Rotational kinetic energy explaination

[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.

 

[Thomas2]

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.

You shouldn't forget that if the central object has a finite mass, the kinetic energy of the weight is not constant anymore but increases/decreases as the wire gets longer/shorter because the central object (e.g. the rocket) loses/gains rotational energy through the pull. It is then dependent on the initial velocity of the weight whether the wire can wind up fully or not. If the initial velocity is not high enough, the kinetic energy of the weight will be used up too quickly and the weight will corotate with the central object before it has fully wound up.
Why don't you run your numerical program simply the other way around by simulating the 'rocket de-spin' problem? Just assume that initially the weight is located at the surface of the central object and corotating with it, and then calculate its velocity as the wire gradually unwinds. This gives you then the velocity needed to reversely fully wrap the wire around the object from a given starting radius. One could also determine this analytically from the energy and momentum conservation equations by solving these for the velocity of the weight at a given radius (assuming again that the weight corotates at the surface of the central object).

 

[Thomas2]

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.

What you said in your post #35 is in fact incorrect. The angular momentum of the ball does not increase but decreases as it moves inwards (the kinetic energy i.e. the velocity v stays constant and hence the angular momentum m*v*r must decrease as r decreases). The difference of angular momentum is being taken up by the pole/earth. It has nothing to do with setting the ball in motion but with the constant non-radial pull of the ball on the pole via the rope. If the pole could rotate it would actually gain rotational kinetic energy through this (and the ball would then lose energy in this case as mentioned in my post just above), but if the pole is fixed to the earth, its mass is effectively infinite and the ball loses only angular momentum but not kinetic energy (this is analogous to elastically bouncing a ball off a wall; the velocity of the ball reverses i.e. the ball loses momentum (which is taken up by the wall) but it does not lose any kinetic energy (unless the wall can move and its mass is not much higher than the mass of the ball; this all follows from the laws of energy and momentum conservation for elastic collisions or more generally speaking from Newton's laws).

 

[pervect]

If you take the limit of an infinitely thin string and pole, the radius of the ball's path never changes, it moves in a circular orbit. This conserves both kinetic energy and angular momentum, but it isn't representative of the original problem where the radius of the ball's path decreases as a function of time as the string "winds up".

With a finite thickness for the pole or the string, the radius of the ball's path does change, it does "wind up", but the force exerted by the string on the ball is not quite central. This implies that angular momentum is not conserved (but to a reasonable degree of approximation, kinetic energy is).

Someone even wrote a Lagrangian down for the problem in the first go-around, coming to the same conclusion:

https://www.physicsforums.com/showpost.php?p=489270&postcount=41

(though it would be a lot more readable if they had used latex).

 

[fantispug]

It's about 30 months too late but...
I'm not convinced energy is conserved for the ball/tether system. We've got a big round post, there's a string coming from it to the ball. The string will be at a tangent to the circle, and hence not radial. The ball is moving around the post and inwards. There are components of force in each of these directions.
Unless someone can give me a good reason the work from these two components must exactly cancel (they will have opposite signs) energy need not be conserved.

In this case the Lagrangian formalism breaks down (d'Alembert's principle of virtual work is not satisfied).

Edit: Never mind me. The string is always at a tangent to the pole and the ball always moves at a tangent to the string (although the latter is not so obvious to me, I now strongly believe it is true). Thus no work is done in a virtual displacement, Lagrange is happy and the world is safe.
As stated many times before me: angular momentum is not conserved, but energy is. The ball thus stays at constant velocity throughout the whole motion as would be expected.

 

[TVP45]

Repost??

I tried to post earlier and apparently failed. If this is indeed my 2nd answer, I apologize.

The question is quite an interesting one with a simple answer.

Assumptions: Heat loss is negligible. The pole is not free to counter-rotate. The laws of thermodynamics still hold.

If initially a mass m, on a tether, has a tangential velocity v, it has linear kinetic energy 0.5 mv^2. If the mass is released at this time, it will fly off on a tangent with all that energy. None will remain behind.

If no work is added or done, the energy of the mass remains the same.

If, at a tighter radius, the mass is then released, it will fly off on a tangent with all the energy. None will remain behind.

That energy has not been dissipated nor supplemented. Therefore, it remains the same as originally. Thus, the tangential velocity has remained constant.

Angular momentum has not been conserved.

If a measurement seemed to show that the linear velocity has increased, that is an error of measurement which is common with small radii. It is easier to measure Omega and calculate v.

Tom

 

[sushruth]

I am very late...but i think that it is pretty simple...angular momentum is conserved in the tetherball system...there is no net torque...
mVR=mvr.
V = vr/R
say at R = r/2
=> V = 2v.
I think the problem u say is in the conservation of energy.
Initial Kinetic energy =Ki= (1/2)m*v^2
Final Kinetic energy = Kf=(1/2)m*V^2= (1/2)m*4v^2=4*(Ki).
Whoa...how did that happen?...
U are forgeting the fact that there is a centripetal force caused by the pulling rope acting on the ball..this force happened to pop up simply because of the velocity we initially applied...
There is some work being done on the ball when it is pulled inward by this force.
We know centripetal force Fc = mV^2/R.
Now V is a function of R as V = m (v^2 r^2/R^3)=(m*(vr)^2/R^3)
Work done by this centripetal force= W = ∫Fc.dR (from r to r/2)
so we will get V = m(vr)^2*∫R^-3dR=-(m(vr)^2)*(1/2(R^2))
applying the limits we get W = (1/2)[(m v^2 r^2 /r^2) - (m v^2 r^2/(r/2)^2)
= -3 ((1/2)*m*v^2) = -3*Ki..
Wow...and it perfectly fits...!
Ki = Kf + Work done by the centripetal force...
So as u see both angular momentum and energy are conserved...You simply were not considering the complete energy equation...The energy conservation is for the net mechanical energy of the system and not just the kinetic energy...
The same caculations would be true even if u did it with angular velocities and Kinetic energies...I just did the simplest way of calculating it...

 

[Michael C]

I am very late...but i think that it is pretty simple...angular momentum is conserved in the tetherball system...there is no net torque...

No, your analysis can't be correct. There is no energy input after the ball has been given the first impulse, so we cannot end with more energy than we started with.

The question has already been correctly answered: the rope does no work on the ball, since there is no motion in the direction of the length of the rope (contrary to what would happen if somebody was pulling the rope inwards through a hole). The ball's KE, and therefore its speed, stays constant. The ball's angular momentum decreases, but the angular momentum of the system ball + pole + Earth stays the same.

If the ball continues spiralling round the pole until it crashes into it. If the ball were a point mass, it would be moving directly towards the centre of the pole at the moment it crashes into it, so at the end of the spiral it has zero angular momentum.

 

[modulus]

Okay, I'm not sure if what I'm about to say here has already been said in this thread, 'cause I kinda skipped through the middle two pages...

I think I've finally found out the answer to the original question asked, about where the energy came from which pulls the volleyball inwards, and the corresponding decrease in energy of the agent.
See, the answer lies in the fact that pole has thickness, and the string which is attached to the volleyball is attached to the pole in a kind of 'ring' around it. When the volleyball is set into motion, the ring rotates around the pole, but the pole offers a resistance to it's motion - through the friction between the string and the pole surface, and this induces an increased tension in the string, which travels all the way up to the point where it is attached to the volleyball.
The increased tension pulls the volleyball inwards, and increases it kinetic energy, and the corresponding decrease in energy is the heat lost to the environment due to the friction acting between the pole and the ring of the string around it.

I think this should answer your original question, Lee...I hope I helped! :)

 

[Doc Al]

Okay, I'm not sure if what I'm about to say here has already been said in this thread, 'cause I kinda skipped through the middle two pages...

I think I've finally found out the answer to the original question asked, about where the energy came from which pulls the volleyball inwards, and the corresponding decrease in energy of the agent.
See, the answer lies in the fact that pole has thickness, and the string which is attached to the volleyball is attached to the pole in a kind of 'ring' around it. When the volleyball is set into motion, the ring rotates around the pole, but the pole offers a resistance to it's motion - through the friction between the string and the pole surface, and this induces an increased tension in the string, which travels all the way up to the point where it is attached to the volleyball.
The increased tension pulls the volleyball inwards, and increases it kinetic energy, and the corresponding decrease in energy is the heat lost to the environment due to the friction acting between the pole and the ring of the string around it.

I think this should answer your original question, Lee...I hope I helped! :)

Please take the time to read through the entire thread before responding. Besides being incorrect--as explained in the thread--your answer is 7 years too late.

 

[sushruth]

No, your analysis can't be correct. There is no energy input after the ball has been given the first impulse, so we cannot end with more energy than we started with.

The question has already been correctly answered: the rope does no work on the ball, since there is no motion in the direction of the length of the rope (contrary to what would happen if somebody was pulling the rope inwards through a hole). The ball's KE, and therefore its speed, stays constant. The ball's angular momentum decreases, but the angular momentum of the system ball + pole + Earth stays the same.

If the ball continues spiralling round the pole until it crashes into it. If the ball were a point mass, it would be moving directly towards the centre of the pole at the moment it crashes into it, so at the end of the spiral it has zero angular momentum.

I have proven that the energy has not increased...ofcourse the sting does work because the centripetal force is acting inwards and the ball is pulled inwards...how does that mean that the string does no work?...
centripetal force does no work in rotating the ball...but since the ball is being pulled in the inward pull is caused by this centripetal force..so it does work...