Question

Bipolar.

Short answer

The values of components of acceleration are mentioned below.

$$\begin{gathered} a_u=a\left[\frac{1}{\cos hu+\cos v}\ddot{u}+\frac{\sin h\cos hu}{{\left(\cos hu+\cos v\right)}^2}{\dot{u}}^2\right]+a\left[\frac{2\sin v}{{\left(\cos hu+\cos v\right)}^2}\dot{v}\dot{u}-\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\dot{v}}^2\right] \\ {a}_v=a\left[\frac{1}{\cos hu+\cos v}\ddot{u}-\frac{\sin v}{{\left(\cos hu+\cos v\right)}^2}{\dot{u}}^2\right]+a\left[\frac{2\sin hu}{{\left(\cos hu+\cos v\right)}^2}\dot{v}\dot{u}-\frac{\sin v}{{\left(\cos hu+\cos v\right)}^2}{\dot{v}}^2\right] \end{gathered}$$

Step-by-step solution

Given Information

The Bipolar.

Definition of cylindrical coordinates.

The coordinate system primarily utilized in three-dimensional systems is the cylindrical coordinates of the system. The cylindrical coordinate system is used to find the surface area in three-dimensional space.

Find the value.

The velocity and scale factors are given below.

$$\begin{gathered} h_u=\frac{a}{\cos hu+\cos v} \\ {h}_v=\frac{a}{\cos hu+\cos v} \\ \dot{r}=\dot{u}\frac{a}{\cos hu+\cos v}{\stackrel{⏜}{e}}_u+\dot{v\frac{a}{\cos hu+\cos v}}{\stackrel{⏜}{e}}_u \end{gathered}$$

The lagrangian in the parabolic coordinates are given below.

$$\begin{gathered} L=\frac{1}{2}m{\dot{r}}^2-V\left(u,v\right) \\ L=\frac{1}{2}m\left[\frac{a^2{\dot{u}}^2}{{\left(\cos hu+\cos v\right)}^2}+\frac{a^2{\dot{v}}^2}{{\left(\cos hu+\cos v\right)}^2}\right]-V\left(u,v\right) \end{gathered}$$

Apply Euler-lagrange equation, the above equation becomes as follows

$$\begin{gathered} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{u}}\right)=\frac{\partial L}{\partial u} \\ m{a}^2\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\ddot{u}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^3}{\dot{u}}^2\right]+m{a}^2\left[\frac{2\sin v}{{\left(\cos hu+\cos v\right)}^3}\dot{v}\dot{u}-\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^3}{\dot{v}}^2\right]=\frac{\partial V}{\partial u} \\ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{u}}\right)=\frac{\partial L}{\partial u} \\ m{a}^2\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\ddot{v}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^3}{\dot{u}}^2\right]+ma\left[\frac{2\sin u}{{\left(\cos hu+\cos v\right)}^3}\dot{v}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^3}{\dot{v}}^2\right]=-\frac{\partial V}{\partial v} \end{gathered}$$

Divide the above equation by respective scalar factors.

$$\begin{gathered} ma\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\ddot{u}+\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\dot{u}}^2\right]+m{a}^2\left[\frac{2\sin v}{{\left(\cos hu+\cos v\right)}^2}\dot{v}\dot{u}-\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\dot{v}}^2\right]=-\frac{1}{h_u}\frac{\partial V}{\partial u} \\ ma\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\ddot{v}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^2}{\dot{u}}^2\right]+ma\left[\frac{2\sin u}{{\left(\cos hu+\cos v\right)}^2}\dot{v}\dot{u}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^2}{\dot{v}}^2\right]=-\frac{1}{h_v}\frac{\partial V}{\partial v} \end{gathered}$$

The components of acceleration are mentioned below.

$$\begin{gathered} a_u=a\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\ddot{u}+\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\dot{u}}^2\right]+m{a}^2\left[\frac{2\sin v}{{\left(\cos hu+\cos v\right)}^2}\dot{v}\dot{u}-\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\dot{v}}^2\right] \\ {a}_v=a\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\ddot{v}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^2}{\dot{u}}^2\right]+ma\left[\frac{2\sin u}{{\left(\cos hu+\cos v\right)}^2}\dot{v}\dot{u}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^2}{\dot{v}}^2\right] \end{gathered}$$