If M=0, theta=arccos(2/3) where theta is the angle between the right, for example and the pearl at the top of the circle (or the ring).

A ring of mass M hangs from a thread, and two beads of mass m slide on it without friction.The beads are released simultaneously from the top of the ring and slide down opposite sides.

Show that the ring will start to rise if m>3M/2, and find the angle at which it occurs

Solution Preview

Let the radius of the ring be R. Let "theta" be the angle mass m makes with the vertical at any time t after they are released from the top.

Since we have a frictionless movement, the force acting on the beads are always pointing along N, the Normal Force acting along the radius, but away from it.

Apply Newton's law for the circular motion of the bead of mass m.

m*g*Cos(theta) - N = m*v^2/R (Resultant Centripetal force) ...

Solution Summary

This solution is provided in 367 words. It uses Newton's law to describe the circular motion of the mass and how this relates to mass. All calculations are included.

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