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.

Question

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Answer 1

Let at a certain instant two particles be at points P and Q, as shown in the following figure.
Consider a point R, which is at a distance $y$ from point Q, i.e.,
$Q R = y$
∴ $P R = d - y$

Angular momentum of the system about point P:
$L_{P} = m v \times 0 + m v \times d = m v d . . . . . . (i)$

Angular momentum of the system about point Q:
$L_{Q} = m v \times d + m v \times 0 = m v d . . . . (i i)$

Angular momentum of the system about point R:

$L_{R} = m v \times (d - y) + m v \times y = m v d . . . . (i i i)$

Comparing equations (i), (ii), and (iii), we get:
$L_{P} = L_{Q} = L_{R} . . . . . . (i v)$

We infer from equation $(i v)$ that the angular momentum of a system does not depend on the point about which it is taken.

Answer 2

Let at a certain instant two particles be at points P and Q, as shown in the following figure.

Consider a point R, which is at a distance

y

from point Q, i.e.,

Q R = y

∴

P R = d - y

Angular momentum of the system about point P:

L_{P} = m v \times 0 + m v \times d = m v d . . . . . . (i)

Angular momentum of the system about point Q:

L_{Q} = m v \times d + m v \times 0 = m v d . . . . (i i)

Angular momentum of the system about point R:

L_{R} = m v \times (d - y) + m v \times y = m v d . . . . (i i i)

Comparing equations (i), (ii), and (iii), we get:

L_{P} = L_{Q} = L_{R} . . . . . . (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.