Air resistance differential equation

In summary, the textbook is teaching you how to solve a linear differential equation for a particle with retarding forces, but when you try to use the actual drag force equation, you get an equation that is not linear. You need to use other methods to solve it.
  • #1
David Koufos
9
4
Hello all, I want to say thank you in advance for any and all advice on my question. My classical mechanics textbook (Marion Thornton) has been taking me through motion for a particle with retarding forces.

The example it keeps giving is:

m dv/dt = -kmv

which can be solved for:
v = v0e-kt and
x = v0/k(1-e-kt)

But out of curiosity I tried using the actual drag force equation "1/2ρCAv2" instead of "kmv." But I can't figure out how to solve the differential:
##\ddot x ## + 1/2ρCA##\dot x ##2 = 0

How do you solve this thing? I'm stuck since it's not the standard
x'' + ax' + bx = 0

My solution yielded:

$$ \int\frac{\mathrm{d}\dot x }{ \dot x^2} = \frac{1}{2m}\rho CA\int \mathrm{dt} $$

which just gives some weird thing:
$$t = e^{\frac{1}{2m}\rho CAx} $$
which can't be right.
 
Last edited:
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  • #2
Having a quadratic air resistance term, your differential equation is no longer linear so you will need to apply other methods. As you have noted, the differential equation you get for ##v## is separable, i.e.,
$$
\frac{dv}{v^2} = K\, dt.
$$
However, your integration of this does not seem correct to me. Please show the details of your computation.
 
  • #3
I treated ##
\int {\frac {\mathrm {d \dot{x}}} {\mathrm {\dot{x^2}}}}## the same as ##\int \frac {\mathrm{dx}} {x^2} = \int x^{-2} \mathrm{dx} = -x^{-1}##.

So then I got ##- \dot x^{-1} = \frac {\mathrm {dt}}{dx} = \frac {1}{2m}\rho CA t##.

Then multiplied and divided: ##\frac {\mathrm {dt}}{t} = \frac {1}{2m}\rho CA \mathrm {dx}##.

Then integrated ##\int {\frac {\mathrm {dt}}{t}} = \frac {1}{2m}\rho CA \int {\mathrm {dx}}##.

So I got ##\ln {t} = \frac {1}{2m}\rho CAx##, ##t = e^{\frac {1}{2m}\rho CAx}##
 
  • #4
David Koufos said:
I treated ##
\int {\frac {\mathrm {d \dot{x}}} {\mathrm {\dot{x^2}}}}## the same as ##\int \frac {\mathrm{dx}} {x^2} = \int x^{-2} \mathrm{dx} = -x^{-1}##.

So then I got ##- \dot x^{-1} = \frac {\mathrm {dt}}{dx} = \frac {1}{2m}\rho CA t##.

Then multiplied and divided: ##\frac {\mathrm {dt}}{t} = \frac {1}{2m}\rho CA \mathrm {dx}##.

Then integrated ##\int {\frac {\mathrm {dt}}{t}} = \frac {1}{2m}\rho CA \int {\mathrm {dx}}##.

So I got ##\ln {t} = \frac {1}{2m}\rho CAx##, ##t = e^{\frac {1}{2m}\rho CAx}##
What about taking into account the initial condition in your integration to get v? You omitted the constant of integration.
 
  • #5
Thank you by the way Orodruin for mentioning that it's nonlinear. I'm doing some research on "homogeneous first-order nonlinear ordinary differential equations." I guess that's the kind of equation this is.
 
  • #6
Chestermiller said:
What about taking into account the initial condition in your integration to get v? You omitted the constant of integration.
Would that make a big difference tho? I could always just set up the constant to be trivial.
 
  • #7
David Koufos said:
Would that make a big difference tho? I could always just set up the constant to be trivial.

Try it for yourself: with a constant of integration, and without it. Do you get mathematically equivalent ##x(t)## formulas?
 
Last edited:
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  • #8
David Koufos said:
Would that make a big difference tho? I could always just set up the constant to be trivial.
Ask yourself this: Do you think I would have responded the way I did if I didn't already know that it would alleviate your difficulty?
 

Related to Air resistance differential equation

What is air resistance differential equation?

The air resistance differential equation, also known as the drag equation, is a mathematical model used to predict the force of air resistance acting on an object moving through a fluid, such as air. It takes into account the velocity of the object, the density of the fluid, and the shape and size of the object.

Why is air resistance differential equation important?

The air resistance differential equation is important because it helps us understand and predict the behavior of objects moving through a fluid. This is crucial in fields such as aerodynamics, where the performance of aircraft and other vehicles is highly dependent on air resistance.

What factors affect the air resistance differential equation?

The air resistance differential equation is affected by several factors, including the velocity of the object, the density of the fluid, and the shape and size of the object. Additionally, factors such as the surface roughness of the object, the fluid's viscosity, and the object's orientation can also impact the equation.

How is the air resistance differential equation derived?

The air resistance differential equation is derived from the Navier-Stokes equations, which describe the motion of fluid particles. Simplifying assumptions are made to account for the behavior of a solid object moving through a fluid, resulting in the drag equation.

Can the air resistance differential equation be solved analytically?

In most cases, the air resistance differential equation cannot be solved analytically and requires numerical methods to obtain a solution. However, for simple cases with specific assumptions, analytical solutions may be possible.

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