JEE Main 2024 — Rotational Motion Question with Solution

Initial Conditions:

Consider a solid disc of mass $M$ and radius $R$ rolling without slipping on a horizontal surface. The center of mass of the disc has an initial speed $v$.

Kinetic Energy of the Rolling Disc:

For an object that rolls without slipping, its total kinetic energy $K$ is the sum of its translational kinetic energy and its rotational kinetic energy.

Translational Kinetic Energy:

$K_{\text{trans}}=\frac{1}{2}M{v}^2.$

Rotational Kinetic Energy:

The moment of inertia $I$ of a solid disc about its center is:

$I=\frac{1}{2}M{R}^2.$

Since the disc rolls without slipping, the angular velocity $\omega$ is related to the linear speed by:

$v=\omega R⟹\omega =\frac{v}{R}.$

Therefore, the rotational kinetic energy is:

$K_{\text{rot}}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)\left(\frac{v^2}{R^2}\right)=\frac{1}{4}M{v}^2.$

Total Initial Kinetic Energy:

Summing these gives:

$K_{\text{total}}=K_{\text{trans}}+K_{\text{rot}}=\frac{1}{2}M{v}^2+\frac{1}{4}M{v}^2=\frac{3}{4}M{v}^2.$

Energy Conservation as the Disc Climbs the Incline:

As the disc moves up the incline, there are no dissipative forces (assuming a smooth incline and no slipping), so energy is conserved. The total kinetic energy at the bottom will be converted entirely into gravitational potential energy at the maximum height $h$:

$K_{\text{total}}=Mgh.$

Substitute $K_{\text{total}}=\frac{3}{4}M{v}^2$:

$\frac{3}{4}M{v}^2=Mgh.$

Canceling $M$ from both sides:

$\frac{3}{4}{v}^2=gh.$

Thus,

$h=\frac{3}{4}\frac{v^2}{g}.$

Final Answer:

$\boxed{\frac{3}{4}\frac{v^2}{g}}$

This corresponds to Option A.