Does only the Tangential momentum count

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The discussion clarifies that when converting linear momentum to angular momentum, only the tangential component of the linear momentum contributes to the angular momentum about a fixed axis. A particle colliding tangentially with a rotating disc transfers all its momentum into angular momentum, while a collision at an angle results in a reduced contribution based on the cosine of that angle. The angular momentum of the system remains constant during the collision, regardless of the direction of the particle's motion. This understanding emphasizes the relationship between linear and angular momentum in rotational dynamics. The conversation effectively highlights the importance of the collision angle in determining the angular momentum generated.
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Does only the Tangential momentum "count"...

When convertering the momentum from linear to angular?
 
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For a particle with linear momentum \vec p, its angular momentum about some origin is \vec L = \vec r \times \vec p. So, if I understand what you're asking, the answer is yes.
 
So a particle traveling into a disc (thats free to rotate) in an inelastic collision that collides on a path tangent to the disc has all of its momentum converted into angular momentum, whearas a path of 30° with respect to the tanget line would have cos(30) times that angular momentum?
 
I'd phrase it a bit differently. (As I don't like saying that linear momentum "converts" to angular momentum.) Let me assume that the axis of the disk is fixed; the disk is free to rotate, but not translate. In any case, prior to the collision the particle has a certain angular momentum with respect to that axis. Yes, the amount of angular momentum it has depends on its direction of motion. During the collision, the angular momentum (of the entire system) does not change. Does this help?
 

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