The combination of an applied force and a constant frictional force produces a constant total torque of 35.8 Nm on a wheel rotating about a fixed axis. The applied force acts for 5.95 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 59.7 s.

AFind the moment of inertia of the wheel. Answer in kgm^2

BFind the magnitude of the frictional torque. Answer in Nm

CFind the total number of revolutions of the wheel.

AIf the total torque (applied + friction)is called L,

L = Lappl + Lfric = I alpha = alpha is the angular acceleration, which is (9.9 rad/s)/5.95 s= 1.664 rad/sec^2

Solve for the moment of inertia, I. b). During deceleration (with the applied force off),

alpha' = (9.9 rad/s)/59.7 s = ? Solve again the equation Lfric = I alpha' for the new torque, using the value of alspha' from part (a)

CCalculate (add) the total number of revolutions accelerating and decelerating, N1 + N2.

N1(2 pi) = (1/2) alpha (t1)^2 N2(2 pi) = (1/2) alpha'(t2)^2 t1 = 5.95 s t2 = 59.7 s