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Exercises · 6.7

Q.Two particles, each of mass mm and speed vv, travel in opposite directions along parallel lines separated by a distance dd. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

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For a system whose total linear momentum is zero the angular momentum is the same about every point. Here the two equal-and-opposite momenta cancel, so L\mathbf{L} is independent of the reference point, with magnitude mvdmvd.

Shift-of-origin identity. Moving the reference point from OO to O′O' (let R\mathbf{R} be the vector from O′O' to OO) changes the total angular momentum by

LO′=∑i(ri−R)×pi=LO−R×Ptotal,Ptotal=∑ipi.\mathbf{L}_{O'} = \sum_i(\mathbf{r}_i - \mathbf{R})\times\mathbf{p}_i = \mathbf{L}_O - \mathbf{R}\times\mathbf{P}_{\text{total}}, \qquad \mathbf{P}_{\text{total}} = \sum_i \mathbf{p}_i.

Total momentum is zero. The two particles carry momenta p1=mv x^\mathbf{p}_1 = mv\,\hat{\mathbf{x}} and p2=−mv x^\mathbf{p}_2 = -mv\,\hat{\mathbf{x}}, so

Ptotal=mv x^−mv x^=0.\mathbf{P}_{\text{total}} = mv\,\hat{\mathbf{x}} - mv\,\hat{\mathbf{x}} = 0.

Hence R×Ptotal=0\mathbf{R}\times\mathbf{P}_{\text{total}} = 0 for every R\mathbf{R}, giving LO′=LO\mathbf{L}_{O'} = \mathbf{L}_O: the angular momentum is the same about any point.

Its value. Taking the two parallel lines as y=+d/2y = +d/2 and y=−d/2y = -d/2, …

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