Question

Prove that the velocity $v$ of translation of a rolling body (like a ring,disc,cylinder or sphere) at the bottom of an inclined plane of a height $h$ is given by $v^2=\frac{2gh}{[1+\frac{k^2}{R^2}]}$

Using dynamical consideration(i.e by consideration of forces and torques).Note $k$ is the radius of gyration of the body about its symmetry axis, and $R$ is the radius of the body.The body starts from rest at the top of the plane.

Answer 1

Let $m$ be the radius of the body,$R$ be the radius ,$K$ be the radius of gyration,$v$ be the translational velocity of the body,$h$ be the height of the inclined plane and $g$ be the acceleration due to gravity.

Total energy at the top of the plane,$E_1=mgh$

Total energy at the bottom of the plane$E_b=K{E}_{rot}+K{E}_{trans}$

$=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$

but $I=m{k}^2$ and $\omega =\frac{v}{R}$

$\therefore E_b=\frac{1}{2}m{k}^2{\left(\frac{v}{R}\right)}^2+\frac{1}{2}m{v}^2$

$\frac{1}{2}m{v}^2(\frac{k^2}{R^2}+1)$

From the law of conservation of energy, we have

$E_T=E_b$

$\Rightarrow mgh=\frac{1}{2}m{v}^2(\frac{k^2}{R^2}+1)$

$\Rightarrow 2gh=v^2(\frac{k^2}{R^2}+1)$

$\therefore v^2=\frac{2gh}{(\frac{k^2}{R^2}+1)}$

Hence proved.