Question

Elliptical cylinder.

Short answer

The values of components of acceleration are mentioned below.

$$\begin{gathered} a_u=a\left[\sqrt{\sin h^{2}u+{\sin }^{2}v}\ddot{u}+\frac{\sin h\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\dot{u}}^2\right]+a\left[\frac{\sin v\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}\dot{u}\dot{v}-\frac{\sin hu\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\dot{v}}^2\right] \\ {a}_v=a\left[\sqrt{\sin h^{2}u+{\sin }^{2}v}\ddot{u}-\frac{\sin v\cos hv}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\dot{u}}^2\right]+a\left[\frac{\sin hu\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}\dot{u}\dot{v}-\frac{\sin vu\cos hu}{\sqrt{\sin h^{2}u+{\sin }^{2}v}}{\dot{v}}^2\right] \\ {a}_z=\ddot{z} \end{gathered}$$

Step-by-step solution

Given Information

The Elliptical cylinder.

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.

$$\begin{gathered} h_u=a\sqrt{\sin h^{2}u+\sin h^{2}v} \\ {h}_v=a\sqrt{\sin h^{2}u+\sin h^{2}v} \\ {h}_z=1 \\ \dot{r}=\dot{u}a\sqrt{\sin h^{2}u+\sin h^{2}v}{\stackrel{⏜}{e}}_u+\dot{v}a\sqrt{\sin h^{2}u+\sin h^{2}v}{\stackrel{⏜}{e}}_v+\dot{z}{\stackrel{⏜}{e}}_z \end{gathered}$$

The lagrangian in the parabolic coordinates are given below.

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

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

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

Solve further.

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

Divide the above equation by respective scalar factors.

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

The components of acceleration are mentioned below.

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

The values of components of acceleration are mentioned below.

$$\begin{gathered} a_u=\ddot{u}\sqrt{u^2+v^2}+\frac{u}{\sqrt{u^2+v^2}}{\dot{u}}^2+\frac{2v}{\sqrt{u^2+v^2}}\dot{u}\dot{v}-\frac{u}{\sqrt{u^2+v^2}}{\dot{v}}^2 \\ {a}_v=\ddot{v}\sqrt{u^2+v^2}-\frac{u}{\sqrt{u^2+v^2}}{\dot{u}}^2+\frac{2u}{\sqrt{u^2+v^2}}\dot{u}\dot{v}-\frac{v}{\sqrt{u^2+v^2}}{\dot{v}}^2 \\ {a}_z=\ddot{z} \end{gathered}$$