Question

Find the rotational kinetic energy of a solid cylinder having moment of inertia of 0.625 kg m^2 moving with angular speed 100 rads^-1

Answer 1

K.E = 3125 J

Explanation:

When the cylinder rotates about its axis , the kinetic energy of rotation can be found by the following relation

K.E = 1/2 I ω²

here I = moment of inertia

ω = angular velocity of cylinder

Thus K.E = 1/2 x 0.625 x ( 100 )² = 3125 J

Thus kinetic energy = 3125 J