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Rotational kinetic energy

by leebenjamin@adelphia
Tags: energy, kinetic, rotational
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Michael C
#73
Feb4-12, 11:50 AM
P: 132
Quote Quote by sushruth View Post
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
#74
Feb5-12, 05:22 AM
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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
#75
Feb5-12, 05:44 PM
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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

Ek = [itex]\frac{1}{2}[/itex]ms2[itex]\dot{θ}[/itex]2

L = ms2[itex]\dot{θ}[/itex]

[itex]\theta[/itex] 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[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{2mEk}[/itex] s

So, if we assume Ek 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
#76
Feb5-12, 11:18 PM
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Quote Quote by Dadface View Post
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
#77
Feb6-12, 02:40 AM
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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
#78
Feb6-12, 12:24 PM
P: 132
Quote Quote by Philip Wood View Post
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 Donald Simanek's example: 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
#79
Feb6-12, 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.
Doc Al
#80
Feb6-12, 02:41 PM
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Quote Quote by geekedsloth View Post
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
#81
Feb8-12, 03:50 AM
P: 1,005
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
#82
Feb8-12, 07:08 AM
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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
#83
Feb9-12, 07:00 PM
P: 1,005
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?
Attached Thumbnails
Unavngivet.jpg  
rcgldr
#84
Feb9-12, 10:34 PM
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Quote Quote by aaaa202 View Post
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
#85
Feb10-12, 03:00 AM
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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.
Attached Thumbnails
ball&pole.jpg  
aaaa202
#86
Feb10-12, 07:01 AM
P: 1,005
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
#87
Feb10-12, 08:16 AM
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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
#88
Feb10-12, 08:28 AM
P: 132
Quote Quote by aaaa202 View Post
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
#89
Feb10-12, 08:42 AM
P: 1,005
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
#90
Feb10-12, 08:45 AM
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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.


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