The combination of an applied force and frictional force produces a constant tot

Question

The combination of an applied force and frictional force produces a constant total torque of 34.2 Nm on a wheel rotating about a fixed axis. The applied force acts for 6.3s during which time, the angular speed of the wheel increases for 1.60 rad/s to 12.20 rad/s. The applied force is then removed. The wheel comes to rest in 66.4s

a.) what is the rad/s^2 the angular acceleration of the wheel before the applied force is removed?

b.) what is in kg m^2 the moment of interia of the wheel?

c.) what is in rad/s^2 the angular acceleration of the wheel after the applied force is removed?

d.) what is in N m the torque due to friction (after the force is removed)?

Explanation / Answer

Part A)

Apply wf = wo + (alpha)t

12.2 = 1.6 + (alpha)(6.3)

alpha = 1.68 rad/s2

Part B)

Torque = I(alpha)

34.2 = I(1.68)

I = 20.3 kg m2

Part C)

wf = wo + (alpha)t)

0 = 12.2 + (alpha)(66.4)

alpha = -.184 rad/s2 (Negative since its slowing down)

Part D)

Torque = I(alpha)

Torque = (20.3)(.184)

Torque = 3.73 Nm (Your answer key may want that as a negative)

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