A horizontal uniform circular disc of mass M and radius R is rotating with an angular velocity ₀ about a frictionless vertical axis passing through its centre. Two small point masses, each of mass m , are gently placed at diametrically opposite edges of the disc. What will be the final angular velocity of the system?

Step-by-step solution

The initial moment of inertia of the disc about its central axis is I_i = MR^2 2 . When two point masses, each of mass m , are placed at the edges (distance R from the axis), the final moment of inertia of the system becomes: I_f = MR^2 2 + 2mR^2 = ( M 2 + 2m )R^2 = ( M + 4m 2 )R^2 Since no external torque acts on the system, the angular momentum is conserved: I_i _i = I_f _f ( MR^2 2 ) ₀ = ( M + 4m 2 )R^2 _f _f = M M + 4m ₀ Answer: M M + 4m ₀