I have been trying this question for so long and I still cannot figure out how to do it...(adsbygoogle = window.adsbygoogle || []).push({});

A solid brass ball of mass 8.2 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 4.9 m, and the ball has radius r << R. (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop? (b) If the ball is released at height h = 6.00R, what is the magnitude of the horizontal force component acting on the ball at point Q?

The picture looks like a straight line angling towards the ground, and when it gets to the ground, it shallows out and does a loop-the-loop. The radius of the loop is R, and the ball is started from height h. Point Q is a line drawn from the center of the loop, directly right to the edge of the loop.

So for this problem, you use the conservation of energy. I have been using M x g x h= M x g x h + 1/2m(v squared) + (1/2 Icom Omega squared). So I know that since the ball is on the verge of leaving at the top of the loop, the kinetic energy must be zero, but my guess is that it still has rotational energy, so how am I supposed to find that out? And I know that for part b, I must use F= m(v squared)/r, but once again, I don't know how to find v, since there is v and omega. But thank you for the help in advance

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# Homework Help: Rotational velocity

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