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Physics LibreTexts

3.12: Torque, Angular Momentum and a Moving Point

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In Figure III.7 I draw the particle mi, which is just one of n particles, n1 of which I haven’t drawn and are scattered around in 3-space. I draw an arbitrary origin O, the centre of mass C of the system, and another point Q, which may (or may not) be moving with respect to O. The question I am going to ask is: Does the equation ˙L=τ apply to the point Q? It obviously does if Q is stationary, just as it applies to O. But what if Q is moving? If it does not apply, just what is the appropriate relation?

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The theorem that we shall prove – and interpret - is

˙LQ=τQ+MrQרrQ.

We start:

LQ=(rirQ)×[mi(vivQ)]

˙LQ=(rirQ)×mi(˙vi˙vQ)+(˙ri˙rQ)×mi(vivQ).

The second term is zero, because ˙r=v

Continue:

˙L=(rirQ)×mi˙vimiri×˙vQ+mirQ×˙vQ

Now mi˙vi=Fi, so that the first term is just τQ

Continue:

˙L=τQmiri×dotvQ+MirQ×dotvQ

=τQM¯rרrQ+MrQרrQ

=τQ+M(rQ¯r)רrQ

˙LQ=τQ+MrQרrQQ.E.D

Thus in general, ˙LQτQ, but ˙LQ=τQ under any of the following three circumstances:

  1. rQ=0 - that is, Q coincides with C.
  2. ¨rQ=0 - that is, Q is not accelerating.
  3. ¨rQ and rQ are parallel, which would happen, for example, if O were a centre of attraction or repulsion and Q were accelerating towards or away from O.

This page titled 3.12: Torque, Angular Momentum and a Moving Point is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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