My Epic Fail at Deriving an Equation with Lagrange

[Arm]

Homework Statement

Two positive charges are placed one meter away from eachother. One is held in place and one is allowed to move. Find an equation that models the velocity of the charge allowed to move after it is released from rest.

Relevant Equations

$$KE = \frac{mv^2}{2}$$
$$U_E= \frac{k q_1 q_2}{r}$$
$$L = KE - PE$$
$$\frac{\partial L}{\partial x} - \frac{d}{dt} ( \frac{\partial L}{\partial \dot x} ) = 0$$

Here is my epic fail at trying to derive the equation using Lagrange (this was my first time trying to use lagrangian mechanics except for when I memorized the derivation for a pendulum)
$$L = \frac{m \dot r^2}{2} - \frac{k q_1 q_2}{r}$$
$$\frac{\partial L}{\partial r} = \frac{k q_1 q_2}{r^2}$$
$$\frac{\partial L}{\partial \dot r} = m \dot r$$
$$\frac{d}{dt} ( m \dot r ) = m \ddot r$$
$$\frac{k q_1 q_2}{r^2} - m \ddot r = 0$$
$$\ddot r = \frac{k q_1 q_2}{m r^2}$$
$$\dot r = \int \frac{k q_1 q_2}{m r^2}dr = \frac{2 k q_1 q_2}{mr}$$

 

[TSny]

Homework Statement:: Two positive charges are placed one meter away from eachother. One is held in place and one is allowed to move. Find an equation that models the velocity of the charge allowed to move after it is released from rest.
Relevant Equations:: $$KE = \frac{mv^2}{2}$$
$$U_E= \frac{k q_1 q_2}{r}$$
$$L = KE - PE$$
$$\frac{\partial L}{\partial x} - \frac{d}{dt} ( \frac{\partial L}{\partial \dot x} )$$

The last line needs to be completed as an equation.

$$L = \frac{m \dot r^2}{2} - \frac{k q_1 q_2}{r}$$
$$\frac{\partial L}{\partial r} = \frac{k q_1 q_2}{r^2}$$
$$\frac{\partial L}{\partial \dot r} = m \dot r$$
$$\frac{d}{dt} ( m \dot r ) = m \ddot r$$
$$\frac{k q_1 q_2}{r^2} + m \ddot r = 0$$
$$\ddot r = \frac{-k q_1 q_2}{m r^2}$$

OK.

$$\dot r = \int \frac{k q_1 q_2}{m r^2} = \frac{2 k q_1 q_2}{mr}$$

What is the variable of integration for the integral shown after the first = sign? It needs to be the same variable of integration that you used to get from $\ddot r$ to $\dot r$.

 

[TSny]

$$\frac{k q_1 q_2}{r^2} + m \ddot r = 0$$
$$\ddot r = \frac{-k q_1 q_2}{m r^2}$$

These two equation have a sign error.

 

[Arm]

The last line needs to be completed as an equation

What is the variable of integration for the integral shown after the first = sign? It needs to be the same variable of integration that you used to get from r¨ to r˙.

fixed and fixed, thank you
But I pulled the dr out of nowhere, I don't know if this is allowed

 

[Arm]

These two equation have a sign error.

also fixed, thank you

 

[TSny]

OK. So you have $\ddot r = \large \frac b {r^2}$, where $b$ is a constant. If I integrate both sides with respect to time, I get $\dot r = \int \frac b {r^2} dt + C$, where $C$ is a constant of integration. The integral on the right is integrated with respect to time. We can't do this integral because we don't know $r$ as a function of time.

Go back to $\ddot r = \large \frac b {r^2}$. There is a trick you can do. Multiply both sides by $\dot r$ to get $$\ddot r \dot r = \large \frac b {r^2} \dot r$$ which may be written as $$v \frac{dv}{dt} = \frac b {r^2} \frac {dr}{dt}$$ where $v = \dot r$. Multiply both sides by $dt$ and proceed.

 

[Arm]

OK. So you have $\ddot r = \large \frac b {r^2}$, where $b$ is a constant. If I integrate both sides with respect to time, I get $\dot r = \int \frac b {r^2} dt + C$, where $C$ is a constant of integration. The integral on the right is integrated with respect to time. We can't do this integral because we don't know $r$ as a function of time.

Go back to $\ddot r = \large \frac b {r^2}$. There is a trick you can do. Multiply both sides by $\dot r$ to get $$\ddot r \dot r = \large \frac b {r^2} \dot r$$ which may be written as $$v \frac{dv}{dt} = \frac b {r^2} \frac {dr}{dt}$$ where $v = \dot r$. Multiply both sides by $dt$ and proceed.

$$v \frac{dv}{dt} = \frac b {r^2} \frac {dr}{dt}$$
$$v {\:} dv = \frac{b}{r^2} \frac{dr}{dt}dt$$
$$\int v {\:} dv = \int \frac{b}{r^2} v {\:} dt$$
Not sure how I would take an integral with position and with velocity in it

 

[TSny]

$$v \frac{dv}{dt} = \frac b {r^2} \frac {dr}{dt}$$
$$v {\:} dv = \frac{b}{r^2} \frac{dr}{dt}dt$$
$$\int v {\:} dv = \int \frac{b}{r^2} v {\:} dt$$
Not sure how I would take an integral with position and with velocity in it

Note that $\frac{dr}{dt}dt = dr$

 

[Arm]

Note that $\frac{dr}{dt}dt = dr$

$$ \int v {\:} dv = \int \frac{b}{r^2} v {\:} dt $$
$$ \int v {\:} dv = \int \frac{b}{r^2} \frac{dr}{dt} {\:} dt $$
$$ \int v {\:} dv = \int \frac{b}{r^2} dr$$
$$ \frac{v^2}{2} = \frac{-b}{r}$$
$$ b = \frac{-k q_1 q_2}{m} $$
$$ \frac{v^2}{2} = \frac{(-)(-)k q_1 q_2}{mr}$$
$$ v^2 = \frac{2k q_1 q_2}{mr}$$
$$ v = \sqrt{\frac{2k q_1 q_2}{mr}}$$
I think this is the correct final answer

 

[TSny]

$$ \int v {\:} dv = \int \frac{b}{r^2} dr$$
$$ \frac{v^2}{2} = \frac{-b}{r}$$

Does the second equation above satisfy the initial condition: $v = 0$ when $r = 1$ m?

$$ b = \frac{-k q_1 q_2}{m} $$

Check to see if the sign is correct.

 

[Arm]

Does the second equation above satisfy the initial condition: $v = 0$ when $r = 1$ m?Check to see if the sign is correct.

I think the sign is correct

 

[TSny]

I think the sign is correct

In post #6 we defined $b$ such that $\ddot r = \large \frac b {r^2}$.

In post #1 you have $\ddot r = \large \frac{k q_1 q_2}{mr^2}$

 

[Arm]

Check to see if the sign is correct.

I think the sign is correct

I see where I made the mistake sign now

Does the second equation above satisfy the initial condition: v=0 when r=1 m?

Ignoring variable of integration and +C for the umpteenth time has gotten me the wrong answer yet again; when will I learn my lesson?
Here's the actual correct answer

$$v^2 = \frac{-2k q_1 q_2}{mr} + C$$
The particle is released from rest $x$ meters away from the other particle. Velocity is 0 at this moment
$$0^2 = \frac{-2k q_1 q_2}{mx} + C$$
$$ C = \frac{2k q_1 q_2}{mx}$$
$$v^2 = \frac{-2k q_1 q_2}{mr} + \frac{2k q_1 q_2}{mx}$$
$$v^2 = \frac{2k q_1 q_2}{mx} - \frac{2k q_1 q_2}{mr}$$

General solution:
$$v = \sqrt{ \frac{2 k q_1 q_2}{m}( \frac{1}{x} - \frac{1}{r} ) }$$

If x = 1 then r also = 1 so
$$v = \sqrt{ \frac{2 k q_1 q_2}{m}( \frac{1}{1} - \frac{1}{1} ) }$$
$$v = \sqrt{ \frac{2 k q_1 q_2}{m}( 1 - 1 ) }$$
$$v = \sqrt{ \frac{2 k q_1 q_2}{m}( 0 ) }$$
$$v = 0$$

Thanks for all the help!

 

[TSny]

Looks good. Of course using energy conservation is a quick way to get the answer. But it’s a good exercise using the Lagrangian approach.

 

[vanhees71]

Well, the Lagrangian approach also tells you to use symmetry principles, which are most effectively formulated using the action principle. Since $\partial_t L=0$ you know that $H=p_r \dot{r}-L=\text{const}$. Using the general principles of the action formalism saves a lot of work!

 

[Arm]

Well, the Lagrangian approach also tells you to use symmetry principles, which are most effectively formulated using the action principle. Since $\partial_t L=0$ you know that $H=p_r \dot{r}-L=\text{const}$. Using the general principles of the action formalism saves a lot of work!

I have no idea what that means because I'm taking the equilivent of general college physics 2 right now but I'll look into it; if you have a recommended resource I'll read/watch it.

 

[vanhees71]

I don't know, what "general college physics 2" includes. You need a good grasp of multivariate calculus to understand variational calculus and the Hamilton-Lagrange formalism of mechanics. A standard introductory textbook is Goldstein, Mechanics, but it must be the older 2nd edition. The newer editions are spoiled by some authors thinking it would need some "modernization" ;-)).