Question

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle ${60}^o$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $\frac{(2-\sqrt{3})}{\sqrt{10}s}$, then the height of the top of the inclined plane, in meters. is. Take $g=10m{s}^{-2}$.

Answer 1

Acceleration of a rolling body (an inclined plane) $a_c=\frac{g\sin \theta }{1+\frac{I_c}{M{R}^2}}$
where $I_c$ is the moment of inertia of the body.
So,
$a_{ring}=\frac{g\sin \theta }{2}$
$a_{disc}=\frac{2g\sin \theta }{3}$
using the formula $S=Ut+\frac{a{t}^2}{2}$

For $t_1,$
$\frac{h}{\sin \theta }=\frac{1}{2}\left(\frac{g\sin \theta }{2}\right)t_1^2$$\Rightarrow t_1=\sqrt{\frac{4h}{g{\sin }^2\theta }}=\sqrt{\frac{16h}{3g}}$

For $t_2,$
$\frac{h}{\sin \theta }=\frac{1}{2}\left(\frac{2g\sin \theta }{3}\right)t_2^2$$\Rightarrow t_2=\sqrt{\frac{3h}{g{\sin }^2\theta }}=\sqrt{\frac{4h}{g}}$

time difference =
$\Delta t=t_1-t_2$
where, $\Delta t=\frac{(2-\sqrt{3})}{\sqrt{10}s}$
Where,
$\Rightarrow \sqrt{\frac{16h}{3g}}-\sqrt{\frac{4h}{g}}=\frac{2-\sqrt{3}}{\sqrt{10}}$

$\sqrt{h}[\frac{4}{\sqrt{3}}-2]=2-\sqrt{3}$

$\sqrt{h}=\frac{(2-\sqrt{3})\sqrt{3}}{(4-2\sqrt{3})}=\frac{\sqrt{3}}{2}\Rightarrow h=\frac{3}{4}=0.75m$