Question

A 28.0 kg wheel, essentially a thin hoop with radius 1.00 m, is rotating at 320 rev/min. It must be brought to a stop in 11 s. How much work must be done to stop it?
What is the required average power?

Answer 1

Moment of inertia of wheel,
I = m*r^2
=28*1^2
= 28 Kgm^2

w= 320 rev/min = 320*2pi rad / 60 s = 33.5 rad/s

Initial kinetic energy = 0.5*I*w^2
= 0.5*28*(33.5)^2
= 15721.2 J

If this wheel has to be brought to rest, final kinetic energy =0
work done = change in kinetic energy
= 15721.2 J
Answer: 15721.2 J

average power= work done/ time
= 15721.2 J / 11s
= 1429.2 W
Answer: 1429.2 W