Newton's second law integration

[Oblio]

I need to integrate

$$\frac{mdv}{F(v)}$$ = dt

to show that

t = m $$\int$$$$^{v}_ {vo}$$ $$\frac{dv '}{F(v ')}$$

Now, I now that velocity is the derivative of position, and the mass is constant, but how does one integrate F(v)?

[Integrating is a weak spot for me] Thanks in advance!

 

[Oblio]

in getting the first equation, since F(v) is the only force, F(v) = mv'.
In rearranging, how does the dt come about?

 

[learningphysics]

in getting the first equation, since F(v) is the only force, F(v) = mv'.
In rearranging, how does the dt come about?

Just integrate both sides of that first equation... the side with the dt becomes t... and the right side gives you the integral...

integral of dt is t.

 

[Oblio]

I'm just wondering how Newtons second law: F(v) = mv' obtained the t...

 

[learningphysics]

I'm just wondering how Newtons second law: F(v) = mv' obtained the t...

oh... v' = dv/dt then rearrange...

 

[Oblio]

I don't understand how i miss things sometimes...

doesn't adding the primes cancel out the integral?

 

[learningphysics]

I don't understand how i miss things sometimes...

doesn't adding the primes cancel out the integral?

Ah I see your confusion... the prime in the equation with the integral has nothing to do with derivative...

They're just using another variable to do the integration...

they could just as well write

$$t = m \int_{v0}^v\frac{du}{F(u)}$$

they just want to keep one variable v for the limit of the integration... ie we're integrating from v0 to the final velocity v... so inside the integral they want a different variable (they used v' which was a bad choice because it can cause confusion with the derivative)

 

[Oblio]

Oohhh. haha

What about the integral and dv?

 

[learningphysics]

Oohhh. haha

What about the integral and dv?

Yeah, when they put in the integral... they just switched from dv to dv'... you can use any variable name you want...

this would also be equally valid:

$$t = m \int_{v0}^{v_{final}}\frac{dv}{F(v)}$$

The only thing is you need a different variable inside the integral than you have at your limits of integration... ie: this would technically be wrong:

$$t = m \int_{v0}^v}\frac{dv}{F(v)}$$

people often use this type of notation... but you shouldn't really have v as the limit of integration, when the variable inside the integral is also v.

 

[Oblio]

Would that just become v' though since its dv' and in an integral?

 

[learningphysics]

Would that just become v' though since its dv' and in an integral?

Not sure what you mean.

 

[Oblio]

same question as before, but instead of prime i mean the d in front of the v.
Derivative of v (dv) and the integral... = v?

I think not because the dt is separated though.

 

[learningphysics]

same question as before, but instead of prime i mean the d in front of the v.
Derivative of v (dv) and the integral... = v?

I think not because the dt is separated though.

I'm sorry... I'm still not following... let's start with this:

$$\frac{mdv}{F(v)} = dt$$

Then I can take the integral of both sides

$$\int_{v_0}^{v_{final}}\frac{mdv}{F(v)} = \int_0^{t_{final}}dt$$ so that gives

$$\int_{v_0}^{v_{final}}\frac{mdv}{F(v)} = t_{final}$$

I can rewrite t_final as t... and take out the m...

$$m\int_{v_0}^{v_{final}}\frac{dv}{F(v)} = t$$

 

[Oblio]

yeah, that is where I'm up to (correctly, compared to your example). I'm basically wondering, what CAN I do with dv (the numerator)?

 

[learningphysics]

yeah, that is where I'm up to (correctly, compared to your example). I'm basically wondering, what CAN I do with dv (the numerator)?

why do you need to do anything? Isn't that the final result? If you want you can switch from v to some other variable... v', u or anything else... but v' is NOT the derivative of v. It's just another variable like u or x etc...

$$\int_3^4{f(t)dt}$$ is exactly the same thing as $$\int_3^4{f(x)dx}$$... just switched the name of the variable... makes no difference.

 

[Oblio]

I'm now supposed to solve for v, now that I have solved this, to get v as a function of t. I think I need to get rid of the integral to rearrange?


(I was partly reading too far into the question again)

 

[learningphysics]

I'm now supposed to solve for v, now that I have solved this, to get v as a function of t. I think I need to get rid of the integral to rearrange?


(I was partly reading too far into the question again)

Can you show exactly you need to solve? I'm confused... is there a formula for F(v) ?

 

[Oblio]

K the whole question is this:

There are certain simple one dimensional problems where the equatin of motion can always be solve,d or at least reduced to the problem of doing an integral. One of these is the motion of a done dimensional particle subject to a force that depends on only the velocity v, that is F=F(v). Write down N2L and separate the variables by rewriting it as mdv/F(v) = dt. Now integrate both sides of this equation and show that

t = m... (we did this)

Provided you can do the integral, this gives t as a function of v. You can also solves to give v as a function of t. Use this method to solve the special case that F(v) = Fo, a constant, and comment on your result.

 

[learningphysics]

K the whole question is this:

There are certain simple one dimensional problems where the equatin of motion can always be solve,d or at least reduced to the problem of doing an integral. One of these is the motion of a done dimensional particle subject to a force that depends on only the velocity v, that is F=F(v). Write down N2L and separate the variables by rewriting it as mdv/F(v) = dt. Now integrate both sides of this equation and show that

t = m... (we did this)

Provided you can do the integral, this gives t as a function of v. You can also solves to give v as a function of t. Use this method to solve the special case that F(v) = Fo, a constant, and comment on your result.

Yes, just plug in Fo into the integral, instead of F(v)... and then solve the integral... since Fo is a constant, it just comes out of the integral... leaving dv inside

 

[Oblio]

so just solve for the integral of dv?

 

[learningphysics]

so just solve for the integral of dv?

yes.

 

[Oblio]

... integral of dv = v?

 

[learningphysics]

... integral of dv = v?

You should use limits in your integral ie $$\int_{v_0}^{v_f}dv = v_f-v_0$$

 

[Oblio]

velocity is constant, is my vo = v or 0?

 

[learningphysics]

velocity is constant, is my vo = v or 0?

why do you say velocity is constant?

 

[Oblio]

doesn't fo = fv?

 

[learningphysics]

doesn't fo = fv?

force is constant. not velocity.

 

[Oblio]

So, I'm unable to take it past vf-vo?

 

[learningphysics]

So, I'm unable to take it past vf-vo?

You should be using this:

$$t = m \int_{v0}^{vf}}\frac{dv}{F(v)}$$

What does the right hand side come out to? What do you get when you solve for vf?

 

[Oblio]

I thought we subbed F(v) out of the integral?

 

[learningphysics]

I thought we subbed F(v) out of the integral?

Yes. what do you get when you do that?

 

[Oblio]

t =mF(v) $$\int$$$$^{v}_{vo}$$ dv

 

[learningphysics]

t =mF(v) $$\int$$$$^{v}_{vo}$$ dv

before you take out F(v), you should replace it with Fo... so what do you get when you solve for v?

 

[Oblio]

t= mF(o) $$\int$$$$^{v}_{vo}$$ dv

t= mF(o) (v-vo)

solve for v from here?

 

[learningphysics]

t= mF(o) $$\int$$$$^{v}_{vo}$$ dv

t= mF(o) (v-vo)

solve for v from here?

Yes, one thing though, it should be m/Fo because the F(v) was in the denominator not the numerator.