Angular Momentum: Disk + Point Mass at Margin

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The angular momentum of a disk with a point mass at its margin can be calculated by adding their individual moments of inertia, expressed as L = (I(disc) + I(point mass))w, where w is the angular velocity. The point mass contributes L = mR²w, while the disk contributes L = 1/2 MR²w. Using the moment of inertia tensor would yield the same result, as the axis of rotation is a principal axis. The calculation involves integrating the densities of the disk and point mass. This approach clarifies the relationship between the components of angular momentum in the system.
Gavroy
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I asked myself,whether the angular momentum of a disk rotating free in horizontal plane when there is a point mass at the margin of the disc, is given by:
L = (I(disc)+I(point mass))w where w is the angular velocity.

or do I have to use the moment of inertia tensor. i am a little bit confused now...
 
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You are right in just adding the two moments of inertia, that is the same as adding their angular momentum together, which you can do: if you imagine the particle was there without the disk, it would have L=mR2w, and the disk would have L=1/2 MR2w, and then to get the total you could just add the two.

Assuming the disk is rotating about its center, if you used the moment of inertia tensor it would come out to the same thing. I think that that axis would be one of the principal axes, and you would have an integral of x2 + y2 multiplied by the corresponding densities, which would be a circle and a delta function.
 
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