A uniform circular ring of mass M and radius r is rotating with an angular speed $\omega$ about an axis passing through its center and perpendicular to the plane of the ring. Two identical beads, each of mass m, some how get attached at two diametrically opposite points. The rotational speed of the ring will become
- A. $\frac{M \omega}{M+2m}$
- B. $\frac{M \omega}{M+m}$
- C. $\frac{2M \omega}{M+2m}$
- D. $\omega$
Answer: A. $\frac{M \omega}{M+2m}$
Explanation: Conservation of angular momentum: $I_1 \omega_1 = I_2 \omega_2$. $I_1 = Mr²$, $I_2 = (M+2m)r²$. $\omega_2 = \frac{M}{M+2m}\omega$.