Physics Help Forum Angular Momentum

 May 9th 2013, 03:44 PM #1 Junior Member   Join Date: Jul 2008 Location: Melbourne, Australia Posts: 11 Angular Momentum So angular momentum of a particle is given by: L = m (r x v) and angular momentum is conserved in the absence of any forces. I want to work in cartesian coordinates and I have a particle rotating around the z axis but not necissarily in the z=0 plane (i.e. dz/dt = 0). How can the angular momentum be conserved when L/m = (r x v) = (y dz/dt - z dy/dt) e1 - (x dz/dt - z dx/dt) e2 + (x dy/dt - y dx/dt) e3 L/m = (r x v) = - z dy/dt e1 + z dx/dt e2 + (x dy/dt - y dx/dt) e3 Ie the angular momentum has components in the x (e1) and y (e2) directions that are non-zero and have variable magnitude.
May 11th 2013, 07:36 AM   #2
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 Originally Posted by Kiwi_Dave So angular momentum of a particle is given by: L = m (r x v) and angular momentum is conserved in the absence of any forces. I want to work in cartesian coordinates and I have a particle rotating around the z axis but not necissarily in the z=0 plane (i.e. dz/dt = 0). How can the angular momentum be conserved when L/m = (r x v) = (y dz/dt - z dy/dt) e1 - (x dz/dt - z dx/dt) e2 + (x dy/dt - y dx/dt) e3 L/m = (r x v) = - z dy/dt e1 + z dx/dt e2 + (x dy/dt - y dx/dt) e3 Ie the angular momentum has components in the x (e1) and y (e2) directions that are non-zero and have variable magnitude.
"angular momentum is conserved in the absence of any forces"

If the particle is rotating about the z-axis, there must be a radial centripetal force acting on it.
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 May 19th 2013, 01:33 AM #3 Junior Member   Join Date: Oct 2012 Posts: 1 Centripetal force State 5 application of centripetal force

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### l m.rxv

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