Question 744: Centre of Mass and Rotational Motion

Question

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 $g=10 \mathrm{~m} \mathrm{~s}^{-2}$.]

Accepted Answer

The angle rotated by disc in $t=\sqrt{\pi} \mathrm{s}$ is
$$
\begin{aligned}
& 	heta=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \Rightarrow 	heta=\frac{1}{2} 	imes \frac{2}{3}(\sqrt{\pi})^2 \\
&=\frac{\pi}{3} \mathrm{rad}
\end{aligned}
$$
and the angular velocity of disc is
$$
\begin{aligned}
\omega & =\omega_0+\alpha t \\
= & \frac{2 \sqrt{\pi}}{3} \mathrm{rad} / \mathrm{s} \\
	ext { and } v_{\mathrm{cm}} & =\omega R=\frac{2 \sqrt{\pi}}{3} 	imes 1 \\
& =\frac{2 \sqrt{\pi}}{3} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
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 $	an 	heta=\frac{\omega R \sin 120^{\circ}}{v_{\mathrm{cm}}+\omega R \cos 120^{\circ}}$
$\Rightarrow 	an 	heta=\sqrt{3}$
$\Rightarrow 	heta=\frac{\pi}{3} \mathrm{rad}$
So, $H_{\max }=\frac{u^2 \sin ^2 	heta}{2 g}$
$=\frac{\left(\frac{2 \sqrt{\pi}}{3}ight)^2 	imes \sin ^2 60^{\circ}}{2 	imes 10}$
$$
\begin{aligned}
& =\frac{4 \pi 	imes 3}{9 	imes 2 	imes 10 	imes 4} \\
& =\frac{\pi}{60} \mathrm{~m}
\end{aligned}
$$
So, height from ground will be
$$

= R\left(1-\cos 60^{\circ}ight)+\frac{\pi}{60}=\frac{1}{2}+\frac{x}{10}$$
$$
\Rightarrow x=\frac{\pi}{6}=0.52
$$

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type