This set of Machine Kinematics Interview Questions and Answers for freshers focuses on “Numericals On Kinetics Of Motion and Loss of Kinetic Energy”.

1. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate kinetic energy of rotation of the wheels and axles at a speed of 9 km/h.

a) 7670 N-m

b) 8670 N-m

c) 9670 N-m

d) 6670 N-m

View Answer

Explanation: Given : m = 12 t = 12 000 kg ;

m

_{1}= 2 t = 2000 kg ; k

_{1}= 0.4 m ; d

_{1}= 1.2 m or r

_{1}= 0.6 m ; m

_{2}= 2.5 t = 2500 kg ; k

_{2}= 0.6 m ; d

_{2}= 1.5 m or r

_{2}= 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,

I_{1} = m_{1}(k_{1})^{2} = 2000 (0.4)^{2} = 320 kg-m^{2}

and mass moment of inertia of the rear axle together with its wheels,

I_{2} = m_{2} (k_{2})^{2} = 2500 (0.6)^{2} = 900 kg -m^{2}

Angular speed of the front roller,

ω_{1} = v/r_{1} = 2.5/0.6 = 4.16 rad/s

and angular speed of rear wheels,

ω_{2} = v/r_{2} = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E_{1} =1/2 I_{1} (ω_{1})^{2} = 1/2 × 320(4.16)^{2} 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E_{2} =1/2 I_{2} (ω_{2})^{2} = 1/2 × 900(3.3)^{2} 4900 N-m

∴ Total kinetic energy of rotation of the wheels,

E = E_{1} + E_{2} = 2770 + 4900 = 7670 N-m

2. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate total kinetic energy of road roller.

a) 25170 N-m

b) 35170 N-m

c) 45170 N-m

d) 55170 N-m

View Answer

Explanation: Given : m = 12 t = 12 000 kg ;

m

_{1}= 2 t = 2000 kg ; k

_{1}= 0.4 m ; d

_{1}= 1.2 m or r

_{1}= 0.6 m ; m

_{2}= 2.5 t = 2500 kg ; k

_{2}= 0.6 m ; d

_{2}= 1.5 m or r

_{2}= 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,

I_{1} = m_{1}(k_{1})^{2} = 2000 (0.4)^{2} = 320 kg-m^{2}

and mass moment of inertia of the rear axle together with its wheels,

I_{2} = m_{2} (k_{2})^{2} = 2500 (0.6)^{2} = 900 kg -m^{2}

Angular speed of the front roller,

ω_{1} = v/r_{1} = 2.5/0.6 = 4.16 rad/s

and angular speed of rear wheels,

ω_{2} = v/r_{2} = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E_{1} =1/2 I_{1} (ω_{1})^{2} = 1/2 × 320(4.16)^{2} 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E_{2} =1/2 I_{2} (ω_{2})^{2} = 1/2 × 900(3.3)^{2} 4900 N-m

∴ Total kinetic energy of rotation of the wheels,

E = E_{1} + E_{2} = 2770 + 4900 = 7670 N-m

We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller,

E_{3} = 1/2 mv^{2} = 1/2 x 1200 (2.5)^{2} = 37500 N-m

This energy includes the kinetic energy of translation of the wheels also, because the total mass (m) has been considered.

∴ Total kinetic energy of road roller,

E_{4} = Kinetic energy of translation + Kinetic energy of rotation

= E_{3} + E = 37 500 + 7670 = 45 170 N-m

3. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate braking force required to bring the roller to rest from 9 km/h in 6 m on the level.

a) 5528.3 N

b) 6528.3 N

c) 7528.3 N

d) 8528.3 N

View Answer

Explanation: Given : m = 12 t = 12 000 kg ;

m

_{1}= 2 t = 2000 kg ; k

_{1}= 0.4 m ; d

_{1}= 1.2 m or r

_{1}= 0.6 m ; m

_{2}= 2.5 t = 2500 kg ; k

_{2}= 0.6 m ; d

_{2}= 1.5 m or r

_{2}= 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,

I_{1} = m_{1}(k_{1})^{2} = 2000 (0.4)^{2} = 320 kg-m^{2}

and mass moment of inertia of the rear axle together with its wheels,

I_{2} = m_{2} (k_{2})^{2} = 2500 (0.6)^{2} = 900 kg -m^{2}

Angular speed of the front roller,

ω_{1} = v/r_{1} = 2.5/0.6 = 4.16 rad/s

and angular speed of rear wheels,

ω_{2} = v/r_{2} = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E_{1} =1/2 I_{1} (ω_{1})^{2} = 1/2 × 320(4.16)^{2} 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E_{2} =1/2 I_{2} (ω_{2})^{2} = 1/2 × 900(3.3)^{2} 4900 N-m

∴ Total kinetic energy of rotation of the wheels,

E = E_{1} + E_{2} = 2770 + 4900 = 7670 N-m

We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller,

E_{3} = 1/2 mv^{2} = 1/2 x 1200 (2.5)^{2} = 37500 N-m

This energy includes the kinetic energy of translation of the wheels also, because the total mass (m) has been considered.

∴ Total kinetic energy of road roller,

E_{4} = Kinetic energy of translation + Kinetic energy of rotation

= E_{3} + E = 37 500 + 7670 = 45 170 N-m

Let F = Braking force required to bring the roller to rest, in newtons.

We know that the distance travelled by the road roller,

s = 6 m … (Given)

∴ Work done by the braking force

= F × s = 6 F N-m

This work done must be equal to the total kinetic energy of road roller to bring the roller to

rest, i.e.

6 F = 45 170 or F = 45 170/6 = 7528.3 N

4. A haulage rope winds on a drum of radius 500 mm, the free end being attached to a truck. The truck has a mass of 500 kg and is initially at rest. The drum is equivalent to a mass of 1250 kg with radius of gyration 450 mm. The rim speed of the drum is 0.75 m/s before the rope tightens. By considering the change in linear momentum of the truck and in the angular momentum of the drum, find the speed of the truck when the motion becomes steady.

a) 0.502 m/s

b) 0.602 m/s

c) 0.702 m/s

d) 0.802 m/s

View Answer

Explanation: Given : r = 500 mm = 0.5 m ; m

_{1}= 500 kg ; m

_{2}= 1250 kg ; k = 450 mm = 0.45 m ; u = 0.75 m/s

We know that mass moment of inertia of drum,

I_{2} = m_{2}.k^{2} = 1250 (0.45)^{2} = 253 kg-m^{2}

Let v = Speed of the truck in m/s, and

F = Impulse in rope in N-s.

We know that the impulse is equal to the change of linear momentum of the truck. Therefore

F = m_{1}.v = 500 v N-s

and moment of impulse = Change in angular momentum of drum

i.e. F x r = I_{1} (ω_{2} − ω_{1}) = I_{2}(u – v/r)

500v x 0.5 = 253(0.75 – v/0.5)

or, 250v = 380 − 506v

∴ 250 v + 506 v = 380

or v = 380/756 = 0.502 m/s

5. An electric motor drives a machine through a speed reducing gear of ratio 9:1. The motor armature, with its shaft and gear wheel, has moment of inertia 0.6 kg-m^{2}. The rotating part of the driven machine has moment of inertia 45 kg-m^{2}. The driven machine has resisting torque of 100 N-m and the efficiency of reduction gear is 95%. Find the power which the motor must develop to drive the machine at a uniform speed of 160 r.p.m.

a) 1764 W

b) 2764 W

c) 3764 W

d) 4764 W

View Answer

Explanation: Given : G = 9; I

_{A}= 0.6 kg-m

^{2}; I

_{B}= 45 kg-m

^{2};

T

_{B}= 100 N-m; η = 95% = 0.95;

N = 160 r.p.m. ; N

_{1}= 0 ; N

_{2}= 60 r.p.m. T

_{A}= 30 N-m

We know that the power which the motor must develop,

P = 2πN T_{B}/60× η

= 2π × 160 × 100/60 x 0.95

= 1764 W

6. The flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant. Determine : Angular acceleration of the flywheel.

a) 0.6 rad/s^{2}

b) 0.8 rad/s^{2}

c) 0.10 rad/s^{2}

d) none of the mentioned

View Answer

Explanation: Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m

Angular acceleration of the flywheel

Let α = Angular acceleration of the flywheel.

We know that mass moment of inertia of the flywheel,

I=m.k

^{2}= 2500×12 = 2500 kg-m

^{2}

We also know that torque ( T ),

1500 = I .α = 2500 × α

or α = 1500 / 2500 = 0.6 rad/s^{2}

7. The flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant. Determine : Kinetic energy of the flywheel after 10 seconds from the start.

a) 50 kJ

b) 60 kJ

c) 45 kJ

d) none of the mentioned

View Answer

Explanation: Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m

Angular acceleration of the flywheel

Let α = Angular acceleration of the flywheel.

We know that mass moment of inertia of the flywheel,

I=m.k

^{2}= 2500×12 = 2500 kg-m

^{2}

We also know that torque ( T ),

1500 = I .α = 2500 × α

or α = 1500 / 2500 = 0.6 rad/s^{2}

Kinetic energy of the flywheel after 10 seconds from start

First of all, let us find the angular speed of the flywheel (ω^{2}) after t = 10 seconds from the start (i.e. ω^{1} = 0 ).

We know that ω^{2} = ω^{1} + α.t = 0 + 0.6 × 10 = 6 rad/s

∴ Kinetic energy of the flywheel,

E = 1/2 I(ω^{2})^{2}

= 1/2 x 2500 x 6^{2}

= 45 000J

= 45 kJ

8. Which of the following objects have momentum?

a) An electron is orbiting the nucleus of an atom.

b) A UPS truck is stopped in front of the school building.

c) The high school building rests in the middle of town.

d) None of the mentioned

View Answer

Explanation: Momentum can be thought of as mass in motion. An object has momentum if it has its mass in motion. It matters not whether the object is of large mass or small mass, moving with constant speed or accelerating; if the object is MOVING, then it has momentum.

9. A truck driving along a highway road has a large quantity of momentum. If it moves at the same speed but has twice as much mass, its momentum is ________________

a) zero

b) quadrupled

c) doubled

d) unchanged

View Answer

Explanation: Momentum is directly related to the mass of the object. So for the same speed, a doubling of mass leads to a doubling of momentum.

10. A ball is dropped from the same height upon various flat surfaces. For the same collision time, impulses are smaller when the most bouncing take place.

a) True

b) False

View Answer

Explanation: Since being dropped from the same height, the balls will be moving with the same pre-collision velocity (assuming negligible air resistance). Upon collision with the ground, the velocity will have to be reduced to zero – that is, the ball will cease moving downwards. This decrease in velocity constitutes the first portion of the velocity change. If the ball bounces, then there is an additional velocity change sending the ball back upwards opposite the original direction. Thus, for the same collision time, bouncing involves a greater velocity change, a greater momentum change, and therefore a greater impulse.

**Sanfoundry Global Education & Learning Series – Machine Kinematics.**

To practice all areas of Machine Kinematics for Interviews, __here is complete set of 1000+ Multiple Choice Questions and Answers__.

**If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]**

**Related Posts:**

- Check Kinematics of Machinery Books
- Apply for Aerospace Engineering Internship
- Practice Mechanical Engineering MCQs
- Practice Aeronautical Engineering MCQs
- Check Mechanical Engineering Books