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

Question

The combination of an applied force and a frictional force produces a constant total torque of 35.9 Nm on a wheel rotating about a fixed axis. The applied force acts for 10.10 s, during which time, the angular speed of the wheel increases from 1.20 rad/s to 15.7 rad/s. The applied force is then removed. The wheel comes to rest in 80.4 s.

a) What is in rad/s2, the angular acceleration of the wheel while the applied force was acting on it?

b) What is the moment of inertia of the wheel? Answer in units of kg m2 .

c) What is in N . m, the frictional torque acting on the wheel?

d) What is in N . m, the torque due to the applied force acting on the wheel?

Explanation / Answer

part A: angular accleration a = (w2-w1)/t

alpha A = (15.7-1.2)/10.10

A = 1.435 rad/s^2

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part B : apply Torque T = I alpha

so moment of inertial I = T/A

I = 35.9/1.435

I = 25 Kgm^2

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Torque T = I alpha

so here I1 A1 = I2 A2

due to Friction, T = 25 * (15.7-1.2)/80.4

Tf = 4.5 Nm

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Torque T = 35.9 - 4.5 = 31.4 Nm

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