Chapter 8, Problem 8.1P

To determine

Prove that the positions of two particles can be expressed in terms of their center of mass and their relative position and thereby its kinetic energy acquires the form $T=\frac{1}{2}M{\dot{R}}^2+\frac{1}{2}\mu {\dot{r}}^2$.

Answer to Problem 8.1P

It is proved that the positions of two particles can be expressed in terms of their center of mass and their relative position and hence its kinetic energy is $T=\frac{1}{2}M{\dot{R}}^2+\frac{1}{2}\mu {\dot{r}}^2$.

Explanation of Solution

Write the general expression connecting individual masses, their position vectors, centre of mass and its position vector.

    $\left(m_1+m_2\right)R=m_1{r}_1+m_2{r}_2$        (I)

Here¸$m_1,m_2$ are the masses, $r_1$ is the position vector of 1^st mass, $r_2$ is the position vector of 2^nd mass and $R$ is the position vector of centre of mass.

Rewrite the above equation by replacing $m_1+m_2$ with M.

    $\begin{array}{c}MR=m_1{r}_1+m_2{r}_2\\ R=\frac{m_1{r}_1+m_2{r}_2}{M}\end{array}$

Express $r_1$ in terms of $r_2$

    $r_1=r+r_2$

Rewrite equation (I) by substituting the above equation.

    $\begin{array}{c}\left(m_1+m_2\right)R=m_1\left(r+r_2\right)+m_2{r}_2\\ \left(m_1+m_2\right)R-m_{1}r=\left(m_1+m_2\right)r_2\\ r_2=\frac{\left(m_1+m_2\right)R-m_{1}r}{m_1+m_2}\\ =R-\frac{m_1}{m_1+m_2}r\end{array}$        (II)

Express $r_2$ in terms of $r_1$.

    $r_2=r_1-r$

Rewrite equation (I) by substituting the above equation.

    $\begin{array}{c}\left(m_1+m_2\right)R=m_1{r}_1+m_2\left(r_1-r\right)\\ \left(m_1+m_2\right)R+m_{2}r=\left(m_1+m_2\right)r_1\\ r_1=\frac{\left(m_1+m_2\right)R+m_{2}r}{m_1+m_2}\\ =R+\frac{m_2}{m_1+m_2}r\end{array}$        (III)

Write the expression for total kinetic energy.

    $T=\frac{1}{2}{m}_1{\dot{r}}_1^2+\frac{1}{2}{m}_2{\dot{r}}_2^2$

Here, ${\dot{r}}_1$ is the linear velocity of 1^st mass and ${\dot{r}}_2$ is the linear velocity f 2^nd mass.

Rewrite the above equation by substituting equations (II) and (III).

    $\begin{array}{c}T=\frac{1}{2}{m}_1{\left[\frac{d}{dt}\left(R+\frac{m_2}{m_1+m_2}r\right)\right]}^2+\frac{1}{2}{m}_2{\left[\frac{d}{dt}\left(R-\frac{m_1}{m_1+m_2}r\right)\right]}^2\\ =\frac{1}{2}{m}_1{\left(\dot{R}+\frac{m_2}{m_1+m_2}\dot{r}\right)}^2+\frac{1}{2}{m}_2{\left(\dot{R}-\frac{m_1}{m_1+m_2}\dot{r}\right)}^2\\ =\frac{1}{2}{m}_1\left({\dot{R}}^2+{\left(\frac{m_2}{m_1+m_2}\right)}^2{\dot{r}}^2+2\frac{m_2}{m_1+m_2}\dot{r}\dot{R}\right)+\\ \frac{1}{2}{m}_2{\left({\dot{R}}^2+{\left(\frac{m_1}{m_1+m_2}\right)}^2{\dot{r}}^2-2\frac{m_1}{m_1+m_2}\dot{r}\dot{R}\right)}^2\\ =\frac{1}{2}\left(\begin{array}{l}{\dot{R}}^2\left(m_1+m_2\right)+\left({\left(\frac{m_2}{m_1+m_2}\right)}^2{\dot{r}}^2+m_2{\left(\frac{m_1}{m_1+m_2}\right)}^2{\dot{r}}^2\right)+\\ 2\frac{m_1{m}_2}{m_1+m_2}\dot{r}\dot{R}-2\frac{m_1{m}_2}{m_1+m_2}\dot{r}\dot{R}\end{array}\right)\\ =\frac{1}{2}\left({\dot{R}}^2\left(m_1+m_2\right)+\left(\frac{m_1{m}_2{\dot{r}}^2}{{\left(m_1+m_2\right)}^2}\right)\left(m_1+m_2\right)\right)\\ =\frac{1}{2}\left({\dot{R}}^2\left(m_1+m_2\right)+\left(\frac{m_1{m}_2{\dot{r}}^2}{m_1+m_2}\right)\right)\end{array}$

Conclusion:

Replace $m_1+m_2$ with M and $\frac{m_1{m}_2}{m_1+m_2}$ with $\mu$.

    $T=\frac{1}{2}\left(M{\dot{R}}^2+\mu {\dot{r}}^2\right)$

Here, $\mu$ is the reduced mass.

Therefore, it is proved that the positions of two particles can be expressed in terms of their center of mass and their relative position and hence its kinetic energy is $T=\frac{1}{2}M{\dot{R}}^2+\frac{1}{2}\mu {\dot{r}}^2$.