A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $\omega$. Two particles having mass $m$ each are now attached at diametrically opposite points. The angular speed of the ring will become:

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

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Rotational Mechanics

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed

ω

. Two particles having mass

m

each are now attached at diametrically opposite points. The angular speed of the ring will become:

(A) $ω \frac{M - 2 m}{M + 2 m}$
(B) $ω \frac{M + 2 m}{M}$
(C) $ω \frac{M}{M + m}$
(D) $ω \frac{M}{M + 2 m}$ ✓ Correct

Answer: (D)

Solution

Attaching the particles exerts no external torque about the axis of rotation, so the angular momentum about that axis is conserved. The ring alone has

I_{i} = M r^{2}

, and the two particles sit on the rim at distance

r

, adding

2 m r^{2}

, so

I_{f} = (M + 2 m) r^{2}

. Conservation gives

M r^{2} ω = (M + 2 m) r^{2} ω^{'}

so the new angular speed is

ω^{'} = \frac{M ω}{M + 2 m}

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option

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