Question

A circular ring of Mass $M$ is rotating about its own axis in horizontal plane at an angular velocity $\omega$. If two point size bodies of mass $m$ are gently attached to the rim of ring at two ends of its diameter, then the angular velocity of ring is:

• A. $\frac{Mw}{M+2m}$
• B. $\frac{2mw}{M+2m}$
• C. $\frac{mw}{M+2m}$
• D. $\frac{2Mw}{M+2m}$

Answer 1

The correct option is A $\frac{Mw}{M+2m}$

Initial moment inertia of the ring $I_1=M{R}^2$

Final moment of inertia when m mass is attached $I_2=M{R}^2+2m{R}^2$

Now using the angular momentum conservation

$I_1\omega =I_2{\omega }_2\Rightarrow {\omega }_2=\frac{M\omega }{M+2m}$