Discussion Forum

A ring and a disc are initially at rest , side by side , at the top of an inclined plane which makes an angle 60° 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=10 m s^-2 ]

Answer:

Explanation:

$a=\frac{g\sin\theta}{1+\frac{1}{MR^{2}}}$

$a_{ring}=\frac{g\sin\theta}{2}$             ($I= MR^{2}$)

$a_{disc}=\frac{2g\sin\theta}{3}$             ($I= \frac{MR^{2}}{2}$)

$s= \frac{h}{\sin\theta}=\frac{1}{2}at^{2}$

    $= \frac{1}{2}(\frac{g\sin\theta}{2})t_{1}^{2}$

    $\Rightarrow$                         $t_{1}=\sqrt{\frac{4h}{g\sin^{2}\theta}}=\sqrt{\frac{16h}{3g}}$

    $s= \frac{h}{\sin \theta}=\frac{1}{2}at^{2}=\frac{1}{2}(\frac{2g\sin\theta}{3})t_{2}^{2}$

$\Rightarrow$                   $t_{2}=\sqrt{\frac{3h}{g\sin^{2}\theta}}=\sqrt{\frac{4h}{g}}$

                       $t_{2}-t_{1}=\sqrt{\frac{16h}{3g}}-\sqrt{\frac{4h}{g}}$

                        $= \frac{2-\sqrt{3}}{\sqrt{10}}$

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

Solving this equation we get, h=0.75m