
#73
Feb412, 11:50 AM

P: 132





#74
Feb512, 05:22 AM

PF Gold
P: 2,002

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. 



#75
Feb512, 05:44 PM

P: 861

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} = [itex]\frac{1}{2}[/itex]ms^{2}[itex]\dot{θ}[/itex]^{2} L = ms^{2}[itex]\dot{θ}[/itex] [itex]\theta[/itex] is the angle of takeoff 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[itex]\theta[/itex], in which b is the full string length, and a is the post radius.] Eliminating [itex]\dot{θ}[/itex] we obtain L = [itex]\sqrt{2mE_{k}}[/itex] 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 offaxis 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. 



#76
Feb512, 11:18 PM

HW Helper
P: 6,925





#77
Feb612, 02:40 AM

P: 861

In a simple ball, string and pole setup 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.] 



#78
Feb612, 12:24 PM

P: 132

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. 



#79
Feb612, 12:52 PM

P: 9

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.




#80
Feb612, 02:41 PM

Mentor
P: 40,889





#81
Feb812, 03:50 AM

P: 992

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 =) 



#82
Feb812, 07:08 AM

P: 861

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



#83
Feb912, 07:00 PM

P: 992

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? 



#84
Feb912, 10:34 PM

HW Helper
P: 6,925

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



#85
Feb1012, 03:00 AM

P: 861

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 [itex]\times[/itex]F in which, in this case, r is the displacement vector from O to the ball. 



#86
Feb1012, 07:01 AM

P: 992

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?




#87
Feb1012, 08:16 AM

P: 861

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 runoff 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. 



#88
Feb1012, 08:28 AM

P: 132

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. 



#89
Feb1012, 08:42 AM

P: 992

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? :)




#90
Feb1012, 08:45 AM

P: 861

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 (noninertial) reference frame. That's why I'm choosing to take our torque and angular momentum about the still centre of the pole. 


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