# Energy dissipated by air resistance through integration

• yoghurt54
In summary: Thanks for the feedback! I think the integral in the form of -1/(1+ax^2) would be the most straightforward to integrate, since a and x are constants.I'll give that a try and report back.
yoghurt54

## Homework Statement

Hey guys, I'm doing a problem on quadratic drag and energy dissipation. Basically, the question asks me to find the energy dissipated when a ball of mass 'm' is thrown directly upwards with a velocity 'v0', during the upwards journey to its maximum height, 'h'. The resistive force is 'av2'. Now, whether or not it is relevant to answering the question (I think it is), I believe I need to find the integral of
'av2' with respect to the displacement upwards, 'x'.

## Homework Equations

I've derived the equation taking upwards as x-positive:

F = mvdv/dx = -mg -av2

ignoring any '-' signs,

E=∫ F dx = ∫( -mg -av2 )dx between x=0 and x=h

## The Attempt at a Solution

Now obviously the energy has to equal 1/2*m*v02 due to the conservation of energy.

And this can be easily shown by integrating the RHS - mvdv/dx.

But when I get to the av2, I tried substituting v=dx/dt, which doesn't make it any clearer.

I tried to do integration by parts, getting the av2 part to be axv2 - ∫2axvdv/dx dx.

Any ideas? Perhaps a knowledge of multi-variable calculus is needed?

EDIT: Thanks for your ideas guys, but I *DON'T* need to find the height 'h' (which is what people are offering suggestions about). Essentially, I'd like to know how to do ∫av2 dx, with v being dx/dt:

i.e. how do you find ∫ (dx/dt)2 dx ?

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You're on the right track--one can use trig identities like found http://www.physics.udel.edu/~szalewic/teach/419/cm08ln_quad-drag.pdf" or break it down by parts where the denominators are a-v and a+v.

Last edited by a moderator:
denverdoc said:
You're on the right track--one can use trig identities like found http://www.physics.udel.edu/~szalewic/teach/419/cm08ln_quad-drag.pdf" or break it down by parts where the denominators are a-v and a+v.

Sorry, but are you suggesting finding v as a function of t? Or are you saying I should find v as a function of x? I've done the latter, it's helped me find the maximum height, but not much else for the energy.

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I was just suggesting a way to find the integral; the solution to your problem if I'm understanding it correctly only requires equating the difference in energy initially less that of potential with the dissipative force of drag. We don't need to take into account whether the retarding force is quadratic or linear. Now to know what the final height is will require integration.

BTW, what I was suggesting was setting up the integral in the form of -1/(1+A*V(t)^2)

as a sum of -1/2 (1/(1+jaV(t))-1/2(1/(1-jaV(t) where a is the sqrt of A and j is sqrt(-1).

Certainly there are other ways to do this including integration by parts and various trig functions. Your call, any way you integrate the function twice, first to get v(t), then x(t) and using initial and final conditions should end up with something like:

h=m/2k*[ln((mg + kVi^2)/mg)

Another approach is just to use an integral table for int(-1/(1+ax^2) the first integral is the inverse tangent and the integral of that is ln.

Ok, thanks for your help denverdoc! I'll give it another shot.

It is easy to integrate your equation

$$mvdv/dx = -mg -av^2$$

by separating the variables v and x.

$$\int_{v_0}^0{\frac{vdv}{mg/a+v^2}}= -\frac{a}{m}\int_0^h{dx}$$

$$\frac{1}{2}\ln{\frac{mg/a}{mg/a+v_0^2}= -\frac{ah}{m}$$

$$h=\frac{m}{2a}\ln(1+\frac{v_0^2*a}{mg})$$

ehild

Thanks a bunch ehild, but that's not what I wanted. It's pretty trivial to get the height by doing what both you (and I) have done, what I need is a way to determine the ENERGY through integration. This means that I have to integrate the LHS wrt x, which is what I am (still!) stuck on.

If you integral between v0 and v(x) on the left side and between 0 and x on the right side, you can get the function v^2(x). It contains a constant term and an exponential function, easy to integrate.

$$\int_{v_0}^{v(x)}{\frac{vdv}{mg/a+v^2}}= -\frac{a}{m}\int_0^x{dx}$$

$$\frac{1}{2}\ln{\frac{mg/a+v(x)^2}{mg/a+v_0^2}= -\frac{ax}{m}$$

$$v(x)^2=-mg+(mg+a*v_0^2)\exp(-2ax/m)$$

$$W=-a\int_0^h{v(x)^2}dx$$

But you need not do this: As Denverdoc pointed out: the change of the mechanical energy is equal to the work of drag:

$$W=mgh- 1/2 mv_0^2$$.

ehild

## What is energy dissipation?

Energy dissipation refers to the process of converting kinetic energy into other forms of energy, such as heat, sound, or light. This can occur through various mechanisms, including friction, air resistance, and fluid turbulence.

## How does air resistance affect energy dissipation?

Air resistance is a type of friction that occurs when an object moves through the air. As the object moves, it collides with air particles, causing them to exert a force on the object in the opposite direction of its motion. This force results in a decrease in the object's kinetic energy, leading to energy dissipation.

## What is integration in relation to energy dissipation?

Integration is a mathematical technique used to calculate the total energy dissipated by air resistance. It involves breaking down the energy dissipated into small, infinitesimal steps and summing them up to get the total energy dissipated over a period of time.

## How does the surface area of an object affect energy dissipation through air resistance?

The surface area of an object affects the amount of air resistance it experiences. Objects with larger surface areas will experience more air resistance, leading to a higher amount of energy dissipated. This is why objects with streamlined shapes, such as airplanes, are designed to minimize surface area and reduce energy dissipation through air resistance.

## What factors can affect the amount of energy dissipated by air resistance through integration?

The amount of energy dissipated by air resistance through integration can be affected by various factors, such as the speed and mass of the object, the density and viscosity of the air, and the shape and surface area of the object. Additionally, the presence of external forces, such as gravity or wind, can also impact the amount of energy dissipated through air resistance.

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