JEE Physics 2023 Question with Solution

Given: A rolling spherical shell has rotational kinetic energy to total kinetic energy ratio $\frac{x}{5}$.

Find: The value of $x$.

For a rolling spherical shell, the angular velocity is

$$\omega = \frac{v}{R}$$

The rotational kinetic energy is given by

$$K_{\text{rot}} = \frac{1}{2} \left( \frac{2}{3} m R^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{2} \left( \frac{2}{3} m R^2 \right) \left( \frac{v^2}{R^2} \right) = \frac{1}{3} m v^2$$

The total kinetic energy is

$$K_{\text{total}} = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{2}{3} m R^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2$$

Therefore,

$$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3} m v^2}{\frac{5}{6} m v^2} = \frac{2}{5}$$

Thus, $\frac{x}{5} = \frac{2}{5}$, so $x = 2$.

Therefore, the required value is $2$.

Given: The body is a spherical shell rolling without slipping.

Find: The value of $x$ in the ratio $\frac{x}{5}$.

For a spherical shell,

$$I = \frac{2}{3} mR^2$$

Using rolling condition,

$$\omega = \frac{v}{R}$$

Now rotational kinetic energy becomes

$$K_{\text{rot}} = \frac{1}{2} I \omega^2$$

Substituting,

$$K_{\text{rot}} = \frac{1}{2} \cdot \frac{2}{3} mR^2 \cdot \left( \frac{v}{R} \right)^2$$ $$K_{\text{rot}} = \frac{1}{3} mv^2$$

Translational kinetic energy is

$$K_{\text{trans}} = \frac{1}{2} mv^2$$

So total kinetic energy is

$$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2$$

Hence,

$$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3} mv^2}{\frac{5}{6} mv^2} = \frac{2}{5}$$

Comparing with $\frac{x}{5}$, we get $x = 2$.

Therefore, the required numerical answer is $2$.