Question

A ring of mass m and radius r is placed on a smooth horizontal surface with its centre at origin a small ball of same mass is given a velocity

Answer 1

Mass per unit length of ring = M/(2×π×R)

A ring of mass m and radius r is placed on a smooth horizontal surface

Mass of the dx element = dm = dx × M/(2×π×R)

For equilibrium,

2T sin θ= dm × v²/ R

Since θ= small, sinθ=θ

2T\θ=dx ×M/(2πR)× v²/ R

2T (dx/R) =dx ×M/(2πR)× v²/ R

Tension in the ring. T= Mv²/ (4πR)