Why is momentum not conserved?

[simphys]

Homework Statement

please check the screenshot.
I have two questions: which don't need any calculation yet:
1. Why is angular momentum of the two disks not conserved?? The friction between is internal no? In the solution it said that there is ang. momentum conserved (neither kinetic of course)

2. What is actually meant with 'an equilibrium situation reached'?

Relevant Equations

thanks in advance.

Here is question + drawing.

 

[phinds]

Can you think of any reason? What, exactly, happens when you smoosh the two disks together.

 

[simphys]

Can you think of any reason? What, exactly, happens when you smoosh the two disks together.

Well, friction will occur. But I can't see how this creates non-conservation of AM
Or.. is it the torque/moment that the force of your hand will apply to the disks when smooshed together that creates an angular impulse?

 

[phinds]

Well, friction will occur. But I can't see how this creates non-conservation of AM
Or.. is it the torque/moment that the force of your hand will apply to the disks when smooshed together that creates an angular impulse?

And what does the friction cause?

 

[simphys]

And what does the friction cause?

well.. a torque? but internally isn't it?

 

[vela]

Well, friction will occur. But I can't see how this creates non-conservation of AM
Or.. is it the torque/moment that the force of your hand will apply to the disks when smooshed together that creates an angular impulse?

Consider the forces exerted on each wheel separately when the disks are in contact. Presumably your hand needs to exert whatever force is necessary so that the wheel can rotate but not undergo translational motion.

 

[phinds]

Do you happen to be a dentist, by chance?

I ask because this is like pulling teeth

When there is friction between two object in motion, be they moving linearly against each other or rotating, what does the friction cause? (I AM assuming slippage)

 

[vela]

well.. a torque? but internally isn't it?

Yes, the friction between the two wheels will produce a torque about the axis on each wheel, and the forces are internal to the system so the resulting torques are internal. But what other forces are acting on each wheel?

 

[jbriggs444]

Yes, the friction between the two wheels will produce a torque about the axis on each wheel, and the forces are internal to the system so the resulting torques are internal. But what other forces are acting on each wheel?

The force of friction on the one wheel produces a torque about that wheel's axis, yes.
The force of friction on the other wheel produces a torque about the other wheel's axis, yes.

But if we are talking about the angular momentum for a system or net torque upon a system, we need to pick one reference axis for the system. Not two. Torques about different axes are incommensurable. You cannot expect to sum them together and obtain a meaningful result.

If one picks only one axis then the net torque from a pair of equal and opposite forces applied at the same point is equal to what, exactly?

That said, a search for other forces acting on the assembly is likely to be fruitful.

 

[bob012345]

well.. a torque? but internally isn't it?

This is not an easy problem. It would help you to draw a free body diagram of all the forces and torques on each disk. That might give you insight as to what is going on.

 

[kuruman]

1. Why is angular momentum of the two disks not conserved?? The friction between is internal no? In the solution it said that there is ang. momentum conserved (neither kinetic of course)

Why are you saying this? Did you calculate the angular momentum before and after and found that it is not conserved contrary to what the solution claims? If so, please post your work.

Note: You can set friction aside and consider that the circumference of each wheel has identical gears in the proper number ratio so that can mesh (see below). Would the gears change anything as far as the angular momentum of the two-gear system before and after meshing is concerned?

2. What is actually meant with 'an equilibrium situation reached'?

It means that one disk rolls on the other without slipping, i.e. the disk that was not spinning has sped up to its final angular velocity and the disk that was spinning has slowed down to its final angular velocity. This equilibrium situation is reached instantly when the the disks have teeth as in the figure above. The teeth should help you see a crucial condition that holds when the equilibrium situation is reached. Just don't let @phinds pull them!

 

[haruspex]

1. Why is angular momentum of the two disks not conserved?? The friction between is internal no? In the solution it said that there is ang. momentum conserved (neither kinetic of course)

Is that what you meant to write? Did you mean that in the solution the angular momentum of the two disc system is not conserved?

 

[simphys]

Is that what you meant to write? Did you mean that in the solution the angular momentum of the two disc system is not conserved?

yes apologies

 

[simphys]

Yes, the friction between the two wheels will produce a torque about the axis on each wheel, and the forces are internal to the system so the resulting torques are internal. But what other forces are acting on each wheel?

It's gravity, isn't it? That the one that creates the torque. If I were to take the 'rotation axis' in f.e. disk2, the gravitational force of disk 1 would still exert an angular impulse

 

[simphys]

The force of friction on the one wheel produces a torque about that wheel's axis, yes.
The force of friction on the other wheel produces a torque about the other wheel's axis, yes.

But if we are talking about the angular momentum for a system or net torque upon a system, we need to pick one reference axis for the system. Not two. Torques about different axes are incommensurable. You cannot expect to sum them together and obtain a meaningful result.

If one picks only one axis then the net torque from a pair of equal and opposite forces applied at the same point is equal to what, exactly?

That said, a search for other forces acting on the assembly is likely to be fruitful.

exactly that is what I didn't do.. should've seen that there still is gravity as well.

 

[haruspex]

It's gravity, isn't it? That the one that creates the torque. If I were to take the 'rotation axis' in f.e. disk2, the gravitational force of disk 1 would still exert an angular impulse

No, it's not to do with gravity.
What keeps these two discs in place? The friction between them would send them off in opposite directions unless there are other forces involved.
What point would you like to take moments about? You can eliminate the contribution from one of these external forces, but how are you going to eliminate both?

 

[simphys]

No, it's not to do with gravity.
What keeps these two discs in place? The friction between them would send them off in opposite directions unless there are other forces involved.
What point would you like to take moments about? You can eliminate the contribution from one of these external forces, but how are you going to eliminate both?

okay right.. it's the support forces then, correct?

 

[haruspex]

okay right.. it's the support forces then, correct?

Yes.

 

[simphys]

Yes.

should've examined it more properly basically.. the problem is not reallly difficult if I would've thought a bit simpler about it. i.e. from the basic principles up.

 

[phinds]

@simphys I clearly had a totally different take on this problem. I though the issue was that there would be initial slippage between the disks as they came together and that this would cause heat, thus converting some of the system energy to heat energy.

Clearly, I'm the only one who looked at it that way, so hard as it is for me to believe I must have had the wrong approach and was leading you down a blind alley, thus my comment in post #7 was uncalled for and I apologize.

 

[wrobel]

Why doesn't one simply write the angular momentum equation for each body and solve them?

 

[Chestermiller]

I haven't seen a single equation yet in this thread. Let F be the tangential friction force between the disks. What is the angular acceleration equation for each disk.

 

[jbriggs444]

I haven't seen a single equation yet in this thread. Let F be the tangential friction force between the disks. What is the angular acceleration equation for each disk.

To be fair, parts a), b) and c) of the quoted question are qualitative. No equations required.

I do like the approach of starting by finding the two angular accelerations. Doing so will reveal a next step.

 

[kuruman]

This is an extension of the problem in which a disk spinning freely in space with angular velocity $\omega_0$ comes in contact with a rough horizontal surface, starts rolling with slipping while the linear velocity of its center $V$ is less than the linear speed at its rim $\omega R$. When the two become equal, the disk starts rolling without slipping. The extension is that instead of the fixed horizontal surface, here we have another disk. However, here too, there is initial slipping until each cylinder rolls without slipping on the other.

In the initial problem, angular momentum about a point on the surface is conserved because friction exerts zero torque about any point on the surface. All it does is convert the initial spin angular momentum about the center of the disk to angular momentum about the center of the disk plus (orbital) angular momentum of the center of mass of the disk about the point of contact, $$I~\omega_0=I~\Omega +M~V_{cm}~R.$$ With $\Omega=V_{cm}/R,$ one can find the final speed $V_{cm}$ in terms of $\omega_0$.

Why is all this relevant here? Because here the axles about which the disks spin are held fixed which means that the disks are not allowed to orbit about each other as they would if they were free. Does this sound like angular momentum conservation?

To solve the problem one must follow the suggestion of @Chestermiller in post #22 and solve the rotational kinematics problem. One disk slows down and the other speeds up until their linear velocities at their rims are equal, $\omega_1R_1=\omega_2R_2$.

I will say no more because this is a "live" problem.

 

[bob012345]

To be fair, parts a), b) and c) of the quoted question are qualitative. No equations required.

I do like the approach of starting by finding the two angular accelerations. Doing so will reveal a next step.

Parts a) where it says imagine one axle in each hand and b) where it asks if you have to apply a torque when you bring them together basically sets up how to think about the problem and leads to the right approach. Full disclosure, I totally missed that myself when I attempted the problem.....

 

[Chestermiller]

To be fair, parts a), b) and c) of the quoted question are qualitative. No equations required.

I do like the approach of starting by finding the two angular accelerations. Doing so will reveal a next step.

Just because the questions are qualitative does not mean that you can't use equations to assist you in answering them. In this case, modeling the problem is very helpful.

 

[wrobel]

To be fair, parts a), b) and c) of the quoted question are qualitative

equations can give qualitative results as well

 

[haruspex]

Let F be the tangential friction force between the disks. What is the angular acceleration equation for each disk.

You can finesse that a bit by thinking in terms of impulsive moment, the time integral of the frictional torque. That avoids worrying about whether the forces are constant.

 

[Chestermiller]

You can finesse that a bit by thinking in terms of impulsive moment, the time integral of the frictional torque. That avoids worrying about whether the forces are constant.

I had realized that too. Thanks.

 

[kuruman]

You can finesse that a bit by thinking in terms of impulsive moment, the time integral of the frictional torque. That avoids worrying about whether the forces are constant.

That avoids going through the kinematics equations as does the shortcut of having disks with meshing gears.

For this method to work one needs the ratio of the moments of inertia. We are told that disk #2 "is in all respects identical to #1, except that its radius is $R_2$." I think it is safe to assume that the disks have the same density and not the same mass.