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Work and energy in rolling (check your understnading)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Objective questions, contained in this module with hidden solutions, help improve understanding of the topics covered under the module "Work and energy in rolling".

The questions have been selected to enhance understanding of the topics covered in the module titled " Work and energy in rolling ". All questions are multiple choice questions with one or more correct answers. There are two sets of the questions. The “understanding level” questions seek to unravel the fundamental concepts involved, whereas “application level” are relatively difficult, which may interlink concepts from other topics.

Each of the questions is provided with solution. However, it is recommended that solutions may be seen only when your answers do not match with the ones given at the end of this module.

Understanding level (Work and energy in rolling)

Exercise 1

A spherical ball rolls without sliding. Then, the fraction of its total mechanical energy associated with translation is :

(a) 1 2 (b) 2 3 (c) 3 4 (d) 5 7 (a) 1 2 (b) 2 3 (c) 3 4 (d) 5 7

Solution

The kinetic energy for translation is distributed between translation and rotation. The ratio of two kinetic energy is given by :

K R K T = I M R 2 K R K T = I M R 2

Adding “1” to either side of the equation and solving, we have :

K K T = I + M R 2 M R 2 K K T = I + M R 2 M R 2

Inverting the ratio,

K T K = M R 2 I + M R 2 K T K = M R 2 I + M R 2

For solid sphere,

I = 2 M R 2 5 I = 2 M R 2 5

Putting MI’s expression in the ratio,

K T K = M R 2 2 M R 2 5 + M R 2 = 5 7 K T K = M R 2 2 M R 2 5 + M R 2 = 5 7

Hence, option (d) is correct.

Note:
Important to note here is that this distribution of kinetic energy between translation and rotation is independent of the path of rolling. It only depends on the MI of rolling body i.e. on mass and its distribution about the axis of rotation. As such, the distribution of kinetic energy between translation and rotation are different for different bodies. In the case of ring and hollow cylinder, the distribution is 50% - 50%.

Exercise 2

Two identical spheres start from top of two incline planes of same geometry. One slides without rolling and other rolls without sliding. If loss of energy in two cases are negligible, then which of the two spheres reachs the bottom first ?

(a) sphere sliding without rolling reaches the bottom first (b) sphere rolling without sliding reaches the bottom first (c) both spheres reaches the bottom at the same time (d) sphere sliding without rolling stops in the middle (a) sphere sliding without rolling reaches the bottom first (b) sphere rolling without sliding reaches the bottom first (c) both spheres reaches the bottom at the same time (d) sphere sliding without rolling stops in the middle

Solution

Here, no loss of energy is involved, so we can employ law of conservation of energy for the two cases. B

Since both spheres move same vertical displacement, they have same gravitational potential energy. Further, there is no loss of energy. As such, spheres have same kinetic energy at the bottom.

In the case of sliding without rolling, the total kinetic energy is translational kinetic energy. On the other hand, kinetic energy is distributed between translation and rotation in the case of rolling without sliding. The sphere in rolling, therefore, has lesser translational kinetic energy.

Therefore, velocity of the sphere sliding without rolling reaches the ground with greater speed. Again, as geometry of the inclines are same, the sphere sliding without rolling reaches the bottom first. With respect to option (d), we note that once sliding starts, the sphere does not stop on an incline.

Hence, option (a) is correct.

Exercise 3

A hollow spherical ball of mass “m” and radius “r” at rest, rolls down the zig-zag track as shown in the figure. If AB = 10 m and CD = 4 m, then speed (m/s) of the sphere at the far right end of the path is (consider g = 10 m / s 2 m / s 2 ) :

Figure 1: A hollow spherical ball from rest, rolls down the zig-zag track.
Rolling motion
Rolling motion (we1a.gif)

(a) 10 (b) 5 √ 2 (c) 6 √ 2 (d) 3 (a) 10 (b) 5 √ 2 (c) 6 √ 2 (d) 3

Solution

The hollow spherical ball is at rest in the beginning. Thus, hollow spherical ball has no kinetic energy, but only gravitational potential energy due to its place at a height from the ground.

There is no loss of energy due to friction and hence, mechanical energy of the ball is conserved. The hollow spherical ball acquires the kinetic energy, which is equal to the change in gravitational potential energy.

K = m g ( AB - CD ) = m g ( 10 - 4 ) = 6 m g K = m g ( AB - CD ) = m g ( 10 - 4 ) = 6 m g

In order to know the speed at the end of the track, we need to know the kinetic energy due to rolling. The kinetic energy of rolling, in terms, of linear velocity is given as :

K = ( I + m r 2 ) x v C 2 2 r 2 K = ( I + m r 2 ) x v C 2 2 r 2

Solving for v C 2 v C 2 and putting value of kinetic energy as derived earlier, we have :

⇒ v C 2 = 2 K r 2 ( I + m r 2 ) ⇒ v C 2 = 2 x 6 m g r 2 ( 2 m r 2 3 + m r 2 ) ⇒ v C 2 = 2 K r 2 ( I + m r 2 ) ⇒ v C 2 = 2 x 6 m g r 2 ( 2 m r 2 3 + m r 2 )

⇒ v C = 6 √ 2 m / s ⇒ v C = 6 √ 2 m / s

Hence, option (c) is correct.

Exercise 4

Two identical solid spheres roll down through the same height along two inclines of different angles. Then

Figure 2: Two identical solid sphere rolls down through the same height along two inclines of different angles.
Rolling motion
Rolling motion (we2.gif)

(a) the speed and time of descent are same in two cases (b) for steeper incline, the speed is greater and time of descent is lesser (c) the speed is same and time of descent for steeper incline is lesser (d) the speed for steeper incline is greater and time of descent is same (a) the speed and time of descent are same in two cases (b) for steeper incline, the speed is greater and time of descent is lesser (c) the speed is same and time of descent for steeper incline is lesser (d) the speed for steeper incline is greater and time of descent is same

Solution

The mechanical energy of the rolling body is conserved in both cases. As such, the rolling of spheres along two inclines through same vertical displacement means that final kinetic energies of the solid spheres are same.

K = M g h K = M g h

We also know that distribution of kinetic energy between translation and rotation is fixed for a given geometry and mass of the body.

K T K R = M R 2 I K T K R = M R 2 I

This means that translational kinetic energy of the solid sphere will be same. In turn, it means that final speeds of the solid sphere in two cases will be same.

However, as the linear distance is less in the case of steeper incline, the sphere will take lesser time.

Hence, option (c) is correct.

Exercise 5

A ring of mass 0.3 kg and a disk of mass 0.4 have equal radii. They are given equal kinetic energy and released on a horizontal surface in such a manner that each of them starts rolling immediately. Then,

(a) ring has greater linear velocity (b) ring has lesser linear velocity (c) ring has greater angular velocity (d) ring and disk have equal linear velocity (a) ring has greater linear velocity (b) ring has lesser linear velocity (c) ring has greater angular velocity (d) ring and disk have equal linear velocity

Solution

The kinetic energy of a rolling body, in terms of linear velocity of center of mass, is given as :

K = ( I + M R 2 ) x V C 2 2 R 2 K = ( I + M R 2 ) x V C 2 2 R 2

⇒ V C 2 = K R 2 I + M R 2 ⇒ V C 2 = K R 2 I + M R 2

For ring, m = 0.3 kg and I = M R 2 I = M R 2

⇒ V C 2 = K R 2 M R 2 + M R 2 = K M = K 0.3 = 10 K 3 ⇒ V C 2 = K R 2 M R 2 + M R 2 = K M = K 0.3 = 10 K 3

For disk, m = 0.4 kg and I = M R 2 2 I = M R 2 2 .

⇒ V C 2 = K R 2 M R 2 2 + M R 2 = 4 K 3 m = 4 K 3 x 0.4 = 10 K 3 ⇒ V C 2 = K R 2 M R 2 2 + M R 2 = 4 K 3 m = 4 K 3 x 0.4 = 10 K 3

Thus, velocities of centers of mass for both ring and disk are same.

Hence, option (d) is correct

Exercise 6

A circular body of mass “M” and radius “R” rolls inside a hemispherical shell of radius “R”. It is released from the top edge of the hemisphere. Then, the angular kinetic energy at the bottom of the shell is maximum, if the circular body is :

(a) a ring (b) a disk (c) a hollow sphere (d) a solid sphere (a) a ring (b) a disk (c) a hollow sphere (d) a solid sphere

Solution

Since mechanical energy is conserved in rolling, the total kinetic energy for a circular body is a constant. Now, ratio of translational and rotational kinetic energy is :

K T K R = M R 2 I K T K R = M R 2 I

This ratio is independent of the position of the rolling body. Thus, rotational kinetic energy (and hence angular velocity) is maximum for the circular body having maximum moment of inertia. Here, ring for same mass and radius has the maximum moment of inertia. As such, angular velocity is maximum for the ring.

Hence, option (a) is correct.

Application level (Work and energy in rolling)

Exercise 7

A small solid sphere of mass “m” and radius “r”, rolls down the incline and then move up the loop of radius “R” as shown in the figure. From what minimum height from the ground, the ball is released so that it does not leave the track at the highest point of the loop?

Figure 3: Sphere starts rolling from a height along the incline.
Rolling of a sphere
Rolling of a sphere (we3.gif)

(a) 2 R 3 (b) 3 R (c) 3 R 10 (d) 7 R 10 (a) 2 R 3 (b) 3 R (c) 3 R 10 (d) 7 R 10

Solution

We know that the minimum speed of the ball at the top (D) of loop, so that it does not leave the highest point, is :

v D = √ ( g R ) v D = √ ( g R )

Corresponding to this requirement, the velocity required at the bottom of the loop is :

v B = √ ( 5 g R ) v B = √ ( 5 g R )

Now, the kinetic energy at the bottom of the loop, in terms of linear velocity, is :

K = ( I + M R 2 ) x V C 2 2 r 2 K = ( I + M R 2 ) x V C 2 2 r 2

⇒ K = ( I + m r 2 ) x 5 g R 2 r 2 ⇒ K = ( I + m r 2 ) x 5 g R 2 r 2

Let “h” be the height, then (assuming r << R) :

⇒ ( I + m r 2 ) x 5 g R 2 r 2 = m g h ⇒ ( I + m r 2 ) x 5 g R 2 r 2 = m g h

⇒ h = ( I + m r 2 ) x 5 g R 2 r 2 m g ⇒ h = ( I + m r 2 ) x 5 g R 2 r 2 m g

For solid sphere,

⇒ h = ( 2 m r 2 5 + m r 2 ) x 5 g R 2 r 2 m g ⇒ h = 7 m r 2 5 x 5 g R 2 r 2 m g ⇒ h = ( 2 m r 2 5 + m r 2 ) x 5 g R 2 r 2 m g ⇒ h = 7 m r 2 5 x 5 g R 2 r 2 m g

⇒ h = 7 R 10 ⇒ h = 7 R 10

Hence, option (c) is correct.

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