A uniform horizontal circular disc of mass M and radius R is rotating freely about a vertical axis passing through its center at an angular speed of 40 rpm . Two small blocks, each of mass M 4 , are initially fixed at diametrically opposite points on the edge of the disc. If the blocks are slowly moved to a distance of R 2 from the center along the diameter, what will be the new angular speed of the system?

Step-by-step solution

The initial moment of inertia of the system is the sum of the moment of inertia of the disc and the two blocks at the edge. I₁ = 1 2 MR^2 + 2 ( M 4 R^2 ) = 1 2 MR^2 + 1 2 MR^2 = MR^2 When the blocks are moved to a distance of R 2 from the center, the new moment of inertia is: I₂ = 1 2 MR^2 + 2 ( M 4 ( R 2 )^2 ) = 1 2 MR^2 + 1 8 MR^2 = 5 8 MR^2 Since no external torque acts on the system, angular momentum is conserved: I₁ ₁ = I₂ ₂ Substituting the given values: MR^2 40 = 5 8 MR^2 ₂ ₂ = 40 8 5 = 64 rpm Answer: 64 rpm