JEE Main 2022 — Rotational Motion Question with Solution

In case of rotational motion of a rigid body like this, the kinetic energy is not just due to the linear motion, but also due to its rotation.

For a solid cylinder, the moment of inertia I is given by $\frac{1}{2}m{r}^2$. The kinetic energy due to rotation is given by $\frac{1}{2}I{\omega }^2$. But we also know that $\omega =\frac{v}{r}$, hence the rotational kinetic energy can be written as $\frac{1}{2}\frac{1}{2}m{v}^2=\frac{1}{4}m{v}^2$, where $\frac{1}{2}$ is from the moment of inertia of the solid cylinder.

Therefore, the total kinetic energy (linear + rotational) when the string snaps is $\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2=\frac{3}{4}m{v}^2$.

Equating this to the potential energy $mgh$ and solving for h gives the result :

$$h=\frac{v^2}{2g}\times \frac{3}{2}=1.2\mathrm{m}=120\mathrm{cm}$$

Alternate Method :

Applying COE, we get

$$mgh=\frac{1}{2}m{v}^2\left(1+\frac{{\mathrm{K}}^2}{r^2}\right)$$

$\mathrm{K}=$ radius of gyration

For a solid cylinder, $\frac{{\mathrm{K}}^2}{r^2}=\frac{1}{2}$

$$\begin{array}{rl}\therefore h& =\frac{v^2}{2g}\left(1+\frac{1}{2}\right)\\ & \\ & =\frac{16}{2\times 10}\times \frac{3}{2}\\ & \\ & =1.2\mathrm{m}\\ & \\ & =120\mathrm{cm}\end{array}$$