Two particles each of mass $\mathbf{m}$are connected by an (inextensible) string of length$\mathbf{l}$. One particle moves on a horizontal table (assume no friction), The string passes through a hole in the table and the particle at the lower end moves up and down along a vertical line.Find the Lagrange equations of motion of the particles. Hint: Let thex coordinates of the particle on the table be$\mathbf{r}$ and$\mathbf{\theta }$, and let the coordinate of the other particle be$\mathbf{z}$. Eliminate one variable from$\mathbf{L}\mathbf{(}\mathbf{using}\mathbf{}\mathbf{r}\mathbf{+}\left|z\right|\mathbf{=}\mathbf{l}\mathbf{)}$and write two Lagrange equations.

The given values are two particles each of mass $\mathrm{m}$are connected by an (inextensible) string of length $\mathrm{l}$. One particle moves on a horizontal table (assume no friction), The string passes through a hole in the table and the particle at the lower end moves up and down along a vertical line.The given equation is $\left(\mathrm{r}+\left|\mathrm{z}\right|=\mathrm{l}\right)$

The Lagrange equations are used to construct the equations of motion of a solid mechanics issue in matrix form, including damping.

For a particle on the flat surface use polar coordinates, and kinetic energy in polar coordinates has the following form:

${\mathrm{T}}_1=\frac{1}{2}\left({\dot{\mathrm{r}}}^2+{\mathrm{r}}^2{\dot{\mathrm{\varphi }}}^2\right)$

To add the kinetic energy of the second particle to the total kinetic energy. Since a constraint that $\mathrm{r}+\left|\mathrm{z}\right|=\mathrm{l}$, take a derivative with respect to time to obtain that $\dot{\mathrm{r}}=\dot{\mathrm{z}}$ (assuming that $\mathrm{z}$direction is facing away from the table and that particle doesn't climb the table which would induce a change in sign in coordinate). From this it follows that the kinetic energy of the hanging particle is given by:

$$\begin{gathered} {\mathrm{T}}_2=\frac{1}{2}\mathrm{m}{\dot{\mathrm{z}}}^2 \\ =\frac{1}{2}\mathrm{m}{\dot{\mathrm{r}}}^2 \end{gathered}$$

So, the total kinetic energy is

$$\begin{gathered} \mathrm{T}={\mathrm{T}}_1+{\mathrm{T}}_2 \\ =\frac{1}{2}\mathrm{m}\left(2{\dot{\mathrm{r}}}^2+{\mathrm{r}}^2{\dot{\mathrm{\varphi }}}^2\right) \end{gathered}$$

Now, for the potential energy, the particle on the table has a constant gravitational potential energy since it is a on a flat surface and in turn would produce a constant term in the Lagrangian (which we can simply ignore since taking derivatives in Euler equations would eliminate a constant term). The potential energy of a hanging particle is

$$\begin{gathered} \mathrm{V}=-\mathrm{mgz} \\ =-\mathrm{mg}\left(\mathrm{l}-\mathrm{r}\right) \end{gathered}$$

Since the $\mathrm{z}$- axis points in the bottom direction (away from the bottom of the table). Therefore, the full Lagrangian is

$$\begin{gathered} \mathrm{L}=\mathrm{T}-\mathrm{V} \\ =\frac{1}{2}\mathrm{m}\left(2{\dot{\mathrm{r}}}^2+{\mathrm{r}}^2{\dot{\mathrm{\varphi }}}^2\right)+\mathrm{mg}\left(\mathrm{l}-\mathrm{r}\right) \end{gathered}$$

First observer the Euler equation for $\mathrm{r}$degree of freedom. The Euler equations reads:

$\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{r}}}-\frac{\partial \mathrm{L}}{\partial \mathrm{r}}=0$

First, calculate the required derivatives.

$$\begin{gathered} \frac{\partial \mathrm{L}}{\partial \dot{\mathrm{r}}}=2\mathrm{m}\dot{\mathrm{r}}\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{r}}} \\ =2\mathrm{m}\ddot{\mathrm{r}}\frac{\partial \mathrm{L}}{\partial \mathrm{r}} \\ =\mathrm{mr}{\dot{\varphi }}^2-\mathrm{mg} \end{gathered}$$

Using all of the equations above,

$2\ddot{\mathrm{r}}-\mathrm{r}{\dot{\varphi }}^2+\mathrm{g}=0$

Next, move onto the $\varphi$degree of freedom. The Euler equations reads:

$\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{\varphi }}}-\frac{\partial \mathrm{L}}{\partial \mathrm{\varphi }}=0$

First, calculate the required derivatives.

role="math" localid="1664351753329" $$\begin{gathered} \frac{\partial \mathrm{L}}{\partial \dot{\mathrm{\varphi }}}={\mathrm{mr}}^2\dot{\mathrm{\varphi }}\frac{\partial \mathrm{L}}{\partial \mathrm{\varphi }} \\ =0 \end{gathered}$$

Calculate the time derivative, since it is a constant with respect to time (it follows from the Euler equation) and it is often written in that form (without taking the derivative with respect to time) because it represents a constant of motion (a quantity which doesn't change with time. So, inserting the calculated derivatives into the Euler equation we obtain (after diving by factors $\mathrm{m}$)

$\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{r}}^2\dot{\mathrm{\varphi }}\right)=0$

Therefore,the two Lagrange equations of motion of the particles is $2\ddot{\mathrm{r}}-\mathrm{r}{\dot{\mathrm{\varphi }}}^2+\mathrm{g}=0$ and $\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{r}}^2\dot{\mathrm{\varphi }}\right)=0$ .