Question

At time $t=0$, a disk of radius $1m$ starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha =\frac{2}{3}rad{s}^{-2}$. 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 $g=10m{s}^{-2}$.]

Answer 1

Answer: 0.52

At $t=0,\omega =0$
at $t=\sqrt{\pi },\omega =\alpha t=\frac{2}{3}\sqrt{\pi },$
$v=\omega r=\frac{2}{3}\sqrt{\pi }$
$\theta =\frac{1}{2}\alpha t^2$
$\theta =\frac{1}{2}\times \frac{2}{3}\times \pi =\frac{\pi }{3}$
$\theta =60^{\circ }$

$v_y=v\sin 60=\frac{\sqrt{3}}{2}V$
$h=\frac{u_y^2}{2g}=\frac{\frac{3}{4}{v}^2}{2g}$
$h=\frac{\frac{3}{4}\times \frac{4}{9}\pi }{2g}$
$h=\frac{3\pi }{9\times 2g}=\frac{\pi }{6g}$
Maximum height from plane, $H=\frac{R}{2}+h$
$H=\frac{1}{2}+\frac{\pi }{6\times 10}$
$x=\frac{\pi }{6};x=0.52$