# Motion with damping involving both $-v^2$ and $-v$ terms

I'm trying to solve for the motion of a particle, accounting for both viscous and drag forces. (There is no potential)

The total resistance by medium is modeled by:

$\vec{F}=-(\mu_1v + \mu_2v^2)\hat{v}$

Where $v$ is the velocity; There is no potential.

I need to set up the Lagrangian and describe motion under the following conditions:

$\frac{\mu_1}{\mu_2} = 10^{-5}$ $\frac{\mu_1}{\mu_2} = 1$ $\frac{\mu_1}{\mu_2} = 500$

I have set up the Lagrangian as follows:

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

$\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} -\frac{\partial L}{\partial x} =- (\mu_1 \dot{x} + \mu_2\dot{x}\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2})$ and similarly for z and y.

Equations of motion turn out to be:

$m\ddot{x}= - (\mu_1 \dot{x} + \mu_2\dot{x}\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2})$ Similarly for z and y.

I'm at a total loss to how these should be solved. Is there even a way of analytically solving such coupled equations?

However, If I orient the x axis along initial direction of velocity, I should get:

$m\ddot{x}= - (\mu_1 \dot{x} + \mu_2\dot{x}^2)$

$y=0$

$z=0$

I'm stuck at this point. Wolfram Alpha suggests:

http://puu.sh/bmj5K/8d1edff9ab.png Where $\frac{\mu_1}{m}=a$ and $\frac{\mu_2}{m}=b$

I'm not aware of the techniques used to get to this. Would appreciate some help/pointing to resources on that.

After this, I'm not really sure of how to apply limits (Since we have the additional constraint of the domain of Log functions), and how to interpret the motion in the 3 regimes mentioned above.

Can someone give me some pointers?

Also what really happens when there is a potential (Say, gravity)? (which really is the real world scenario)

• Your differential equation involves $v$ and $dv/dt$. It can be solved by separation of variables. See, e.g., en.wikipedia.org/wiki/… Sep 5, 2014 at 7:45

However, If I orient the x axis along initial direction of velocity, I should get: $$m\ddot{x}= - (\mu_1 \dot{x} + \mu_2\dot{x}^2)$$

That's not quite true. You implicitly assumed that $\sqrt{\dot x^2} = \dot x$. That is only true if $\dot x\ge 0$. Otherwise you need to use the more general $\sqrt{\dot x^2} = |\dot x|$.

You don't want to go there.

You want to convince yourself (and your grader!) that the simplification $\dot y = \dot z = 0$ and $\dot x \ge 0$ doesn't discard any physics. (It doesn't, but you need to show this.)

I'm stuck at this point. Wolfram Alpha suggests: ...

You are biting off too much at once by trying to attack this second order non-linear ODE in one chunk. Break the problem down into smaller, more manageable parts.

First, let's rewrite your expression for acceleration as $\ddot x + \omega \dot x + k \dot x^2 = 0$. (I'm using $\omega$ and $k$ in instead of your $a$ and $b$.) There's no dependence on $x$ here, just velocity. You can rewrite the above as $\dot v + \omega v + kv^2 = 0$. This is a first order ODE, which is good, but it's also a non-linear first order ODE thanks to that $v^2$ term.

While this is a first order non-linear ODE, it is trivially separable. You could use separation of variables to solve this as higgsss suggested.

Another approach is divide through by $v^2$ yielding $\frac {\dot v}{v^2} + \frac {\omega} v + k = 0$. This should strongly suggest a change of variables $v(t) = 1/u(t)$. (In fact, the expression $\dot v + \omega v + kv^2 = 0$ should strongly suggest that change of variables.) Make this change of variables and you'll find that you have a very simple first order linear ODE.

Either way, you'll get an explicit representation of $v(t)$, and then you can integrate that to yield $x(t)$.

Also what really happens when there is a potential (Say, gravity)? (which really is the real world scenario).

Most real world scenarios don't have nice clean analytic solutions. This is one of them.