How much tangential force must be exerted to stop a rotating

How much tangential force must be exerted to stop a rotating solid disk (radius = 1.50 feet weighing 480 lbs in 12.0 seconds? The disk is initially rotating at 600 RPM.

Solution

First we have to find the Torque, now Torque = F * r (r=radius of the wheel)

= F * 1.5 ft = 1.5 F foot pounds

Mass = 480/32.2 = 14.9 slugs

I = (1/2) m r^2 = .5 (14.9)(2.25)
= 16.8 slug ft^2

Now finding K.E(kinetic energy)

K.E = (1/2) I w^2
value of w in radians/sec
w = omega = 600 rev/60 s * 2 pi rad/rev = 62.8 rad/s
so
K.E = (1/2)(16.8)(62.8)^2 = 33,054 slug ft^2/s^2

work done by force = torque * total angle in radians to stop
= 1.5 F (theta)
Average speed in rad/s times time to stop is expressed in theta
Now Average speed = 62.8/2 = 31.4 rad/s
so
theta = (31.4 * 12 )
= 377 radians to stop
so
work to stop = 1.5 F (377)
= 566 F ft pounds

so
566 F = 33,054
Tangential force exerted to stop the rotating disc = 58.5 pounds