The combination of an applied force and a friction force produces a constant tot
ID: 1474157 • Letter: T
Question
The combination of an applied force and a friction force produces a constant total torque of 36.7 N m on a wheel rotating about a fixed axis. The applied force acts for 5.90 s. During this time, the angular speed of the wheel increases from 0 to 9.9 rad/s. The applied force is then removed, and the wheel comes to rest in 59.1 s. Find the moment of inertia of the wheel, Find the magnitude of the torque due to friction. Find the total number of revolutions of the wheel during the entire interval of 65.0 s.Explanation / Answer
a)
Angular acceleration = (2- 1)/t = 9.9/ 5.90 = 1.677 rad/s^2
I = I *1.677 = 36.7
I = 36.7/ 1.677 = 21.884 kg m/s^2
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b)
The wheel comes to rest only due to the frictional torque (friction)
(friction) = I ' = 21.884 '
' = - 9.9/59.1 = - 0.167 minus to show that the speed reduces.
(friction) = - 21.884*0.167 = 3.654 N.m
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c)
in time 5.90s
= 0.5 t^2 = 0.5*1.677*5.9^2 = 29.18 radians.
Or from average angular velocity *time = (9.9/2)*5.90 =29.20 radians
In time 59.1 s
(9.9/2)*59.1 = 292.54 radians
Total angle traversed
292.54 +29.20 = 321.74 radians
321.74 / (2) revolutions = 51.20 revolutions
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