Having conceptual trouble with this aspect of centripetal force. Say we have a puck on a frictionless table attached to a string that I am holding through a small hole, so that the puck moves in a circular path. So $F=m\;\dfrac{v^{2}}{r}=m\;\omega^{2}\; r$. If I pull on the string to change the radius of the puck's path, does angular velocity change, or linear? The math makes it look like it could be either, or both, which seems like it can't be right.
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$\begingroup$ Please can you show your reasoning : What do you expect to happen and why? What do you mean that "the math makes it look like it could be either, or both"? I am asking you to show that you have made some attempt to work out what happens. $\endgroup$– sammy gerbilCommented Apr 26, 2016 at 21:24
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$\begingroup$ You must consider conservation ideas and determine what, if any, quantities might remain constant. Include that in the work that @sammygerbil asked for. $\endgroup$– Bill NCommented Apr 26, 2016 at 21:32
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$\begingroup$ As the puck is moving in uniform circular motion the centripetal force is normal to the velocity vector, no work is produced and the total energy of the puck, which is kinetic$=(1/2)mv^{2}$, remains constant. Pulling the string, even slowly, the string tension becomes oblique to the velocity vector during the transition, work is produced transfered to the puck as kinetic energy. So the speed $v$ is increased, the radius is decreased and so from $\omega=v/r$ the angular velocity is increased also. $\endgroup$– VoulkosCommented Apr 26, 2016 at 21:37
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$\begingroup$ @Frobenius, shouldn't that be an answer? $\endgroup$– Alfred CentauriCommented Apr 27, 2016 at 0:26
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$\begingroup$ @Alfred Centauri If the meaning of your comment is that we must not post complete answers to homework like questions then I apologize and I'll delete my comment. If you mean to post it as an anwser then I'll do it with pleasure. Please confirm. $\endgroup$– VoulkosCommented Apr 27, 2016 at 5:55
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2 Answers
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In your scenario, angular momentum $m v r$ is preserved (because your pulling force is radial, with no tangential component). So if you reduce $r$ by half, $v$ must double, and since $\omega = v/r$, it increases by a factor of four.
Note this means in a small amount of time that the area swept out by the string is proportional to $v$ and $r$. Since they change by opposite factors the area remains the same. It's a basic rule of orbits:
Note also that the kinetic energy increases by a factor of four, because $v$ doubles. That energy comes from the work done in pulling the string.
If the pulling had not been radial, but had a tangential component, that would have confused the issue.
By the way: Doing work against centripetal force is (to me) an interesting topic. For example, if you have a pendulum hanging quietly from a string, making the string shorter or longer has no interesting effect. But if the pendulum is already swinging, you can pull it up when it is in the center of the swing (against large centripetal force), and drop it down when it is at either end of the swing (small centripetal force), moving the weight in a figure-8. This imparts energy to the pendulum so that it swings higher.
I suspect this is also a possible way to look at other forms of oscillatory motion, such as traveling uphill on a skateboard in a series of S-turns. It might also be relevant to flapping wings, where the downstroke works against a higher force because the bird is in a slight upward curve, and the upstroke works against lesser force because the bird is in a slight downward curve. The net work comes out as kinetic energy. I'm sure this explanation will get flak, but I think it could be one way to understand what's going on.
answered Apr 26, 2016 at 21:33
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$\begingroup$ Do we have to say something like "the string is pulled uniformly" so that the puck has no radial acceleration? Also, since the speed of the puck is increasing, the velocity vector and force vector cannot be perpendicular can they? $\endgroup$ Commented Apr 27, 2016 at 0:40
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$\begingroup$ @Al: I doubt your first point. On the second point, I've been wondering that myself. I'm supposing some kind of infinitessimal argument would clear it up, but I haven't been able to give it enough thought. $\endgroup$ Commented Apr 27, 2016 at 12:04
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As the puck is moving in uniform circular motion the centripetal force is normal to the velocity vector, no work is produced and the total energy of the puck, which is kinetic $\mathcal K=(1/2)mv^{2}$, remains constant.
Pulling the string, even slowly, the string tension becomes oblique to the velocity vector during the transition, work is produced transfered to the puck as kinetic energy. So the speed $v$ is increased, the radius is decreased and from $\omega=v/r$ the angular velocity is increased also.
