Question

Two unequal point masses $M$ and $m$, are a distance $d$ apart and revolve with same angular speed about their common center of mass. The ratio of their angular momentum about their center of mass $\frac{L_M}{L_m}$ is :

• A. ${\left(\frac{M}{m}\right)}^2$
• B. $\frac{M}{m}$
• C. $\frac{m}{M}$
• D. ${\left(\frac{m}{M}\right)}^2$

Answer 1

The correct option is C $\frac{m}{M}$

$\begin{array}{c}\text{Two particles}A\text{and}B\text{having mass}M\text{and}m\text{,}\\ \text{respectively.}\end{array}$

$\begin{array}{c}\text{Let, both}A\text{and}B\text{revolves about}C\text{with}\\ \text{angular speed}{\omega }_{\text{.}}\end{array}$

$\begin{array}{c}\text{Coordinates of}B=(d,0)\\ \text{Thus, coordinates of centre of mass (C) is}(\frac{M\times 0+m\times d}{M+m},0)\\ =(\frac{m}{M+m}d,0)\end{array}$

$\begin{array}{cc}& \text{Therefore,}\frac{L_M}{L_m}=\frac{I_M\cdot C_M}{I_m\cdot C_m}=\frac{I_M}{I_m}(\because C_m={\omega }_m=\omega )\\ & \therefore \frac{L_M}{Lm}=\frac{[M\cdot {\left(\frac{m}{m+M}\right)}^2{d}^2]}{[m\cdot {(d-\frac{md}{m+M})}^2]}=\frac{m}{M}\end{array}$