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Angular momentum confusion 
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#1
Dec712, 12:53 AM

P: 113

Imagine i have a particle in the xy plane rotating in a circle around the zaxis at a constant angular velocity. The angular momentum is r cross p which would give me a constant value at all times, there is no change and it's conserved. However if I was to move the rotating particle higher up and then draw my r vector out from the origin the direction of angular momentum would constantly change as the particle rotated.
So at one chosen origin the angular momentum is constant, but at another chosen origin it changes? If this is true, wouldn't that imply there is a torque and also that there's no torque? 


#2
Dec712, 02:32 AM

PF Gold
P: 963

Interesting. I agree with everything you say, including your conclusion. But there is nothing paradoxical about the conclusion. It's well known that torque (like angular momentum) depends upon the chosen point about which we calculate it.
Only when we have a system of particles acting on each other by central forces can we say that the sum of angular momenta about any point is constant, and that the resultant torque about any point is zero. 


#3
Dec712, 08:29 AM

P: 789




#4
Dec712, 06:15 PM

P: 1,991

Angular momentum confusion
There is nothing unrealistic.
In order to have uniform circular motion you need some centripetal force. The torque of this force is zero in respect to the center of the trajectory but is nonzero in respect to a different point along the axis. 


#5
Dec712, 09:36 PM

P: 789




#6
Dec712, 09:43 PM

P: 1,991

I suppose we should make sure that we mean the same thing by "unrealistic".
How will this make it unrealistic? What parts are not conform with reality? 


#7
Dec712, 10:00 PM

P: 789

The angular momentum of a closed system (not subject to outside forces) is constant in time and independent of the coordinate system you choose. If you ever calculate the angular momentum of a system to be changing in time, then you will know that it is not closed  there is some outside force causing it. If you expand the system to include the source of that force, and there are no other forces on that system, then its angular momentum (and linear momentum, and energy) will be constant in time and independent of the coordinate system that you choose. 


#8
Dec712, 10:16 PM

P: 1,991

But who implied that you will? Who said that the particle is free? I did not for sure. If you just look at the angular momentum of one particle, what you find out is not less realistic than what you get when looking at the system. As long as you know what are you looking at. 


#9
Dec712, 10:58 PM

P: 789




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