Question

A thin, 90.0 g disk with a diameter of 9.00 cm rotates about an axis through its center with 0.270 J of kinetic energy. What is the speed of a point on the rim?

Answer 1

Angular kinetic energy of the disk KE = 1/2 Iw^2; where I = 1/2 mr^2 for a thin disk, m = 0.09 kg, r = 0.09 m.

Tangential velocity (on the rim) = v = wr where w is the angular velocity of the spinning disk;

so that KE = 1/2 (1/2 mr^2) (v^2/r^2) = 1/4 mv^2 = 0.27

Solve for v = sqrt(v^2) = sqrt[4KE/m] = sqrt[4*0.27/0.09] = 3.46 m/s

The physics is this...as the disk is fixed but spinning, the only KE is angular, there is no linear kinetic energy. Angular KE is based on inertia moment I = kmr^2 and angular velocity w, where k = 1/2 for a thin disk.