Calculating Time to Reach Point O with Force F = kx^-2

[Hafid Iqbal]

[PLAIN]http://img204.imageshack.us/img204/9204/fisika.jpg
How much time will elapse until this particle arrive at point O?

Edit : that force is F = kx^-2 not kx²


Thanks

 

[Dick]

It looks to me like you want to solve a differential equation. The force tells you the acceleration at any time. Integrate that to get the velocity and displacement. Please try it.

 

[Hafid Iqbal]

Yes, i have had velocity. But the time is my problem, i still can't get the variable t

 

[Dick]

Yes, i have had velocity. But the time is my problem, i still can't get the variable t

Can you show what you have done?

 

[Hafid Iqbal]

Here it is

[URL]http://latex.codecogs.com/gif.latex?\&space;\&space;\&space;\&space;\&space;F=m.a=m.\frac{dv}{dt}&space;\\&space;\\&space;.&space;\&space;\&space;kx^{-2}=m.\frac{dv}{dt}&space;\&space;.&space;\&space;\frac{dr}{dr}&space;=&space;m.v.\frac{dv}{dr}&space;\\&space;\\&space;.&space;\&space;\&space;\&space;\&space;k(a-r)^{-2}&space;\&space;dr=m.v&space;\&space;\&space;dv&space;\\&space;\\&space;\int_0^a&space;k(a-r)^{-2}&space;\&space;dr=&space;\int_0^{v_O}&space;m.v&space;\&space;\&space;dv&space;\\&space;\\&space;.&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\&space;\fbox{\textit{v}_O&space;=&space;\infty}​

[/URL]

 

[Ray Vickson]

This is incorrect: F is a^n attractive force (pointing toward the origin), so you need x''(t)=-k/x(t)^2. Alternatively, you can work out the potential energy V associated with the force, then use conservation of (total) energy.

RGV

 

[BruceW]

Actually, I think Hafid's equation is correct (Since dr=-dx).
To get further on this problem, you should leave the limits as v and r, and then integrate the right-hand side (easy), and integrate the left hand side (using trigonometric substitution).
So then you will have a function of r on the left and a function of v on the right. Then you need rearrange it to the form v.f(r)=constant
So then, you integrate both sides with respect to t (you will need to use the trigonometric substitution again), and use the limit r=a, and this will give you the time.

 

[Hafid Iqbal]

Thanks for the reply

This is incorrect: F is a^n attractive force (pointing toward the origin), so you need x''(t)=-k/x(t)^2. Alternatively, you can work out the potential energy V associated with the force, then use conservation of (total) energy.

RGV

Actually, I think Hafid's equation is correct (Since dr=-dx).
To get further on this problem, you should leave the limits as v and r, and then integrate the right-hand side (easy), and integrate the left hand side (using trigonometric substitution).
So then you will have a function of r on the left and a function of v on the right. Then you need rearrange it to the form v.f(r)=constant
So then, you integrate both sides with respect to t (you will need to use the trigonometric substitution again), and use the limit r=a, and this will give you the time.

Which one is true?

 

[BruceW]

Well, its true that $F=-k x^{-2}$ and $F = m \frac{d^2x}{dt^2}$ so that:
$$m \frac{d^2x}{dt^2} = -k x^{-2}$$
But now if we use the substitution x=a-r, we get:
$$m \frac{d^2r}{dt^2} = k {(a-r)}^{-2}$$
So your equation is correct, if we define $v = \frac{dr}{dt}$.

 

[BruceW]

Also, your equation:
$$\int_0^r \ k(a-r)^{-2} \ dr= \ \int_0^{v} \ m.v \ dv$$
IS the equation for energy conservation. On the left, you have the integral of the force (which equals the negative change in potential energy), and on the right, you have the change in kinetic energy.

You now need to do the integration, then rewrite for v and integrate again to get the time it takes to get to r=a.

 

[Hafid Iqbal]

Ok, thanks sir for the help... :)

 

[BruceW]

no worries, it is quite a tricky problem