For constant particle masses define the centre of mass by , with . Summing Newton's second law over all particles gives
Each pair of internal forces cancels by Newton's third law. Hence
Only the action-reaction condition is needed for this translation law; centrality of the internal forces is needed for the angular momentum result in part (ii).
About a fixed inertial origin, the total angular momentum is . Differentiation gives
because . Separate the external and internal contributions. The combined torque of an internal pair is
The first equality uses Newton's third law, and the second uses the given central-force condition. All internal torques therefore cancel, leaving
An equal-and-opposite pair which is not central need not have zero combined torque; this explains why both internal-force hypotheses matter.
Write and , where . The definition of the centre of mass and its derivative give
Expand the total angular momentum:
The two middle terms vanish. Thus the angular momentum decomposition about the centre of mass is
The first term is orbital angular momentum of the centre; the second is angular momentum about the centre of mass.
Let and . A differentiable central potential energy gives the internal forces
There are no external forces. Differentiating the total kinetic energy and using Newton's second law therefore gives
Since , the last expression is . Thus
This conserves total mechanical energy, including the kinetic energy of centre-of-mass motion, not merely the relative kinetic energy.

Articles by others on the same topic (0)

There are currently no matching articles.