19.12 Work and energy in rolling (check your understnading)

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

$$\begin{array}{l}\frac{K_R}{K_T}=\frac{I}{M{R}^2}\end{array}$$

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

$$\begin{array}{l}\frac{K}{K_T}=\frac{I+M{R}^2}{M{R}^2}\end{array}$$

Inverting the ratio,

$$\begin{array}{l}\frac{K_T}{K}=\frac{M{R}^2}{I+M{R}^2}\end{array}$$

For solid sphere,

$$\begin{array}{l}I=\frac{2M{R}^2}{5}\end{array}$$

Putting MI’s expression in the ratio,

$$\begin{array}{l}\frac{K_T}{K}=\frac{M{R}^2}{\frac{2M{R}^2}{5}+M{R}^2}=\frac{5}{7}\end{array}$$

Hence, option (d) is correct.

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.