Chapter 8, Problem 8.2P

(a)

To determine

Decompose the Lagrange equation into equation 8.13.

Answer to Problem 8.2P

Lagrange equation is decomposed into equation 8.13.

Explanation of Solution

Write the general expression for position coordinate of centre of mass of system.

    $R=\frac{m_1{r}_1+m_2{r}_2}{m_1+m_2}$        (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.

Write the equation for the total mass of the system.

    $M=m_1+m_2$

  $\begin{array}{l}R=\frac{m_1{r}_1+m_2{r}_2}{M}\\ MR=m_1{r}_1+m_2{r}_2\end{array}$        (II)

Draw the vector diagram showing the position vectors of individual masses and centre of mass.

Express $r_1$ in terms of $r_2$

    $r_1=r+r_2$

Here, $r$ is the displacement vector.

Rewrite equation (II) by adding $m_2{r}_1$ on both sides.

    $\begin{array}{c}MR+m_2{r}_1=m_1{r}_1+m_2{r}_2+m_2{r}_1\\ MR+m_2{r}_1-m_2{r}_2=m_1{r}_1+m_2{r}_1\\ MR+m_2\left(r_1-r_2\right)=\left(m_1+m_2\right)r_1\\ R+m_{2}r=M{r}_1\\ r_1=\frac{R+m_{2}r}{M}\end{array}$

Write the equation for linear velocity of 1^st mass from the above equation.

      ${\dot{r}}_1=\dot{R}+\frac{m_2}{M}\dot{r}$        (III)

Here, ${\dot{r}}_1$ is the linear velocity of 1^st mass.

Rewrite equation (II) by adding $m_1{r}_2$ on both sides.

    $\begin{array}{c}MR+m_1{r}_2=m_1{r}_1+m_2{r}_2+m_1{r}_2\\ MR+m_1{r}_2-m_1{r}_1=m_2{r}_2+m_1{r}_2\\ MR-m_1\left(r_1-r_2\right)=\left(m_1+m_2\right)r_2\\ R-m_{1}r=M{r}_2\\ r_2=\frac{MR-m_{1}r}{M}\\ =R-\frac{m_1}{M}r\end{array}$

Write the equation for linear velocity of 2^nd mass from the above equation.

    ${\dot{r}}_2=\dot{R}-\frac{m_1}{M}\dot{r}$        (IV)

Here, ${\dot{r}}_2$ is the linear velocity f 2^nd mass.

Write the expression for total kinetic energy.

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

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

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

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

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

Here, $\mu$ is the reduced mass.

Write the expression for total potential energy. It is the sum of interaction potential energy due to two masses and the gravitational potential energy.

    $U=U\left(r\right)+U_g$

Here, $U$ is the total potential energy, $U\left(r\right)$ is the interaction potential energy due to two masses and $U_g$ is the gravitational potential energy.

Write the equation for $U_g$.

    $U_g=mgZ$

Rewrite the equation for $U$ by substituting the above equation.

    $U=U\left(r\right)+mgZ$        (VI)

Write the general form of LaGrange of a system.

    $L=T-U$

Rewrite the above equation by substituting equations (V) and (VI).

     $\begin{array}{c}L=\frac{1}{2}\left(M{\dot{R}}^2+\mu {\dot{r}}^2\right)-\left(U\left(r\right)+mgZ\right)\\ =\frac{1}{2}\left(M{\dot{R}}^2+\mu {\dot{r}}^2\right)-U\left(r\right)-mgZ\end{array}$

Express $\dot{R}$ using cartesian coordinates.

    $\dot{R}={\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2$

Express $\dot{r}$ using cartesian coordinates.

    $\dot{r}={\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2$

Here, x, y, z are relative coordinates.

Rewrite the equation for $L$ by substituting the above two equations.

    $L=\frac{1}{2}\left[M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2+\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2\right]-U\left(x,y,z\right)-mgZ$

Split the Lagrange into two parts-Lagrange of mass $M$ moving with the speed of centre of mass and Lagrange of mass $\mu$ moving with speed of relative position $r$.

Write the Lagrange of mass $M$ moving with the speed of centre of mass from the above equation.

    $L_{cm}=\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ$

Here, $L_{cm}$ is the Lagrange of mass $M$ moving with the speed of centre of mass from the above equation.

Write the Lagrange of mass $\mu$ moving with speed of relative position $r$.

    $L_{rel}=\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)$

Here, $L_{rel}$ is the Lagrange of mass $\mu$ moving with speed of relative position $r$.

So, the total Lagrange can be expressed as follows.

    $L=L_{cm}+L_{rel}$

Conclusion:

Therefore, the Lagrange equation is decomposed into equation 8.13.

(b)

To determine

Write the Lagrange equation in cartesian coordinates, three Lagrange equation for relative coordinates and prove that the motion is equivalent to that of a single particle of mass equal to that of reduced mass of the system.

Answer to Problem 8.2P

Lagrange equation in cartesian coordinates are $\dot{X}=\text{constant}$, $\dot{Y}=\text{constant}$,  $\ddot{Z}=-g$ and Lagrange equations for relative coordinates are $-\frac{\partial U}{\partial x}=\mu \ddot{x}$, $-\frac{\partial U}{\partial y}=\mu \ddot{y}$, $-\frac{\partial U}{\partial z}=\mu \ddot{z}$ and hence proved that motion of system is equivalent to that of a single particle of mass equal to that of reduced mass of the system.

Explanation of Solution

Write the Lagrange of mass $M$ moving with the speed of centre of mass from the above equation.

    $L_{cm}=\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ$

Write the Lagrange equation in x-direction of centre of mass.

    $\frac{\partial L_{cm}}{\partial X}=\frac{d}{dt}\left(\frac{\partial L_{cm}}{\partial \dot{X}}\right)$

Rewrite the above equation by substituting the previous equation.

    $\begin{array}{c}\frac{\partial \left[\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ\right]}{\partial X}=\frac{d}{dt}\left(\frac{\partial \left[\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ\right]}{\partial \dot{X}}\right)\\ 0=\frac{d}{dt}\left(\frac{1}{2}M\left(2\dot{X}\right)-0\right)\\ 0=M\frac{d\dot{X}}{dt}\\ M\ddot{X}=0\Rightarrow \dot{X}=\text{constant}\end{array}$

Write the Lagrange equation in y-direction of centre of mass.

    $\frac{\partial L_{cm}}{\partial Y}=\frac{d}{dt}\left(\frac{\partial L_{cm}}{\partial \dot{Y}}\right)$

Rewrite the above equation by substituting $\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ$ for $L_{cm}$.

    $\begin{array}{c}\frac{\partial \left[\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ\right]}{\partial Y}=\frac{d}{dt}\left(\frac{\partial \left[\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ\right]}{\partial \dot{Y}}\right)\\ 0=\frac{d}{dt}\left(\frac{1}{2}M\left(2\dot{Y}\right)-0\right)\\ 0=M\frac{d\dot{Y}}{dt}\\ M\ddot{Y}=0\Rightarrow \dot{Y}=\text{constant}\end{array}$

Write the Lagrange equation in z-direction of centre of mass.

    $\frac{\partial L_{cm}}{\partial Z}=\frac{d}{dt}\left(\frac{\partial L_{cm}}{\partial \dot{Z}}\right)$

Rewrite the above equation by substituting $\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ$ for $L_{cm}$.

    $\begin{array}{c}\frac{\partial \left[\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-mgZ\right]}{\partial Z}=\frac{d}{dt}\left(\frac{\partial \left[\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2+{\dot{Z}}^2\right)}^2-MgZ\right]}{\partial \dot{Z}}\right)\\ 0=\frac{d}{dt}\left(\frac{1}{2}M\left(2\dot{Z}\right)-Mg\right)\\ -Mg=M\ddot{Z}\\ \ddot{Z}=-g\end{array}$

Above results show that the system is moving at constant speed in x and y direction and is moving downwards with gravitational acceleration.

Write the Lagrange equation in relative coordinate x.

    $\frac{\partial L_{cm}}{\partial x}=\frac{d}{dt}\left(\frac{\partial L_{cm}}{\partial \dot{x}}\right)$

Rewrite the above equation by substituting $\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)$ for $L_{cm}$.

  $\begin{array}{c}\frac{\partial \left[\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)\right]}{\partial x}=\frac{d}{dt}\left(\frac{\partial \left[\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)\right]}{\partial \dot{x}}\right)\\ -\frac{\partial U}{\partial x}=\frac{d}{dt}\left(\frac{1}{2}\mu \left(2\dot{x}+0+0\right)-0\right)\\ -\frac{\partial U}{\partial x}=\mu \frac{d\dot{x}}{dt}\\ -\frac{\partial U}{\partial x}=\mu \ddot{x}\end{array}$

Write the Lagrange equation in relative coordinate y.

    $\frac{\partial L_{cm}}{\partial y}=\frac{d}{dt}\left(\frac{\partial L_{cm}}{\partial \dot{y}}\right)$

Rewrite the above equation by substituting $\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)$ for $L_{cm}$.

  $\begin{array}{c}\frac{\partial \left[\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)\right]}{\partial y}=\frac{d}{dt}\left(\frac{\partial \left[\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)\right]}{\partial \dot{y}}\right)\\ -\frac{\partial U}{\partial y}=\frac{d}{dt}\left(\frac{1}{2}\mu \left(0+2\dot{y}+0\right)-0\right)\\ -\frac{\partial U}{\partial y}=\mu \frac{d\dot{y}}{dt}\\ -\frac{\partial U}{\partial y}=\mu \ddot{y}\end{array}$

Write the Lagrange equation in relative coordinate z.

    $\frac{\partial L_{cm}}{\partial z}=\frac{d}{dt}\left(\frac{\partial L_{cm}}{\partial \dot{z}}\right)$

Rewrite the above equation by substituting $\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)$ for $L_{cm}$.

$\begin{array}{c}\frac{\partial \left[\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)\right]}{\partial z}=\frac{d}{dt}\left(\frac{\partial \left[\frac{1}{2}\mu {\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)}^2-U\left(x,y,z\right)\right]}{\partial \dot{z}}\right)\\ -\frac{\partial U}{\partial z}=\frac{d}{dt}\left(\frac{1}{2}\mu \left(0+0+2\dot{z}\right)-0\right)\\ -\frac{\partial U}{\partial z}=\mu \frac{d\dot{z}}{dt}\\ -\frac{\partial U}{\partial z}=\mu \ddot{z}\end{array}$

Conclusion:

Therefore, the Lagrange equation in cartesian coordinates are $\dot{X}=\text{constant}$, $\dot{Y}=\text{constant}$,  $\ddot{Z}=-g$ and Lagrange equations for relative coordinates are $-\frac{\partial U}{\partial x}=\mu \ddot{x}$, $-\frac{\partial U}{\partial y}=\mu \ddot{y}$, $-\frac{\partial U}{\partial z}=\mu \ddot{z}$ and hence proved that motion of system is equivalent to that of a single particle of mass equal to that of reduced mass of the system.