Question

A ring of mass \(M\) and radius \(R\) is rotating about its axis with angular velocity \(\omega\). Two identical bodies each of mass \(m\) are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be :

• A. \(\frac{m(M+2m)}{M}\omega^2R^2\)
• B. \(\frac{Mm}{(M+m)}\omega^2R^2\)
• C. \(\frac{Mm}{(M+2m)}\omega^2R^2\)
• D. \(\frac{(M+m)M}{(M+2m)}\omega^2R^2\)

Hint:

Conserve angular momentum about the axis; the two bodies at the ends of a diameter are each at distance R from the axis.

Answer 1

The Correct Option is C

Initial moment of inertia \(I_1=MR^2\). After attaching the two masses at distance \(R\) from the axis: \(I_2=MR^2+2mR^2=(M+2m)R^2\).
Angular momentum conservation: \(I_1\omega=I_2\omega'\Rightarrow\omega'=\frac{M\omega}{M+2m}\).
\(\Delta KE=\frac12I_1\omega^2-\frac12I_2\omega'^2=\frac12MR^2\omega^2\left[1-\frac{M}{M+2m}\right]=\frac12MR^2\omega^2\cdot\frac{2m}{M+2m}\).
\(\Delta KE=\frac{Mm}{M+2m}\omega^2R^2\).
Answer: (3) \(\frac{Mm}{(M+2m)}\omega^2R^2\)