# Angular motion

1. Jan 21, 2009

### tommyhakinen

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$\omega_{f} = \omega_{i} + \alpha t$$
$$\theta_{f} = \theta_{i} + \omega t + 0.5 \alpha t^{2}$$
$$\tau = I \alpha$$

3. 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
$$\omega_{f} = \omega_{i} + \alpha t$$
$$0 = 10 + \alpha (10)$$
$$\alpha = -1/6 rad/s^{2}$$

The answer I get is negative. However if I substitute it into
$$\theta_{f} = \theta_{i} + \omega t + 0.5 \alpha t^{2}$$

it becomes
$$\theta_{f} = 10 + 0.5 (-1/6) (60)^{2}$$
$$\theta_{f} = -290 rad$$

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

Regards,

tommy

2. Jan 21, 2009

### psykatic

well, you've mistakenly put the values as,
whereas it should be,
$$10 = 0 + \alpha (6)$$,
because, according to the given question, $$\omega_ f$$ is 10 whereas $$\omega_ i$$ is 0. This is evident from the line,

Now, try working on it!