A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at t = 0, the wheel turns through 8.20 revolutions in 12.0 s. At t = 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia of the wheel?

Solution 38 E Step 1: Data given Velocity = d = 8.20 rev Time t = 12 s Kinetic energy 36.0 J We need to find moment of inertia We shall use s = u + 1/2 (at ) Here initial velocity is = 0 Hence we have 2 8.20 rev × 2 = 1/2 × a × (12 s) Rearrange to find acceleration a we get 8.20 rev×2 a = (1/2)×(12 s) a = 0.71 rev/s 2 Step 2 : We need to find velocity it is obtained using 2 = 2ad 2 2 = 2 × 0.71 rev/s × 8.20 rev × 2 2 =73.16 rad/s