A 240 kg wheel essentially a thin hoop with radius 140 m is

A 24.0 kg wheel, essentially a thin hoop with radius 1.40 m, is rotating at 461 rev/min. It must be brought to a stop in 17.0 s. (a) How much work must be done to stop it? (b) What is the required average power?

Solution

The work needed to stop the wheel will be the same as the rotational kinetic energy of the wheel it needs to lose.

W = I*^2/2

Since the mass is concentrated on the outside of a loop, its moment of inertia would be I = mr^2

W = m*r^2*^2/2
W = (24kg)(1.40 m)^2(461*2/60 rad/s)^2/2
W = 5.48*10^4J

(b) P = W/t
P = 5.48*10^4 J / 17.0 s
P = 3224.38W