A uniform solid cylinder of mass M and radius R has a wedge-shaped slice corresponding to a 60^ sector removed along its entire length. The moment of inertia of the remaining portion of the cylinder about its natural geometric axis is

Step-by-step solution

The angle of the removed sector is 60^ . The angle of the remaining portion of the cylinder is 360^ - 60^ = 300^ . Since the cylinder is uniform, the mass is proportional to the sector angle. The mass of the remaining portion M' is: M' = 300^ 360^ M = 5 6 M The moment of inertia of a solid cylinder (or any sector of it) about its central geometric axis is given by 1 2 mass radius ^2 . Therefore, the moment of inertia of the remaining portion is: I' = 1 2 M' R^2 = 1 2 ( 5 6 M ) R^2 = 5 12 M R^2 Answer: 5 12 MR^2