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Rotation of a body about a point with decreasing radius 
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#1
Jun2811, 06:17 AM

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I have problem to see clearly the case of rotation of a body about a point with decreasing radius.this can be viewed as a body being rotated about a point with the help of a thread which passes through a hole in a disc and the thread is being pulled to apply a tension and gradually to decrease the length of the thread.I have attached a word doc. to have a clear pic. about it.It can seen that here that the instantaneous velocity has two components ,one perpendicular to the radius & other along radius. The tension is along the thread or radial.
there are some points to be noted. 1 .the tension being perpendicular cannot do any work on the perpendicular comp. of velocity & cannot change its magnitude. 2.there is no torque about the center as tension is radial. So, angular momentum should be conserved. But the problem is angl. momen. is rXp . or " r*mv_{perpendicular}". as angu. momen. is constant so r*mv_{perpendicular} should be constant. Then if r decreases so then v_{perpendicular} must change. And it means that the perpendicular comp. of velocity changes ,which violate the 1st point. 


#2
Jun2811, 01:14 PM

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#3
Jun2811, 01:17 PM

P: 617

Angular momentum is conserved and the perpendicular component of the velocity will increase. That is why figure skaters spin faster when they bring their arms in and why planets closer to their host star have faster orbits. Your error is in the first point.



#4
Jun2811, 04:09 PM

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Rotation of a body about a point with decreasing radius



#5
Jun2911, 05:16 AM

P: 12

thanks for your replies .All the statements made by you are absolutely right. Surely the speed of the revolving body will increase as the tension is doing work on it. The attachment given by "rcgldr" is OK. but have a look in this attachment below. the instantaneous velocity is (here in RED) tangential to the spiral path that the body makes. Now we resolve it into two comp.. one along the radius(in BLUE color) and the other perpendicular(in BLUE color) to it. ofcourse the ,the tension is acting along the radial direction and it doesnot have any comp. along the perpendicular component of the instantaneous velocity. So, the tension will do work on the radial velocity , as usually the net speed will increase .But .... the perpendicular component of the velocity will remain same. that is what I meant to say. The perpen. one will remain constant.
there is no torque about the center as tension is radial. So, angular momentum should be conserved. But the problem is angl. momen. is rXp. or " r*mv_{perpendicular}". as angu. momen. is constant so r*mv_{perpendicular} should be constant. Then if r decreases so then v_{perpendicular} must change. And it means that the perpendicular comp. of velocity changes. 


#6
Jun2911, 05:32 AM

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


#7
Jun2911, 06:26 AM

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#8
Jun2911, 09:07 AM

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#9
Jun2911, 12:18 PM

P: 617

What exactly is the problem? You are right, no work is done in the tangential direction. That is why angular momentum is conserved and the speed must increase as the radius decreases. What part of this process is causing confusion.



#10
Jun3011, 02:30 AM

P: 12

So, tention , being perpendicular to this perpendicular component, can change its(perpendicular component's) direction but not the magnitude ofcourse. So, clearly 1.Angular momentum is constant . 2.the V_{perpendicular} is constant in magnitude. Now my confussion is: Since, Angular momentum is constant,its direction and magnitude both must remain constant. The expression for angular momentum is rxV . ignoring its direction which always remains same,it can be written simply as ........................magnitude of angular momentum = r.V_{perpendicular}............. Since,its magnitude is also constant , if 'r' decreases then V_{perpendicular} must also change. This is my confussion because V_{perpendicular} is constant (as you also agreed) . This miraculas result is really disturbing me. 


#11
Jun3011, 03:29 AM

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For example, there's a moment in time when that direction of the perpendicular component of velocity is horizontal. Prior to that moment in time, there was a period of time when there was acceleration in the horizontal direction. Only at the instant of that moment in time is there no acceleration in the horizontal direction. Immediately afterwards, there will be a decleration in the horizontal direction because the direction is being changed towards vertical (assuming a spiral like path). As long as the string is being pulled inwards or released outwards, the velocity and the components of that velocity will be continously changing. 


#12
Jun3011, 05:29 AM

P: 12

But now take the similar case when a body is in just a simple circular motion. The instantaneous velocity and tangential velocity is same here.This velocity is always perpendicular to the tension. Now going with your same logic, suppose there was a moment in time when the string is vertical and the instantaneous velocity is horizontal. Before this ,there was a period of time, it had acceleration along horizontal direction because of tension . So,this component of tension should also change the magnitude this velocity .But as we know that the tangential velocity ,which is also always perpendicular to the radius, is constant in magnitude. 


#13
Jun3011, 03:45 PM

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P: 7,171

The main thing to consider is if there is any component of acceleration in the direction of (or opposing) velocity, then the magnitude of velocity, V will change. Unless the path is a straight line in the direction of the string, the string is rotating, and as I mentioned before since the direction is changing, then a radial component later ends up as part of the future perpendicular component. 


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