Question

The combination of an applied force and a constant frictional force produces a constant total torque of 35.8 N·m on a wheel rotating about a fixed axis. The applied force acts for 6.04 s. During this time the angular speed of the wheel increases from 0 to 10.2 rad/s. The applied force is then removed, and the wheel comes to rest in 59.7 s.

(a) Find the moment of inertia of the wheel.
_______ kg

Answer 1

a)
Angular acceleration ? = (?2- ?1)/t = 10.2/6.04 = 1.688 rad/s^2

I ? = I *1.688 = 35.8

I = 35.8/ 1.688 = 21.208 kg m/s^2
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b)
The wheel comes to rest only due to the frictional torque ? (friction)

? (friction) = I ?' = 21.208 ?'
?' = - 10.2/59.7 = - 0.170 minus to show that the speed reduces.
? (friction) = - 21.208*0.170 = 3.605 N.m
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c)
in time 6.04s
?= 0.5 ? t^2 = 0.5*1.688*6.04^2 =30.79 radians.

Or from average angular velocity *time = (10.2/2)*6.04 =30.08 radians

In time 59.7 s
(10.2/2)*59.7 = 281.2 radians

Total angle traversed
281.2 +27.6 = 304.47 radians
304.47 / (2?) revolutions = 48.45 revolutions