Question

A net torque of 36 N.m acts on a wheel rotating about a fixed axis for 6 s. During this time the angular speed of the wheel increases from 0 to 12 rad/s. The applied force is then removed, and the wheel comes to rest in 75 s. a. What is the moment of inertia of the wheel? b. What is the magnitude of the frictional torque? c. How many revolutions does the wheel make

Answer 1

a.

$$\tau=I\alpha\to \tau=I\frac{\omega-0}{t}$$

$$I=\frac{\tau t}{\omega}=\frac{36 \cdot 6}{12}=18\ kgm^2$$

b.

$$I=\frac{\tau_f t'}{\omega}$$

$$18=\frac{75\tau_f }{12}$$

$$\tau_f=2.88\ Nm$$

c.

$$n=\frac{t}{T}=\frac{\omega t}{2\pi}=\frac{(12)(75)}{2\pi}=143$$