# Angular momentum sanding disk

1. Oct 24, 2006

### bearhug

A sanding disk with rotational inertia 1.2E-3 kg m^2 is attached to an electric drill whose motor delivers a torque of magnitude 16 N m about the central axis of the disk. About that axis, and with the torque applied for 33 ms,
(a) What is the magnitude of the angular momentum of the disk?

I have τ= 16 N m and t= 0.033 s
I= 1.2e-3 kg m^2

In my book first of all there are two equations given for angular momentum (L)
Those are L= Iw and L=rp and I'm not sure which one I'm suppose to use. I have I but not w and I don't have r or p and p=mv correct? So I feel like this is actually a lot easier than I'm making it out to be but I can't get it. Any help will be greatly appreciated.

2. Oct 24, 2006

Use the fact that M*t = Iw2 - Iw1 = change in angular momentum, assuming that the disk started from rest, where M is the torque.

3. Oct 24, 2006

### Staff: Mentor

One way is to use Newton's 2nd law for rotation to find the angular acceleration of the disk while the torque is applied:
$$\tau = I \alpha$$

Once you have the angular acceleration, treat it as a kinematics problem.

(Even easier is to use angular impulse, as radou suggests, if you've covered that.)

4. Oct 25, 2006

### bearhug

Thanks for the help. There's a second part of the question that asks for the angular speed in rev/min. The angular speed I calculated was 438.9 rev/s and I just converted that. Just wanted to know if this is correct.

5. Oct 26, 2006

### Staff: Mentor

Standard units for angular speed are radians/sec. How did you calculate your value of rev/s?

6. Mar 31, 2007

### cardioid

if you treat it as a kinematics problem, how do you actually find the rotational momentum?

7. Apr 1, 2007

### Staff: Mentor

Once you've found the angular acceleration, as I describe in post #3, use it to find the angular speed at the end of the given time interval:
$$\omega = \alpha \Delta t$$

The angular momentum is just $I \omega$.

8. Apr 1, 2007

### cardioid

thank you Doc Al