# Energy dissipated by air resistance through integration

1. Dec 23, 2009

### yoghurt54

1. The problem statement, all variables and given/known data
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'.

2. Relevant 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

3. 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 ??????

Last edited: Dec 23, 2009
2. Dec 23, 2009

### denverdoc

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" [Broken] or break it down by parts where the denominators are a-v and a+v.

Last edited by a moderator: May 4, 2017
3. Dec 23, 2009

### yoghurt54

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.

Last edited by a moderator: May 4, 2017
4. Dec 23, 2009

### denverdoc

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.

5. Dec 23, 2009

### denverdoc

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.

6. Dec 23, 2009

### yoghurt54

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

7. Dec 23, 2009

### ehild

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

8. Dec 23, 2009

### yoghurt54

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.

9. Dec 23, 2009

### ehild

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