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.

Answer 1

aarrgh, English units

torque = F * .75 ft = .75 F foot pounds

mass = 480/32.2 = 14.9 slugs

I = (1/2) m r^2 = .5 (14.9)(.5625)
= 4.19 slug ft^2

kinetic energy = (1/2) I w^2
find w in radians/sec
w = omega = 600 rev/60 s * 2 pi rad/rev = 62.8 rad/s
so
ke = (1/2)(4.19)(62.8)^2 = 8271 slug ft^2/s^2

work done by force = torque * total angle in radians to stop
= .75 F (theta)

theta is average speed in rad/s times time to stop
average speed = 62.8/2 = 31.4 rad/s
so
theta = 31.4 rad/s * 12 s
= 377 radians to stop
so
work to stop = .75 F (377)
= 283 F ft pounds

so
283 F = 8271
F = 29.2 pounds

Answer 2

I used radius = .75, should be 1.5

Answer 3

aarrgh, English units

torque = 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

kinetic energy = (1/2) I w^2
find w in radians/sec
w = omega = 600 rev/60 s * 2 pi rad/rev = 62.8 rad/s
so
ke = (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)

theta is average speed in rad/s times time to stop
average speed = 62.8/2 = 31.4 rad/s
so
theta = 31.4 rad/s * 12 s
= 377 radians to stop
so
work to stop = 1.5 F (377)
= 566 F ft pounds

so
566 F = 33,054
F = 58.5 pounds