A thin circular ring of mass $m$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $\omega^{\prime}=$

(C) $\frac{\omega(m-2 M)}{(m+2 M)}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Rotational Mechanics

A thin circular ring of mass

m

and radius

R

is rotating about its axis with a constant angular velocity

ω

. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity

ω^{'} =

Answer: (A)

Angular momentum about the axis is conserved, since the two masses are placed on gently and exert no external torque. Initially

L = I ω = m R^{2} ω

. Once the masses

M

are attached at opposite ends of a diameter, the moment of inertia becomes

(m + 2 M) R^{2}

. Equating,

m R^{2} ω = (m + 2 M) R^{2} ω^{'}

, so the new angular velocity is

ω^{'} = \frac{m ω}{m + 2 M}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option