Bug on a bowling ball, Physics

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Bug on a Bowling Ball

A bug sitting on top of a bowling ball begins to slide o? with negligible friction. Conclude the angle at which the bug leaves the surface.

This problem is easily solved with the aid of the diagram

1978_Bug on a Bowling Ball.png

The forces on the bug include the centripetal force due to gravity, the centrifugal force due to motion along the curve and the resulting normal force.

Fn = mv2/r - mg cos θ

The get done in kinetic energy as the bug slides is provided by the loss in potential energy. Since h = r - r cos θ we have

993_Bug on a Bowling Ball1.png

At the instant the bug leaves the ball the normal force Fn becomes zero. We are able to now substitute the value of the centrifugal force

Fn =(mv2/r)- mgcos θ

The get in kinetic energy as the bug slides is provided by the loss in potential energy. As h = r - r cos θ we have

mv2/2 = mgh

mv2/2 = mgr(1 - cos θ)

mv2/r = 2mgr(1 - cos θ)

At the moment the bug leaves the ball the normal force Fn becomes zero. We are able to now substitute the value of the centrifugal force

285_Bug on a Bowling Ball2.png


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