# Rotational kinetic energy explaination

ZapperZ
Staff Emeritus
krab said:
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
Homework Helper
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

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ZapperZ
Staff Emeritus
Andrew Mason said:
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 transfered 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 transfered 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
Homework Helper
ZapperZ said:
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
Staff Emeritus
Andrew Mason said:
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
Homework Helper
ZapperZ said:
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
Staff Emeritus
Andrew Mason said:
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 transfered 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.

ZapperZ said:
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 :rofl: ) 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
Homework Helper
ZapperZ said:
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

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.

krab said:
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).

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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.

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ZapperZ
Staff Emeritus
krab said:
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.

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ZapperZ said:
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 text book, but what book I don't recall

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.

reilly said:
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

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

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).

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reilly--

reilly said:
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 in to 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 text book and look at it again.

reilly said:
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