A thin circular ring having mass M and radius r rotates about its axis with an angular speed . Subsequently, two particles, each of mass m , are attached at diametrically opposite points. The resulting angular speed of the ring will be

Step-by-step solution

Initial moment of inertia of the ring is I₁ = M r^2 . Initial angular speed is ₁ = . When two particles each of mass m are attached at diametrically opposite points on the ring, the new moment of inertia becomes I₂ = M r^2 + 2m r^2 = (M + 2m) r^2 . Since no external torque acts on the system, the angular momentum is conserved. I₁ ₁ = I₂ ₂ M r^2 = (M + 2m) r^2 ₂ ₂ = M M + 2m