The Angular Momentum of a System of Particles — Question 7

Question 7

Relative motion can be represented with reduced mass

Two particles have masses $m_1=2$ and $m_2=3\,\mathrm{kg}$. Their relative position is $\vec r=\vec r_1-\vec r_2=(4,0,0)\,\mathrm m$ and relative velocity is $\vec v=\vec v_1-\vec v_2=(0,5,0)\,\mathrm{m/s}$.

1. Derive and evaluate angular momentum about the center of mass.

2. Find the same value using each particle separately in the center frame.

Question 7 - Solution

Relative motion can be represented with reduced mass

(a) Derivation and calculation

With $M=m_1+m_2$, $$\vec\rho_1=\frac{m_2}{M}\vec r,\quad\vec\rho_2=-\frac{m_1}{M}\vec r,\quad \vec u_1=\frac{m_2}{M}\vec v,\quad\vec u_2=-\frac{m_1}{M}\vec v.$$ Substitution into the particle sum gives $$\vec L_C=\frac{m_1m_2(m_1+m_2)}{M^2}\vec r\times\vec v =\mu\vec r\times\vec v,\quad\mu=\frac{2(3)}5=1.20\ \mathrm{kg}.$$ Thus $\boxed{\vec L_C=1.2(4)(5)\hat z=24\hat z\ \mathrm{kg\,m^2/s}}$.

(b) Result and interpretation

The individual center-frame variables are $$\vec\rho_1=2.4\hat x,\quad\vec\rho_2=-1.6\hat x,\qquad \vec u_1=3\hat y,\quad\vec u_2=-2\hat y.$$ Their angular momenta are $$L_{1,z}=2.4(2)(3)=14.4,\qquad L_{2,z}=(-1.6)(3)(-2)=9.6,$$ $$\boxed{L_{C,z}=14.4+9.6=24\ \mathrm{kg\,m^2/s}}.$$

Physical check

The relative velocity is $3-(-2)=5$ m/s and the center-frame momenta sum to zero.