Solving Complex Velocity from Particle Falling in Gravity

[fobos3]

I'm trying to derive the equations of motion for a particle falling in a uniform gravitational fill with air drag proportional to the square of velocity. However I'm getting the velocity as a complex number. Here is what I've done

The force of friction is F=-k\left(\dfrac{dx}{dt}\right)^2\dfrac{\textbf{v}}{||\textbf{v}||}=-k\left(\dfrac{dx}{dt}\right)^2

We put the particle stationary at x=0

The Lagrangian is \mathcal{L}=\dfrac{1}{2}m\left(\dfrac{dx}{dt}\right)^2-mgx

\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{x}}\right)-\dfrac{\partial \mathcal{L}}{\partial x}=-k\left(\dfrac{dx}{dt}\right)^2

m\dfrac{d^2 x}{dt^2}+mg=-k\left(\dfrac{dx}{dt}\right)^2

If we put c=\dfrac{k}{m}

\dfrac{d^2x}{dt^2}+c\left(\dfrac{dx}{dt}\right)^2+g=0

We put p=\dfrac{dx}{dt}

We have \dfrac{d^2x}{dt^2}=\dfrac{dp}{dt}=\dfrac{dp}{dx}\dfrac{dx}{dt}=\dfrac{dp}{dx}p

The differential equation becomes

\dfrac{dp}{dx}p+cp^2+g=0

\dfrac{p}{cp^2+g}\dfrac{dp}{dx}=-1

\int\dfrac{p}{cp^2+g}\,dp=-x

To solve the integral we put u=cp^2+g

\dfrac{du}{dp}=2cp

p=\dfrac{1}{2c}\dfrac{du}{dp}

\dfrac{1}{2c}\int \dfrac{1}{u}\,du=-x

\int \dfrac{1}{u}\,du=\ln |u|=\ln u because u>0

\dfrac{1}{2c}\ln (cp^2+g)+A=-x

A(cp^2+g)=e^{-2cx}

At t=0,x=0,p=0

Ag=1

A=\dfrac{1}{g}

\dfrac{c}{g}p^2+1=e^{-2cx}

p^2=\dfrac{g(e^{-2cx}-1)}{c}

Now the sign of \dfrac{g(e^{-2cx}-1)}{c} is determined by e^{-2cx}-1 which is not necessary positive definite.

In fact if we put c=1,x=1 we get e^{-2}-1<0 which means that p\in \mathbb{C}

But p=\dfrac{dx}{dt} which makes no sense at all. Did I do something wrong?

 

[hikaru1221]

\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{x}}\right)-\dfrac{\partial \mathcal{L}}{\partial x}=-k\left(\dfrac{dx}{dt}\right)^2

The sign of the drag force in the above equation is wrong.
By the way, I don't see why we have to use Lagrangian here, since Newtonian method is much more simple.

EDIT: Actually you may change either the sign of the drag force or the sign of the potential energy, but only one of them. Both yield the same correct equation.

 

[RandomGuy88]

m\dfrac{d^2 x}{dt^2}+mg=-k\left(\dfrac{dx}{dt}\right)^2

I think you have a sign issue here. Shouldn't it be:

m\dfrac{d^2 x}{dt^2}=mg-k\left(\dfrac{dx}{dt}\right)^2

Mass*acceleration equals the sum of the forces. The net force is gravity minus drag.

 

[fobos3]

The sign of the drag force in the above equation is wrong.
By the way, I don't see why we have to use Lagrangian here, since Newtonian method is much more simple.

EDIT: Actually you may change either the sign of the drag force or the sign of the potential energy, but only one of them. Both yield the same correct equation.

Yes I see that now. Can you explain where I went wrong in my derivation of the friction.This is what I thought

F=-k\dot{x}^2 \hat{\textbf{v}}

\hat{\textbf{v}}=\dfrac{\textbf{v}}{||\textbf{v}||}=\dfrac{(\dot{x})}{\dot{x}}=(1)

Where the brackets denote a vector in the x direction.

Now obviously \hat{\textbf{v}}=(1) is wrong, because the particle accelerates in the negative direction, but why?

Edit
Never mind. I find out on my own.

 

[hikaru1221]

There is nothing wrong with \vec{F}=-kv^2\hat{v}. But you must be careful when jotting down F = -kv^2 (*) (that means F<0). Since the ball is falling, the force must act upwards. So when you write (*), that means the positive direction of x-axis is downward.
Now in order that potential energy V = mgx, the x-axis must point upwards, which is inconsistent with (*) as explained above.