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

$\theta=\frac{1}{2}\alpha t^2$
$=\frac{1}{2}\times\frac{2}{3}\pi=\frac{\pi}{3}=60\degree$
$V_{cm}=\alpha t$
The resultant velocity of point P is represented as $V = αt$, making an angle of 60° with the horizontal, where $u_y = αt\ sin 60°.$

$y_{max}=\frac{1}{2}+\frac{u_y^2}{2g}$
$=\frac{1}{2}+\frac{\alpha^{2}t^23}{20\times4}$

$=\frac{1}{2}+\frac{\pi}{60}$
$=0.52$

Answer: 0.52