Horizontal Motion with Quadratic Drag

[Lonely Lemon]

Homework Statement

Consider an object that is coasting horizontally (positive x direction) subject to a drag force f = -bv - cv^2 . Write down Newton's second law for this object and solve for v by separating variables. Sketc the behaviour of v as a function of t. Explain the time dependence for t large. (Which force term is dominant when t is large?)

Homework Equations

The Attempt at a Solution

I start with Newton's second law:

m(dv/dt) = -bv -cv^2

m(dv/dt) = -v(cv + b)

(-m/v)(dv/dt) -cv = b

dv/v + (cv/m)dt = (-b/m)dt

Then... do I integrate with respect to v for the dv terms and t for the dt terms? I can't figure out what this will give me with the second term on the right...

 

[nickjer]

You need all the 'v' terms together with 'dv'. But you subtracted 'cv' in line 3 which is a mistake. Instead divide the whole equation in line 2 by 'v(cv+b)'. Then solve for it.

 

[Lonely Lemon]

I get to the point where all my v terms are on one side so that I have:

(m dv)/[v(cv+b)] = -dt

From here when I try and integrate with respect to v on the left and t on the right, I'm doing a definite integral setting the limits between v and v0 and t and t0 so I get:

m[ln(cv^2 + bv) - ln(cv0^2 - bv0^2)] = -t

ln(cv^2 + bv) = -t/m + ln(cv0^2 - bv0^2)

Then if I take the exponential of both sides I still end up in a situation where I can't solve for v because there will be v on both sides... i.e

cv^2 + bv = e^(-t/m) + cv0^2 - bv0^2

How can I solve this for v?

 

[nickjer]

You need to check your integration, it is incorrect.

 

[Lonely Lemon]

How exactly do I integrate m(dv)/[v(cv+b)] ? The way I do it I consistently get natural log expressions which I am guessing is wrong because the function should tend towards 0 when I plot it...

 

[gabbagabbahey]

How exactly do I integrate m(dv)/[v(cv+b)] ? The way I do it I consistently get natural log expressions which I am guessing is wrong because the function should tend towards 0 when I plot it...

\frac{d}{dv}\ln|cv^2+bv|=\frac{2cv+b}{cv^2+bv}\neq\frac{1}{cv^2+bv}

To integrate your expression properly, try completing the square on the denominator and then making an appropriate substitution.

 

[nickjer]

Another trick is to break the fraction up into to fractions that look like this:

\frac{1}{v(cv+b)} = \frac{A}{v} + \frac{B}{cv+b}

Then solve for A and B. Integrating the two fractions separately will be much easier.

 

[Lonely Lemon]

Ah that clears things up. I see where I went wrong there, thanks for your help!