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 down 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

$a_{ring}=\frac{g\sin \theta }{2}$

$a_{disc}=\frac{2g\sin \theta }{3}$

using the formula $S=Ut+\frac{a{t}^2}{2}$
$\frac{h}{\sin \theta }=\frac{1}{2}\left(\frac{g\sin \theta }{2}\right)t_1^2\Rightarrow t_1=\sqrt{\frac{4h}{g{\sin }^2\theta }}=\sqrt{\frac{16h}{3g}}$
$\frac{h}{\sin \theta }=\frac{1}{2}\left(\frac{2g\sin \theta }{3}\right)t_2^2\Rightarrow t_2=\sqrt{\frac{3h}{g{\sin }^2\theta }}=\sqrt{\frac{4h}{g}}$
$\Rightarrow \sqrt{\frac{16h}{3g}}-\sqrt{\frac{4h}{g}}=\frac{2-\sqrt{3}}{\sqrt{10}}$
$\sqrt{h}[\frac{4}{\sqrt{3}}-2]=2-\sqrt{3}$
$\sqrt{h}=\frac{(2-\sqrt{3})\sqrt{3}}{(4-2\sqrt{3})}=\frac{\sqrt{3}}{2}\Rightarrow h=\frac{3}{4}=0.75m$