Rotational kinetic energy explaination

[leebenjamin@adelphia]

I'm afraid I need some help on this one. We are studying rotational dynamics and I read in my textbook, the explanation of a common classroom deonstration. A person sits on a rotating stool and holds heavy weights in their outstretched hands. Someone sets the individual into rotational motion and when they pull their hands in, the speed of rotation increases. This is easily understood with the conservation of angular momentum. We are asked to imagine that we actually perform this demonstration and that pulling our arms in results in halving our total moment of inertia. This obviously doubles our angular speed according to the law of conservation of angular momentum. If we consider the rotational kinetic energy, we quickly see that we have doubled the energy associated with that rotation. Again, not a difficult concept because we have done work on the weights (The source of energy being food that we ate earlier that day, or perhaps even days ago.) But here's the problem I have. Suppose we do a DIFFERENT experiment. Suppose we tether a volleyball on a pole. Now we strike the volleyball perpendicularly to the string that holds it on the pole. We have given the ball a certain amount of energy and it begins to rotate at a given angular speed. The string wraps around the pole and in a period of time, the radius will have decreased just enough to make the moment of inertia half of its orignal value just as in the previous demonstration. When that happens, again the angular speed will double and the kinetic energy will also double. But this time, NO ADDITIONAL ENERGY WAS ADDED AND THE POLE DIDN'T HAVE LUNCH TODAY... OR YESTERDAY EITHER! How do we reconcile the fact that the ball seems to have gained energy with no additional energy input here? I really need help here. Thanks, in advance, for whatever you can supply.

 

[ZapperZ]

I'm afraid I need some help on this one. We are studying rotational dynamics and I read in my textbook, the explanation of a common classroom deonstration. A person sits on a rotating stool and holds heavy weights in their outstretched hands. Someone sets the individual into rotational motion and when they pull their hands in, the speed of rotation increases. This is easily understood with the conservation of angular momentum. We are asked to imagine that we actually perform this demonstration and that pulling our arms in results in halving our total moment of inertia. This obviously doubles our angular speed according to the law of conservation of angular momentum. If we consider the rotational kinetic energy, we quickly see that we have doubled the energy associated with that rotation. Again, not a difficult concept because we have done work on the weights (The source of energy being food that we ate earlier that day, or perhaps even days ago.) But here's the problem I have. Suppose we do a DIFFERENT experiment. Suppose we tether a volleyball on a pole. Now we strike the volleyball perpendicularly to the string that holds it on the pole. We have given the ball a certain amount of energy and it begins to rotate at a given angular speed. The string wraps around the pole and in a period of time, the radius will have decreased just enough to make the moment of inertia half of its orignal value just as in the previous demonstration. When that happens, again the angular speed will double and the kinetic energy will also double. But this time, NO ADDITIONAL ENERGY WAS ADDED AND THE POLE DIDN'T HAVE LUNCH TODAY... OR YESTERDAY EITHER! How do we reconcile the fact that the ball seems to have gained energy with no additional energy input here? I really need help here. Thanks, in advance, for whatever you can supply.

First of all, let me just say that I WISH, when I was teaching, that all of my students are like you. It is a JOY to teach students who are are thinking like this, and stumble upon things that don't quite jell with what they understand.

Back to you question. The thing with both moving your arm inwards and with the string coiling in is that there is a force within the system that is doing work. When the centripetal force is applied to move the object either inwards or outwards, work is either done onto the rotational system, or done by the rotational system (remember that work requires a force AND a net displacement of the object in question).

In the case of a person pulling in the weights, the centripetal force is provided by the arm/hand, while the rope+pole provide this force in the other example. This "external" force (external to the rotation motion) does change the angular momentum because it is always perpendicular to the motion (orthorgonal to the direction of the angular momentum vector), and therefore adds no net torque to the system to affect angular momentum conservation.

So moral of the story: the source of the force can be anything. As long as they act the same way, the result will be the same.

Zz.

 

[leebenjamin@adelphia]

Mr. Zapper...

I think you misinterpreted my question. I understand fully that work is done by the tension in the string. And that the work done is Force times the distance. My question is, WHAT IS THE SOURCE OF THE ENERGY GIVEN TO THE VOLLEYBALL? I only gave it so much energy and the pole or the string don't have any food (or fuel of any kind) to provide more energy, yet at a later point in time, the ball now has twice the energy it did after I hit it. Where did this extra ENERGY come from?

Lee

 

[ZapperZ]

Mr. Zapper...

I think you misinterpreted my question. I understand fully that work is done by the tension in the string. And that the work done is Force times the distance. My question is, WHAT IS THE SOURCE OF THE ENERGY GIVEN TO THE VOLLEYBALL? I only gave it so much energy and the pole or the string don't have any food (or fuel of any kind) to provide more energy, yet at a later point in time, the ball now has twice the energy it did after I hit it. Where did this extra ENERGY come from?

Lee

The energy came from the tension in the string - the EM force that's holding the molecules of the string together. This energy is external to the rotational energy.

Zz.

 

[leebenjamin@adelphia]

But I don't think that works. This string applied its force without outside influence. In every case that I can think of, some source of energy needs to be responsible for causing a rope to exert tension. For example, an electric or a gas motor can cause a tension in a rope which can then lift a load or give it kinetic energy. I am not comfortable with a rope spontaneously exerting energy and furthermore, even if it did, wouldn't it have to decrease in its own energy? I, for example, use up some of the energy from the food I ate, when I exert the force on the masses. And if I give the masses kinetic energy using a motor, the motor has to use up fuel or utilize electric energy which results in a battery having less energy than before, etc., etc. I don't see the reduction in the rope's energy like these other examples. I still fail to see the source of the energy.

Can you be more specific?

Thanks!
Lee

 

[ZapperZ]

But I don't think that works. This string applied its force without outside influence. In every case that I can think of, some source of energy needs to be responsible for causing a rope to exert tension. For example, an electric or a gas motor can cause a tension in a rope which can then lift a load or give it kinetic energy. I am not comfortable with a rope spontaneously exerting energy and furthermore, even if it did, wouldn't it have to decrease in its own energy? I, for example, use up some of the energy from the food I ate, when I exert the force on the masses. And if I give the masses kinetic energy using a motor, the motor has to use up fuel or utilize electric energy which results in a battery having less energy than before, etc., etc. I don't see the reduction in the rope's energy like these other examples. I still fail to see the source of the energy.

Can you be more specific?

Thanks!
Lee

OK, let's go back to the human example. Forget about the details for the moment. All you care about is that SOMETHING is causing the radius of rotation to become smaller, ya? I don't really have to tell you what is causing this. All I need to tell you is that the radius got smaller without any application of an external torque on the rotational system.

When we account for both L and KE, we see that L is conserved, but KE is not. How do we account for the non-conservation of KE? We say that, well, there is a radial force from SOMEWHERE that is doing work on the system (the relevant system here is a mass at in a rotational motion). Work (and energy) is transferred between the origin of this force and the rotational system, when this radial force does something that causes a radial displacement of that object.

Now so far, I still have not invoked any need for explaining the origin of this force. All I have done is explain this purely via mechanics. At this point, mechanics stops here and the explanation (at least as far as mechanics is concerned) is complete. Up to this point, I can easily use the same explanation for the ball+string system.

But since we are trying to fully account of what comes from where, we then invoke biological explanation for the human pulling the weight in - we say the muscles burn some energy that he ate, and this allows him to use this energy to pull the object in, etc... You want an explanation similar to this for the ball+string system. OK, let's try this:

As the string is wrapping around the pole, it's radius is of rotation is getting smaller. It means that the rope is PULLING THE ball inwards radially. What is doing the work?

Recall that when you put a charged particle in an electrostatic field, and the charged particle moves due to the electrostatic force, the field is doing the work on to the particle. The energy of the field is transferred into the translational energy of the particle. I can invoke the same thing with the string pulling the ball inwards. The immovable pole (that is attached to the "immovable" earth) transfers the EM force to the string, which then transfers the force to pull the ball in. That's why I said that it is the EM field that's doing the work. One can easily argue that it is the whole Earth that's providing the support for the pole+string+ball, so it is the Earth that's doing the work. I would have no argument with that. It depends on how far up the chain of events you want to pursue.

Zz.

 

[Doc Al]

I think the tetherball problem is a bit subtle. It is different from the example of a person pulling in the weights, which you seem to understand. In that example, there is a force doing work on the weights, so you can see that mechanical energy is not conserved. And you recognize that the energy must come from the person's internal energy (food!). Also, there is no external torque on the system, so angular momentum is conserved.

So what's different with the tetherball winding around the pole? For one thing: the rope is not pulling in the ball (although it is pulling on the ball), it is merely winding around the pole. I believe the tension in the rope does zero work on the ball and thus there is no need for an energy source: Mechanical energy is conserved.

Note: The tension in the rope exerts a torque on the ball. (The pole has thickness.) Angular momentum is not conserved!

(If I'm wrong, Zapper, please correct me. )

 

[ZapperZ]

I think the tetherball problem is a bit subtle. It is different from the example of a person pulling in the weights, which you seem to understand. In that example, there is a force doing work on the weights, so you can see that mechanical energy is not conserved. And you recognize that the energy must come from the person's internal energy (food!). Also, there is no external torque on the system, so angular momentum is conserved.

So what's different with the tetherball winding around the pole? For one thing: the rope is not pulling in the ball (although it is pulling on the ball), it is merely winding around the pole. I believe the tension in the rope does zero work on the ball and thus there is no need for an energy source: Mechanical energy is conserved.

Note: The tension in the rope exerts a torque on the ball. (The pole has thickness.) Angular momentum is not conserved!

(If I'm wrong, Zapper, please correct me. )

But mechanical energy cannot be conserved if L is conserved. This is because L has a linear dependence on angular velocity, while KE has a quadratic dependence on angular velocity. They cannot be satisfied simultaneously unless w=0 or 1.

There is an analogous experiment to this, which is a rope attached to a ball that is rotating on a frictionless table, but the other end of the rope passes through a small hole in the table and is attached to a weight. Here, the gravitational force is doing work by pulling onto the weight, and this is transferred to the tension in the string to pull in the ball. I see no difference between these two other than the origin/source of the work done.

Zz.

 

[leebenjamin@adelphia]

Well, okay, but in the case of the person pulling the weights inward, the internal energy of the person is DECREASED by an amount equal to the increase of KE for the mases at the ends of his arms. In the case of the volleyball, where is the corresponding DECREASE of energy of the pole/string system that would account for the increase of energy of the ball?

Lee

 

[ZapperZ]

Well, okay, but in the case of the person pulling the weights inward, the internal energy of the person is DECREASED by an amount equal to the increase of KE for the mases at the ends of his arms. In the case of the volleyball, where is the corresponding DECREASE of energy of the pole/string system that would account for the increase of energy of the ball?

Lee

That's difficult to account for. I mean, when you drop a mass in a gravitational field, and the field is doing work and transfering energy to the mass, where is the corresponding reduction in energy? We can do this ad nauseum and consider the source of the gravitatioinal field and say that it is also being pulled in (miniscule) etc. etc. But at some point this no longer becomes practical.

Zz.

 

[leebenjamin@adelphia]

To Dr. Al...

If there is no need for a mechanical energy source, then how do we account for the fact that at a later point in time, the kinetic energy of the volleyball is TWICE as much as it was at the beginning of the problem? That cannot happen without an energy source. (Or if it can, everything I learned about conservation of energy was false.)

Lee

 

[Doc Al]

But mechanical energy cannot be conserved if L is conserved. This is because L has a linear dependence on angular velocity, while KE has a quadratic dependence on angular velocity. They cannot be satisfied simultaneously unless w=0 or 1.

Right. Angular momentum is not conserved.

There is an analogous experiment to this, which is a rope attached to a ball that is rotating on a frictionless table, but the other end of the rope passes through a small hole in the table and is attached to a weight. Here, the gravitational force is doing work by pulling onto the weight, and this is transferred to the tension in the string to pull in the ball. I see no difference between these two other than the origin/source of the work done.

Here the rope pulls in the ball, performing work on the ball. (That energy comes from the change in gravitational energy of the descending weight, of course.) In this case the force is radial, so it exerts no torque on the ball; Angular momentum is conserved.

 

[Doc Al]

If there is no need for a mechanical energy source, then how do we account for the fact that at a later point in time, the kinetic energy of the volleyball is TWICE as much as it was at the beginning of the problem? That cannot happen without an energy source. (Or if it can, everything I learned about conservation of energy was false.)

What makes you say that the kinetic energy is twice as much at a later point in time? (Neglect any change in height of the ball.) You are assuming conservation of angular momentum. I don't believe that's true in this case: the rope exerts a torque on the ball.

 

[ZapperZ]

Right. Angular momentum is not conserved.


Here the rope pulls in the ball, performing work on the ball. (That energy comes from the change in gravitational energy of the descending weight, of course.) In this case the force is radial, so it exerts no torque on the ball; Angular momentum is conserved.

But where is the torque in this set up? Ideally, I can make the string as thin, and the pole as thin as I want to (after all, we neglect the string's mass). So I don't see how you can insert an angular component to the force from just the string.

Zz.

 

[Doc Al]

But where is the torque in this set up? Ideally, I can make the string as thin, and the pole as thin as I want to (after all, we neglect the string's mass). So I don't see how you can insert an angular component to the force from just the string.

The problem assumes the rope winds around the pole. You cannot neglect the thickness of the pole, else the problem disappears. The rope is not aligned in the radial direction; it acts at an angle, thus exerting a torque on the ball (with respect to the center of the pole).

 

[ZapperZ]

The problem assumes the rope winds around the pole. You cannot neglect the thickness of the pole, else the problem disappears. The rope is not aligned in the radial direction; it acts at an angle, thus exerting a torque on the ball (with respect to the center of the pole).

Yes, but this is enough to truly violate conservation of angular momentum, and produce a conservation of energy? I also cannot make the "hole in the table" case arbitrarily small and frictionless, yet even rudimentary intro physics lab on this can produce quite a good agreement with conservation of angular momentum principle, which implies that this conservation is actually quite robust.

Note that due to the quadratic dependence on w, to be able to conserve KE, L has to be severely violated no matter how large (or small) we make the pole to be. So if I give the same initial condition to "mass attached to a string+pole" and "mass on frictionless table pulled by a vertically hanging mass", I should see severe difference in the angular velocity between the two if L is violated by KE isn't. I find this difficult to accept especially since I can make the rope VERY long when compared to the diameter of the pole.

Zz.

 

[leebenjamin@adelphia]

In response to Dr. Al's answer:

The reason why the kinetic energy is twice as great is this: The moment of inertia becomes one half of the original while (according to the law of conservation of angular momentum) the angular velocity doubles. Now KE (rotational) is 1/2 I (omega)^2 So since I becomes 1/2 as great and omega becomes twice as great, omega squared becomes 4 times as great and 4 times 1/2 = 2. Kinetic energy doubles.

All of these responses are not giving me what I need. The example in the book said that the reason why the kinetic energy could double like this was that the person spent energy from food previously eaten to increase the KE. I understand that. And the person's energy would certainly decrease by an amount equal to the increase in kinetic energy (That's easy to see and understand.) We could draw an analogy by cranking the masses in with an electric motor instead of a person. Clearly, the amount of energy in the battery would be reduced by an amount equal to the increase in kinetic energy of the masses. I do not see such a decrease in the energy of anything related to the problem if we do the volleyball thing. It seems as if we are CREATING energy here and I don't think that's possible. I haven't heard anything so far, however, that says it isn't so.

lee

 

[Doc Al]

I'll have to give this one some more thought to give you a good answer. But here's something for you to ponder: If you think the energy is increasing, where's it coming from? (This was the original question.) I think we can both agree that the pole isn't moving--so it can't be doing any work on the rope. And the rope itself doesn't do work (unless it contracts); it merely transfers force.

 

[Doc Al]

In response to Dr. Al's answer:

The reason why the kinetic energy is twice as great is this: The moment of inertia becomes one half of the original while (according to the law of conservation of angular momentum) the angular velocity doubles. Now KE (rotational) is 1/2 I (omega)^2 So since I becomes 1/2 as great and omega becomes twice as great, omega squared becomes 4 times as great and 4 times 1/2 = 2. Kinetic energy doubles.

My response to this remains the same: You are assuming conservation of angular momentum. I say it's not conserved in this case, because of the torque exerted by the rope.

All of these responses are not giving me what I need. The example in the book said that the reason why the kinetic energy could double like this was that the person spent energy from food previously eaten to increase the KE. I understand that. And the person's energy would certainly decrease by an amount equal to the increase in kinetic energy (That's easy to see and understand.) We could draw an analogy by cranking the masses in with an electric motor instead of a person. Clearly, the amount of energy in the battery would be reduced by an amount equal to the increase in kinetic energy of the masses. I do not see such a decrease in the energy of anything related to the problem if we do the volleyball thing. It seems as if we are CREATING energy here and I don't think that's possible. I haven't heard anything so far, however, that says it isn't so.

I agree with you! If the energy increases, then it must have a source. In my understanding of the problem, the energy does not increase.

I think this is an excellent and subtle problem.

 

[leebenjamin@adelphia]

Mr. Zapper...

You said:

But mechanical energy cannot be conserved if L is conserved. This is because L has a linear dependence on angular velocity, while KE has a quadratic dependence on angular velocity. They cannot be satisfied simultaneously unless w=0 or 1.

There is an analogous experiment to this, which is a rope attached to a ball that is rotating on a frictionless table, but the other end of the rope passes through a small hole in the table and is attached to a weight. Here, the gravitational force is doing work by pulling onto the weight, and this is transferred to the tension in the string to pull in the ball. I see no difference between these two other than the origin/source of the work done.

As I see it, there is a HUGE difference between these two experiments and it underlies the problem I have been asking. If you do the one with the mass hanging through a hole in the table, the increase in kinetic energy of the spinning mass is EXACTLY balanced by a decrease in gravitational potential energy of the falling mass (at the center of the table) It is just that kind of a reduction in energy that I am looking for. It was the origin of my original question.

Lee

 

[leebenjamin@adelphia]

To Dr. Al:

You said:

My response to this remains the same: You are assuming conservation of angular momentum. I say it's not conserved in this case, because of the torque exerted by the rope.


But if you apply the law of conservation of angular momentum to the situation and if you use high speed photography, the numbers bear out very nicely that angular momentum is conserved. And as Zapper said in an earlier message, suppose we think of the diameter of the rod in millimeters and the length of the string in METERS. Then clearly, making the approximation that this is circular motion and that the tension in the string is providing a centripetal force + a force to do the work to increase the kinetic energy of the spinning volleyball doesn't seem like it should be much of a stretch of the imagination. I have never questioned the conservation of momentum. I am sure that it is true. The problem still remains that I cannot find a source of energy for the increase in KE of the volleyball.

Lee

 

[Doc Al]

But if you apply the law of conservation of angular momentum to the situation and if you use high speed photography, the numbers bear out very nicely that angular momentum is conserved.

Well, now we're getting somewhere! An experiment will settle things. Do you have a reference for this experiment?

 

[leebenjamin@adelphia]

No. This is just an experiment you can do with a video camera and a volleyball on a pole. And you don't really have to DO it anyway! Just do it as a thought experiment like Einstein would have done. Use a VERY thin pole and strike the volleyball perpendicularly to the outstretched string. The ball now has energy mv^2/2.

The string continues to wrap around the pole until its length is such that the moment of inertia is half of what it was at the beginning. Since the pole is SOOO thin, the force the tension applies is radially inward and the tension does work on the ball bringing it closer to the center of the circle. The result is that while momentum is conserved (The angular velocity is now twice as great as it was, while the moment of inertia is half of its original value) the kinetic energy (since it depends on the SQUARE of omega) is now twice its original value. Well work had to be done, so that had to cost some energy. But where did that energy come from? What energy has been reduced as a result of the kinetic energy of the ball going up? That's the essence of my problem, but I don't seem to be getting any answers here. I REALLY hope that you can help!

Thanks!
Lee

 

[PBRMEASAP]

Wow, this is a neat problem! I wish I had thought of stuff like this when I was taking physics. I had trouble keeping up with the homework, so I didn't have much free time to ponder.

This thread is going so fast that I can't read all of every post, so forgive me if I'm repeating someone.

Here's my two cents:

I agree with Doc Al that no work is being done on the ball by the string or the pole. I also agree that imagining an infinitely thin pole doesn't help much, because the string would not wrap around it. Instead, imagine a very fat pole. At all times (after the initial punch given to the ball) the string is tangent to the pole, and the ball is accelerating toward that instantaneous point of tangency. But it is moving at right angles to that direction of acceleration, so no work can be done (edit: by the string).

Therefore, I would say that the angular velocity changes in such a way that energy is conserved. Also, as Doc Al pointed out, the string exerts a torque on the ball (about the center of the fat pole), so angular momentum of the ball need not be conserved. In fact, if you look at the whole system of the earth, the pole, and the ball, I believe that some angular momentum is transferred to the earth.

 

[krab]

I think Doc Al is right. If you put the assembly of ball, rope, pole on an air table, (actually you'll need 2 balls, one on either side, to avoid wobble), then as the balls wind up, the assembly will begin to turn as well. The string does no work. In a tether-ball situation, the ball does not speed up; it has 1/2mv^2 energy at the start and has the same 1/2mv^2 at the end just before it is all wound up and crashes into the pole. It seems to go faster at the end, but in fact is only going a faster angular speed, not a faster linear speed.

ZZ and lee are arguing in the limit of zero pole thickness. In that limit, there is no torque of the ball on the pole, but OTOH, neither does the string wind up.

edit: Looks like PBRMEASAP and I gave substatially the same answer simultaneously.

 

[ZapperZ]

I think Doc Al is right. If you put the assembly of ball, rope, pole on an air table, (actually you'll need 2 balls, one on either side, to avoid wobble), then as the balls wind up, the assembly will begin to turn as well. The string does no work. In a tether-ball situation, the ball does not speed up; it has 1/2mv^2 energy at the start and has the same 1/2mv^2 at the end just before it is all wound up and crashes into the pole. It seems to go faster at the end, but in fact is only going a faster angular speed, not a faster linear speed.

ZZ and lee are arguing in the limit of zero pole thickness. In that limit, there is no torque of the ball on the pole, but OTOH, neither does the string wind up.

Actually, I don't have to have an infinitely thin pole. All I need is that the length of the string >> the diameter of the pole. So I could have the string at 2 m in length, while a pole that's 1 cm in diamter. With such geometry, it is very hard for me to be convinced that there's any significant angular component to the tension in the string.

It is also puzzling to me that no work is thought of being done, especially when the mass is moving in radially. A centripetal force alone will simply cause a uniform circular motion. It requires an additional force to pull the ball in. This additional force translates to an additional tension in the string beyond just having it move in a circular motion. Again, relating this to a similar observation, this is what most science museum has where a ball rolls on the outer edged of a "funnel" and people see it moving faster and faster it it reaches the center of the funnel. Here, instead of a string pulling the ball in, it is the side of the funnel that is pushing the ball in with the help of gravity. Here, it should be even more complicated than the ball+string+pole, since the contact point here is not on the vertical side of the ball, but rather at an angle below the horizontal. Yet, all of these demonstrations claim to show conservation of angular momentum.

Zz.

 

[Andrew Mason]

No. This is just an experiment you can do with a video camera and a volleyball on a pole. And you don't really have to DO it anyway! Just do it as a thought experiment like Einstein would have done. Use a VERY thin pole and strike the volleyball perpendicularly to the outstretched string. The ball now has energy mv^2/2.

The string continues to wrap around the pole until its length is such that the moment of inertia is half of what it was at the beginning. Since the pole is SOOO thin, the force the tension applies is radially inward and the tension does work on the ball bringing it closer to the center of the circle. The result is that while momentum is conserved (The angular velocity is now twice as great as it was, while the moment of inertia is half of its original value) the kinetic energy (since it depends on the SQUARE of omega) is now twice its original value. Well work had to be done, so that had to cost some energy. But where did that energy come from?

I have just looked at the posts in this thread. It is an interesting issue, but I think we have to be careful, as Doc Al points out, when we assume things. We know that energy is conserved. We also know that angular momentum is conserved in any isolated system. So, if this is an isolated system then both E and L must be conserved.

But if it is an isolated system and the ball winds around the pole AND conserves angular momentum, it has to speed up at a rate that is greater than its energy would permit. So that tells us there is something wrong with our assumptions. The only thing that can be wrong is our assumption that this is an isolated system.

And, of course, one can see that is right. The pole is connected to the earth. The torque on the pole caused by the winding of the rope, is transferred to the earth. So, if you include the earth, angular momentum is conserved without the ball having to speed up (since the Earth is so massive). Angular momentum and energy are both conserved. The universe and the laws of physics are saved.

AM

 

[ZapperZ]

I have just looked at the posts in this thread. It is an interesting issue, but I think we have to be careful, as Doc Al points out, when we assume things. We know that energy is conserved. We also know that angular momentum is conserved in any isolated system. So, if this is an isolated system then both E and L must be conserved.

No, I disagree. We need to know what FORM of energy is conserved. When you change the moment of inertia via changing the radius of the object, without any external force, this is analogous to an explosion where two particles go off in opposite direction - the linear momentum may be conserved, but KE isn't, typical inelasic collision.

The "system" here is the rotational motion of a mass. An "external" force can come from within the object in which the energy accounting did not include in the beginning. This is because the rotational KE only includes I, the moment of inertia of the body, and w, the angular velocity. It did not include the energy from the string, the pole, the muscles, the organs, etc., only the rotational motion. So when a radial force is applied to the system, this does not affect L, since such radial forces are perpendicular to w.

However, the rotational KE (which is the energy in question) doesn't care if it is perpendicular, parallel, at an angle, etc. If such energy is being transferred either from or to the system (system still object m in a rotational motion), then it will change the KE, and there's no reason to expect the rotational KE is conserved.

Now the issue here is two separate things:

1. Is the fact that the string is wrapping around a post contributes a significant angular component of the force so much so that it causes L to not be conserved, but rotational KE will? I do not see this. The fact that I can make the radius of motion large compared to the radius of the pole means that even if L is not entirely conserved, it will only be weakly unconserved. But it is certainly not enough to cause w to increase THAT quickly to be able to save rotational KE (which depends on the square of w). If we have learned anything from linear motion, it is that momentum conservation is more ROBUST than KE conservation.

2. If KE isn't conserved, where did it go? Here, the question then is what force is doing the work of causing dr/dt of the system to be non-zero? It is why I brought up the hole-in-the-table example. The gravitational pull on the hanging mass is doing the work, which is transferred to the EM interaction in the string that is pulling in the ball, and this, in principle, changes the gravitational energy of the earth. But how is this different than the pole? Instead of gravitational interaction, replaces the Earth having EM interaction with the pole, and it having EM interaction with the string that is pulling in the ball. I see no difference here, only different flavor of mechanism.

Zz.

 

[Andrew Mason]

The "system" here is the rotational motion of a mass. An "external" force can come from within the object in which the energy accounting did not include in the beginning. This is because the rotational KE only includes I, the moment of inertia of the body, and w, the angular velocity. It did not include the energy from the string, the pole, the muscles, the organs, etc., only the rotational motion. So when a radial force is applied to the system, this does not affect L, since such radial forces are perpendicular to w.

When I referred to conservation of energy, I simply meant that the energy of the system was constant - no energy was being added. In the problem as posed, there is no source of additional energy.

I agree that when a radial force is applied, angular momentum is conserved. This is simply because the rate of change of angular momentum is torque. Since torque is 0 with a perfectly radial force (\vec \tau = \vec F \times \vec r = Frsin(0) = 0), angular momentum is constant.

But if there is no torque (ie. only central force), the rope/ball cannot wrap itself around the pole. If the ball wraps itself around the pole, the pole must exert a torque that is equal and opposite to the torque is applied to the pole by the rope. The torque on the pole is: \vec \tau = \vec T \times \vec r where \vec T is the tension force in a direction from the ball to the point of contact on the pole surface and \vec r is the radius vector from the point of contact on the pole to the centre of the pole. These vectors are at 90 degrees to each other so \tau = Trsin(90) = Tr.

If the pole remains fixed (no angular acceleration) so the rope wraps around it, then an equal and opposite torque must be applied to the pole. This means there is an external torque applied to the pole which means that angular momentum of the ball/rope/pole is not conserved (I\dot \omega \ne 0) if the ball/rope wraps around the pole.

AM

 

[ZapperZ]

But if there is no torque (ie. only central force), the rope/ball cannot wrap itself around the pole. If the ball wraps itself around the pole, the pole must exert a torque that is equal and opposite to the torque is applied to the pole by the rope. The torque on the pole is: \vec \tau = \vec T \times \vec r where \vec T is the tension force in a direction from the ball to the point of contact on the pole surface and \vec r is the radius vector from the point of contact on the pole to the centre of the pole. These vectors are at 90 degrees to each other so \tau = Trsin(90) = Tr.

If the pole remains fixed (no angular acceleration) so the rope wraps around it, then an equal and opposite torque must be applied to the pole. This means there is an external torque applied to the pole which means that angular momentum of the ball/rope/pole is not conserved (I\dot \omega \ne 0) if the ball/rope wraps around the pole.

AM

Again, I disagree. The "torque" on the system here only comes in because the string isn't exactly at the center of the central force motion. So the slight angular deviation produces an angular component of the force from the tension on the string. But that is why I said that I can easily make this so that the diameter of the pole is << than the radius of the rotation of the ball! The component of the tension along the the angular direction is practically miniscule! It is why I said if L is not conserved, it will only be so very weakly. In any case, rotational KE certainly would not be conserved either since w certainly cannot increase THAT significantly to preserve it.

Zz.

 

[Andrew Mason]

Again, I disagree. The "torque" on the system here only comes in because the string isn't exactly at the center of the central force motion. So the slight angular deviation produces an angular component of the force from the tension on the string. But that is why I said that I can easily make this so that the diameter of the pole is << than the radius of the rotation of the ball! The component of the tension along the the angular direction is practically miniscule!

You can't make the wrapping radius smaller than 1/2 of the rope's thickness, so the torque cannot be made arbitrarily small.

It is why I said if L is not conserved, it will only be so very weakly. In any case, rotational KE certainly would not be conserved either since w certainly cannot increase THAT significantly to preserve it.

So are you saying that the ball would lose KE? If so, where does it go?

AM

 

[ZapperZ]

You can't make the wrapping radius smaller than 1/2 of the rope's thickness, so the torque cannot be made arbitrarily small.

Come again?

All I said was that I can make the diameter of the pole << the radius that the ball is rotating. I didn't realize I also have to specify the rope's thickness, because I didn't realize this also needs to come into play. Are we making this arbitrarily difficult to also include the rope's mass?

And I also said that I can make it so that the angular component of the tension to not be THAT significant to severely change L. Thus, I said this non-conserving factor is weak, and chances are, if you try to measure it, you can't get it accurate enough to detect it. It is why we can do many of these conservation of angular momentum experiment in simple elementary labs, despite of friction, non-ideal conditions, etc.

So are you saying that the ball would lose KE? If so, where does it go?

AM

Er.. I have explained this several times!

"2. If KE isn't conserved, where did it go? Here, the question then is what force is doing the work of causing dr/dt of the system to be non-zero? It is why I brought up the hole-in-the-table example. The gravitational pull on the hanging mass is doing the work, which is transferred to the EM interaction in the string that is pulling in the ball, and this, in principle, changes the gravitational energy of the earth. But how is this different than the pole? Instead of gravitational interaction, replaces the Earth having EM interaction with the pole, and it having EM interaction with the string that is pulling in the ball. I see no difference here, only different flavor of mechanism."

It is always the work done by the the "structure", be it human muscles, EM forces in the string connected to the pole that is connected to the earth, etc. etc. I believe, if you looked through the beginning of this thread, I have mentioned this a few times already.

Zz.

 

[PBRMEASAP]

1. Is the fact that the string is wrapping around a post contributes a significant angular component of the force so much so that it causes L to not be conserved, but rotational KE will? I do not see this.

Yep, you are right. That would not make sense and I completely overlooked that.


So there was something obviously wrong with my first post. Since angular momentum is transferred to the Earth via the pole, then (duh ) so is energy. The entire system of earth, pole, and ball must be considered if you want to apply either conservation of energy or momentum. For the ball by itself, neither is conserved.

The ball does work on the earth+pole by exerting a torque on it through a nonzero angle. Of course, you can also look at that as the earth+pole doing work on the ball, with oposite sign. But since the ball's rotation is slowing down and the earth+pole's is speeding up, the tension in the string steadily decreases until the ball smacks into the pole. So the work being done doesn't last forever--there's no perpetual motion or anything weird like that.

This holds true even if the tether is much longer than the radius of the pole. All that means is that it will take a lot longer for the transfer of energy/momentum to be complete.

I am going to think about some more for the case that krab suggested, where there are two balls diametrically opposed to balance out the wobble. Maybe I can get some more insight that way.

Props to the OP for coming up with an awesome question.

 

[Andrew Mason]

And I also said that I can make it so that the angular component of the tension to not be THAT significant to severely change L.

I would agree with you if you meant 'severely change L' in a given time interval \Delta t where \Delta t is much less than the total time it takes to completely wrap around the pole. I don't think you can avoid a change in angular momentum by making the pole diameter smaller. All you do is make the torque smaller so it takes longer for the change in angular momentum to occur.

The change in angular momentum \Delta L = \tau \Delta t is always the same where \Delta t is the time to wrap around the pole. For a given ball speed, total wrapping time is inversely proportional to the diameter, d, of the pole (t \propto 1/d). Since \Delta L = \tau \Delta t = T\frac{d}{2}\Delta t \propto Td/d = T, the total change in momentum is a function only of tension T, not the pole diameter. It is just that if you make pole diameter smaller, it takes longer to achieve the change in momentum.

Thus, I said this non-conserving factor is weak, and chances are, if you try to measure it, you can't get it accurate enough to detect it.

You can if you take enough time - the time it takes to wrap the rope around the pole.

Er.. I have explained this several times!

Maybe. But Leebenjamin was talking about the ball speeding up - gaining kinetic energy. I thought the point of this exercise was to explain why the ball did not speed up. He points out that the ball has to speed up to avoid changing angular momentum. And the solution has to be: the ball can't speed up because there is no source of additional energy, so angular momentum has to change.

If the ball simply wraps around the pole, the ball/rope does almost no work because the pole does not turn (ok, the Earth might turn an infinitesimal angle). So there is virtually no way for the ball to lose or gain kinetic energy (\Delta KE = Work = \tau \Delta \theta) The ball maintains its tangental speed until it crashes into the pole.

AM

 

[gerben]

I think all the energy involved is provided by:

The source of energy being food that we ate earlier that day, or perhaps even days ago.

When you throw the ball you are moving the ball in one direction and the Earth in the opposite direction. The two masses that you have set in motion (the ball and the earth) are connected via a pole, and the force on the ball and the Earth are both exerted at a distance from the pole. So, relative to the axis of the pole you exert a torque on both the Earth and the ball. As the ball is moving inwards its angular momentum relative to the pole increases, but at the same time the angular momentum of the Earth relative to the pole decreases.

The angular momentum of the ball increases at the cost of the angular momentum of the earth, both were initially provided by muscle power.

 

[krab]

PBRMEASAP: The transfer of angular momentum is significant, but the transfer of energy to the Earth is negligible. To understand this, consider a car coming to a stop at the equator. The conservation of momentum means that mv of the car is transferred to the earth, where it is (2/5)M\Delta V (the 2/5 comes from the moment of inertia of the earth, a sphere), where M is the Earth's mass and V the equatorial speed. But this means the extra kinetic energy of the Earth is (1/2)I\omega^2, which is (1/5)M(\Delta V)^2=(5/2)(m/M)(mv^2/2). So of the energy (1/2)mv^2 lost by the car, only an incredibly tiny fraction (5/2)(m/M) goes to speeding up the earth. Essentially, all the lost energy is dissipated in the brakes, and car engineers need not worry about the Earth's change in speed when designing braking ststems. The upshot is that in problems like these, the momentum transfer to the Earth figures into conservation of momentum, but the energy transfer practically does not figure into the conservation of energy.

 

[PBRMEASAP]

PBRMEASAP: The transfer of angular momentum is significant, but the transfer of energy to the Earth is negligible.

Good point. As Andrew Mason pointed out, the extra angle that the Earth turns through as a result of the ball is very, very small. And the torque is on the order of a few lb-ft (or is it ft-lbs?). So it is true that the ball has lost very little kinetic energy, even neglecting sources of dissipation. I liked your car example, though. It really put the different magnitudes involved in perspective.

edit: I forgot to say, this means the part in my last post about the decreasing tension is not only not true, but also irrelevant (which simplifies things).

 

[krab]

In thinking the problem through, I've found a few variants that are helpful: Imagine the pole is attached top and bottom to something fixed, by means of frictionless bearings. IOW, the pole can spin about its axis.

1. Swing the ball around like a tether ball. The pole will spin so that the point of attachment always faces the ball. There is no winding up and in the absence of friction, the ball revolves around the pole centre forever. Not very interesting. Imagine you are holding the pole. The work you do is torque times angle and since the torque is zero, work is zero and the ball continues at constant speed.

2. Now hold the pole still. The string winds around the pole. But in this case also you do no work. There is torque, but the angle you allow the pole to spin is zero, so torque times angle is zero. This means the ball, even though it is winding up the string, maintains a constant speed. You can also look at it from the point of view of circular motion. The kinetic energy is (1/2)I\omega^2, and though angular speed \omega is increasing, I is decreasing and this exactly offsets it so the kinetic energy is nevertheless constant. The string length is R, and the pole radius is r. The distance from ball to centre of pole is \sqrt{r^2+R^2}, so the moment of inertia is M(r^2+R^2) where M is the mass of the ball (all other components being massless). But the angular speed is \omega=v/\sqrt{r^2+R^2}, so the kinetic energy is (1/2)I\omega^2=(1/2)Mv^2, which is constant.

3. Now turn the pole steadily so that the radius form the pole centre to the attachment point is always at right angles to the string. In this case, the torque is Fr=Mv^2(r/R). Turning the pole in this way through an angle theta is really doing work on the ball, so the ball steadily speeds up. But everyone is familiar with this situation: it's exactly what you do when you twirl something like a sling over your head, when you use a skipping rope, or the basic design of a weedeater, or whatever. These examples of course have friction, so they level off to a constant speed after initial acceleration.

 

[ZapperZ]

2. Now hold the pole still. The string winds around the pole. But in this case also you do no work. There is torque, but the angle you allow the pole to spin is zero, so torque times angle is zero. This means the ball, even though it is winding up the string, maintains a constant speed. You can also look at it from the point of view of circular motion. The kinetic energy is (1/2)I\omega^2, and though angular speed \omega is increasing, I is decreasing and this exactly offsets it so the kinetic energy is nevertheless constant. The string length is R, and the pole radius is r. The distance from ball to centre of pole is \sqrt{r^2+R^2}, so the moment of inertia is M(r^2+R^2) where M is the mass of the ball (all other components being massless). But the angular speed is \omega=v/\sqrt{r^2+R^2}, so the kinetic energy is (1/2)I\omega^2=(1/2)Mv^2, which is constant.

But krab, in all cases where you do have a conservation of L, there are no "work done angularly" either, which is what you are describing above (torque x angular displacement). Yet, in those cases, you still have L being conserved, while rotational KE is not. Unlike angular momentum, the energy being added or taken out of the rotational system need not be just non-orthorgonal. While you can impose force radially to the rotating system without affecting L, rotational KE is not immune to this. Thus, as soon as you have a dr/dt term not being zero, some thing somewhere is doing work and this changes the rotational KE of the system. So the added energy to the system is not just from the angular work done, but also radial work done.

The fact that dr/dt is not zero is a clear displacement due to some additional force beyond the centripetal force to maintain uniform circular motion. I don't see how this can be explained away as not changing the rotational KE of the system. I will also say that if I have a pole of let's say 10 cm in diameter, and I start this with a string that's 2 m and then remeasure everything with the radius has shrunk to 1 m, I would be hard pressed to be convinced that the finite diameter of the pole changes L THAT much that one can clearly detect this. Certainly the angular velocity would not have increase that much to preserve rotational KE considering its quadratic relationship.

Zz.

 

[PBRMEASAP]

I will also say that if I have a pole of let's say 10 cm in diameter, and I start this with a string that's 2 m and then remeasure everything with the radius has shrunk to 1 m, I would be hard pressed to be convinced that the finite diameter of the pole changes L THAT much that one can clearly detect this.

This sounds like a reasonable experiment. Anyone want to do it?

 

[reilly]

Difficult problem. Perfectly suited to a Lagrangian approach, mainly due to the constraints.

Let r be the distance from the center of the pole to the tethered ball, which has mass m. The length of the tether will be SQRT( r*r + a*a), where a is the radius of the pole. (The cord must be tangential to the pole, thus you get a right triangle with s the hypotenuse and sides of length a and r -- the ball is at the end of r.) The motion has both radial and angular components. So the KE is given by

(m/2){ (dr/dt)*(dr/dt) + r*r(d@/dt)*d@/dt)}, where @ denotes the angle swept out by the radius vector.

The key constraint concerns the time change in s, which is given by ds/dt=-a*d@/dt. You an convince yourself of this by drawing a picture of s at t1, and of s at t1+dt. The angular displacement will be (d@/dt)*dt*a -- the cord wraps around the circumference.Hence, the constraint.

Thus it's possible to rewrite the KE in terms of s and d@/dt, and eliminate r. But, the next step is to eliminate d@/dt with the constraint ds/dt=-a d@/dt, and the KE, Lagrangian, becomes

L = m/2a*a(M(s) ds/dt*ds/dt)

where M(s) = s2-a2 +s2 a2/(s2-a2) here s2=s squared, a2= a squared

So, the system is a conservative one -- no explicit time dependence, no odd powers of ds/dt. As it clearly must be, energy is conserved. Angular momentum is not conserved. The tension has a component along the r direction, and the @ direction, so there's a torque involved.

The equation of motion is a bear, and is identical to the equation stemming from
d KE/dt = 0.

The ball has both negative work and positive work done on it, and the two cancel out.

I'm quite certain that what I've written is correct. But, it's been a very long time (pushing 40 years) since I've done a Lagrangian with constraints problem, so I'd appreciate someone taking an independent crack at deriving the equations of motion.

Regards,
Reilly Atkinson

PS. I recall seeing this problem in a textbook, but what book I don't recall

 

[PBRMEASAP]

reilly:

I agree with your conclusions. From the picture though, it looks to me like r^2 = a^2 \ + \ s^2, so that the radius is the hypotenuse of the right triangle instead of the tether. But then that makes the constraint relation between \theta (angle swept out by radius) and s a little more tricky. One way around this is to let \phi be the angle swept out by the tangent, so that we have s(t) = s_{0} - a \phi (t), where s_0 is the original length of the tether.

Then the kinetic energy is just

T = \frac{1}{2} m s^2 (\frac{d \phi}{dt})^2
= \frac{1}{2} m (s_{0} - a\phi)^2 (\frac{d \phi}{dt})^2

The conclusions are the same--no explicit time dependence. I have forgotten though, what does an odd power of a velocity mean? Can that even happen? I could really use a refresher on mechanics, and it's only been a coupla years for me.

 

[PBRMEASAP]

so I'd appreciate someone taking an independent crack at deriving the equations of motion.

Equation of motion:

\frac {d}{dt} \frac {\partial T}{\partial \dot{\phi}} \ = \ \frac {\partial T} {\partial \phi}

m \frac {d}{dt} ((s_0 - a \phi)^2 \dot{\phi}) = -am(s_0 - a \phi) \dot{\phi}^2

-2am(s_0 - a \phi) \dot{\phi}^2 + m(s_0 - a \phi)^2 \ddot{\phi} = -am(s_0 - a \phi) \dot{\phi}^2

m(s_0 - a\phi)^2 \ddot{\phi} - am(s_0 - a \phi) \dot{\phi}^2 = 0

If you multiply through by \dot{\phi}, you get the same equation you would by requiring \frac {dT}{dt} = 0, just as in your derivation.

 

[reilly]

PBMEASAP -- You are right about r2=s2+a2 (See my post, #41 for notation). I should have noted that the limit as s->a should not be singular as it is in my first formulation.

Where I differ, after a bit of work, is in the connection between @ (theta, angular location of the radius) and phi, the anglular loation of the tangent point. If you draw the diagram, let phi be the angle from reference to the radius to the tangent point -- just as you've defined it. Then let beta be the angle from that radius to r, the radial vector to the mass, the "lower" angle of the triangle, s,r,a. Then I claim that

@ = phi + beta, and beta = arcsin(a/r) in which case

d@/dt = d(phi)/dt - a (dr/dt)/ { r SQRT(r2 - a2)}

Your idea to go with phi is a good one indeed. With s= s0 -a (phi) you get your expression for the KE, due to d@/dt. But there's also the 1/2m(dr/dt)*(dr/dt) term, which i find to be

KE1=
m/2 a*a (d phi/dt*d phi/dt)/( [s0-a(phi)]*[s0-a(phi)]/ [a2 +[s0-a(phi)]*[s0-a(phi)]

and KE2 = = \frac{1}{2} m (s_{0} - a\phi)^2 (\frac{d \phi}{dt})^2

and L = KE1 + KE2.


And thank you for checking my work, with appropriate math rather than with a bunch of words.

Regards,
Reilly Atkinson

 

[PBRMEASAP]

reilly:

Okay I went back and looked at the drawing again, and I agree with you except that I have beta = arccos(a/r) instead of arcsin(a/r). That simply introduces a sign difference between my d (beta)/dt and yours.

Here's what I arrived at from my drawing:

r^2 = a^2 + s^2 (as before), which means

\dot{r} = \frac{s}{\sqrt{a^2+s^2}} \dot{s}

We also have
\theta = \phi + \beta = \phi + \arctan{\frac{s}{a}}

\dot{\theta} = \dot{\phi} + \frac{a \dot{s}}{a^2+s^2}

Plug these values into the formula for kinetic energy:

T = \frac{1}{2} m(\dot{r}^2 + r^2 \dot{\theta}^2)

\ \ \ \ = \frac{1}{2} m\{ \frac{s^2 \dot{s}^2}{a^2 + s^2} + (a^2 + s^2)(\dot{\phi} + \frac{a \dot{s}}{a^2+s^2})^2 \}

now we apply the constraint ds/dt = -a d(phi)/dt:

T = \frac{1}{2} m \{\frac{a^2 s^2 \dot{\phi}^2}{a^2+s^2} + (a^2+s^2)(\frac{s^2 \dot{\phi}}{a^2+s^2})^2 \} = \frac{1}{2} m (\frac{a^2 s^2 \dot{\phi}^2}{a^2+s^2} + \frac{s^4 \dot{\phi}^2}{a^2+s^2}) = \frac{1}{2}ms^2 \dot{\phi}^2

so it seems to check out according to my drawing.

everyone:

What was the conclusion of the energy/momentum conservation debate? It should be noted that by writing down the lagrungian as L = KE, where KE does not depend on time explicitly, we have already assumed that energy is conserved. So that doesn't prove anything really.

 

[reilly]

PBRMEASAP -- I should go back and study geometry and trig. But,...

Your results look good to go.

Since there is no potential, and the system is isolated, energy must be conserved, as we have both shown.

Angular momentum, on the other hand, is not conserved. Effectively, the problem involves a non-central force, which means no rotational symmetry. (Think about an electron scattering from a charged cube, vs. an eletron scattering from a charged sphere.)

Regards,
Reilly

 

[Thomas2]

The kinetic energy is unchanged for the tetherball because the rope is freely winding itself up on the pole, i.e. the angular momentum of the ball must decrease as the radius decreases. This angular momentum is being taken up by the pole (which one won't notice if the latter is fixed to the earth).

The same principle is actually used in reverse to de-spin rockets and satellites. These are usually launched with a spin in order to stabilize them, but at the observation stage one does not necessarily want the spin any more. So in this case, basically two long wires with weights are wrapped around them and the latter then released and allowed to wind themselves off the rocket. This causes the angular momentum to be transferred from the rocket to the weights (the latter are then allowed to fly away after they have spun off).

Both the cases of the radial shortening of the rope (which leads to an increase of kinetic energy as work is being done against the centrifugal force) as well as the tetherball example are actually problems in the Berkeley Physics Course Vol.1 (Chpt. 6), although the full solution is not given for the latter case (but it is mentioned that the kinetic energy stays constant).

 

[PBRMEASAP]

reilly--

Since there is no potential, and the system is isolated, energy must be conserved, as we have both shown.

Yep, that is definitely true. I don't know if you could say that we've shown it (mathematically) though, since energy conservation was built into our formalism from the get go. What you and I did is exactly equivalent to obtaining the equation of motion by writing KE = constant and going from there. Not that there's anything wrong with that. I'm glad you brought up the Lagrangian approach though--it caused me to hunt down a textbook and look at it again.

Angular momentum, on the other hand, is not conserved. Effectively, the problem involves a non-central force, which means no rotational symmetry. (Think about an electron scattering from a charged cube, vs. an eletron scattering from a charged sphere.)

Yes! This, on the other hand, could be shown from the Lagrangian by writing it in terms of @. We know that the Lagrangian at least contains the kinetic energy term we came up with (and maybe some mysterious time-dependent field due to the string). This term explicitly contains @, which means its conjugate momentum (which is the angular momentum) is not constant. This is equivalent to what you said about lack of rotational symmetry.



Thomas2--

I didn't know they did that with spinning satellites/rockets. That is very cool.
Is that vol 1 of Berkeley Physics Course any good? I have "Waves" and "Electricity and Magnetism" (vols 2 and 3, i think). They are great.

-pbr

 

[krab]

The same principle is actually used in reverse to de-spin rockets and satellites. These are usually launched with a spin in order to stabilize them, but at the observation stage one does not necessarily want the spin any more. So in this case, basically two long wires with weights are wrapped around them and the latter then released and allowed to wind themselves off the rocket. This causes the angular momentum to be transferred from the rocket to the weights (the latter are then allowed to fly away after they have spun off).

Is that ever cool! It's like my thought experiment of post 25. I was actually trying to solve this case where the central device has a moment of inertia I, and then show how the behaviour changes with I. But the equations are not integrable, so I'd have to do it by a Runge Kutta technique, and the physics gets obscured. My starting condition was that the weights are winding up. But after a while, the central device (let's call it the rocket) will start turning in the same direction, thus stopping the winding up process. I now see that at some point, the rocket reaches a maximum spin speed and then will again slow down as the weights go off to infinity.

 

[gerben]

The kinetic energy is unchanged for the tetherball because the rope is freely winding itself up on the pole, i.e. the angular momentum of the ball must decrease as the radius decreases. This angular momentum is being taken up by the pole (which one won't notice if the latter is fixed to the earth).

This is what I said in post #35, the rotation of the Earth with reference to the pole (induced by throwing the ball) will be counterbalanced by the ball moving inwards.