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
