Integration help on physics problem

  • #1
Ashley1nOnly
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Homework Statement


upload_2018-9-22_6-10-14.png

Homework Equations


F=ma
F/m = a

The Attempt at a Solution


Fc = constant force
fquad= cv'^2
[/B]


(Fc-cv'^2)/m = a

(Fc-cv'^2)/m = dv' / dt' * using the primes to differentiate between v and v' during integration

dt '(Fc-cv'^2) dv'*(m)

dt' = (m/ (Fc-cv'^2)) dv'


dt' = m / [ c ((Fc/c) - v'^2) ] dv'

∫c dt' = ∫ m / ((Fc/c) - v'^2) dv'

I need a method to integrate the right hand side that will allow me to get v by itself.
*u-substitution won't work
*I can't integrate it straight through
* I can use ln
* I can't use partial fractions
* I don't think I can use any trig identities here either
*I'm not sure where to take this.

***More work added on,

right hand side
(m*c)/Fc = ∫1/ (1- ((c*v^2)/Fc)) dv'

so I found that

∫ 1/ (A^2 - x^2) dx = 1/ (2A) * ln | (x+A)/(x-a)| + C

I can rewrite my right hand side as the below to fit the equation but what do I do with the constant (c/Fc)? Once I figure that out I can use e to get rid of ln and solve for v

∫1/ (1^2- (c/Fc)*(v^2) dv'





 

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Answers and Replies

  • #2
kuruman
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* I can't use partial fractions
Yes you can. For problems like this, I find it useful to express the differential equation in terms of the constant terminal velocity vT and eliminate the force. Do that and see what you get.

Hint on edit: What is dv/dt when terminal velocity is reached?
 
  • #3
haruspex
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using the primes to differentiate between v and v' during integration
Then you should be using v' throughout the differential equation. v only comes into play after integration.
 
  • #4
Ashley1nOnly
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So this is my initial equation
(Fc-cv^2) = am

the terminal speed is reach when Fc=cv^2

vter= sqrt( Fc/c)

solve for c= Fc/ (vter)^2

m* dv'/dt' = Fc- (Fc/ (vter)^2)* v^2)

m dv'/dt' = Fc(1-V^2/(vter)^2)

∫ (1-V^2/(vter)^2) dv' =∫(Fc/mc)* dt' t = t0 to t v= v0 to v

t= ((vter * Fc)/ m)* tanh^-1 (v/ vter)

v= vter*tanh( (m/Fc)/vter)

* what can I do to get rid of tanh so that the integration will be easier?
 
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  • #5
kuruman
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Eliminate ##F_c~##; you can always put it back in later. Then the integrand becomes$$\frac{1}{c} \left( \frac{1}{v_T^2-{v'}^2} \right). $$ You can easily write this as the sum of two fractions. Also, when you integrate remember to use limits of integration. When t' = 0, v' = 0; when t' = t, v' = v.
 
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  • #6
kuruman
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I thought we eliminated the c when we plugged in the value for c so we only have Fc=Force constant
In post #2, I suggested eliminating the constant force ##F_c## not the constant ##c## in ##cv^2.## Sorry if I was not clear. There are too many letters ##c## that cause confusion. Perhaps you should call the constant driving force just ##F##. The problem doesn't call it ##F_c.##
 
  • #7
Ashley1nOnly
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Changed the equation to use only F
(F-cv^2) = am

the terminal speed is reach when F=cv^2

vter= sqrt( F/c)

solve for c= F/ (vter)^2

m* dv'/dt' = F- (F/ (vter)^2)* v^2)

m dv'/dt' = F(1-V^2/(vter)^2)

∫ 1/ [(1-(V/vter)^2)] dv' =(F/m)∫ dt'
 
  • #8
kuruman
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Changed the equation to use only F
"Eliminate ##F_c##" is another way of saying "Get rid of ##F_c##". This means "Replace all occurrences of ##F_c## with ##cv_T^2.##" Then you will get the integrand in post #5 which you can split into fractions. Please read this and my previous posts carefully.
 
  • #9
Charles Link
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"Eliminate ##F_c##" is another way of saying "Get rid of ##F_c##". This means "Replace all occurrences of ##F_c## with ##cv_T^2.##" Then you will get the integrand in post #5 which you can split into fractions. Please read this and my previous posts carefully.
If I can make a comment: I think this one can be solved by partial fractions as @kuruman has suggested in post 5, (where you ultimately get two terms that both integrate as natural logs), but inverse hyperbolic tangent also works. The OP's algebra was very sloppy in post 4, and the OP even dropped the ## t ## in their expression, or the approach they were taking should have worked.
 
  • #10
kuruman
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If I can make a comment: I think this one can be solved by partial fractions as @kuruman has suggested in post 5, (where you ultimately get two terms that both integrate as natural logs), but inverse hyperbolic tangent also works. The OP's algebra was very sloppy in post 4, and the OP even dropped the ## t ## in their expression, or the approach they were taking should have worked.
I agree. However in post #4 OP is asking in the end
what can I do to get rid of tanh so that the integration will be easier?
Well, the answer can be cast as a hyperbolic tangent, at which point no further integration is needed. I assumed OP was asking for a way to do the integral bypassing hyperbolic tangents and that's why I pushed for partial fractions which is a mehtod that OP eliminated in post #1.
 
  • #11
Ashley1nOnly
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Changed the equation to use only F
(F-cv^2) = am

the terminal speed is reach when F=cv^2

vter= sqrt( F/c)

solve for F= c* (vter)^2

m* dv'/dt' = c* (vter)^2- cv^2)

m*dv'/dt'= c (vter)^2- v^2)

(m/c)* ∫1/ (vter^2 -v^2) dv' = ∫ dt

working with the integral only
∫1/ (vter^2 -v^2) dv' = A/ (vter -v) + B/ (vter+v)

= [A /(vter+v) +B /(vter-v)] / [ (vter+v) (vter-v)]

v=vter A = 1/(2*vter)
v=-vter B= 1/(2*vter)

[1/(2*vter) [- ∫1/(v'-vter) dv' + ∫(vter+v') dv' ]]

applying the limits of integration v0= 0 v=v
first integral

-ln(v-vter)+ln(0-vter)= ln(v-vter)+ln(vter) = ln(v-vter * vter) = ln(v*vter -vter^2)
Second integral
ln (vter+v) +ln(vter+0)= ln (vter+v) +ln(vter) = ln((vter+v)*(vter))= ln(vter^2+ vter*v)

first integral + second integral
ln(v*vter -vter^2)+ ln(vter^2+ vter*v)





now adding in the t part

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

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

e^((2*vter*c*t/m) )= vter^2 - v^2

v^2= vter^2 - e^((2*vter*c*t/m) )

v=sqrt( vter^2 - e^((2*vter*c*t/m) ) )
 
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  • #12
kuruman
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(m/c)[1/(2*vter)[ln ((vter-v)) + ln((vter+v))]
This is incorrect. You did not use the limits of integration as I indicated in post #5. Ignoring the constants up front, you have integrals of the form$$\int_0^v\frac{dv'}{v_T \pm v'}$$Each integral gives you two terms when you evaluate at the upper and lower limits, four terms overall.
 
  • #13
Ashley1nOnly
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This is incorrect. You did not use the limits of integration as I indicated in post #5. Ignoring the constants up front, you have integrals of the form$$\int_0^v\frac{dv'}{v_T \pm v'}$$Each integral gives you two terms when you evaluate at the upper and lower limits, four terms overall.
I added in the limits and made the changes. I got
t=1/(2*vter)[ln (vter^2 -v^2)-2ln(vter)]
 
  • #14
kuruman
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You must have messed up a minus sign in front of one of the integrals. Don't forget that ##ln[x+y]-ln[y] = ln[\frac{x+y}{y}]##. Use this when you evaluate each integral separately then combine the two results which have a relative negative sign between them.
 
  • #15
Ashley1nOnly
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You must have messed up a minus sign in front of one of the integrals. Don't forget that ##ln[x+y]-ln[y] = ln[\frac{x+y}{y}]##. Use this when you evaluate each integral separately then combine the two results which have a relative negative sign between them.

So I went back to my post #11 and made some changes to how I did the integration with the limits applied and got

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

after simplifying

t=(m/(c*2*vter))*ln(vter^2 - v^2)
 
  • #16
kuruman
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This is still incorrect. The argument of the logarithm must be a dimensionless quantity. There is a mistake somewhere and I suspect a minus sign error. Forum rules prevent me from posting my solution, but if you post yours, I might be able to find out where you went wrong.
 
  • #17
Ashley1nOnly
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This is still incorrect. The argument of the logarithm must be a dimensionless quantity. There is a mistake somewhere and I suspect a minus sign error. Forum rules prevent me from posting my solution, but if you post yours, I might be able to find out where you went wrong.
can you message it to me and I'll post it?

I have my steps showed in post #11
 
  • #18
Charles Link
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applying the limits of integration v0= 0 v=v
first integral

ln(vter-v)-ln(vter-0)= ln(vter-v)-ln(vter) = ln ( (vter-v)/(vter)) = ln(1- (v/vter))
Right here (post 11) is where the minus sign is getting omitted. The denominator needs to be changed to ## v'-vter ## before integrating: ## \int \frac{1}{v'-vter} \, dv'=\ln|v'-vter| +C ##. If the sign of the denominator is reversed, you get a minus in front of ## ln ##.
 
  • #19
Charles Link
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## \frac{1}{vter-v'}=-\frac{1}{v'-vter} ##. Here ## v'## is your ## x ##. ## \int \frac{1}{a-x} \neq ln|x-a| ## . Instead ## \int \frac{1}{a-x}=-ln|x-a| ##.
 
  • #20
kuruman
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Right here (post 11) is where the minus sign is getting omitted. The denominator needs to be changed to ## v'-vter ## before integrating: ## \int \frac{1}{v'-vter} \, dv'=\ln|v'-vter| +C ##. If the sign of the denominator is reversed, you get a minus in front of ## ln ##.
Yes, but what is ##C##? The integral is not indefinite. $$\int_0^v\frac{dv'}{v_T - v'}=\ln \left(\frac{v_T-v}{v_T} \right)$$
BTW, the absolute values for the argument of the logarithm are superfluous. We are doing physics here and the result (if the math is done correctly) should exist. Here ##v## cannot be larger than the terminal velocity ##v_T##.

On edit: There should be a negative sign in front of the result as noted by @Charles Link in post #31.
 
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  • #21
Charles Link
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∫v0dv′vT−v′=ln(vT−v−VT)
The equation in post 20 needs a minus sign in front of the right side. It also needs absolute value signs: ## \\ ## ## \int\limits_{0}^{v} \frac{dv'}{v_T-v'} =-\int\limits_{0}^{v} \frac{dv'}{v'-v_T} =-[\ln|v-v_T|-\ln|0-v_T|]=-ln|\frac{v-v_T}{-v_T}| ##. ## \\ ## I believe I have this correct.
 
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  • #22
SammyS
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This is incorrect. You did not use the limits of integration as I indicated in post #5. Ignoring the constants up front, you have integrals of the form$$\int_0^v\frac{dv'}{v_T \pm v'}$$Each integral gives you two terms when you evaluate at the upper and lower limits, four terms overall.

I added in the limits and made the changes. I got
t=1/(2*vter)[ln (vter^2 -v^2)-2ln(vter)]
@Ashley1nOnly ,

What @kuruman was telling you was, (in addition to integration limits) , you need to do each of those integrations, one with a denominator of ##\ v_{T} + v' \,,\ ## the other with denominator ##\ v_{T} - v' \,.\ ##

That second one has a subtle twist.
 
  • #23
Ashley1nOnly
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so the integral now becomes

[1/(2*vter) [- ∫1/(v'-vter) dv' - ∫(v'+vter) dv' ]]

what's the purpose of switching them?
 
  • #24
kuruman
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@Ashley1nOnly ,

What @kuruman was telling you was, (in addition to integration limits) , you need to do each of those integrations, one with a denominator of ##\ v_{T} + v' \,,\ ## the other with denominator ##\ v_{T} - v' \,.\ ##

That second one has a subtle twist.
Please read post#20 about absolute signs; it came up in the previous page. Yes, there is a subtle twist in the second term.
 
  • #25
Ashley1nOnly
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Please post#20 about absolute signs; it came up in the previous page. Yes, there is a subtle twist in the second term.

1/(2*vter) [- ∫1/(v'-vter) dv' - ∫(-v'-vter) dv' ]
 
  • #26
Charles Link
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1/(2*vter) [- ∫1/(v'-vter) dv' - ∫(-v'-vter) dv' ]
I spelled it out for you in post 21=but do work it yourself, and you will see that minus sign out front is necessary.
 
  • #27
Ashley1nOnly
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The sign in front of the first integral should be + (positive) not - (negative).

1/(2*vter) [ ∫1/(v'-vter) dv' - ∫1/(-v'-vter) dv' ]

first integral
-ln (-v/vter - 1)

second integral
ln(-v-vter)- ln(0-vter)= ln(-v-vter)+ln(vter) = ln (-v/vter -1)

is this right so far?
 
  • #28
kuruman
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OK, here it is spelled out again.
$$\frac{1}{v_T^2-v'^2}=-\frac{1}{2v_T} \left( \frac{1}{v'-v_T} -\frac{1}{v'+v_T} \right).$$Add the fractions inside the parentheses to see why this is so. The positive sign in front of the integration variable ##v'## makes it easier to keep track of negative signs when you integrate.
 
  • #29
Charles Link
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From my quote in post 18, it appears you (the OP) have been editing post 11. If this is the case, it is making things harder, because I think you have changed things that only needed a minus sign so that they would be correct, and now they need much more work.
 
  • #30
Ashley1nOnly
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From my quote in post 18, it appears you (the OP) have been editing post 11. If this is the case, it is making things harder, because I think you have changed things that only needed a minus sign so that they would be correct, and now they need much more work.
yes I was updating post #11
 
  • #31
Charles Link
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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).
 
  • #32
SammyS
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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.
 
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  • #33
kuruman
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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.
 
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  • #34
Ashley1nOnly
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first integral

-∫ 1/(v' -vter) dv'


second integral

∫ 1/(v'+ vter) dv'


with limit from 0 to v
 
  • #35
Charles Link
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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.
 

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