JEE Advanced 2022 Paper 1, Physics Section 1 Q5: Rolling Motion

Kinematics at detachment ($t=\sqrt{\pi }$ s):

Angle rotated: $\theta =\frac{1}{2}\alpha t^2=\frac{1}{2}\cdot \frac{2}{3}\cdot \pi =\frac{\pi }{3}$ rad.

Angular velocity: $\omega =\alpha t=\frac{2\sqrt{\pi }}{3}$ rad/s.   Velocity of CM: $v_{cm}=\omega R=\frac{2\sqrt{\pi }}{3}$ m/s.

Velocity of stone: The stone is at angular position $\pi /3$ from the bottom. Its velocity is the vector sum of $v_{cm}$ (horizontal) and the tangential velocity $\omega R$ at angle $120^{\circ }$ relative to $v_{cm}$.

$v=\sqrt{v_{cm}^2+(\omega R{)}^2+2v_{cm}(\omega R)\cos 120^{\circ }}=v_{cm}=\frac{2\sqrt{\pi }}{3}$ m/s.

$\tan {\theta }_{\text{proj}}=\frac{\omega R\sin 120^{\circ }}{v_{cm}+\omega R\cos 120^{\circ }}=\sqrt{3}\Rightarrow {\theta }_{\text{proj}}=\frac{\pi }{3}$.

Maximum height of trajectory above release point:

$h_1=\frac{v^2{\sin }^{2}60^{\circ }}{2g}=\frac{(4\pi /9)(3/4)}{20}=\frac{\pi }{60}$ m.

Height of release point above the plane: $R(1-\cos 60^{\circ })=\frac{1}{2}$ m.

Total: $H_{\max }=\frac{1}{2}+\frac{\pi }{60}=\frac{1}{2}+\frac{x}{10}$, giving $x=\frac{\pi }{6}\approx 0.52$.