- #1

tommyhakinen

- 36

- 0

## Homework Statement

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.

## Homework Equations

[tex]\omega_{f} = \omega_{i} + \alpha t[/tex]

[tex]\theta_{f} = \theta_{i} + \omega t + 0.5 \alpha t^{2}[/tex]

[tex]\tau = I \alpha[/tex]

## The Attempt at a Solution

I am able to get part (a). However, for part (b) and (c), in order to get the frictional torque and total number of revolutions, I need to get the second angular acceleration from

[tex]\omega_{f} = \omega_{i} + \alpha t[/tex]

[tex]0 = 10 + \alpha (10)[/tex]

[tex]\alpha = -1/6 rad/s^{2}[/tex]

The answer I get is negative. However if I substitute it into

[tex]\theta_{f} = \theta_{i} + \omega t + 0.5 \alpha t^{2}[/tex]

it becomes

[tex]\theta_{f} = 10 + 0.5 (-1/6) (60)^{2}[/tex]

[tex]\theta_{f} = -290 rad[/tex]

it doesn't make sense for it to be negative. advice please. thanks.

Regards,

tommy