Question

Prove the result 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
$v^2=\frac{2gh}{\frac{1+k^2}{R^2}}$
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

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

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

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

$I=m{k}^2$, $v=R\omega$

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

$⟹v^2=\frac{2gh}{\frac{1+k^2}{R^2}}$