Question

A spherical object (with non-uniform density) of mass 16 kg and radius 0.22 m rolls along a horizontal surface at a constant linear speed without slipping. The moment of inertia of the object about a diameter is 0.59 M R2 . The object’s rotational kinetic energy about its own center is what fraction of the object’s total kinetic energy

Answer 1

Answer:

Kr = 0.7618K

Explanation:

Suppose that the object's velocity is V, then his kinetic energy is:

K = $\frac{mv^{2} }{2}$

K = $\frac{(16)v^{2} }{2}$

K = 8$v^{2}$

The rotational kinetic energy is

Kr = $\frac{Iw^{2} }{2}$

           where I: The moment of inertia

                      ω: angular velocity

Kr =$\frac{(0.59)w^{2} }{2}$

Kr = 0.295$w^{2}$

How the movement is without slipping, then  

ω = $\frac{v}{r}$

ω = $\frac{v}{0.22}$

Thus

Kr = $\frac{0.295v^{2} }{0.22^{2} }$

Kr = 6.095$v^{2}$

8$v^{2}$  ---->  1

6.095$v^{2}$----->?

Kr = 0.7618K