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.

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.

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.

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.

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

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.

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.