Question

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.

Answer 1

Answer:

(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.