Irodov 1.196: Angular Momentum Shifts with the Origin

Problem Irodov 1.196

A system of particles has a total momentum $\vect p$ and an angular momentum $\vect M$ measured about a reference point $O$ . A second reference point $O'$ is located relative to $O$ by the radius vector $\vect r_0$ . Express the system’s angular momentum $\vect M'$ about $O'$ in terms of $\vect M$ , $\vect p$ , and $\vect r_0$ , and determine the condition under which the system’s angular momentum does not depend on which reference point is chosen.

Two reference points O and O' separated by r_0. The system carries one total momentum p; the angular momenta M and M' about the two points are the two axial marks. — Two reference points $O$ and $O'$ separated by $\vect r_0$ . The system carries one total momentum $\vect p$ ; the angular momenta $\vect M$ and $\vect M'$ about the two points are the two axial marks.

Given

$\vect p$ , $\vect M$ about $O$
$\vect r_0 = \overrightarrow{OO'}$

Find

Find

\vect M'

about

O'

, and the condition under which

\vect M

does not depend on the choice of

O

Solution

Step 1Rewrite each particle’s position about the new point

Let $\vect r_i$ be the position of particle $i$ relative to $O$ and $\vect r_i'$ its position relative to $O'$ . Since $O'$ sits at $\vect r_0$ from $O$ , every particle’s position shifts by the same constant vector,

\vect r_i' = \vect r_i - \vect r_0. \tag{1}

Step 2Sum the individual angular momenta about

O'

The angular momentum about $O'$ is $\vect M' = \sum_i [\vect r_i' \vect p_i]$ , with $\vect p_i$ the momentum of particle $i$ (unchanged by the choice of origin). Substituting (1) and expanding the cross product,

\begin{align*} \vect M' &= \sum_i \big[(\vect r_i - \vect r_0)\,\vect p_i\big] \\ &= \sum_i [\vect r_i \vect p_i] - \Big[\vect r_0 \sum_i \vect p_i\Big]. \tag{2}\end{align*}

The first sum is $\vect M$ by definition, and the second sum is the total momentum $\vect p$ , giving

\vect M' = \vect M - [\vect r_0\vect p]

(3)

Step 3Find when the choice of point does not matter

By the boxed relation, $\vect M'=\vect M$ for every possible $\vect r_0$ exactly when the correction term vanishes identically, which forces the vector it is built from to vanish,

[\vect r_0\vect p] = \vect 0 \text{ for every } \vect r_0 \quad\Longleftrightarrow\quad \vect p = \vect 0.

Since the total momentum of a system is zero precisely in the reference frame moving with its center of inertia, the angular momentum of a system of particles is independent of the reference point exactly in that frame.

Check

At $\vect r_0 = \vect 0$ (so $O' = O$ ) the boxed relation gives $\vect M' = \vect M$ identically, as it must for two labels on the same point.

Remark

This is why the angular momentum measured in the center-of-inertia frame is called the system’s proper angular momentum: it is the one value every observer in that frame agrees on, whatever point of it they choose.