Set up Lagrange’s equations in cylindrical coordinates for a particle of mass $\mathbf{m}$in a potential field $\mathbf{V}\left(r,\theta ,z\right)$. Hint: $\mathit{v}\mathbf{=}\frac{\mathbf{d}\mathbf{s}}{\mathbf{d}\mathbf{t}}$; writein cylindrical coordinates.

It is given that mass is $\mathrm{m}$and potential field is $V\left(r,\theta ,z\right)$.

In the same way that geometry is the study of shape and algebra is the study of generalisations of arithmetic operations, calculus, sometimes known as infinitesimal calculus or "the calculus of infinitesimals," is the mathematical study of continuous change.. Differentiation and integration are the two main branches.

Let the velocity vector be $\overline{v}$, kinetic energy be $\mathrm{T}$and Lagrangian be $\mathrm{L}$. Write the velocity vector and its kinetic energy in cylindrical coordinates.

$\overline{v}=\dot{r}\hat{r}+r\dot{\theta }\hat{\theta }+\dot{z}\hat{z}$

$T=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2\right)$

Now define the Lagrangian from kinetic energy and potential field.

$$\begin{gathered} L=T-V \\ L=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2\right)-V\left(r,\theta ,z\right) \end{gathered}$$

Write the Euler’s equation with $\mathrm{r}$as its degree of freedom, and find it’s derivatives.

$$\begin{gathered} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right)-\frac{\partial L}{\partial r}=0 \\ \frac{\partial L}{\partial \dot{r}}=m\dot{r} \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}=m\ddot{r} \\ \frac{\partial L}{\partial r}=0 \end{gathered}$$

Use the equations above and obtain, $m\left(\ddot{r}-r{\dot{\theta }}^2\right)=-\frac{\partial V}{\partial r}$.

Write the Euler’s equation with $\mathrm{\theta }$as its degree of freedom, and find it’s derivatives.

$$\begin{gathered} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)-\frac{\partial L}{\partial \theta }=0 \\ \frac{\partial L}{\partial \dot{\theta }}=m{r}^2\dot{\theta } \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}=m\left(2r\dot{r}\dot{\theta }+r^2\ddot{\theta }\right) \\ \frac{\partial L}{\partial \theta }=-\frac{\partial V}{\partial \theta } \end{gathered}$$

Use the equations above and obtain, $m\left(2\dot{r}\dot{\theta }+r\ddot{\theta }\right)=-\frac{1}{r}\frac{\partial V}{\partial \theta }$.

Write the Euler’s equation with $\mathrm{z}$as its degree of freedom, and find it’s derivatives.

$$\begin{gathered} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{z}}\right)-\frac{\partial L}{\partial z}=0 \\ \frac{\partial L}{\partial \dot{z}}=m\dot{z} \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{z}}=m\ddot{z} \\ \frac{\partial L}{\partial z}=-\frac{\partial V}{\partial z} \end{gathered}$$

Use the equations above and obtain, $m\ddot{z}=-\frac{\partial V}{\partial z}$.

Therefore, Lagrange’s equations in cylindrical coordinates for a particle of mass $\mathrm{m}$in a potential field $V\left(r,\theta ,z\right)$, are $m\left(\ddot{r}-r{\dot{\theta }}^2\right)=-\frac{\partial V}{\partial r}$,$m\left(2\dot{r}\dot{\theta }+r\ddot{\theta }\right)=-\frac{1}{r}\frac{\partial V}{\partial \theta }$, and $m\ddot{z}=-\frac{\partial V}{\partial z}$, respectively.