A 30.0 kg wheel, essentially a thin hoop with radius 0.620 m, is rotating at 202 rev/min. It must be brought to a stop in 26.0 s. (a) How much work must be done to stop it? (b) What is the required average power? Give absolute values for both parts.

(a) 2579.99 J.

(b) 99.23 W

Explanation:

(a)

The work done  by the wheel = Kinetic energy of the wheel.

W = 1/2mv² Equation 1

Where W = work done, m = mass of the wheel, v = velocity of the wheel.

But,

v = rω Equation 2

Where r = radius of the wheel, ω = angular velocity

Substitute equation 2 into equation 1

W = 1/2mr²ω² Equation 3

Given: m = 30 kg, r = 0.620 m, ω = 202 rev/min = (202×0.10472) rad/s = 21.153 rad/s

Substitute into equation 3

W = 1/2(30)(0.62²)(21.153²)

W = 2579.99 J.

(b)

Power = work done/time

P = W/t Equation 4

where P = power, W = work done, t = time

Given: W = 2579.99 J, t = 26 s

Substitute into equation 4

P = 2579.99/26

P = 99.23 W.