
#19
Feb413, 12:28 AM

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#20
Feb413, 12:59 AM

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A coordinate system is not the same as the frame of reference.
Describing a motion with a polar coordinate system, there are both radial and azimuthal components of acceleration. The azimuthal component includes both the "Coriolis term" and the angular acceleration. As the string can exert only radial force, the azimuthal component of acceleration is zero. That does not mean that the acceleration or the force has constant direction as you said in #11. It means that the force is "central", acts along the radius, and the acceleration is radial. ehild 



#21
Feb413, 01:13 AM

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The expression given for the acceleration vector implies its direction is constant. Which should not be the case for a rotating mass in an inertial frame. Consequently, the expression is an noninertial frame. The article that you linked has the correct expression for acceleration at the end. It should just be taken from there. 



#22
Feb413, 03:08 AM

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PhizKid said that the azimuthal component of acceleration was zero, that was, why he did not include that term in the integral.
ehild 



#23
Feb413, 03:45 AM

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ehild 



#24
Feb413, 05:28 AM

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#25
Feb413, 08:00 AM

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ehild 



#26
Feb413, 09:06 AM

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"Pulling slowly" means that the motion is circular and uniform at any instant, which automatically proves the assumption. Whether this is the case with any other kind of pulling must be proved. 



#27
Feb413, 09:36 AM

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



#28
Feb413, 11:28 AM

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#29
Feb413, 11:43 AM

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The tangential and the centripetal acceleration are defined in the instantaneous system of coordinates, moving together with the particle along its trajectory. The velocity points in the direction tangent to the trajectory, and the acceleration can have both tangential and normal components. The tangential acceleration is equal to the time derivative of the speed. The normal (centripetal) component of acceleration is v^{2} /R where R is the curvature of the trajectory. In this problem, the trajectory is a spiral. The acceleration has components both tangential and normal to the path . In polar coordinates, the velocity has both radial and azimuthal components, but the acceleration has only radial component, as the force is radial. ehild 



#30
Feb413, 11:56 AM

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Oh gosh I'm so confused. I don't know if my solution is right or not. I never took into account the slow approximation so does that void it's correctness? Or does applying the slow approximation simply get rid of the radial velocity terms in my final solution? I'm not seeing any reason to introduce Coriolis forces into this and I have no idea where voko got the idea that I was working in a frame co  rotating with the mass so that its trajectory in such a frame would be purely radial (this is what would happen in such a frame correct? If someone could answer that even though its unrelated =D). Thanks a ton!




#31
Feb413, 02:19 PM

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Getting back to the issue of the work related to radial acceleration, there should be a way to show that since the initial and final state of the whirling mass has no radial speed, the net work related to radial acceleration is zero. This could be done by showing that angular momentum is conserved, since the time derivative of angular momentum is torque, and in this case, the torque is zero. Once it's shown that angular momentum is conserved (and that the initial and final state of the whirling mass has no radial speed), then it doesn't matter if the string is pulled slowly or quickly, the net work done by radial acceleration and deceleration is zero. wiki article on angular momentum (this may help): http://en.wikipedia.org/wiki/Angular_momentum 



#32
Feb413, 03:31 PM

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Your derivation is correct, the work of the net force is equal to the change of KE, and you just arrived to that result. But I think you should expand the expression for KE and give it in terms of the initial and final speeds. And ignore radial speeds as that was said in the problem. You are right, you work in inertial frame of reference, using polar coordinates. No centrifugal force, no Coriolis force, no other pseudoforces. In a frame of reference, rotating together with the mass it would move in radial direction. ehild 



#33
Feb513, 03:45 AM

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The statement, by you and ehild, that the associated fictitious forces can be ignored entirely, is manifestly wrong, because the ## r\dot{\theta}^2 ## component, which is retained in the analysis, is the centrifugal term. 



#34
Feb513, 05:14 AM

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Also, note that in his expression for the radial component of the acceleration, the ##r\dot{\theta}^2## term appears with a minus sign. It corresponds to the centripetal acceleration, not the centrifugal acceleration. In a rotating frame, that term moves to the other side of F=ma and corresponds to the centrifugal force. See, for example, http://en.wikipedia.org/wiki/Polar_c...Coriolis_terms. 



#35
Feb513, 08:11 AM

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#36
Feb613, 02:20 AM

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Anyway, it seems the OP has long lost interest in this discussion, so I won't debate this any longer. 


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