Question

a thin, 50.0 g disk with a diameter of 9.00 cm rotates about an axis through its center with 0.210 j of kinetic energy. you may want to review (pages 299 - 301) .

Answer 1

KE = 1/2 I ω^2 kinetic energy of rotating disk with inertia I

I = 1/2 M R^2 = 1/2 * .050 kg * (.09 m)^2 = .0002025 kg m^2

ω^2 = 2 * .210 / .0002025 = 2074 sec^2

ω = 45.5 / sec