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

Step-by-step explanation:

The kinetic energy of translation

`E_1=(1)/(2)mv^2`

m= mass v= linear velocity

The kinetic energy of rotation

`E_2=(1)/(2)I\omega^2`

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

where ω= v/R= angular velocity

`E_2=(1)/(2)kmR^2(v)/(R)^2`

Then

`(E_1)/(E_2) = (0.5mv^2)/(0.5kmR^2(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