Question

The combination of an applied force and a friction force produces a constant total torque of 36.0 N,m on a wheel rotating about a fixed axis. The applied force acts for 6.00 s. During this time the angular speed of the wheel increases from 0 to 10.0 rad/s. The applied force is then removed, and the wheel comes to rest in 60.0 s. Find (a) the moment of inertia of the wheel, (b) the magnitude of the frictional torque, and (c) the total number of revolutions of the wheel.

Answer 1

Answer:

Part a)

$I = 21.6 kg m^2$

Part b)

$\tau_f= -3.6 Nm$

Part c)

$N = 52.5 rev$

Explanation:

As we know that the net torque on the wheel is given as

$\tau = 36 Nm$

now we know that its angular speed changes from 0 to 10 rad/s in time interval of 6.00 s

so we have

$\alpha = \frac{10 - 0}{6} = \frac{5}{3} rad/s^2$

Part a)

We know that the relation between torque and angular acceleration is given as

$\tau = I\alpha$

$36 = I(\frac{5}{3})$

$I = 21.6 kg m^2$

Part b)

Now the angular speed will reduce to zero after removing external torque in t = 60 s

now we have

$\alpha = \frac{0 - 10}{60} = -\frac{1}{6}$

so frictional torque is given as

$\tau_f = I\alpha$

$\tau_f = (21.6)(-\frac{1}{6}) = -3.6 Nm$

Part c)

Number of revolutions are given by the equation

$N = \frac{0 + \omega}{4\pi}t_1 + \frac{\omega + 0}{4\pi}t_2$

$N = \frac{10}{4\pi}6 + \frac{10}{4\pi}60 = 52.5 rev$