
#1
Dec1305, 01:24 PM

P: 34

Hi
we just studied motion under central force. we got the following question... is this possible trajectory(see attachment) under central force and force source is outside the loop? (my answer is that it is possible if force source is repulsive) whatever the answer is how can i explain it using physical arguments ? 



#2
Dec1305, 09:29 PM

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What quantities are conserved in motion about a central force?




#3
Dec1305, 10:45 PM

P: 34

i know that one of the angular momentum vector components is conserved so the entire motion is in the plane(perpendicular to that component)
also usual energy is conserved... 



#4
Dec1405, 12:56 AM

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possible trajectories under central force and... 



#5
Dec1405, 03:12 PM

P: 34

i already know that i was wrong...
this trajectory is not possible under central force but i don't see how can i say that by the picture. 



#6
Dec1405, 03:16 PM

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#7
Dec1605, 02:24 PM

P: 34

L=rmv\sin(\alpha) [/tex] so if one wants to maintain L constant then if r increases then [itex]v\sin(\alpha)[/itex] decreases and if r closer to force center the [itex]v\sin(\alpha)[/itex] is larger. i don't see any contradiction to this on the picture(or maybe the direction of L is what matters?) when particle is moving from (a) > (b) its r vector becomes shorter and v increases and vica versa when it moves from (c) > (d) by the way, what exactly makes such a trajectory impossible under central force? is it because two lines cross each other? 



#8
Dec1605, 06:31 PM

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#9
Dec1705, 05:03 AM

P: 34

ok...
so at when particle moves from (a) to (b) its velocity vector point to left while when particle moves from (c) to (d) its velocity vector points to right. so according to righthand rule [itex]\vec{L}[/itex] changes direction... right? 



#10
Dec1705, 05:46 AM

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#11
Dec1705, 03:03 PM

P: 34

thanks a lot:)



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