Conservation of spin, energy and momentum

[k4ff3]

I have some problems understanding when a system is isolated or not. I know the definition:

An isolated system is a system which is free from the influence of a net external force which alters the momentum of the system.

Well, my confiusion is better shown by an example.

Consider an Merry-go-round (MGR):

A person runs towards the MGR while the MGR stand still. Call this situation 1). He the jumps on it. This obviously makes the MGR rotate. Call this situation 2). Define that for the spin, the center of rotation is in the middle of the MGR.

Im confused regarding what properties is conserved. Angular momentum, energy and (linear momentum). If you don't care about my ramblings, you can stop read now ;p

I would think that momentum (linear, p=mv) is not conserved. In 1) the person running makes the system have a p>0, but in 2) the system will only have angular momentum, i.e p=0. Is this correct?

I would say that energy is not conserved due to the fact that it is friction that makes the person stick to the MGR. But I am not sure about this. I can think of situations where one would not need friction to stick to the MGR, for example if one run into a wall placed on the MGR perpedicular to it. This makes me confused.

And then you have angular momentum (spin). The person has angular momentum around the axis of rotation just before he hits the MGR in 1). in 2) the MGR and person both have Angular momentum. I would say that the spin is conserved, because there is no external net torque acting on the person-MGR system. However, you have that friction again.. One can perhaps look at the friction as a torque.

 

[Doc Al]

I would think that momentum (linear, p=mv) is not conserved. In 1) the person running makes the system have a p>0, but in 2) the system will only have angular momentum, i.e p=0. Is this correct?

Linear momentum of the system is not conserved. The MGR is fixed by its axis, which exerts an external force on it.

I would say that energy is not conserved due to the fact that it is friction that makes the person stick to the MGR. But I am not sure about this. I can think of situations where one would not need friction to stick to the MGR, for example if one run into a wall placed on the MGR perpedicular to it. This makes me confused.

However the person 'sticks' to the MGR, the person essentially makes an inelastic collision with the MGR. Mechanical energy is not conserved. (The same would be true if the person just ran and jumped onto a wagon.)

And then you have angular momentum (spin). The person has angular momentum around the axis of rotation just before he hits the MGR in 1). in 2) the MGR and person both have Angular momentum. I would say that the spin is conserved, because there is no external net torque acting on the person-MGR system. However, you have that friction again.. One can perhaps look at the friction as a torque.

There are no external torques acting on the system. (Presumably, the axle of the MGR is frictionless and exerts no torque.) Whatever forces the MGR and person exert on each other (whether friction or other), those forces produce no net torque on the system. Angular momentum is conserved.

 

[k4ff3]

Thanks for the quick reply!
Can you explain to me why energy is not conserved in inelastic collisions?
And why can you neglect the force acting from the axis on the system in the angular momentum case, but not in the linear momentum case?

Do I got it right:
-Isolated system for angular momentum <=> no external net *torque* action on the system
-Isolated system for linear momentum <=> no external net *force* acting on the system

Why is the force from the axis considered external anyway? It is a part of the carousel, and thus a part of the system, and thus internal, not external.

 

[Doc Al]

Can you explain to me why energy is not conserved in inelastic collisions?

Whenever things collide and stick together, mechanical energy is 'lost'. You can prove this for yourself by setting up a few simple examples of colliding objects and applying conservation of momentum to find the final speed after the collision.

And why can you neglect the force acting from the axis on the system in the angular momentum case, but not in the linear momentum case?

The reason is just as you state next:

Do I got it right:
-Isolated system for angular momentum <=> no external net *torque* action on the system
-Isolated system for linear momentum <=> no external net *force* acting on the system

Right!

Why is the force from the axis considered external anyway? It is a part of the carousel, and thus a part of the system, and thus internal, not external.

It's perfectly OK to consider the axle as part of the system if you like, but then realize that something must be exerting a force (an external force) to hold the axle in place.

 

[k4ff3]

What if I consider the axis AND the ground that holds it AND the Earth that carries the ground as a part of the system? Will there still be an external force acting on the system?

 

[Doc Al]

What if I consider the axis AND the ground that holds it AND the Earth that carries the ground as a part of the system? Will there still be an external force acting on the system?

No. You can always expand your system so that any force is then part of the system. Of course, that's not always a helpful thing to do.

Considering the entire Earth and all its parts as a system, then sure, linear momentum of that system is conserved when the person jumps onto the MGR.

 

[k4ff3]

Interesting... Yet confusing. =)

Thank you for your help, Doc!

 

[k4ff3]

Btw, what happens to the "excess" energy after an inelastic collision?
Edit:
I know the basic answer probably is "heat, deforming the colliding bodies and vibrational waves (sound)". However, why is there is no heat loss in elastic collisions? And would you hear a sound when you are standing close to an elastic collision? If yes, then energy has been lost.. I can't understand why there should be more heat loss in an inelastic collision than in an elastic one..

 

[Doc Al]

Btw, what happens to the "excess" energy after an inelastic collision?
Edit:
I know the basic answer probably is "heat, deforming the colliding bodies and vibrational waves (sound)". However, why is there is no heat loss in elastic collisions? And would you hear a sound when you are standing close to an elastic collision? If yes, then energy has been lost.. I can't understand why there should be more heat loss in an inelastic collision than in an elastic one..

It's a matter of degree. All real collisions of macroscopic bodies lose energy to some degree. But some collisions are approximately elastic to a high degree--for example, a collision between steel balls may well be close to elastic. (But you're right, you'll still lose a bit of energy.) On the other hand, a collision between two lumps of clay that stick together is far from elastic.

What makes the difference is the ability of the bodies to store and release elastic energy. (Thus the name elastic collision.) Steel is highly elastic--it springs back after you deform it; clay is not--it just gets mushed and stays that way.

 

[rkmurtyp]

If you are running in a st. line that does not passthrough the center of MGR, yourself and MGR together (system) has an angular momentum. When you jump on to the MGR and come to a dead stop, MGR and yourself together rotate with an angular momentum equal to that before you jumped. Angular momentum is conserved. No friction is assumed in this case. The kinetic energy is conserved. If friction is assumed to be present, heat is produced to the extent of loss of kinetic energy.

If the line along which you run passes through the center of MGR, then when you jump on to MGR, you will simply slide along the surface and land on the ground . If you come to a stop on the ground, KE is transformed into heat. If you don't come to a stop and keep moving, KE is lost to the extent that your speed decreases. If your speed does not decrease, both linear momentum, angular momentum and KE are conserved. If your speed decreases, the linear momentum of yourself and the Earth together is conserved, so is the angular momentum wrt MGR.

 

[Doc Al]

If you are running in a st. line that does not passthrough the center of MGR, yourself and MGR together (system) has an angular momentum. When you jump on to the MGR and come to a dead stop, MGR and yourself together rotate with an angular momentum equal to that before you jumped. Angular momentum is conserved. No friction is assumed in this case. The kinetic energy is conserved.

Kinetic energy is not conserved. (You and the MGR exert force on each other, else how can you end up rotating together.)

 

[rkmurtyp]

In kinematics, you don't invoke the concept of force at all. You rotate together because of the angular momentum.

 

[Delta2]

Isnt it possible to construct a device or make an experiment that converts linear momentum to angular momentum or vice versa?
I guess the answer is no, since conservation of linear and angular momentum hold together. It is somehow confusing because conservation of momentum holds but not as total (angular+linear) but only for each kind separately.

 

[Doc Al]

In kinematics, you don't invoke the concept of force at all. You rotate together because of the angular momentum.

Using conservation of angular momentum alone you can show that kinetic energy is not conserved. (Nonetheless, you and the MGR do exert forces on each other!)

 

[rkmurtyp]

Isnt it possible to construct a device or make an experiment that converts linear momentum to angular momentum or vice versa?
I guess the answer is no, since conservation of linear and angular momentum hold together. It is somehow confusing because conservation of momentum holds but not as total (angular+linear) but only for each kind separately.

You are right. Linear momentum and angular momentum are not inter convertible.
Since you know that each of them is converted individually, it goes without saying that they are conserved together too!

 

[rkmurtyp]

Using conservation of angular momentum alone you can show that kinetic energy is not conserved. (Nonetheless, you and the MGR do exert forces on each other!)

I don't understand exactly what you are saying. But, I can perhaps say something more if you can show (as you say) that KE is not conserved, from conservation of angular momentum alone.

 

[Doc Al]

I guess the answer is no, since conservation of linear and angular momentum hold together. It is somehow confusing because conservation of momentum holds but not as total (angular+linear) but only for each kind separately.

Linear and angular momentum are conserved (or not) independently. They are different physical quantities with different units, so adding them in a single expression is not physically meaningful.

 

[Doc Al]

I don't understand exactly what you are saying. But, I can perhaps say something more if you can show (as you say) that KE is not conserved, from conservation of angular momentum alone.

Here you go:

m ≡ mass of person
p ≡ initial momentum of person
I ≡ rotational inertia of MGR

Assume that the person is moving tangentially to the MGR.

Conservation of angular momentum:
Rp = (I + mR²)ω

Initial KE: ½p²/m

Final KE: ½(I + mR²)ω² = ½R²p²/(I + mR²)

As long as I > 0, the Final KE will be less than the Initial KE.

 

[Delta2]

Linear and angular momentum are conserved (or not) independently. They are different physical quantities with different units, so adding them in a single expression is not physically meaningful.

Yes you are right about this, however conceptually they are both momentums and one could argue mathematically that P_{total}=P_{angular}+P_{linear}=C holds but only because P_{angular}=C_1, P_{linear}=C_2 hold.

 

[Doc Al]

Yes you are right about this, however conceptually they are both momentums and one could argue mathematically that P_{total}=P_{angular}+P_{linear}=C holds

I'd say that was a meaningless expression. What would be the units of P_total? How do you add quantities with different dimensions?

 

[Delta2]

I'd say that was a meaningless expression. What would be the units of P_total? How do you add quantities with different dimensions?

You add them as real numbers

Btw, on your other post u treat the inertia of the person as that of a point particle having mass m, doesn't it depends on the inertia of the person if KE_{final}&lt;KE_{initial} (for example if the inertia of the person is taken as \frac{1}{2}mR^2 it doesn't seem to hold for every I&gt;0

 

[Doc Al]

You add them as real numbers

As long as you're willing to ignore units, you can make all sorts of silly expressions.

Btw, on your other post u treat the inertia of the person as that of a point particle having mass m, doesn't it depends on the inertia of the person if KE_{final}&lt;KE_{initial} (for example if the inertia of the person is taken as \frac{1}{2}mR^2 it doesn't seem to hold for every I&gt;0

I do treat the person as a point particle with mass m. The rotational inertia of the person is mR², assuming they jump onto the edge of the MGR.

If the person jumps onto the MGR at some other point, the rotational inertia will be less. But the final KE will still be less than the initial for any perfectly inelastic collision.

 

[Delta2]

As long as you're willing to ignore units, you can make all sorts of silly expressions.

I don't think they are silly, at least not mathematically. But anyway.

I do treat the person as a point particle with mass m. The rotational inertia of the person is mR², assuming they jump onto the edge of the MGR.

If the person jumps onto the MGR at some other point, the rotational inertia will be less. But the final KE will still be less than the initial for any perfectly inelastic collision.

I guess it only makes the math more complex to consider different rotational inertia for the person. Since the collision is inelastic the conclusion holds, still it is kind of surprising that without making any assumptions about friction forces or other internal forces, not even a reference to them, just from conservation of momentum it follows that the final KE will be reduced.

 

[rkmurtyp]

You are quite right. We cannot and should not add linear and angular momenta for the simple reason, as you said, that they have different units.

More interesting is your derivation. I stand corrected; the person and MGR do not rotate together after collision, as I stated earlier.

Assume masses of the person and MGR to be equal. In this case, the person comes to a stop and the MGR starts rotating with an angular velocity that gives angular momentum equal to the initial angular momentum of the person, thus satisfying the conservation of angular momentum. Because we assumed no friction, the person keeps slipping at the same location where he landed on MGR, as MGR keeps rotating. This is similar to 1-D elastic collision, in linear motion.

When the masses are unequal, and MGR has greater mass, the person transfers a portion of his momentum to MGR and himself reverses his direction of motion and move with a slower speed. MGR rotates with angular momentum received from the person. When the mass of MGR is so high that the mass of the person is negligible in comparison, the person simply reverses his velocity, i.e. retraces his path with same speed as he had before jumping; MGR remains still - does not rotate.

The essential point is that the sign of the relative angular momentum changes its sign after collision.

If, on the other hand, the mass of MGR is lower, then the person continues his motion after the collision, along the same line but with a lower value of speed. MGR rotates with an angular momentum that has same sign as the angular momentum of the person, and a speed that gives the reversal of angular velocity. This satisfies the conservation of angular momentum.

If you represent angular momentum vector by an arrow, then those vectors corresponding to the person and MGR can be manipulated just as in the case of elastic collisions in linear motion.

Reversal of relative angular velocity due to elastic collision is the key to the solution of the problem, just as in the case of elastic collisions in linear motion.

 

[k4ff3]

When the masses are comparable, the axis of rotation exerts a force on the MGR under the collision, so this is not similar to 1-D linear elastic collision, as you state.

But when the mass of the MGR is so big that the mass of the "person" bouncing off the MGR again is negligible, then the axis of rotation does not exert a force on the MGR, and thus the collision is similar to 1-D linear elastic collision.

.. I think :)

 

[Doc Al]

More interesting is your derivation. I stand corrected; the person and MGR do not rotate together after collision, as I stated earlier.

Assume masses of the person and MGR to be equal. In this case, the person comes to a stop and the MGR starts rotating with an angular velocity that gives angular momentum equal to the initial angular momentum of the person, thus satisfying the conservation of angular momentum. Because we assumed no friction, the person keeps slipping at the same location where he landed on MGR, as MGR keeps rotating. This is similar to 1-D elastic collision, in linear motion.

You're confusing this with the 1-D linear elastic collision, in which a mass colliding elastically with an equal one at rest will end up with the first one coming rest and the other moving with the original speed. Not so here.

 

[rkmurtyp]

When the masses are comparable, the axis of rotation exerts a force on the MGR under the collision, so this is not similar to 1-D linear elastic collision, as you state.

But when the mass of the MGR is so big that the mass of the "person" bouncing off the MGR again is negligible, then the axis of rotation does not exert a force on the MGR, and thus the collision is similar to 1-D linear elastic collision.

.. I think :)

For your first sentence: If you like to understand only through the concept of force, perhaps the following may help.
When the axis exerts a force on MGR, MGR reacts by exerting an equal and opposite force on the axis. Since the motion of axis does not change, and MGR being movable frictionlessly, it starts rotating.
Your second sentence is not understandable. Are you saying that the person continues his motion with the same velocity after the collision OR are you saying that he bounces off with the same speed in the opposite direction, ie. with reversed velocity?

 

[rkmurtyp]

You're confusing this with the 1-D linear elastic collision, in which a mass colliding elastically with an equal one at rest will end up with the first one coming rest and the other moving with the original speed. Not so here.

I think I left something important unstated earlier; and that is: Imagine the person is running along a straight line as in your problem.The person has an angular momentum. Now also assume that MGR is rotating with an equal and opposite momentum. The total angular momentum of the system (The person and the MGR) is zero. This is the situation before collision. When the person jumps on to the MGR, he will be thrown back to move with his velocity reversed and the MGR starts rotating with its angular velocity reversed. The total momentum is zero again. Symmetry is satisfied.
In reply to your comment I may say, you may try to use an arrow to represent angular velocity of the person before collision. This vector also represents the relative velocity before collision. Then, after collision, you will have two arrows, corresponding to the velocities of the person and the MGR. They are parallel to the same straight line as the original arrow was. Reverse the direction of the arrow that corresponds to the angular velocity of the MGR. Add this to the angular velocity arrow of the person. The resulting arrow represents the relative velocity after collision. This arrow will have the same length but opposite direction to the original arrow that represented the angular velocity of the person before collision. Since all arrows are parallel to one straight line the vectors are algebraic quantities, in this case. The momenta are obtained by multiplying the lengths of the arrows by the respective masses of the object (either the person or MGR as the case may be).
When we use arrows for vectors, we'll find that angular velocities as well as linear velocities will both be arrows and become indistinguishable.

 

[k4ff3]

Your second sentence is not understandable. Are you saying that the person continues his motion with the same velocity after the collision OR are you saying that he bounces off with the same speed in the opposite direction, ie. with reversed velocity?

The absolute value of the velocity is the same, but in the opposite direction.

 

[Doc Al]

I think I left something important unstated earlier; and that is: Imagine the person is running along a straight line as in your problem.The person has an angular momentum. Now also assume that MGR is rotating with an equal and opposite momentum. The total angular momentum of the system (The person and the MGR) is zero. This is the situation before collision. When the person jumps on to the MGR, he will be thrown back to move with his velocity reversed and the MGR starts rotating with its angular velocity reversed. The total momentum is zero again. Symmetry is satisfied.

OK. That's certainly possible for an elastic collision. It's not the scenario we've been discussing, but OK.

In reply to your comment I may say, you may try to use an arrow to represent angular velocity of the person before collision.

Yes, you can represent the person's angular momentum and angular velocity by vectors.

This vector also represents the relative velocity before collision.

I don't understand. Relative to what?

Then, after collision, you will have two arrows, corresponding to the velocities of the person and the MGR. They are parallel to the same straight line as the original arrow was. Reverse the direction of the arrow that corresponds to the angular velocity of the MGR. Add this to the angular velocity arrow of the person. The resulting arrow represents the relative velocity after collision. This arrow will have the same length but opposite direction to the original arrow that represented the angular velocity of the person before collision.

Seems like you're trying to say that angular velocities can be represented by vectors. OK.

Since all arrows are parallel to one straight line the vectors are algebraic quantities, in this case. The momenta are obtained by multiplying the lengths of the arrows by the respective masses of the object (either the person or MGR as the case may be).

To find the angular momenta, multiply the angular velocity by the rotational inertia, not the mass.

When we use arrows for vectors, we'll find that angular velocities as well as linear velocities will both be arrows and become indistinguishable.

Huh??

 

[rkmurtyp]

OK. That's certainly possible for an elastic collision. It's not the scenario we've been discussing, but OK.


Yes, you can represent the person's angular momentum and angular velocity by vectors.

I don't understand. Relative to what?

Seems like you're trying to say that angular velocities can be represented by vectors. OK.


To find the angular momenta, multiply the angular velocity by the rotational inertia, not the mass.

Huh??

There was a slip. It should read as relative angular velocity and not relative velocity. Relative angular velocity = (angular velocity of the person - angular velocity of MGR)
Relative angular velocity is the angular velocity of the person relative to the angular velocity of MGR.

I stand corrected. You are right. To find the angular momenta, we must multiply the angular velocity by the rotational inertia, not the mass.