A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 5.0 m. The cylinder arrives at the bottom of the plane 2.6 s after the sphere. Determine the angle between the inclined plane and the horizontal.

Answer 1

hollow cylinder I = m r^2

m a = m g sin A - torque/r
m a = m g sin A - m r^2 alpha /r
a = g sin alpha - r alpha
but r alpha = a
so
a = (1/2) g sin A

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solid sphere

I = (2/5) m r^2
force down = m g sin A again
force up * r = I alpha
ma = m g sin A - (2/5)mr^2 alpha/r
m a = m g sin A - (2/5) m a
(7/5)a = g sin A
a = (5/7) g sin A

5 = (1/2) a t^2
two different a values dependent only on sin A
solve for sin A :)