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.
a) find the moment of inertia of the wheel. answer in kg*m^2
(b) find the magnitude of the frictional torque. answer in n*m
(c) find the total number of revolutions of the wheel.
(a) if 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)
(c) calculate (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