Posted by **John** on Sunday, March 4, 2007 at 11:09pm.

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

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