The combination of an applied force and a constant frictional forceproduces a co
ID: 1748425 • Letter: T
Question
The combination of an applied force and a constant frictional forceproduces a constant total torque of 36.2N·m on a wheel rotating about a fixed axis.The applied force acts for 5.93 s. Duringthis time the angular speed of the wheel increases from 0 to10.4 rad/s. The applied force is thenremoved, and the wheel comes to rest in 60.9 s. (a) Find the moment of inertia of thewheel.1kg·m2
(b) Find the magnitude of the frictional torque.
2N·m
(c) Find the total number of revolutions of the wheel.
3 rev (a) Find the moment of inertia of thewheel.
1kg·m2
(b) Find the magnitude of the frictional torque.
2N·m
(c) Find the total number of revolutions of the wheel.
3 rev
Explanation / Answer
a)The combination of an applied force and a constant frictional forceproduces a constant total torque of 36.2N·m on a wheel rotating about a fixedaxis. So = 36.2 N.m During the time interval 5.93s, the angular speed of the wheelincreases from 0 to 10.4rad/s So angular acceleration () of the wheel in this timeis = (10.4rad/s - 0) / (5.93s) =1.754 rad/s2 We know the formula = I I = / = (36.2 N.m) / (1.754 rad/s2) = 20.64 kg.m2 b) During the deceleration of the wheel the wheel comes to restfrom 10.4 rad/s to 0 in the time 60.9s. So deceleration of the wheel in this case is = (0 - 10.4rad/s)/(60.9s) = - 0.17 rad/s2 Now magnitude of the frictional torque is f = I = (20.64 kg.m2)(0.17 rad/s2) = 3.525 N.m c) In the accelerating phase, total number of revolutions made bythe wheel is 1 = ot +(1/2)t2 = 0 + (1/2)(1.754 rad/s2)(5.93s)2 = 30.84 rad In the decelerating phase, total number of revolutions made bythe wheel is 2 = ot +(1/2)t2 = (10.4rad/s)(60.9s) + (1/2)(- 0.17rad/s2)(60.9s)2 = 318.1 rad/s Then = 1 + 2 =30.84 rad + 318.1 rad/s =348.94 rad =55.5 rev = 318.1 rad/s Then = 1 + 2 =30.84 rad + 318.1 rad/s =348.94 rad =55.5 rev