Rotational kinetic energy explaination

[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...

 

[Michael C]

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...

First of all, work needs energy. If there was no energy input to the system after the initial given state (which is the case here), then no work can have been done.

Note that the force of the string is not directly towards the centre, it is at an angle to a line between the ball and the centre of the pole. The end of the string attached to the ball does indeed get closer to the centre of the pole, but not because of motion in the direction of the string, only because of motion perpendicular to the direction of the string. Since the ball only moves perpendicular to the force of the string, this force can do no work on the ball.

 

[AlchemistK]

We started with rotational mechanics not to long ago, and sure enough a similar problem arose in our class, though our teacher simply told us the reason, without letting us work our brains on it.

The thing is, even though the tension in the string is perpendicular, the hinge reaction force by the pole is not, and hence this "non-conservation" of energy, again bringing us to the same conclusion as the rest of the people here, the pole AND the Earth too have to be considered in the system.

 

[sushruth]

First of all, work needs energy. If there was no energy input to the system after the initial given state (which is the case here), then no work can have been done.

Note that the force of the string is not directly towards the centre, it is at an angle to a line between the ball and the centre of the pole. The end of the string attached to the ball does indeed get closer to the centre of the pole, but not because of motion in the direction of the string, only because of motion perpendicular to the direction of the string. Since the ball only moves perpendicular to the force of the string, this force can do no work on the ball.

Agreed...that is in causing rotational motion the centripetal force does no work...but now we are talking about a tether ball with its radius decreasing because of each rotation...the force which pulls this tether ball inwards is the centripetal force...there is this change in displacement in the direction of the decrease in radius...

 

[sushruth]

First of all, work needs energy. If there was no energy input to the system after the initial given state (which is the case here), then no work can have been done.

Note that the force of the string is not directly towards the centre, it is at an angle to a line between the ball and the centre of the pole. The end of the string attached to the ball does indeed get closer to the centre of the pole, but not because of motion in the direction of the string, only because of motion perpendicular to the direction of the string. Since the ball only moves perpendicular to the force of the string, this force can do no work on the ball.

The centripetal force in this case work in pulling the ball inwards...and if u want to know how the centripetal force came into existence it would be from the electrostatic attractions..

 

[Michael C]

We started with rotational mechanics not to long ago, and sure enough a similar problem arose in our class, though our teacher simply told us the reason, without letting us work our brains on it.

The thing is, even though the tension in the string is perpendicular, the hinge reaction force by the pole is not, and hence leads to this increase in energy, again bringing us to the same conclusion as the rest of the people here, the pole AND the Earth too have to be considered in the system.

There can be no increase in energy! Consider the whole system: ball, pole, Earth. We are using a frame of reference in which the Earth is stationary. By definition, the Earth has no KE in this frame. Nor does the pole. The ball cannot gain KE, since there is nowhere to gain it from.

When considering the whole system, it's clear that:
1. Angular momentum is conserved. The ball loses angular momentum, the Earth gains the same amount. Since the Earth is so massive, we don't notice its change of momentum.
2. Energy is conserved. The ball loses an immeasurably tiny amount of KE to the Earth. I'm not going to do the calculations: if somebody else wants to they're welcome! For all reasonable limits of accuracy, the ball's KE remains constant. This means that the speed of the ball (not its velocity!) remains constant.

Donald Simanek correctly analyses this problem http://www.lhup.edu/~dsimanek/scenario/insight.htm#skater, using the example of a skater spiralling around a pillar.

 

[sushruth]

The ball loses an immeasurably tiny amount of KE to the Earth. I'm not going to do the calculations: if somebody else wants to they're welcome! For all reasonable limits of accuracy, the ball's KE remains constant. This means that the speed of the ball (not its velocity!) remains constant.

Donald Simanek correctly analyses this problem http://www.lhup.edu/~dsimanek/scenario/insight.htm#skater, using the example of a skater spiralling around a pillar.

I am sorry but the kinetic energy does increase...but the mechanical energy of the net system is conserved..
yes..the calculation is simple...i have posted it above...if u do find any error in those calculations please do tell me..

 

[Michael C]

I am sorry but the kinetic energy does increase...but the mechanical energy of the net system is conserved..

How is the mechanical net energy conserved if the KE of the ball increases? At the start we have:
KE of Earth: zero
KE of pole: zero
KE of ball: 1/2 m v^2
That's all the mechanical energy we have. The string is massless and inelastic, so it has no KE and no stored energy. There is no energy input to this system, so no way to increase the KE of the ball.

yes..the calculation is simple...i have posted it above...if u do find any error in those calculations please do tell me..

The error is not mathematical, it's an incorrect assumption. In your calculations you make the assumption that the centripetal force does work on the ball. It doesn't, as already explained.

Your calculations would apply correctly to the following case:
Imagine that the string is coming out of a hole at the top of the pole and rotating around this fixed point. Here there is no torque and angular momentum is conserved. As long as the string stays the same length, the ball keeps rotating with the same angular velocity. If the string is pulled through the hole, making the rotating part shorter, the KE of the ball will increase. In this case, work is being done on the system: the work done by the force that is pulling the string through the hole is equal to the increase in KE of the ball.

 

[AlchemistK]

There can be no increase in energy!

Yes, I did realize that I had written it wrong, which is why I edited it later, a bit too late though.

 

[rcgldr]

The two dimensional versions of this problem have been discussed in other threads. Below is a link to one of them that includes images for the string pulled through a hole case and string wraps around a pole case. Short summary: In the hole case, during the time the string is pulled inwards or allowed to move outwards, part of the force is in the direction (path) of the spiraling ball, changing the ball's velocity and kinetic energy. In the pole case, the path is involute of circle, and the tension in the string is always perpendicular to the path of the ball, so it's velocity remains constant.

https://www.physicsforums.com/showthread.php?t=328121

 

[sushruth]

H
The error is not mathematical, it's an incorrect assumption. In your calculations you make the assumption that the centripetal force does work on the ball. It doesn't, as already explained.

Then what is the force pulling the ball inwards?...voodoo?...:|...there is an obvious decrease in the radius...this displacement according to ur theory is caused by no force?..initially there is no inward velocity...so all of a sudden it moves in?...sounds spooky...So much for Newton's 1st law...!..

Can u explain Why i can't use the same analogy for this problem ?

Also...according to u if angular momentum is conserved..
mVR=mvr...
intially v is the velocity we provide...
at R = r/2
V = 2v
So obviously kinetic energy for the ball 4 times the initial (1/2 m v^2 becomes 4 (1/2 m v^2))...
And there is no torque acting...ofcourse angular momentum conservation hold true..

P.S.. i am sorry if i sound too harsh...I am just in a bad mood...:/

 

[Michael C]

Then what is the force pulling the ball inwards?...voodoo?...:|...there is an obvious decrease in the radius...this displacement according to ur theory is caused by no force?..initially there is no inward velocity...so all of a sudden it moves in?...sounds spooky...So much for Newton's 1st law...!..

At all times, also initially, there is an inward velocity, because the centre of rotation of the string is not the centre of the pole. The centre of rotation of the string is always at a point on the circumference of the pole, so that the velocity of the end of the string with the ball always has a component towards the centre of the pole.

Also...according to u if angular momentum is conserved..
mVR=mvr...
intially v is the velocity we provide...
at R = r/2
V = 2v
So obviously kinetic energy for the ball 4 times the initial (1/2 m v^2 becomes 4 (1/2 m v^2))...
And there is no torque acting...ofcourse angular momentum conservation hold true..

P.S.. i am sorry if i sound too harsh...I am just in a bad mood...:/

once more:

1. Angular momentum of the system Earth + pole + ball is conserved. Ball transfers angular momentum to Earth: ball's angular momentum decreases.

2. Torque is acting since the centre of rotation of the string is not the centre of the pole.

3. KE of the ball cannot increase because there is no energy source external to the system.

Look at the link to http://www.lhup.edu/~dsimanek/scenario/insight.htm#skater that I already posted.

 

[sushruth]

At all times, also initially, there is an inward velocity, because the centre of rotation of the string is not the centre of the pole. The centre of rotation of the string is always at a point on the circumference of the pole, so that the velocity of the end of the string with the ball always has a component towards the centre of the pole.
once more:

1. Angular momentum of the system Earth + pole + ball is conserved. Ball transfers angular momentum to Earth: ball's angular momentum decreases.

2. Torque is acting since the centre of rotation of the string is not the centre of the pole.

3. KE of the ball cannot increase because there is no energy source external to the system.

Look at the link to http://www.lhup.edu/~dsimanek/scenario/insight.htm#skater that I already posted.

http://ocw.mit.edu/courses/physics/...of-angular-momentum/MIT8_01SC_quiz25_sols.pdf
check out the problem 5...

 

[Michael C]

http://ocw.mit.edu/courses/physics/...of-angular-momentum/MIT8_01SC_quiz25_sols.pdf
check out the problem 5...

Yes, that is the case where the string is pulled through a hole. That is not the same as the case presented here, where the string winds around a pole.

 

[Dadface]

Imagine the ball being set into circular motion but let there be two main differences to the original set up.

1.The initial circle can be horizontal or at any angle(including 90 degrees)to the horizontal.

2.The string does not wind itself around the support point so that its effective length remains constant.

When analysing the resulting motion we would take into account,amongst other things, the tension in the string and the weight of the ball.In terms of energy changes we would observe that there are changes between gravitational potential energy and kinetic energy.
It seems that in this thread gravity has been overlooked,but I think that's a mistake. Gravity does not disappear just because the string wraps itself around a post and I don't think that the effects of gravity can be considered as negligible.
Suppose that gravity could truly be considered as negligible,for example let the tethering post and ball be in a region of deep space and with the string taut.The way I see any circular motion now is that both post and ball rotate about a common centre of mass.

 

[Philip Wood]

I doubt if this will help very much, but at least it's not speculative...

I formed a vector expression for the position of the whirling particle relative to the centre of the post. I then differentiated it wrt time to find the particle velocity. Using these, and the relevant dot and cross products, I found expressions for the KE and angular momentum. Here they are

E_k = \frac{1}{2}ms²\dot{θ}²

L = ms²\dot{θ}

\theta is the angle of take-off of the string from the post (not the angular position of the particle!). s is the length of string still to wind on to the post. [s = b - a\theta, in which b is the full string length, and a is the post radius.]

Eliminating \dot{θ} we obtain

L = \sqrt{2<i>mE</i>_k} s

So, if we assume E_k to be constant, as I do, then L diminishes as the 'free' string shortens. I see no problem with the effect being due to the off-axis force from the string, but perhaps I'm being stupid. [And, as AM points out, the Law of conservation of momentum isn't being violated, because the law applies not to the ball and string, but to the Earth, post, ball, string system.]

The result, expressed in terms of s, is independent of the pole diameter.

 

[rcgldr]

Gravity

That's one of the issues with this problem. Depending on the intial conditions, the ball height could increase or decrease over time, which would affect the speed of the ball, exchanging gravitational potential energy and kinetic energy. I'm not sure if there's an initial condition where the ball's center of mass spirals in a horizontal plane as the rope winds around the pole, in which case speed should be constant until impact.

 

[Philip Wood]

In a simple ball, string and pole set-up the ball will indeed spiral outwards and – under gravity – downwards, and kinetic energy will increase.

For me, though, the problem is more interesting without gravity. We might imagine the unwinding to take place in a freely falling environment, or, less exotically, with the ball moving on a smooth horizontal platform (so the pull of gravity does no work, and is balanced by the normal contact force). While we're about it, I'd like the string very thin compared with the diameter of the pole.

In this way we're back (aren't we?) to the original paradox. [It's certainly the situation which I analysed - post hash 75.]

 

[Michael C]

In a simple ball, string and pole set-up the ball will indeed spiral downwards under gravity, and kinetic energy will increase.

I think that the ball could move upwards or downwards, depending upon the initial impulse.

First imagine the case where the string doesn't change length: one end is at a fixed point. In this case, there is a combination of pendular motion and rotational motion. Starting with the string at a certain angle to the horizontal, it must be possible to give the ball a horizontal impulse so that the weight of the ball is equal to the vertical component of the force on the ball from the string, in which case the ball would rotate indefinitely at a constant speed in a horizontal plane.

Now go back to the original problem. Let's say we start with the string not far from horizontal, and we give the ball an impulse such that it would continue in a horizontal circle if the string didn't change length. As soon as the ball starts moving, the radius of rotation starts decreasing. If the ball's speed stays constant, the tension in the string increases as the radius of rotation decreases, so the angle of the string with the horizontal will decrease: at least initially, the ball rises. I think the ball will end up higher than it started in this case, but I'm not completely sure.

With other initial conditions the ball could certainly move downwards as it moves inwards.

For me, though, the problem is more interesting without gravity. We might imagine the unwinding to take place in a freely falling environment, or, less exotically, with the ball moving on a smooth horizontal platform. While we're about it, I'd like the string very thin compared with the diameter of the pole.

In this way we're back (aren't we?) to the original paradox. [It's certainly the situation which I analysed - post hash 75.]

Yes, that's the way the problem has been stated in some other threads, or in http://www.lhup.edu/~dsimanek/scenario/insight.htm#skater: a skater or a puck on ice. In that case your analysis is surely correct. It makes sense: when the ball hits the pole, its angular velocity is zero, since its direction of movement is directly towards the centre of the pole.

 

[geekedsloth]

I believe this may have been over complicated. I don't know a whole lot about physics (that's why I am here) but put more simply I believe the original question was where does the energy come from that increases kinetic energy and velocity. First, I don't think we can assume that kinetic energy actually increases at any point in this case because (as Mr. Lee pointed out) there is no input of energy after the ball is struck. Rather,as the radius decreases shouldn't the amount of energy required for the ball to make a rotation decrease as well? If my understanding is correct, then with that being said kinetic energy does not increase, but the velocity increases as a result of the decreasing radius. In summary, shorter distance to rotate around pole = less energy required to rotate = higher velocity, and kinetic energy remains the same as it was with initial input. If I am mistaken please correct me as I am here to learn.

 

[Doc Al]

If my understanding is correct, then with that being said kinetic energy does not increase, but the velocity increases as a result of the decreasing radius.

Velocity and kinetic energy are related. You cannot increase the ball's velocity without increasing its kinetic energy. (KE = 1/2mv²)

 

[aaaa202]

I'm not so good at rotational dynamics - if the pole has a thickness why is angular momentum for the ball not conserved? Please explain carefully?
And btw: Nice thread =)

 

[Philip Wood]

Do you recall how to calculate the torque (or moment) of a force about an axis? [Force x perpendicular distance from axis to line-of-action of force]?

Now, unbalanced torques acting on things make them change their angular momentum. If this is intuitively ok, then fine. If not, then we'd need to explain exactly what angular momentum means. [Actually, it's defined very much like torque, but with velocity instead of torque.] And then we'd need to do some vector algebra. Let's assume it's intuitive!

We need to choose a fixed axis about which to take the torque. The centre of the pole is an obvious axis to choose. Now. because the string is coming off the side of the pole, it's not pulling the ball straight towards our axis. Indeed, there'd a torque on the ball, according to our definition, of Fa about the axis, if F is the force and a is the pole radius. So therefore the ball's angular momentum keeps changing (decreasing).

 

[aaaa202]

hmm okay I think I see it, when I imagine the pole to be very thick. Though I'm not sure, can you draw how the torque vector looks?
Just put it on the figure I attached.
Does this mean that the angular momentum is conserved in the frame of the point of application on the line? And that then means that it is not conserved for the ball?
What is the criteria that angular momentum is conserved for the ball? Must the rotation be around the center of mass for the pole (which its not) and why?

 

[rcgldr]

What is the criteria that angular momentum is conserved for the ball?

That there are no torques related to the tension in the string. Even if it's not clear how a string applies a torque on a ball following a spiral path (in this case involute of circle), it should be clear that there is a torque applied to the pole by the tension in the string, and applying Newton's 3rd law to torques, torques only exist in equal and opposing pairs, so if there's a torque on the pole due to the string, then there's and equal and opposing torque on the ball due to the string.

Angular momentum is conserved only if you consider the angular momentum of whatever the pole is attached to, usually the earth.

 

[Philip Wood]

Hallo aaaa202. Does this make sense?

The torque vector (since you ask) is into the page; if you're into vector algebra, torque, G is defined by G = r \timesF in which, in this case, r is the displacement vector from O to the ball.

 

[aaaa202]

Okay thanks! Very helpful! Though one question: Why is it that the rotation MUST be around the center of pole, if the balls angular momentum is to be conserved?

 

[Philip Wood]

There is rotation about the centre in the case we're considering, isn't there? Angular momentum is not conserved, though. Admittedly it's not a pure rotation,because the ball is also moving radially inwards. But this isn't relevant, either!

When you talk about the ball's angular momentum (that is the angular momentum of a body acted upon by an external force) you have to specify about what point you're calculating that angular momentum. You need a fixed point (certainly not, for example, the point of run-off of the string from the circumference of the pole, because that point keeps moving round the pole, and is therefore accelerating towards the centre).

Exactly the same goes for torque. If you choose the same fixed point about which to calculate torque, G and angular momentum, L, a very simple law applies: G = dL/dt.

 

[Michael C]

Okay thanks! Very helpful! Though one question: Why is it that the rotation MUST be around the center of pole, if the balls angular momentum is to be conserved?

Angular momentum is always conserved for the whole system (ball + pole + Earth), no matter which axis is chosen to measure it.

In this case angular momentum is transferred from the ball to the Earth via the pole: there is a torque on the pole, and therefore on the Earth. This means that if we consider only the pole and the ball we come to the conclusion that angular momentum is not conserved: in fact the momentum has been transferred elsewhere.

The ball would have a constant angular momentum if it did not exert a torque on something else. This would be the case if the centre of rotation of the string always stayed at the same point, for instance in the examples where the string is being pulled through a hole.

 

[aaaa202]

hmm it's just that when you see the ball for the point of contact between string and pole it makes a uniform circular motion. So can't you say that the angular momentum is conserved in this frame for the ball? And why does that not qualify to the ball's angular momentum being conserved like if the rotation was around the center of mass? :)

 

[Philip Wood]

Michael C. Agreed. Though, of course, there are interesting cases, such as a system of two charges moving at an angle to each other. [That's not the complete system, I hear someone say.]

aaaa202 As I said, the point of contact of string and pole is accelerating. The laws of Physics need modifying somewhat for use in an accelerating (non-inertial) reference frame. That's why I'm choosing to take our torque and angular momentum about the still centre of the pole.

 

[Michael C]

hmm it's just that when you see the ball for the point of contact between string and pole it makes a uniform circular motion. So can't you say that the angular momentum is conserved in this frame for the ball? And why does that not qualify to the ball's angular momentum being conserved like if the rotation was around the center of mass? :)

The point of contact is changing all the time: it's turning in a circle around the pole, so (as Philip pointed out) it's constantly accelerating. If we fix one point on the surface of the pole and measure the angular momentum around this point, we'll see that the momentum of the ball must be changing, since there is only one instant when the ball exerts no torque in this frame: the instant when the centre of rotation is at the point we have fixed. For the rest of the time, the centre of rotation is not at the point we have fixed, so there is torque around this point.

 

[aaaa202]

yes okay, I should have realized that. But doesn't there exist conservation laws in non inertial reference frames?

 

[Philip Wood]

I expect so. Indeed I expect we could easily find such a frame in which angular momentum is conserved for the ball. But that would not be anything to be especially pleased about. The ball's angular momentum would be conserved simply because we've chosen a special frame in which it is conserved. In this frame, things whose angular momentum we'd normally expect to be conserved won't have it conserved... The laws of Physics are usually easier in inertial frames.