A solid cylinder of radius $R$ is at rest at a height $h$ on an inclined plane. If it rolls down then its velocity on reaching the ground is

Question

Aicharya · Accepted Answer

The given situation is shown below

Let $v$ be the velocity of solid cylinder on reaching the ground. According to conservation of energy, Total energy at point $A=$ Total energy at point $B$ i. e $m g h+0=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2+0$ $\Rightarrow \quad m g h=\frac{1}{2}\left(\frac{1}{2} M R^2\right)\left(\frac{v}{R}\right)^2+\frac{1}{2} m v^2$ $\Rightarrow$ (Where, $I=\frac{1}{2} M R^2$ and $\omega=\frac{v}{R}$ ) $\begin{aligned} & \Rightarrow \quad g h=\frac{v^2}{4}+\frac{v^2}{2} \Rightarrow g h=\frac{3 v^2}{4} \\ & \Rightarrow \quad v^2=\frac{4 g h}{3} \Rightarrow v=\sqrt{\frac{4 g h}{3}}\end{aligned}$

A solid cylinder of radius $R$ is at rest at a height $h$ on an inclined plane. If it rolls down then its…

Solution