Question

$\begin{array}{l}\mathbf{x}\mathbf{=}\mathbf{uv}\\ \mathbf{y}\mathbf{=}\mathbf{u}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{v}}^{\mathbf{2}}}\end{array}$

Short answer

The required values are mentioned below.

$$\begin{gathered} \stackrel{.}{s}=\stackrel{.}{u}{\hat{e}}_u+\frac{u}{\sqrt{1-v^2}}\stackrel{.}{v}{\hat{e}}_v \\ L=\frac{1}{2}m\left({\stackrel{.}{u}}^2+\frac{u^2}{1-v^2}\stackrel{.}{v^2}\right)-V\left(u,v\right) \\ -\frac{1}{h_u}\frac{\partial V}{\partial u}=m\left\{\stackrel{..}{u}-\frac{u}{1-v^2}\stackrel{.}{v^2}\right\} \\ -\frac{1}{h_v}\frac{\partial V}{\partial v}=\left\{\frac{u}{\sqrt{1-v^2}}\stackrel{..}{v}+\frac{2}{\sqrt{1-v^2}}\stackrel{.}{u}\stackrel{.}{v}+\frac{uv}{{\left(1-v^2\right)}^{3/2}}\stackrel{.}{v^2}\right\} \end{gathered}$$

Step-by-step solution

Given Information

The equations are mentioned below.

$$\begin{gathered} x=uv \\ y=u\sqrt{1-v^2} \end{gathered}$$

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 equations are mentioned below.

$$\begin{gathered} x=uv \\ y=u\sqrt{1-v^2} \end{gathered}$$

The formula for position vector s is$s=uv{\hat{e}}_x+u\sqrt{1-v^2{\hat{e}}_y}$.

Compute the velocity.

$$\begin{gathered} \stackrel{.}{s}=\frac{\partial s}{\partial u}\stackrel{.}{u}\frac{\partial s}{\partial v}\stackrel{.}{v} \\ \frac{\partial s}{\partial u}=v{\hat{e}}_x+\sqrt{1-v^2{\hat{e}}_y} \\ \frac{\partial s}{\partial v}=v{\hat{e}}_x-\frac{uv}{\sqrt{1-v^2}}{\hat{e}}_y \end{gathered}$$

Orthogonal metric tensors are diagonal hence, the scalar vectors are given below.

$$\begin{gathered} h_u=1 \\ {h}_v=\frac{u}{\sqrt{1-v^2}} \end{gathered}$$

Express the velocity in terms of u and v.

$\stackrel{.}{s}=\stackrel{.}{u}{\hat{e}}_u+\frac{u}{\sqrt{1-v^2}}\stackrel{.}{v}{\hat{e}}_v$

Express the Lagrangian in terms of u and v.

$$\begin{gathered} L=\frac{1}{2}m{\stackrel{.}{s}}^2-V\left(u,v\right) \\ L=\frac{1}{2}m\left(\stackrel{.}{u^2}+\frac{u^2}{1-v^2}\stackrel{.}{v^2}\right)-V\left(u,v\right) \end{gathered}$$

Apply Euler-lagrange equation.

$$\begin{gathered} \frac{d}{dt}\left(\frac{\partial L}{\partial {\displaystyle \stackrel{.}{u}}}\right)=\frac{\partial L}{\partial u} \\ m\left\{u-\frac{u}{1-v^2}\stackrel{.}{v^2}\right\}=-\frac{\partial V}{\partial u} \\ \frac{d}{dt}\left(\frac{\partial L}{\partial {\displaystyle \stackrel{.}{v}}}\right)=\frac{\partial L}{\partial v} \\ m\left\{\frac{u^2}{1-v^2}v+\frac{2u}{1-v^2}\stackrel{.}{u}\stackrel{.}{v}+\frac{u^{2}v}{\left(1-v\right)}\stackrel{.}{v^2}\right\}=-\frac{\partial V}{\partial v} \end{gathered}$$

Find the components of acceleration.

$$\begin{gathered} -\frac{1}{h_u}\frac{\partial V}{\partial u}=m\left\{\stackrel{..}{u}-\frac{u}{1-v^2}\stackrel{.}{v^2}\right\} \\ -\frac{1}{h_u}\frac{\partial V}{\partial v}=\left\{\frac{u}{\sqrt{1-v^2}}\stackrel{..}{v}+\frac{2}{\sqrt{1-v^2}}\stackrel{.}{u}\stackrel{.}{v}+\frac{uv}{{\left(1-v^2\right)}^{3/2}}\stackrel{.}{v^2}\right\} \end{gathered}$$

Hence, the required values are mentioned below.

$$\begin{gathered} \stackrel{.}{s}=\stackrel{.}{u}{\hat{e}}_u+\frac{u}{\sqrt{1-v^2}}\hat{v}{\hat{e}}_v \\ L=\frac{1}{2}m\left(\stackrel{.}{u^2}+\frac{u^2}{1-v^2}\stackrel{.}{v^2}\right)-V\left(u,v\right) \\ -\frac{1}{h_u}\frac{\partial V}{\partial u}m\left\{\stackrel{..}{u}-\frac{u}{1-v^2}\stackrel{.}{v^2}\right\} \\ -\frac{1}{h_v}\frac{\partial V}{\partial v}=\left\{\frac{u}{\sqrt{1-v^2}}\stackrel{..}{v}+\frac{2}{\sqrt{1-v^2}}\stackrel{..}{u}\stackrel{.}{v}+\frac{uv}{{\left(1-v^2\right)}^{3/2}}\stackrel{.}{v^2}\right\} \end{gathered}$$