Reduce angular momentum from Iw to mvr

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SUMMARY

The discussion centers on the relationship between angular momentum (L), moment of inertia (I), and angular velocity (ω) for a point particle. The equation L=Iω simplifies to L=rxp, but the user encounters an extra factor of r². For a point particle, I is defined as Mr², and ω is expressed as rxv. The correct formulation reveals that the simplification to L=rxp holds only when r=1, which is impractical in real scenarios.

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  • Understanding of angular momentum and its vector representation
  • Familiarity with moment of inertia (I) for point particles
  • Knowledge of angular velocity (ω) and its relation to linear velocity (v)
  • Basic grasp of vector cross product operations
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  • Explore the implications of moment of inertia in rotational dynamics
  • Learn about the vector cross product and its applications in physics
  • Investigate the conditions under which simplifications in physics equations are valid
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Physics students, mechanical engineers, and anyone studying rotational dynamics and angular momentum in classical mechanics.

Aziza
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In vector form,

L=Iω.

I am trying to show that for a point particle, this reduces to L=rxp, but am getting an extra factor of r2 :

For the case of a point particle, I=Mr2.
Also, ω=rxv.
Plugging these into L=Iω gives:

L=Mr2(rxv) = r2(rxp).

Thus this reduces to a point particle only for r=1..but that is silly.

I appreciate any help!:)
 
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