Angular Rotation with Innertia, Work, and Power

A 27.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 270 rev/min. It must be brought to a stop in 22.0 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) How much work must be done to stop it?

The moment of inertia I of a thin circular hoop of radius r and mass m is given by (ref 1):

(1) I = m * r^2

The angular kinetic energy is given by (ref 2):

(2) E = (1/2) * I * w^2, where

w is the angular speed in rads/s. Given that the hoop is rotating at 270 rev/min,

(3) w = 2π * f

= 2π * 270 [rev/min] / 60s/min = 28.3rad/s

Substituting (1) and (3) in (2)

(4) E = (1/2) * I * w^2

= (1/2) * (m * r^2) * 28.3^2

= (1/2) * (27kg * 1.8m^2) * 28.3^2 = 3.50x10^4J of work to stop


(b) What is the required average power?

(5) Power = Work / t

= 3.50x10^4J / 22s = 1.59x10^3 W
.

KE(rot) = 1/2 * I * (omega)^2

Moment of inertia for a hoop is m * r^2

KE(rot) = 1/2 * (m*r^2) * (omega)^2
KE(rot) = 1/2 * (27.0kg * (1.8m)^2) * (270rev/min * 2 pi radians/revolution * 1 min/60 seconds)^2
KE(rot) = 1/2 * 87.5 kg m^2 * (28.27 radians/second)^2
KE(rot) = 3.50 x 10^4 J

Power = E / time
Power = 3.50 x 10^4 J / 22.0 s
Power = 1590 W