Why Does the Derivative of Momentum Appear in Angular Momentum Calculations?

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SUMMARY

The discussion centers on the derivation of the angular momentum theorem for a system of particles, specifically addressing the relationship between the derivative of momentum and angular momentum calculations. The user initially struggles with understanding why the difference in velocities leads to the derivative of momentum (dp/dt) rather than simply momentum (p). After further analysis and differentiation, the user realizes the additional term that appears in the angular momentum expression, indicating a deeper understanding of the concept.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly angular momentum.
  • Familiarity with calculus, specifically differentiation.
  • Knowledge of momentum concepts in physics.
  • Ability to interpret vector equations in physics.
NEXT STEPS
  • Study the derivation of the angular momentum theorem in detail.
  • Learn about the relationship between linear momentum and angular momentum.
  • Explore advanced calculus techniques related to physics applications.
  • Investigate the implications of additional terms in physics equations.
USEFUL FOR

Physics students, educators, and anyone interested in advanced mechanics, particularly those studying angular momentum and its applications in particle systems.

Xyius
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I do not understand this one step in deriving the angular momentum theorem for a system of particles.

The vector angular momentum about the point Q, not necessarily the origin is..
[PLAIN]http://img27.imageshack.us/img27/8196/pfquestion.gif

I do not understand why the difference in the velocities of the point Q and k equals the DERIVATIVE of the momentum. If you bring the mass term into the expression, then shouldn't it be equal to just p? Why is it dp/dt??
 
Last edited by a moderator:
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Never mind, I just got it by differentiating. Now what I don't get is the next part. I will post a pic soon.

EDIT:
Okay so, when I differentiate the expression for angular momentum, I get the expression above. When THEY differentiate it, they get what I have plus this additional term.

[PLAIN]http://img24.imageshack.us/img24/9514/pfquestion1.gif

Maybe I am being oblivious again, but I cannot get this term to make sense!
 
Last edited by a moderator:
Yup! being oblivious again! I got it. I think on that note, I should finally stop studying haha.
 

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