Question

A thin-walled sphere rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy about an axis through its center of mass?

Answer 1

Explanation:

The kinetic energy of translation

$E_1=\frac{1}{2}mv^2$

m= mass v= linear velocity

The kinetic energy of rotation

$E_2=\frac{1}{2}I\omega^2$

I= MOI of the thin walled sphere =kmR^2

where ω= v/R= angular velocity

$E_2=\frac{1}{2}kmR^2\frac{v}{R}^2$

Then

$\frac{E_1}{E_2} = \frac{0.5mv^2}{0.5kmR^2\frac{v}{R}^2 }$

=1/k

solid sphere: k=0.4;   E1/E2 =1/0.4 = 2.5;  

 hollow sphere: k=2/3;   E1/E2 = 1.5