A solid sphere of weight 36.0 N rolls up an incline at an angle of 30.0o. At the bottom of

Question

the incline the center of mass of the sphere has a translational speed of 4.90 m s-1.
Isphere = 2/5mr^2.

What is the kinetic energy of the sphere at the bottom of the incline?

How far does the sphere travel up along the incline?

Answer 1

It rolls up the hill until the initial kinetic energy, which is both translational and rotational, equals the potential energy gained.

Call the initial speed V and the maximum height H.

Initial KE = (1/2)MV^2 + (1/2)Iw^2
and, since I = (2/5)MR^2 and w = V/R,
Initial KE = (1/2)MV^2 + (1/5)MV^2 = ?

Set that sum equal to M g H and solve for H. The distance it travels along the incline is H/sin 30

Answer 2

Good