A solid sphere of radius r is rolling without slipping on a track with a loop th

Question

A solid sphere of radius r is rolling without slipping on a track with a loop the loop of radius R. It starts from a height h, which is less than that required to just make it through the loop starting at rest. Therefore, it already has to start at some initial speed, in order to reach the required speed of       (sqrt{?}gR)    at the top of the loop. Employ conservation of mechanical energy and find an expression for the initial speed. Simplify the result as much as possible.  

Please do this step by step.. i dont understand :(  

Explanation / Answer

KE = 1/2 m v^2 + 1/2 I w^2 = 1/2 m v^2 + 1/2 (2 m r^2 / 5) v^2 / r^2

KE = 1/2 m v^2 + 1/5 m v^2 = 7/10 m v^2     KE of rolling ball

m g = m vt^2 / R     for ball to stay in contact at top of loop

vt = (R g)^1/2      required speed at top

E(top) = 7/10 m vt^2 + 2 g R m     energy required at top

E(bottom) = 7/10 m vb^2 = m g h + 7/10 m vs^2

where vs = starting speed and PE = 0 at bottom

m g h + 7/10 m vs^2 = 7/10 m vt^2 + 2 g R m

7/10 (vs^2 - vt^2) = (2 R - h) g

Note that if h = 2 R then vs = vt   as it should be

Now substitute vt = (R g)^1/2 and solve for vs as a function of h

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