**A solid sphere has a radius of 0.2 m and a mass of 150 kg. How much work is required to get the sphere rolling with an angular speed of 50.0 rad/s on a horizontal surface**?

**Answer:**

Since work equals the change in kinetic energy and the initial kinetic energy is zero since this sphere starts from rest, we can say that:

Work=the total kinetic energy

let’s determine the kinetic energy:

The moment of inertia of a solid sphere is

I = (2/5)MR^2

I = (2/5) (150 kg) (0.2 m)^2

I = 2.4 kg m^2

While it rolls along, it has kinetic energy of translation K_tr and kinetic energy of rotation K_rot

K_Tot = K_tr + K_rot = (1/2) mv^2 + (1/2)I^2

v = r w= (0.2 m) (50 rad/s) = 10 m/s

K_Tot = (1/2) mv^2 + (1/2)I^2

K_Tot = (1/2) (150 kg) (10 m/s)2 + (1/2) (2.4 kg m2) (50 rad/s)2

K_Tot = 7 500 J + 3 000 J

KTot = 10 500 J

conclude that W= 10 500 J

W = 10 500 J