Integrating quadratic drag force equation to solve for velocity

[Charles Link]

yes I was updating post #11

The one simple correction as seen in post 21 compared to post 20 has this problem solved. (Post 20 needs a minus sign, and the correct form is in post 21).

 

[SammyS]

Please read post#20 about absolute signs; it came up in the previous page. Yes, there is a subtle twist in the second term.

I am very reluctant to join such a busy thread, so I was very very reluctant to join this.

Assuming that @kuruman was posting the quoted post (#24) as a reply to my post (#22), I'll respond as follows. This is more for OP (Ashley) than anyone else.

$\displaystyle \int_0^v\frac{dv'}{v_T -v'} = -\int_0^v\frac{-dv'}{v_T -v'}$

$\displaystyle\ \ \ = \int_v^0\frac{-dv'}{v_T -v'}$​

Added in Edit:
Of course, the corresponding indefinite integral evaluates to : $\displaystyle \ \ln |v_T-v'|+C \,. \$ and the absolute value signs here are superfluous.

 

[kuruman]

The one simple correction as seen in post 21 compared to post 20 has this problem solved. (Post 20 needs a minus sign).

I agree.

 

[Ashley1nOnly]

first integral

-∫ 1/(v' -vter) dv' second integral

∫ 1/(v'+ vter) dv'with limit from 0 to v
​

 

[Charles Link]

If you look at post 28, which is correct, you can see the integral with $\frac{1}{v'+vter}$ does not need a minus sign in front of it. If you put one there, and then proceed to do several lines of algebra using that incorrect minus sign, it will be an error that propagates.

 

[Ashley1nOnly]

what's the purpose of switching the two values?

first integral

-∫ 1/(v' -vter) dv' second integral

∫ 1/(v'+ vter) dv'

 

[SammyS]

what's the purpose of switching the two values?

first integral

-∫ 1/(v' -vter) dv'second integral

∫ 1/(v'+ vter) dv'

Switch what two values?

 

[Ashley1nOnly]

first integral

-∫ 1/(v' -vter) dv'

-ln(v'-vter) } with limit from 0 to V

-ln(v-vter) +ln(0-vter)
-ln(v-vter)+ln(-vter)
-ln(v-vter / -vter)
-ln(-v/vter +1)

second integral

∫ 1/(v'+ vter) dv'

ln(v'+vter) } with limit from 0 to V

ln(v+vter) -ln (0+vter)
ln (v+vter / vter)
ln( v/vter +1)

first + second

ln( v/vter +1) +( -ln(-v/vter +1) )

ln( v/vter +1)-ln(-v/vter +1)

=ln [ ( v/vter +1) /(-v/vter +1) ]

 

[Charles Link]

The mistake you originally made, and otherwise you had the calculation correct, was to integrate $\frac{1}{vter-v' }$. In this case, $v'$ is the variable of integration, and it needs to be in the form $\int \frac{dx}{x+a}=ln|x+a|$. When you have it in the form $\int \frac{dx}{a-x}$, it is not in a standard form where you can use the result $\ln|a-x|$. That is incorrect. It is $-\ln|a-x|$. $\\$ Just because $\int \frac{dx}{x+a}=\ln|x+a|$, that does not mean $\int \frac{dx}{a-x}=\ln|a-x|$. This last equation here is incorrect. This second integral needs to get converted to $\int \frac{dx}{a-x}=-\int \frac{dx}{x-a}=-\ln|x-a|$.

 

[Ashley1nOnly]

The mistake you originally made, and otherwise you had the calculation correct, was to integrate $\frac{1}{vter-v' }$. In this case, $v'$ is the variable of integration, and it needs to be in the form $\int \frac{dx}{x+a}=ln|x+a|$. When you have it in the form $\int \frac{dx}{a-x}$, it is not in a standard form where you can use the result $\ln|a-x|$. That is incorrect. It is $-\ln|a-x|$. $\\$ Just because $\int \frac{dx}{x+a}=\ln|x+a|$, that does not mean $\int \frac{dx}{a-x}=\ln|a-x|$.

so this
ln(vter-v)-ln(vter-0)= ln(vter-v)-ln(vter) = ln ( (vter-v)/(vter)) = ln(1- (v/vter))

became this

ln(vter-v)-ln(vter-0)= ln(vter-v)-ln(vter) = ln ( (vter-v)/(vter)) = -ln(1- (v/vter))

 

[Charles Link]

Please read post 39 again very carefully. I added a line or two at the end that should help.

 

[SammyS]

first integral

-∫ 1/(v' -vter) dv'

-ln(v'-vter) } with limit from 0 to V

-ln(v-vter) +ln(0-vter)
-ln(v-vter)+ln(-vter)
-ln(v-vter / -vter)
-ln(-v/vter +1)
...

Technically you have two errors here, which cancel (by coincidence).,
The second, and third highlighted lines should have absolute values inside the logs. After all, $\ v < v_T\,.\$ Right?

And $\ | v-v_T | = (v_T - v ) \$ for $\ v < v_T\,.\$

(Also, the need for absolute value is especially important for the first highlighted line. )​

 

[Ashley1nOnly]

so when you integrate the first one you get

-ln|vt-v|

and the second one

ln|vt+v|

without applying the limits

 

[Ashley1nOnly]

applying limits

first integral
-ln|vt -v| - [-ln |vt-0|]
-ln|vt -v| +ln|vt|
ln| vt/ (vt-v) |
ln| (1-vt/v)|

second integral
ln|vt+v| -ln|vt+0|
ln|vt+v|-ln|vt|
ln| (vt + v )/(vt) |
ln| (1+ v/vt)

first integral + second integral
ln| (1-vt/v)| + ln| (1+ v/vt)|

ln | (1-vt/v) / (1+ v/vt)|

 

[SammyS]

so when you integrate the first one you get

-ln|vt-v|

and the second one

ln|vt+v|

without applying the limits

Yes, if you are referring to the 'first integral' and 'second integral' of your post (#28) .

 

[SammyS]

applying limits

first integral
-ln|vt -v| - [-ln |vt-0|]
-ln|vt -v| +ln|vt|
ln| vt/ (vt-v) |
ln| (1-vt/v)|

second integral
ln|vt+v| -ln|vt+0|
ln|vt+v|-ln|vt|
ln| (vt + v )/(vt) |
ln| (1+ v/vt)

Fine.

If you hold off on combining the logs, you will have

−ln|vt −v| + ln|vt| + ln|vt+v| − ln|vt|

Also, no need for abs. val. Why?Added in Edit:

That would look much beter in LaTeX.

$\displaystyle −\ln(v_t −v) + \ln(v_t) + \ln(v_t+v) − \ln(v_t)$

 

[Ashley1nOnly]

so final answer for the integration part is

ln(1- vt^2 / v^2)

 

[Ashley1nOnly]

t= m/(c*2*vt) * ln(1- vt^2 / v^2)

v= sqrt ( e^((t*c*2*vt )/m) *(1-vt^2)

 

[kuruman]

−ln|vt −v| + ln|vt| + ln|vt+v| − ln|vt|

−ln|vt −v| + ln|vt| + ln|vt+v| − ln|vt| =ln|vt+v|−ln|vt −v|=ln[(vt+v)/(vt-v)].

 

[Ashley1nOnly]

But now I have two v's I just need one

 

[Charles Link]

ln| vt/ (vt-v) |
ln| (1-vt/v)|

Out of post 44, this step is incorrect. It is also really an unnecessary step, but if you do want to proceed that way, you do need to get the algebra correct.

 

[Ashley1nOnly]

Right I totally skipped over that. I can't divide like that.

first integral

ln[ vt/ (vt-v)]

second integral

ln[ 1 + v/vt]

ln[ vt/ (vt-v)]+ln[ 1 + v/vt]

=ln [ (vt+v )/ (vt-v)]

 

[Ashley1nOnly]

t= m/(c*2*vt) *ln [ (vt+v )/ (vt-v)]

 

[Charles Link]

Without checking it carefully, I believe what you have is correct. Now as $t \rightarrow +\infty$, what must $v$ do to have the right side give $+\infty$?

 

[SammyS]

Right I totally skipped over that. I can't divide like that.

first integral

ln[ vt/ (vt-v)]

second integral

ln[ 1 + v/vt]

ln[ vt/ (vt-v)]+ln[ 1 + v/vt]

=ln [ (vt+v )/ (vt-v)]

Please read posts (#46) and (#49), and save yourself some work.

It's fine to practice your algebra, but ...

 

[Ashley1nOnly]

v= vt*e^((t*c*2*vt)/m) -vt) / (e^((t*c*2*vt)/m) +1)

where x = (t*c*2*vt)/mv must go to 0

 

[Ashley1nOnly]

How do I calculate the F?

 

[SammyS]

How do I calculate the F?

Look at the OP (original post):
line 3 or 4 under

The Attempt at a Solution

,
assuming F is what you initially had as F_C .

 

[Charles Link]

v= vt*e^((t*c*2*vt)/m) -vt) / (e^((t*c*2*vt)/m) +1)

where x = (t*c*2*vt)/mv must go to 0

Looks like you might have gone back to your hyperbolic tangent =tanh solution, but as $t \rightarrow +\infty$ in the numerator and denominator, the limit will be 1 for that fraction and not zero. So this answer is incorrect. $\\$ From post 53, you should also be able to get the correct answer very quickly by looking at the denominator $v_T-v$ of the natural log function argument there. $\\$ Here's a hint: Suppose we let $v \rightarrow v_T$, so that $v_T-v \approx 0$, what happens to that natural log function?

 

[Ashley1nOnly]

Right because they were all constants, I kept looking at that but the answer is usually always infinity or zero so I was hesitant to pick that.