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.
Exercise
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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 $P$,
$L_p=m v \times 0+m v \times d=m v d \ldots$...(i) Angular momentum of the system about point $Q L_Q=m v \times d+m v \times 0=m v d \ldots$
(ii) Consider a point $R$, which is at a distance $y$ from point $Q$ i.e., $Q R=y \therefore P R=d-y$ Angular momentum of the system about point $R, L_R=m v \times(d-y)+m v \times y m v d-m v y+m v y=m v d$...(iii) Comparing equations (i), (ii), and (iii), we get, $L_p=L_Q=L_R \ldots$..(iv) We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.
Angular momentum of the system about point $P$,
$L_p=m v \times 0+m v \times d=m v d \ldots$...(i) Angular momentum of the system about point $Q L_Q=m v \times d+m v \times 0=m v d \ldots$
(ii) Consider a point $R$, which is at a distance $y$ from point $Q$ i.e., $Q R=y \therefore P R=d-y$ Angular momentum of the system about point $R, L_R=m v \times(d-y)+m v \times y m v d-m v y+m v y=m v d$...(iii) Comparing equations (i), (ii), and (iii), we get, $L_p=L_Q=L_R \ldots$..(iv) We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.
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