JEE Advance - Physics (2022 - Paper 1 Online - No. 5)
At time $t=0$, a disk of radius $1 \mathrm{~m}$ starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha=\frac{2}{3} \mathrm{rad} \,\mathrm{s}^{-2}$. A small stone is stuck to the disk. At $t=0$, it is at the contact point of the disk and the plane. Later, at time $t=\sqrt{\pi} \,s$, the stone detaches itself and flies off tangentially from the disk. The maximum height (in $m$ ) reached by the stone measured from the plane is $\frac{1}{2}+\frac{x}{10}$. The value of $x$ is ____________ , [Take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$.]
Answer
0.52
Explanation
The angle rotated by disc in $t=\sqrt{\pi} \mathrm{s}$ is
$$ \begin{aligned} & \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\\\ & \Rightarrow \theta=\frac{1}{2} \times \frac{2}{3}(\sqrt{\pi})^2 \\\\ &=\frac{\pi}{3} \mathrm{rad} \end{aligned} $$
and the angular velocity of disc is
$$ \begin{aligned} \omega & =\omega_0+\alpha t \\\\ = & \frac{2 \sqrt{\pi}}{3} \mathrm{rad} / \mathrm{s} \\\\ \text { and } v_{\mathrm{cm}} & =\omega R=\frac{2 \sqrt{\pi}}{3} \times 1 \\\\ & =\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s} \end{aligned} $$
So, at the moment it detaches the situation is
$v=\sqrt{(\omega R)^2+v_{\mathrm{cm}}^2+2(\omega R) v_{\mathrm{cm}} \cos 120^{\circ}}$
$=v_{\mathrm{cm}}=\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s}$
and $\tan \theta=\frac{\omega R \sin 120^{\circ}}{v_{\mathrm{cm}}+\omega R \cos 120^{\circ}}$
$\Rightarrow \tan \theta=\sqrt{3}$
$\Rightarrow \theta=\frac{\pi}{3} \mathrm{rad}$
So, $H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}$
$=\frac{\left(\frac{2 \sqrt{\pi}}{3}\right)^2 \times \sin ^2 60^{\circ}}{2 \times 10}$
$$ \begin{aligned} & =\frac{4 \pi \times 3}{9 \times 2 \times 10 \times 4} \\\\ & =\frac{\pi}{60} \mathrm{~m} \end{aligned} $$
So, height from ground will be
$$ = R\left(1-\cos 60^{\circ}\right)+\frac{\pi}{60}=\frac{1}{2}+\frac{x}{10}$$
$$ \Rightarrow x=\frac{\pi}{6}=0.52 $$
$$ \begin{aligned} & \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\\\ & \Rightarrow \theta=\frac{1}{2} \times \frac{2}{3}(\sqrt{\pi})^2 \\\\ &=\frac{\pi}{3} \mathrm{rad} \end{aligned} $$
and the angular velocity of disc is
$$ \begin{aligned} \omega & =\omega_0+\alpha t \\\\ = & \frac{2 \sqrt{\pi}}{3} \mathrm{rad} / \mathrm{s} \\\\ \text { and } v_{\mathrm{cm}} & =\omega R=\frac{2 \sqrt{\pi}}{3} \times 1 \\\\ & =\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s} \end{aligned} $$
So, at the moment it detaches the situation is
$v=\sqrt{(\omega R)^2+v_{\mathrm{cm}}^2+2(\omega R) v_{\mathrm{cm}} \cos 120^{\circ}}$
$=v_{\mathrm{cm}}=\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s}$
and $\tan \theta=\frac{\omega R \sin 120^{\circ}}{v_{\mathrm{cm}}+\omega R \cos 120^{\circ}}$
$\Rightarrow \tan \theta=\sqrt{3}$
$\Rightarrow \theta=\frac{\pi}{3} \mathrm{rad}$
So, $H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}$
$=\frac{\left(\frac{2 \sqrt{\pi}}{3}\right)^2 \times \sin ^2 60^{\circ}}{2 \times 10}$
$$ \begin{aligned} & =\frac{4 \pi \times 3}{9 \times 2 \times 10 \times 4} \\\\ & =\frac{\pi}{60} \mathrm{~m} \end{aligned} $$
So, height from ground will be
$$ = R\left(1-\cos 60^{\circ}\right)+\frac{\pi}{60}=\frac{1}{2}+\frac{x}{10}$$
$$ \Rightarrow x=\frac{\pi}{6}=0.52 $$
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