# Homework Help: Conserving Angular Momentum

1. Apr 13, 2017

### Game_Of_Physics

• Member advised to use the homework template for posts in the homework sections of PF.
1. A moving rough cylinder of radius a, and mass m collides with an identical cylinder, on a smooth horizontal surface. Its centre of mass moves with linear velocity v0, and its angular velocity is ω0. What is the motion of the cylinders after the collision?

I have be told that the answer to this problem does not involve using conservation of angular momentum because there is an external torque during the collision, and the way to do it is using conservation of rotational energy. I do not understand why angular momentum is not conserved if you consider that it is a closed system.

2. Apr 14, 2017

### scottdave

Since they both have rough surface, it would seem that they stick together at one point. Can you draw out a diagram of what that system would look like? Can you see how the external forces (horizontal surface and gravity) work to create a torque on the new system?

3. Apr 14, 2017

### Game_Of_Physics

Okay so the new system is now made to rotate because of the torque, which therefore conserves angular momentum?

4. Apr 14, 2017

### scottdave

Imagine these two cylinders floating out in space. One (call it A) is moving toward B (which is stationary and not rotating) and also A is rotating. When they hit, and if the surfaces are able to interact, then A and B could spin together for an amount of time. The same angular momentum, but now a new moment of inertia is the two cylinders together. If they come apart, then both could be spinning, but the sum of angular momentums of both will equal the initial. But in actuality, there is the surface, which will push a force if the angular momentum is 'wanting' to spin down. And if something is wanting to spin up, then gravity will work against that. Will they stick? Will the first cylinder try to 'roll over' the second? Will the first one just stop, and the second one starts rolling at the same speed as the first? I am not sure of the answer, but these are things to look at.
What I am saying, is that you might be able to use angular momentum, if you know how the external torques are contributing to the change in momentum.

5. Apr 15, 2017

### vela

Staff Emeritus
What is "it"? The first cylinder or the second cylinder?

I'm inclined to agree with you. What's this supposed external torque acting on the system?

I don't think you can assume they'll stick together. The two surfaces will slide against each other and exert kinetic friction forces.

6. Apr 15, 2017

### haruspex

It seems unlikely energy would be conserved. In linear impulses there are three circumstances:
- elastic, so energy is conserved
- partly elastic, in which case you need to know the coefficient of restitution
- completely inelastic, in which case the maximum energy is lost consistent with conservation of momentum and constraints such as impenetrable surfaces.
The same applies here, but the elastic cases get nasty because you get a rotational bounce as well as a linear one. Unless the question specifies otherwise I would assume completely inelastic.

It depends what axis one uses. The first cylinder will exert a frictional impulse down on the stationary one, resulting in a vertically upward impulse from the ground on it (in addition to the ongoing normal force supporting the weight). So ok if the point of contact of the second cylinder with the ground is the axis.
Correspondingly, there will be a vertically upward frictional impulse on the first cylinder, which is guaranteed to make it airborne.
Not necessarily. We are told the surface is rough, but no actual coefficient, so we should take as rough enough to prevent slipping, as long as that is possible with a finite coefficient. And it is. The horizontal impulse between them is an unbounded force, permitting a correspondingly unbounded frictional impulse.

@Game_Of_Physics, let the impulse between them on collision be J, at some angle θ to the horizontal. Can you think how to find θ? What equations can you write relating the horizontal and vertical components of J to the subsequent motions?

Last edited: Apr 15, 2017
7. Apr 15, 2017

### Game_Of_Physics

I am sorry I forgot to mention that neither of the cylinders leave the surface. What I find puzzling about this is why an external torque is created. I would have thought the only way to achieve this would be for the 2nd cylinder to be on a rough surface, (or more precisely, for the cylinders to be on surfaces with unequal coefficients of friction), or for one of the two cylinders to be inelastic? Could someone explain to me if this is incorrect, and if so, why?

8. Apr 15, 2017

### Game_Of_Physics

The question does not specify otherwise, but how can you use energy conservation to answer the question if it is an inelastic collision? Surely it must be elastic?

9. Apr 15, 2017

### zwierz

The notations

By superscripts $+,-$ we denote quantities after the collision and before the collision respectively.
So that
$\boldsymbol v^{\pm}_1=v^{\pm}_1\boldsymbol e_x$ is the velocity of the center $S_1$ before and after the collision;
$\boldsymbol v^{\pm}_2=v^{\pm}_2\boldsymbol e_x$ is the velocity of the center $S_2$ before and after the collision;
$\boldsymbol \omega_1^\pm=\omega_1^\pm\boldsymbol e_z$ is the angular velocities of the first cylinder before and after the collision;
$\boldsymbol \omega_2^\pm=\omega_2^\pm\boldsymbol e_z$ -- the same for the second cylinder;
$\boldsymbol R_{1,2}=R_{1,2}\boldsymbol e_y$ are the reactions of the floor for the first and the second cylinder respectively;
$\boldsymbol T=T_x\boldsymbol e_x+T_y\boldsymbol e_y$ is the reaction that the second cylinder exerts the first one.

The equations

$$J(\boldsymbol \omega_2^+-\boldsymbol \omega_2^-)=-\boldsymbol{S_2A}\times \boldsymbol T,\quad J(\boldsymbol \omega_1^+-\boldsymbol \omega_1^-)=\boldsymbol{S_1A}\times \boldsymbol T,$$
$$m(\boldsymbol v^+_1-\boldsymbol v^-_1)=\boldsymbol R_1+\boldsymbol T,\quad m(\boldsymbol v^+_2-\boldsymbol v^-_2)=\boldsymbol R_2-\boldsymbol T.$$
Here we have 6 scalar equations for 8 unknowns: $\omega_{1,2}^+,T_x,T_y,R_{1,2}, v^+_{1,2}$.
To close this system add kinematic equations: $\omega_1^+=-\omega_2^+,\quad v_1^+=v_2^+$

Last edited: Apr 15, 2017
10. Apr 15, 2017

### Game_Of_Physics

Right okay. So you have used conservation of angular momentum and conservation of linear momentum, without having to use conservation of angular/linear energy? I have been told the answer involves using conservation of energy because the creation of the torque shown in your diagram invalidates the use of conservation of angular momentum....

11. Apr 15, 2017

### zwierz

I used the equations of the impact theory and the hypothesis by scottdave
If you wish to have the conservation of energy then you should provide another hypothesis about the collision and replace two equations in the last line of #9 with equations of your hypothesis.
For example you can replace those equations with equation of conservation of energy and $\omega_1^+=-\omega^+_2$

12. Apr 15, 2017

### vela

Staff Emeritus
How do you know it's going to go airborne?

I don't agree that "rough" means "enough friction to prevent slipping." I would take "rough" to mean "there's friction between the two cylinders." But it sounds like your interpretation is correct from what the OP has subsequently said.

13. Apr 15, 2017

### haruspex

I mostly agree with that. In particular, it was not immediately obvious to me that the two cylinders should (on the basis of complete inelasticity) have the same horizontal velocity immediately after impact, just as they would with no rotations or vertical motion, but I concluded that is the case.
However, in your definitions you do not seem to have allowed for vertical motion of the first cylinder, even though you have Ty.

14. Apr 15, 2017

### haruspex

There was a more fundamental point I should have made. There is no law of conservation of rotational energy, just as there is no law of conservation of linear energy in a specific coordinate direction. There is just conservation of energy.
Edit: This statement was wrong; it is not conserved: But, curiously, in this question, if by rotational energy you mean that embodied in the spin of each cylinder on its own axis, it is conserved.
Unfortunately that is not physically possible without some other external impulse, so adding that "information" creates an internal contradiction. When a question statement contains an internal contradiction different answers may be validily obtained by different methods. You must overrule that statement.
As I posted, torque depends on the choice of axis.
Because of the downward (frictional) impulse of the first cylinder on the second, there is a corresponding impulsive normal force from the ground on the second cylinder.
If you take the centre of the first cylinder, at moment of impact, as the axis then that normal impulse has a torque about your axis.
You can't.
Because the upward frictional impulse of the second cylinder on the first has no opposing vertical impulse. A change in vertical velocity is inevitable.
In general, no, but since we are not given a coefficient of friction I see no alternative.
By the way, I believe when scottdave writes about the surfaces "sticking" he means the same thing.

Last edited: Apr 16, 2017
15. Apr 15, 2017

### zwierz

yes I considered biliteral constraint:

16. Apr 15, 2017

### haruspex

As I just posted, that is not possible and must be rejected. Your own equations prove it will.

17. Apr 15, 2017

### zwierz

I do not see problems with biliteral constraints

18. Apr 15, 2017

### haruspex

If you mean the impulse T shown in the diagram, that is internal to the system consisting of the two cylinders. The only external impulse is from the ground on the second cylinder. This can be eliminated from angular momentum equations by using an axis in the vertical line through it, such as the centre of the second cylinder.

19. Apr 15, 2017

### haruspex

Not sure what you mean. Are you saying to add a smooth ceiling to hold both cylinders down?

20. Apr 15, 2017

### zwierz

Yes it is exactly the statement I studied. I understood OP in such a way. But surely another statements are also possible. For example as I mentioned above one can consider perfectly elastic collision with conservation of energy. Surely we can study the case when cylinder can fly up. Then one must bring additional hypotheses about interaction between cylinders and the floor

21. Apr 15, 2017

### haruspex

Then the question statement should have specified that ceiling. It is not enough to say "it does not leave the surface"; the reason must be given. This is because it will change the equations, producing a different answer, and the way it changes the equations may depend on the reason it does not leave the surface.
I repeat, your equations are not consistent as they stand, since you have a vertical component in T with no consequent momentum change.
Of course, but it will make things quite complicated. For one thing, the downward frictional impulse on the second cylinder would lead to a subsequent bounce, so both would tend to leave the surface.

22. Apr 15, 2017

### vela

Staff Emeritus
Unless the upward frictional force is greater than the weight of the cylinder, the cylinder will stay on the surface. The only difference will be that the normal force on it is reduced.

Perhaps you worked out the problem and found that to be the case, but I don't think it's obvious that the cylinder will leave the surface. Consider the case where it's barely spinning when it runs into the other cylinder. My intuition tells me it's not going to leave the surface in that scenario.

23. Apr 15, 2017

### zwierz

perhaps it is not enough for engineers, in fundamental science it is enough to impose biliteral constraint mathematically. Concrete realisation of this constraint is another question. Discussing of such a question is not of interest for me.
They are consistent. Solve them and make sure that they have a unique solution

Last edited: Apr 15, 2017
24. Apr 16, 2017

### haruspex

A collision is generally presumed to take a very short time, involving unknown large forces, but combining to form a moderate inpulse, i.e. ∫F.dt. Modest forces, such as gravity and the corresponding normal force for that, produce no impulse worth mentioning in that time.
Just as a normal force N supports a frictional force up to Nμs, a normal impulse J supports a frictional impulse Jμs. That easily overwhelms mgΔt.

25. Apr 16, 2017

### haruspex

You have an equation $m(v_1^+-v_1^-)=R_1+T$. Your R1 and T each have an ey component, but the v1 terms do not. You can make R1 zero, but not Ty.

I did solve what were effectively the same equations, but allowing for a vertical motion of the first cylinder. As I mentioned somewhere, it does turn out that the rotational KE (measured in terms of each cylinder's rotation about its centre) is conserved, though I could not think of any simple reason for that. The first cylinder stops spinning and the second rotates at the same speed the first cylinder had been rotating, but in the opposite sense.

I have not tried to solve with an added force holding the first cylinder down, but I believe a lot will change.