Question

At time $t=0$, a disk of radius $1 \,m$ 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 $g=10 \,m s ^{-2}$.]

Answer 1

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}{2 g }=\frac{\frac{3}{4} v ^2}{2 g } $
$ h =\frac{\frac{3}{4} \times \frac{4}{9} \pi}{2 g } $
$h =\frac{3 \pi}{9 \times 2 g }=\frac{\pi}{6 g }$
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$