Question

A wheel of mass M and radius R is released from rest and rolls to the bottom of an inclined plane of height h without slipping. What is its velocity at the bottom of the incline? The moment of inertia of a wheel of mass M and radius R rotating about an axis through its center of mass is $1/2M{R}^2$.

Answer 1

From conservation of energy,

Loss in gravitational potential energy=Gain in translational kinetic energy+Gain in rotational kinetic energy

$⟹Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$

From the no slipping condition,

$v=R\omega$

$⟹Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2$

$⟹Mgh=\frac{3}{4}M{v}^2$

$⟹v=\sqrt{\frac{4}{3}gh}$