
#1
Dec1112, 10:30 AM

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1. The problem statement, all variables and given/known data
A cone of height h and base radius R is free to rotate about a fixed vertical axis. It has a thin groove cut in the surface. The cone is set rotating freely with angular speed ω0 and a small block of mass m is released in the top of the frictionless groove and allowed to slide under gravity. Assume that the block stays in the groove. Take the moment of inertia of the cone about the vertical axis to be I0. (a) What is the angular velocity of the cone when the block reaches the bottom? (b) Find the speed of the block in inertial space when it reaches the bottom. 3. The attempt at a solution My issue isn't really with solving the two parts because that part is quite straightforward. Just apply conservation of angular momentum about the axis of rotation for part (a) and then conservation of energy for part (b). My problem is more with this: since there is never a presence of a net torque about the axis of rotation, [itex]\partial _{t}L_{y} = 0[/itex] (taking the y axis to be the vertical axis of rotation) so, since the block and cone become one combined system much like in a perfectly inelastic collision, [itex]I_{0}w_{0} = (I_{0} + mR^{2})w_{f}[/itex] and we can easily solve for the final angular velocity of the combined system which is equivalent to the final angular velocity of the block. My question is simply: where does the angular velocity of the initial cone without the block go? At the end when the block reaches the ground there is obviously a loss in the angular velocity of the system from before the mass was placed on the rotating cone. I'm not sure where the angular velocity actually goes. Using conservation of energy I got that [itex]v_{f} = \sqrt{\frac{I_{0}}{m}(w_{0}^{2}  w_{f}^{2}) + 2gh}[/itex] and since [itex]w_{0} >= w_{f} >= 0[/itex], [itex]w_{0}^{2} >= w_{f}^{2} >= 0[/itex] so unless the rotation was zero to begin with (in which case vf is just sqrt(2gh)), there is a positive term that adds on to the 2gh term so am I right in saying that the loss in angular velocity simply gets added to the translational velocity of the block? It seemed too obvious so I don't know if there is some shady business here (=p). Thanks! 



#2
Dec1112, 10:51 AM

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#3
Dec1112, 11:45 AM

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#4
Dec1112, 12:20 PM

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Conceptual Problem in rotating cone problem
The block can rotate quicker or slower than the cone. Actually, it has to do this in order to get down.




#5
Dec1112, 12:32 PM

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#6
Dec1112, 04:42 PM

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Maybe that solution assumed that the velocity of the block relative to the cone is negligible. In general, it is not.




#7
Dec1112, 04:49 PM

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#8
Dec1112, 05:20 PM

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Looking again at how it wrote down the energy conservation equation, it totally disregarded the angular velocity of the block and considered only the new angular velocity of the cone after the block is placed on it (it included the translational velocity of the block of course) but I can see no justification for this based on the problem statement. Do you?




#9
Dec1112, 06:18 PM

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#10
Dec1112, 06:24 PM

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#11
Dec1112, 06:40 PM

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#12
Dec1112, 06:54 PM

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#13
Dec1112, 07:02 PM

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#14
Dec1112, 07:05 PM

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#15
Dec1112, 07:24 PM

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#16
Dec1112, 07:34 PM

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#17
Dec1112, 08:33 PM

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Yep.



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