Question

Prove that the center of masses of two particles divides the line joining the particles in the inverse ratio of their masses.

Answer 1

Use the general equation to find the center of mass of any object under consideration by considering the two particles to be joined by an imaginary line. Shift the origin of the coordinate axes to the center of mass of the system. Evaluate the equation and study the obtained equation.

Complete step by step solution:
Let the mass of the two particles be ${{m}_{1}}$ and ${{m}_{2}}$ . Let the center of mass of the system be ${{\overset{\to }{\mathop{R}}\,}_{cm}}$ . Let ${{\overset{\to }{\mathop{r}}\,}_{1}}$ and ${{\overset{\to }{\mathop{r}}\,}_{2}}$ be the position vectors of ${{m}_{1}}$ and ${{m}_{2}}$ respectively. The center of mass of any object is given by
${{\overset{\to }{\mathop{R}}\,}_{cm}}=\dfrac{{{m}_{1}}{{\overset{\to }{\mathop{r}}\,}_{1}}+{{m}_{2}}{{\overset{\to }{\mathop{r}}\,}_{2}}}{{{m}_{1}}+{{m}_{2}}}$
Now, let us shift the origin of the coordinate axes to the center of mass of the two-body system.
This gives,
$0=\dfrac{{{m}_{1}}{{\overset{\to }{\mathop{r}}\,}_{1}}+{{m}_{2}}{{\overset{\to }{\mathop{r}}\,}_{2}}}{{{m}_{1}}+{{m}_{2}}}$
Cross multiplying the above equation gives us
${{m}_{1}}{{\overset{\to }{\mathop{r}}\,}_{1}}+{{m}_{2}}\overset{\to }{\mathop{{{r}_{2}}}}\,=0$
Taking one term to the other side gives,
${{m}_{1}}{{\overset{\to }{\mathop{r}}\,}_{1}}=-{{m}_{2}}{{\overset{\to }{\mathop{r}}\,}_{2}}$
We can remove the negative sign in the equation by taking purely the magnitude of the position vector. That further gives us
${{m}_{1}}{{r}_{1}}={{m}_{2}}{{r}_{2}}$
By cross multiplying, we get
$\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{r}_{2}}}{{{r}_{1}}}$
That is, the ratio of the line joining the center of mass of the two particles is equal to the inverse ratio of their masses.

Note:
The center of mass of an object or a system is the point where one can imagine the whole mass of the object or the system( mass points in a system) to be concentrated at a single point. For a system of $n$ mass points, the equation is given as
${{\overset{\to }{\mathop{R}}\,}_{cm}}=\dfrac{{{m}_{1}}{{\overset{\to }{\mathop{r}}\,}_{1}}+{{m}_{2}}{{\overset{\to }{\mathop{r}}\,}_{2}}+...+m{}_{n}{{\overset{\to }{\mathop{r}}\,}_{n}}}{{{m}_{1}}+{{m}_{2}}+...+{{m}_{n}}}$