# Homework Help: Angular Momentum of a Disk

1. Aug 18, 2013

### postfan

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

A solid disc of mass m = 5.0 kg and radius r = 0.45 m is spinning clockwise about the axis passing through the center of the disc. At t = 0, three forces are applied to the disc as shown.

Part 1. What is the magnitude of the net torque acting on the disc about its center?
Part 2. Immediately after the three forces are applied, is the magnitude of the angular momentum of the disc increasing, decreasing or staying the same?
Part 3. Assuming that the net torque stays constant, what is the change in the magnitude of the angular momentum of the disc between t = 0 and t = 1.2 s?
Part 4. What is the change in the magnitude of the angular speed between t = 0 and the moment described in Part 3?

2. Relevant equations

3. The attempt at a solution

For part 1 (taking counterclockwise to be positive) I found the torque by calculating 14*r/2+9*r-sin(30)*r*8=7*r+9r-4r=12r=12*.45=5.4

For part 2, since the disk is rotating clockwise, adding counterclockwise torques will cause it to reduce its angular momentum.

For part 3, I multiplied the time interval by the change in torque: 5.4*1.2=6.48

What am I doing wrong?

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2. Aug 18, 2013

### rude man

Check the signs of the 9r and 8rsin30 terms?

3. Aug 18, 2013

### postfan

So is it 7r-9r+4r=2r? Why?

4. Aug 18, 2013

### rude man

Yes. Look at the way the forces tend to turn the disc!

5. Aug 18, 2013

### postfan

OK I figured out parts 1 and 2 from above and common sense, on part 3 I just took the integral over the time period of the torque, but I am not sure about part 4.

I used the equation 1.08L=MR^2/2*omega, m=5, r=.45 and I got omega=2.13, is that right?

6. Aug 18, 2013

### rude man

For part 2, the disc is spinning cw and your net trque is also in that direction. Again, it asks for magnitude of change.
3. part 3 your approach is right but you need to make the correction in 1. I frankly don't remember the formula for I for a disc. BTW it's not 'change in torque' but just 'torque'.
part 4: you know how much angular momentum has changed (= I*Δω), surely deriving Δω is a snap?

7. Aug 18, 2013

### postfan

The part of part 4 that I don't understand is on the LHS of the equation do I use 1.08L or L+1.08?

8. Aug 18, 2013

### rude man

Where does 1.08 come from? What is L?

You have determined the change in angular momentum in part 3. Angular momentum change = I*Δω. Solve for Δω.

9. Aug 18, 2013

### postfan

1.08 is the change of angular momentum, and L is the symbol of angular momentum.

10. Aug 18, 2013

### haruspex

You said you were taking counterclockwise as positive. The above is taking clockwise as positive. In the OP it was the 14r/2 term that had the wrong sign. Anyway, you got 2r clockwise, which is correct.
As rude man asks, what is L and where does 1.08 come from? I will add, what did you get for part 3? (It is not 1.08.)

11. Aug 18, 2013

### rude man

OK, so what is L(t=0)? BTW you have I = mR^2/2 right, I looked it up.
Then ΔL = L(t=1.2) - L(t=0). And I*Δω = ΔL.

@haruspex, direction is irrelevant since they're asking for magnitudes thruout.