QJEEM22Phedbd: JEE Main 2022 Physics - Class 11 JEEM22Phedbd Solution: A solid cylinder length is suspended symmetrically through two…

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

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

$\mathrm{K}=$ radius of gyration

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

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