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?

Question

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?

anonymous · Answer

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 ?

jamesj · Answer

A is the top of the loop?

anonymous · Answer

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