JEE Main 2023 — Rotational Motion Question with Solution

For a rolling spherical shell, we must consider the fact that it has both translational and rotational kinetic energy. The total kinetic energy ($K_{total}$) can be expressed as the sum of the translational kinetic energy ($K_{trans}$) and the rotational kinetic energy ($K_{rot}$):

$$K_{total}=K_{trans}+K_{rot}$$

The translational kinetic energy of an object with mass (m) and linear velocity (v) is given by:

$$K_{trans}=\frac{1}{2}m{v}^2$$

The rotational kinetic energy of a rolling spherical shell with moment of inertia (I) and angular velocity (ω) is given by:

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

For a rolling object without slipping, the relationship between linear velocity (v) and angular velocity (ω) is:

$$v=R\omega$$

Where R is the radius of the spherical shell.

The moment of inertia for a spherical shell is given by:

$$I=\frac{2}{3}m{R}^2$$

Now, we can substitute the moment of inertia and the relationship between linear and angular velocity into the equation for rotational kinetic energy:

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

Simplifying the equation:

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

$$K_{rot}=\frac{1}{3}m{v}^2$$

Now, we can find the ratio of rotational kinetic energy to total kinetic energy:

$$\frac{K_{rot}}{K_{total}}=\frac{\frac{1}{3}m{v}^2}{\frac{1}{2}m{v}^2+\frac{1}{3}m{v}^2}$$

Simplifying the equation:

$$\frac{K_{rot}}{K_{total}}=\frac{\frac{1}{3}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{5}{6}}$$

Multiplying both the numerator and the denominator by 6:

$$\frac{K_{rot}}{K_{total}}=\frac{2}{5}$$

Comparing this to the given ratio of $$\frac{x}{5}$$, we can determine that the value of $$x$$ is 2.