Velocity relation of a particle

[Wiz]

I have a relation for velocity as v= sqrt [(1-x)/x)]
I need to find the time the particle takes to reach x= 0.25
Initially it is at x=1
I had problems integrating furthur...to find time
Can you help??..
Wiz

 

[Jameson]

I did this integral on my calulator and it's gross. Integration by parts perhaps? See what you can do with that..

\int \sqrt{\frac{1-x}{x}} dx = \int (\sqrt{(1-x)}) * (\sqrt{x})^{-1} dx


Jameson

 

[dextercioby]

From the definition of velocity
v(t)=:\frac{dx}{dt} Separation of variables and corresponding limits of integration give:

\int_{0}^{\tau} dt=\int_{1}^{0.25} \frac{dx}{\sqrt{\frac{1-x}{x}}}

The integral in the first member is really simple,it is the time you need,denoted by me with " \tau"...

For the integral in the RHS,use 2 substitutions

First
x=t^{2}
The second
t=\sin y

Pay attention to:
1.Signs.
2.Transformations of the limits of integration.

Daniel.



EDIT:I'm getting a "tau" smaller than 0.It's weird.Are u sure the problem is stated correctly...?

 

[Wiz]

prob

I'm not sure...here's the original prob from which i derived tht velocity relation.

There is a particle initially on x = 1
It is acted upon by a force which varies as F(x)= -0.5 * x^-2
What is the time it takes to reach x= 0.25

Thanks,
Wiz

 

[Wiz]

forgot

I forgot...the mass of the particle is 10^-2 kg...

 

[dextercioby]

That's another problem,man,totally different.

The advice is the same.Separate variables & integrate with corresponding limits.(EDIT 1:It doesn't work,it's second order).

Daniel.

EDIT 2:See next post for further comments.

 

[dextercioby]

Nope,i'm afraid this problem (the second u posted) is not completely integrable

It requires solving the ODE

\frac{d^{2}x(t)}{dt^{2}}=-|C|\frac{1}{x^{2}(t)}

,but subject only to the initial condition:

x(0)=1

Unfortunately,the closure of the Cauchy problem requires 2 independent initial conditions to determine the 2 coefficients (<============ a II-nd order ODE).

Daniel.

P.S.Look it up again.

 

[Wiz]

The prob i posted is the same,
All i did was...i found out acc. from the force..
Then integrated it and i found velocity...the constant in tht case was c=-1

Btw i hv mentioned tht the particle is initially at x=1
The prob is same as given in the book,
This is an iit-jee question meant for students who hv cleared the 12th grade...and i am in 11th rite now,
Thanks for the help but i think u are getting a bit too advanced for me,,,

Wiz

 

[dextercioby]

Could you please post your work.I'm really curious of what you've done...

Daniel.

 

[Wiz]

Oh,my god...i am sooooo sorry
the correct eq is F(x)= -k * 0.5 * x^-2
where k= 10^-2 Nm^2
the mass of the body is 10^-2 kg
Now itz 100% correct.

Wht i did was, i got a= -0.5 * x^-2 as k and m canceled out
so i got dv/dt=(-0.5 * x^-2)
hence dv/dx*dx/dt=(-0.5 * x^-2)
hence dv*v=(-0.5 * x^-2)dx
integrating i got c=-1 and simplification led to the vel eq i posted...
Cud u please check it now?

 

[dextercioby]

It looks correct.(In general,the constant are less worrying).

Daniel.

P.S.Can u uniquely determine v=v(x)...?

 

[Yegor]

You can be sure. I also got: v= sqrt [(1-x)/x)].
It's time just to integrate. Substitution sqrt [(1-x)/x)]=t leads to rather simple integral.

 

[dextercioby]

Are u sure it's not a

\int a(x) \ dx=\int v \ dv

Plug the values & integrate.The RHS is simple

\int v \ dv =\frac{1}{2}v^{2}+C



Daniel.

 

[Yegor]

I hope we shall see it after integration. There must be "tau">0. If we'll have opposite we can just change the sign. Am i right?

 

[Yegor]

Sorry. i didn't notice edition. I think You are right in both posts

 

[dextercioby]

The acceleration is force divided by mass:

a(x)=\frac{F(x)}{m}=\frac{-5\cdot 10^{-3} x^{-2}}{10^{-2}}=-0.5\cdot x^{-2}

Therefore:

\int a(x) \ dx=-0.5\int x^{-2} \ dx =0.5 x^{-1}

What initial condition does the velocity satisfy...?

Daniel.

 

[Wiz]

Hey i solved the prob,
The fact we forgot was tht since the body is moving towards the orgin due to -ve value of acc, the correct relation wud be v= - sqrt[(1-x)/x] ----(1)
Now put x = sin^2($)
hence dx/d$ = sin2$
hence dx = sin2$*d$ --------(2)

By (1),
- dx/sqrt[(1-x)/x] = dt
=> -dx /cot($) = dt
By (2)

dt = - sin2$*d$/cot($)
=> dt = -2sin^2($)*d$
Now integrate lhs from 0 to T and rhs from pi/2 to pi/6
So the final ans I got was
T= [pi/3 + sqrt(3)/4 ]
which happens to be correct

I hope I haven't made any typing mistakes this time..
Thanks for replying...