A solid cylinder of mass 4 ~kg and radius 0.5 ~m is rotating about its axis with an angular speed of 20 ~rad/s . The work required to bring the cylinder to rest is:

Step-by-step solution

The moment of inertia of a solid cylinder about its axis is given by: I = 1 2 MR^2 I = 1 2 4 (0.5)^2 = 0.5 ~kg m^2 According to the work-energy theorem, the work required to stop the cylinder is equal to its initial rotational kinetic energy: W = K = 1 2 I ₀^2 W = 1 2 0.5 (20)^2 W = 1 2 0.5 400 = 100 ~J Using I = MR^2 (as for a ring) would incorrectly give 200 ~J , and omitting the 1 2 factor in kinetic energy would incorrectly give 200 ~J or 400 ~J . Answer: 100 ~J