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A bead slides without friction around a loopthe-loop. The bead is released from a height 20.3 m from the bottom of the loop-the-loop which has a radius 7 m. The acceleration of gravity is 9.8 m/s^2. How large is the normal force on it at point A if its mass is 3 g?

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I already calculated speed. 11.112 m/s. To find the answer to this problem do I use the formula. F=mv^2/R = ((.005kg)(11.112 m/s)^2)/5m = 0.123N ?
A is the top of the loop?
if point "A" in top normal force is N=(mv^2/r)-mg if "A" in bottom N equal to:|dw:1329244817817:dw|(mv^2/r)+mg

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hence in bottom 2nd law of newton is N-mg=mv^2/r & on top is N+mg=mv^2/r
top... thainks guys.
.02347N ?

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