Question

A ring of radius R lies in the vertical plane. A bead of mass ${}^{\prime }{m}^{\prime }$ can move along the ring without friction. Initially, the bead is at rest at the bottom-most point on the ring. Find the minimum constant horizontal speed $v$ with which the ring must be pulled such that the bead completes the vertical circle.

Answer 1

When the ring is pulled, due to the relative motion of the bead and the ring, the bead starts moving up on the ring. The force acting on the bead is the normal force and the force of gravity. Since, the ring is pulled with velocity $V$, the bead also acquires a velocity $V$ relative to ground at its bottom most point. Now, since normal force is zero.

At the top:

$\frac{m{V}_B^2}{R}=mg$

$V_B^2=\sqrt{gR}$

From energy conservation:

$\frac{1}{2}m{V}_A^2=\frac{1}{2}m{V}_B^2+2mgR$

$⟹\frac{V^2}{2}=\frac{gR}{2}+2gR$

$⟹V^2=5gR$

$⟹V=\sqrt{5gR}$