Point transformation for a constrained particle

[Lambda96]

Homework Statement

Lagrangian mechanics for a mass particle

Relevant Equations

none

Hi,

unfortunately, I'm not that fit concerning the Lagrangian formalism, so I'm not sure if I solved the problem 1a correctly.

I have now proceeded as follows

the Lagrangian is

$$L=T-U$$

Since there are no constraining or other forces acting on the point mass, I assume that the potential energy is 0 and thus the system has only kinetic energy, i.e.

$$T=\frac{1}{2}m*(\dot{x}^2+\dot{y}^2)$$

I would now represent the entire equation in the x coordinate only, so.

$$x=x$$
$$y=f(x)$$

Insert into the Lagrangian $$T=\frac{1}{2}m*(\dot{x}^2+\dot{f}^2(x))$$

Thus, I would be done with task 1a, or did I do something wrong?

 

[pasmith]

That is correct, except that if y = f(x) then \dot y = f&#039;(x)\dot x, so <br /> T = \frac12 m (1 + f&#039;(x)^2)\dot x^2.

 

[Lambda96]

Thank you pasmith for your help , I had completely forgotten the chain rule.

Concerning task b, the Euler-Lagrange equation is as follows

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

For $$\frac{\partial L}{\partial x}$$

I am not quite sure since f(x) depends on x, would the following then apply

$$\frac{\partial L}{\partial x}=m f''(x)$$

and

$$\frac{\partial L}{\partial \dot{x}}=m\dot{x}+m f'(x)^2\dot{x}$$

$$\frac{d}{dt}m\dot{x}+m f'(x)^2\dot{x}=m\ddot{x}+m f''(x)^2\ddot{x}\dot{x}$$

Then the Euler-Lagrange equation is

$$m f''(x)-m\ddot{x}-m f''(x)^2\ddot{x}\dot{x}=0$$

 

[pasmith]

Your lagrangian is L = \frac12 m g(x)\dot x^2 where g(x) = 1 + f&#039;(x)^2. Hence <br /> \begin{split}<br /> \frac{d}{dt}\left( \frac{\partial L}{\partial \dot x}\right) - \frac{\partial L}{\partial x} &amp;= \frac{d}{dt}(mg(x)\dot x)<br /> - \tfrac12 m g&#039;(x) \dot x^2 \\<br /> &amp;= mg(x) \ddot x + mg&#039;(x) \dot x^2 - \tfrac12 m g&#039;(x) \dot x^2 \\<br /> &amp;= mg(x) \ddot x + \tfrac12 m g&#039;(x) \dot x^2. \end{split} So what is g&#039;(x) using the chain rule?

 

[vela]

$$\frac{\partial L}{\partial \dot{x}}=m\dot{x}+m f'(x)^2\dot{x}$$

$$\frac{d}{dt}m\dot{x}+m f'(x)^2\dot{x}=m\ddot{x}+m f''(x)^2\ddot{x}\dot{x}$$

Don't forget the product rule.

 

[PeroK]

$$\frac{\partial L}{\partial \dot{x}}=m\dot{x}+m f'(x)^2\dot{x}$$

$$\frac{d}{dt}m\dot{x}+m f'(x)^2\dot{x}=m\ddot{x}+m f''(x)^2\ddot{x}\dot{x}$$

This is not good!

I suggest you take $f(x) = \sin x$ or something like that and check your differentiation against a definite function.

 

[LCSphysicist]

I don't want to be rude, but you are missing trivial points about elementary calculus, as chain rule and product rule.

May i ask, have you take those class? Don't try to skip steps in your learning process.

 

[Lambda96]

Thanks to all for the help, especially pasmith

I have now formed the derivative of $g(x)=1+f'(x)^2$ as follows.

The outer derivative is $2f'(x)$and the inner derivative would be $f''(x)$

Thus, the derivative of $g'(x)$ would be $g'(x)=2f'(x)f''(x)$

The Euler-Lagrange equation is $m\ddot{x}+mf'(x)^2\ddot{x}+mf'(x)f''(x)\dot{x}^2$Concerning task part c and d

$\textbf{task c}$
$|l(x)|=\int_{x_0}^{x} \sqrt{1+f'(x)^2}dx$

$\textbf{task d}$
I would now simply parameterize the curve $l(x)=\left(\begin{array}{c} x \\ f(x) \end{array}\right)$ of $[x_0,x]$ with $l_1=x$ and $l_2=f(x)$

Then the Lagrange function would look like $L=\frac{1}{2}m(\dot{l_1}^2+\dot{l_2}^2)$