Question

A ball of mass m and radius R is both sliding and spinning on a horizontal surface so that its rotational kinetic energy equals its translational kinetic energy. What is the ratio of the ball’s center-ofmass speed to the speed due to rotation only of a point on the ball’s surface? The moment of inertia of the ball is 0.63 m R2 .

Answer 1

Answer:

$\frac{v_{cm}}{\omega} = 1.122\cdot R$

Explanation:

According to the statement of the problems, the following identity exists:

$K_{t} = K_{r}$

$\frac{1}{2}\cdot m \cdot v_{cm}^{2} = 0.63\cdot m \cdot R^{2} \cdot \omega^{2}$

After some algebraic handling, the ratio is obtained:

$\frac{v_{cm}^{2}}{\omega^{2}}=1.26\cdot R^{2}$

$\frac{v_{cm}}{\omega} = 1.122\cdot R$