# Angular momentum of gum help

1. Oct 15, 2007

### snoggerT

The figure below shows an overhead view of a rectangular slab that can spin like a merry-go-round about its center at O. Also shown are seven paths along which wads of bubble gum can be thrown (all with the same speed and mass) to stick onto the stationary slab.

(a) Rank the paths according to the angular speed that the slab (and gum) will have after the gum sticks, greatest first (use only the symbols > or =, for example 1=2=3>4>5>6=7).

L=Iw

3. The attempt at a solution
- I'm really not sure how to start on this problem. I'm struggling with angular momentum as a whole, but I think if I can get a better understanding of this problem, I'll have a better understanding of angular momentum as a whole. So I guess I just need some pointers on how to view this problem

2. Oct 15, 2007

### learningphysics

Use torque... the magnitude of the torque is the force times the perpendicular distance from the line of the force to the pivot...

The magnitude of the force exerted is the same in all direction... call the long side L, the short side s...

call the force F...

What is the magnitude of the torque due to number 2?

3. Oct 15, 2007

### snoggerT

- The magnitude of #2 would be 0.

4. Oct 15, 2007

### learningphysics

exactly... same way #3 and #5 are 0. So 2, 3 and 5 are the lowest...

how about 4,6,7 and 1 ?

5. Oct 15, 2007

### snoggerT

- I would say that 6 is the greatest. Followed by 4 and then 7 and then 1 being the least. I'm not sure if thats correct though.

6. Oct 15, 2007

### learningphysics

Don't guess.

What is the torque of 6... call the force F... long side L... short side s...

7. Oct 15, 2007

### snoggerT

My problem is with L and s. I know that torque is rXF and that in this case it would be rF(perpendicular)sin(theta). For 6, it's perpendicular to the origin, so sin90=1. Therefor to solve 6 it would just be rF(perpendicular). I just don't know how to find r since I don't know the position of the origin (whether it's in the middle of the block or not).

8. Oct 15, 2007

### learningphysics

the magnitude of torque is |r X F| = rFsin(theta) which is rperpendicular*F or r*Fperpendicular (meaning that rsin(theta) is rperpencular... Fsin(theta) is Fperpendicular)

here I think rperpendicular*F is more convenient...

so for 6 rperpendicular is L/2. so magnitude of torque is FL/2.

9. Oct 15, 2007

### snoggerT

- I just realized I thought the "O" in the picture was the origin and didn't see the dot next to it (that is really the origin). So the rest would be:
4> F*(sqrt((L/2)^2+(s/2)^2))
7> Fs/2
1> I'm unsure on this, but from looking at the picture. It appears that it would be Fs/4.

10. Oct 15, 2007

### learningphysics

Everything looks right.

I agree with you about 1 having Fs/4

also note that (sqrt((L/2)^2+(s/2)^2)) is greater than both L/2 and s/2.

So now what is the final answer from greatest to least.

11. Oct 15, 2007

### snoggerT

- So the torque's would be the same as the angular speed that the question is asking? also, how would that relate to angular momentum? The 2nd part of the question is which of them have angular momentum of zero.

12. Oct 15, 2007

### learningphysics

Torque gives the rate of change of angular momentum. Assuming the slab is initially at rest... the greater the torque, the greater the rate of change of angular momentum...

torque*(delta t) = final angular momentum - initial angular momentum

torque*(delta t) = final angular momentum

delta t is the time of impact... (time during which the force is exerted on the slab). we assume the delta t is the same for all 7 cases...

So the greater the torque, the greater the angular momentum and hence the angular speed...

If torque is 0... that means that the angular momentum doesn't change at all... so it stays at an angular momentum of 0.