Question
Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum 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.
= mvd ...(i)
Angular momentum of the system about point Q,
LQ = mv × d + mv × 0
= mvd ...(ii)
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,
LR = mv × (d - y) + mv × y
= mvd - mvy + mvy
= mvd ...(iii)
Comparing equations (i), (ii), and (iii), we get
LP = LQ = LR ...(iv)
From equation (iv), we infer that 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,
Lp = mv × 0 + mv × d = mvd ...(i)
Angular momentum of the system about point Q,
LQ = mv × d + mv × 0
= mvd ...(ii)
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,
LR = mv × (d - y) + mv × y
= mvd - mvy + mvy
= mvd ...(iii)
Comparing equations (i), (ii), and (iii), we get
LP = LQ = LR ...(iv)
From equation (iv), we infer that that the angular momentum of a system does not depend on the point about which it is taken.
