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A mass is released from rest from the edge of a large hemispherical, frictionless bowl as...

A mass is released from rest from the edge of a large hemispherical, frictionless bowl as shown. When the mass slides through the bottom of the bowl, what is the ratio of the magnitude of the normal force to the magnitude of the gravitational force acting on the mass?

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Answer #1

centripetal force = mv2/r. Thus at the lowest point:

Fn - mg = mv2/r.

Picture not posted, so I assume the mass slides from height r where its potential energy relative to the lowest point is mgr. Therefore energy conservation implies

1/2 m v2 = m gr

v2 = 2 gr

Thus

Fn - mg = m (2gr)/r = 2mg

Fn = 3mg

Fn/(mg) = 3

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