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Old May 9th 2013, 03:44 PM   #1
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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.
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Old May 11th 2013, 07:36 AM   #2
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Originally Posted by Kiwi_Dave View Post
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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Old May 19th 2013, 01:33 AM   #3
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Cool Centripetal force

State 5 application of centripetal force
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