Question

A soccer ball of diameter d and mass m rolls up a hill without slipping, reaching a maximum height of h above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then?

Answer 1

Answer:

Convert the units to standard units.

r = d / 2 (cm)

r = r in cm [ 1 m / 100 cm ]

m = m in g * [ 1 kg / 1000 g] = m in kg

Write the general equation for this problem.

The KE at the bottom of the hill = the PE at the top.

1/2 I w^2 + 1/2 m v^2 = mgh

I for a hollow sphere = 2/3xmxr^2

1/2 (2/3 x mr^2 x w^2) + 1/2 x m x v^2 = mgh

Now we need to relate w with v

v = w x r

w = v / r put this result in for w

1/2 (2/3 x mxr^2 z v^2 / r^2) + 1/2 m x v^2 = mgh

1/2 (2/3 x m v^2) + 1/2 x m x v^2 = mgh

1/3 m x v^2 + 1/2 mv^2 = mgh

5/6 m x v^2 = mgh Notice that the m's are all the same on both sides. They cancel.

5/6 v^2 = gx h

Solve for v

v^2 = 6/5 x g x h

Find the Rotational Kinetic Energy (KE_r)

PE - KE_translational = KE_r

mgh - 1/2 x m x v^2 = KE_r

Find the rotational rate at the bottom w

KE_r = 1/2 I x w^2

I = 2/3 m R^2