Why Is the Total Angular Momentum of a Particle System Constant?

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SUMMARY

The total angular momentum of a system of n distinct particles, defined as the vector sum of the angular momentum of each particle, remains constant when only internal forces act between the particles. This conclusion is derived from Newton's second law, which states that the rate of change of momentum is equal to the net force acting on the system. By applying the definition of angular momentum, \vec L = \vec r \times \vec p , and taking the time derivative, it is shown that \frac{d\vec L}{dt} = 0 , confirming the constancy of angular momentum in the absence of external forces.

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  • Understanding of Newton's second law of motion
  • Familiarity with the concept of angular momentum
  • Knowledge of vector calculus and cross products
  • Basic principles of classical mechanics
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Students of physics, particularly those studying classical mechanics, educators teaching angular momentum concepts, and anyone interested in the principles governing particle systems and their motion.

ElDavidas
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hey everyone, I've been working through a past paper and I'm stuck on a question:

"(a) Consider a system consisting of n distinct particles P1, P2, : : :, Pn, with masses m1, m2, ..., mn and position vectors r1, r2, : : :, rn, relative to the origin O of an inertial frame, respectively.

For each i = 1, 2, : : :, n, suppose that the only forces acting on Pi are internal forces Fij , which always act along the line joining Pi and Pj , for j = 1, 2, : : :, n and j doesn't equal i.

Assume that Fij = ¡Fji for i = 1, 2, : : :, n, j = 1, 2, : : :, n and i doesn't equal j.

Define what is meant by the total angular momentum of the system about O.

Show, as a consequence of Newton's second law, that the total angular momentum of the system about O is constant."

I can define the total angular momentum of the system, it's the 2nd part of the Q I have trouble with. :confused:

Thanks
 
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Use the definition of angular momentum:

[tex]\vec L = \vec r \times \vec p[/tex]

Take the time derivative and sum over all particles noting that

[tex]\frac {d \vec p}{dt} = \vec F[/tex]

You will find [itex]d\vec L /dt = 0[/itex].
 

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