Irodov 1.197: Angular Momentum, Split at the Center – Amit & Deepali Sinha Foundation

Problem Irodov 1.197

A system of particles is observed in a reference frame $K$ , and a second frame moves translationally alongside it, always centered on the system’s center of inertia. Let $\vect r_C$ be the radius vector of the center of inertia relative to a point $O$ of frame $K$ , let $\vect p$ be the system’s total momentum as measured in $K$ , and let $\tilde{\vect M}$ be the system’s angular momentum about the center of inertia as measured in the comoving frame (its proper angular momentum). Show that the system’s angular momentum $\vect M$ about the point $O$ , as measured in $K$ , equals

\vect M = \tilde{\vect M} + [\vect r_C\vect p].

Particle m_i reached from O directly by r_i, or in two steps through the center of inertia C: r_C then r_i. M and M are the angular momenta about O and about C in the frame that translates with C. — Particle $m_i$ reached from $O$ directly by $\vect r_i$ , or in two steps through the center of inertia $C$ : $\vect r_C$ then $\tilde{\vect r}_i$ . $\vect M$ and $\tilde{\vect M}$ are the angular momenta about $O$ and about $C$ in the frame that translates with $C$ .

Given

System of particles, frame $K$ , point $O$

Find

Show

\vect M = \tilde{\vect M} + [\vect r_C\vect p]

Solution

Step 1Split every particle’s motion into a center-of-inertia part and a relative part

Let $\vect r_C$ and $\vect v_C$ be the position and velocity of the center of inertia in $K$ , and let $\tilde{\vect r}_i = \vect r_i - \vect r_C$ , $\tilde{\vect v}_i = \vect v_i - \vect v_C$ be particle $i$ ’s position and velocity relative to the center of inertia. Because the comoving frame only translates, $\tilde{\vect v}_i$ is exactly the time derivative of $\tilde{\vect r}_i$ , so

\vect r_i = \vect r_C + \tilde{\vect r}_i, \qquad \vect v_i = \vect v_C + \tilde{\vect v}_i. \tag{1}

Step 2Substitute into the definition of

\vect M

and expand

\begin{align*} \vect M &= \sum_i m_i [\vect r_i \vect v_i] = \sum_i m_i \big[(\vect r_C+\tilde{\vect r}_i)(\vect v_C+\tilde{\vect v}_i)\big] \\ &= \Big(\sum_i m_i\Big)[\vect r_C \vect v_C] + \Big[\vect r_C \sum_i m_i\tilde{\vect v}_i\Big] + \Big[\Big(\sum_i m_i\tilde{\vect r}_i\Big) \vect v_C\Big] + \sum_i m_i [\tilde{\vect r}_i \tilde{\vect v}_i]. \tag{2}\end{align*}

Step 3Drop the two cross terms by the definition of the center of inertia

The center of inertia satisfies $\sum_i m_i \tilde{\vect r}_i = \vect 0$ at every instant, by definition. Differentiating that identity in time, and using that the comoving frame only translates so it contributes no extra term,

\frac{d}{dt}\sum_i m_i\tilde{\vect r}_i = \sum_i m_i\tilde{\vect v}_i = \vect 0. \tag{3}

Both sums vanish identically, so the second and third terms of (2) drop out.

Step 4Identify the two surviving terms

Write the total mass as $m=\sum_i m_i$ and the total momentum as $\vect p = m\vect v_C$ , so the first term of (2) becomes

\Big(\sum_i m_i\Big)[\vect r_C\vect v_C] = m[\vect r_C\vect v_C] = [\vect r_C\,(m\vect v_C)] = [\vect r_C\vect p].

The last term of (2) is, by definition, the proper angular momentum of the system, $\tilde{\vect M} = \sum_i m_i [\tilde{\vect r}_i \tilde{\vect v}_i]$ , built entirely from positions and velocities relative to the center of inertia. Collecting both surviving terms,

\vect M = \tilde{\vect M} + [\vect r_C\vect p]

(4)

Check

If $O$ is placed at the center of inertia itself, $\vect r_C = \vect 0$ and the identity collapses to $\vect M = \tilde{\vect M}$ , as it must: the two angular momenta about the same point can only differ by the term the point’s offset carries.

Remark

The sums that vanish in (3) are also why the angular momentum about the center of inertia comes out the same whether it is built from the frame $K$ ’s velocities or from the comoving frame’s: the term that would separate the two constructions is exactly the third term of (2), $\big[\big(\sum_i m_i\tilde{\vect r}_i\big)\vect v_C\big]$ , and it vanishes identically, for any $\vect r_C$ , by the very definition of the center of inertia.