it is sent to an error detector which detects any error in the output from the required value thus adjusting the input in a way to get the required output.

5. Which of the following is an open loop control system ?

a) Field controlled D.C. motor

b) Ward leonard control

c) Metadyne

d) Stroboscope

Answer: a

Explanation: In field control D.C. motor, the input is dependent of the output. So it is an open loop control system.

6. Which of the following statements is not necessarily correct for open control system ?

a) Input command is the sole factor responsible for providing the control action

b) Presence of non-linearities causes malfunctioning

c) Less expensive

d) Generally free from problems of non-linearities

Answer: b

Explanation: When the input to a system is independent of the output from the system, then the system is called an open-loop or unmonitored system. It is also called as a calibrated system. Most measuring instruments are open-loop control systems, as for the same input signal, the readings will depend upon things like ambient temperature and pressure.

7. In open loop system

a) the control action depends on the size of the system

b) the control action depends on system variables

c) the control action depends on the input signal

d) the control action is independent of the output

Answer: d

Explanation: When the input to a system is independent of the output from the system, then the system is called an open-loop or unmonitored system.

8. The following has tendency to oscillate.

a) Open loop system

b) Closed loop system

c) Both  and 

d) Neither  nor 

Answer: b

Explanation: Both open loop system and closed loop system have the tendency to oscillate.

9. A good control system has all the following features except

a) good stability

b) slow response

c) good accuracy

d) sufficient power handling capacity

Answer: b

Explanation: Repose is not included in a good control system.

Answer: c

Explanation: The needle of the speedometer is only the indicator of the speed and to keep the speed constant, the driver has to maintain the speed of 50 km/h.

This set of Machine Dynamics Problems focuses on “Closed-Loop Transfer Function”.

1. The initial response when the output is not equal to input is called

a) Transient response

b) Error response

c) Dynamic response

d) None of the mentioned

Answer: a

Explanation: Transient response of control system means changing so, this occurs mainly after two conditions and these two conditions are

1. Just after switching ‘on’ the system that means at the time of application of an input signal to the system.

2. Just after any abnormal conditions. Abnormal conditions may include sudden change in the load, short circuiting etc.

2. A control system working under unknown random actions is called

a) computer control system

b) digital data system

c) stochastic control system

d) adaptive control system

Answer: c

Explanation: Adaptive control is the control method used by a controller which must adapt to a controlled system with parameters which vary, or are initially uncertain. Stochastic control system is a control system working under unknown random actions.

3. An automatic toaster is a __________ loop control system.

a) open

b) closed

c) partially closed

d) none of the mentioned

Answer: a

Explanation: The output of a toaster is not dependent on the input of the toaster. Hence, it is a open loop control system.

4. Any externally introduced signal affecting the controlled output is called a

a) feedback

b) stimulus

c) signal

d) gain control

Answer: b

Explanation: The instrument measuring the output of the machine for comparison with the input to the machine is known as feedback. Any externally introduced signal affecting the controlled output is called a stimulus.

5. A closed loop system is distinguished from open loop system by which of the following ?

a) Servomechanism

b) Feedback

c) Output pattern

d) Input pattern

Answer: b

Explanation: In open loop system we do not have any feedback but in closed loop system we have feedback.

6. ______________ is a part of the human temperature control system.

a) Digestive system

b) Perspiration system

c) Ear

d) Leg movement

Answer: b

Explanation: Digestive system controls the digestion of food, ear is to get the sound and leg movement is controlled by brain. So, perspiration system only controls human body temperature.

7. By which of the following the control action is determined when a man walks along a path ?

a) Brain

b) Hands

c) Legs

d) Eyes

Answer: d

Explanation: Walking id done through legs, so legs are only responsible.

8. _______________ is a closed loop system.

a) Auto-pilot for an aircraft

b) Direct current generator

c) Car starter

d) Electric switch

Answer: a

Explanation: Auto-pilot for an aircraft is a closed loop system as the input is totally dependent on the output whichever is required.

9. Which of the following devices are commonly used as error detectors in instruments ?

a) Vernistats

b) Microsyns

c) Resolvers

d) None of the mentioned

Answer: d

Explanation: The commonly used error detectors are self-balancing recording potentiometer, hot wire anemometer, eloctro-magnetic flow meter, torque sensor etc.

Answer: b

Explanation: By increasing the gain of the system the unstable system can be stabalized.