Question

what is the rotational inertia of a solid iron disk of mass 41.0 kg with a thickness of 5.00 cm and radius of 30.0 cm, about an axis perpendicular to the disk and passing through its center?

Answer 1

I = 1/2 M R^2 moment of inertia of solid disk

I = 1/2 * 41.0 kg * (.3 m)^2

I = 20.5 kg * .09 m^2 = 1.85 kg-m^2