Four identical particles, each of mass $m$, are positioned at the corners of a square with side leng

Question

Four identical particles, each of mass $m$, are positioned at the corners of a square with side length $L$. At time $t=0$, they are simultaneously given initial velocities. Particle 1 (at corner A) has velocity $\vec{V}_A=V_0 \hat{u}_{A B}$, Particle 2 (at corner B) has velocity $\vec{V}_B=V_0 \hat{u}_{B C}$, Particle 3 (at corner C) has velocity $\vec{V}_C=V_0 \hat{u}_{C D}$, and Particle 4 (at corner D) has velocity $\vec{V}_D=V_0 \hat{u}_{D A}$. Here, $\hat{u}_{X Y}$ represents the unit vector along the side from X to Y . If these particles interact only through mutual gravitational forces, what is the magnitude of the total angular momentum of the system about its center of mass at the instant they are about to collide?

Professor PC Thomas Classes · Answer

Place a coordinate system with origin at the square's center, where the four particles of mass $m$ are positioned at $(-L / 2, L / 2),(L / 2, L / 2),(L / 2,-L / 2)$, and $(-L / 2,-L / 2)$. Their initial velocities are $V_0 \hat{i},-V_0 \hat{j},-V_0 \hat{i}$, and $V_0 \hat{j}$, respectively.
The center of mass remains at the origin due to symmetry, as confirmed by $\boldsymbol{X}_{C M}=\mathbf{0}$ and $Y_{C M}=\mathbf{0}$.
Angular momentum about the center of mass is computed for each particle using $\vec{L}=\vec{r} 	imes m \vec{v}$.
Each calculation yields the same result: $-\frac{1}{2} m L V_0 \hat{k}$.
$$
\vec{L}_{	ext {total }}=4 	imes\left(-\frac{1}{2} m L V_0 \hat{k}ight)=-2 m L V_0 \hat{k}
$$With no external torques, angular momentum is conserved.
The magnitude is $\left|\vec{L}_{	ext {total }}ight|=2 m L V_0$ initially and at collision.