Question

A solid uniform sphere of radius 2.0 m starts from rest and rolls without slipping down an inclined plane of vertical height 12 m. What is the angular speed of the sphere at the bottom of the inclined plane? (I = 2/5 Mr2)

Answer 1

Let us use the law of conservation of energy. At the top of the inclined plane, energy consists only of potential energy $m g h$, and at the bottom, of linear and rotational kinetic energy $\frac{m v^2}{2} + \frac{I}{2} \omega^2$. Since the energy is conserved, $m g h = \frac{m v^2}{2} + \frac{I}{2} \omega^2$.

The connection between linear velocity and angular velocity is $v = \omega r$, hence substituting this into linear part of the kinetic energy in the previous expression, obtain:

$m g h = \frac{m \omega^2 r^2}{2} + \frac{I}{2} \omega^2$

Solving for $\omega$, obtain: $\omega = \sqrt{\frac{2 m g h}{I + m r^2}} = [I = \frac{2}{5} m r^2] = \sqrt{\frac{2 m g h}{\frac{2}{5} m r^2 + m r^2}} = \sqrt{\frac{10}{7}\frac{g h}{r^2}} = \frac{1}{r}\sqrt{\frac{10}{7} g h} \approx 6.48 s^{-1}$