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Two particles, each of mass $m$ and speed $v$, travel in opposite directions along parallel lines separated by a distance $d$. 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.
Solution
Let at a certain instant two particles be at points $P$ and $Q$, as shown in the following figure.
Angular momentum of the system about point $P$ :
$\vec{L}_{p}=m v \times 0+m v \times d$
$=m v d.. .(i)$
Angular momentum of the system about point $Q:$ $\vec{L}_{\Omega}=m v \times d+m v \times 0$
$=m v d…(i i)$
Consider a point $R,$ which is at a distance $y$ from point $Q,$ i.e.
$QR =y$
$PR =d-y$
Angular momentum of the system about point $R:$ $\vec{L}_{R}=m v \times(d-y)+m v \times y$
$=m v d-m v y+m v y$
$=m v d\ldots(i i i)$
Comparing equations $(i),(i i),$ and $(iii)$, we get:
$\vec{L}_{P}=\vec{L}_{0}=\vec{L}_{R}\ldots(i v)$
We infer from equation ($i v$) that the angular momentum of a system does not depend on the point about which it is taken.
