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A solid brass ball of mass .280g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 14.0 cm, 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? If the ball is released at height h = 6.00R, what are the (b) magnitude and (c) direction of the horizontal force component acting on the ball a point Q?

Solution and with diagram on first page, #11.8

www.monmsci.net/~fasano/phys1/Chapter_11_09.pdf

first what does r<<R mean?? is that the same as r<R?

a) so the initial E is mgh when the ball is at the very top.

Ei = Ef

mgh = 1/2mv^2 + 1/2Iω^2 + mg(2R)

So what is mg(2R)?? how can we substitute 2R for h? Also where did 2R come from, the problem never mention anything about 2R.

b) mg(6R) = mg (R) + 1/2mv^2 + 1/2Iω^2

so i understand that the problem states that h = 6R on the left hand side, but why is the right hand side only R (mgR). What does that R mean?

thanks!

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# Energy, Angular Momentum, Torque, solid ball rolling down loop track? help?

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