Q. A thin circular ring of mass M and radius r is rotating about its axis with 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 angular velocity given by
A) (M + 2m)ω/(2m)
B) 2Mω/(M + 2m)
C) (M + 2m)ω/M
D) Mω/(M + 2m)
Explanation
[Solution from source] No external torque acts, so angular momentum is conserved: Mr²ω = (M + 2m)r²ω′, giving ω′ = Mω/(M + 2m). Adding mass increases the moment of inertia, so the angular velocity must fall; expressions bigger than ω such as (M + 2m)ω/M are wrong. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.