Variable acceleration problem.

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  • #1
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A particle of mass m is released from rest at a distance b from a fixed origin of force that attracts the particle according to the inverse square law:

F = -kx^-2

Show that the time required for the particle to reach the origin is:

[pi](mb^3/8k)^1/2

I have no idea where the pi came from.

This is what I've done.

F=dp/dt
dp/dt=mdv/dt
dv/dt=d^2x/dt^2
m*d^2x/dt^2=m*dv/dt
m*d^2x/dt^2=-k/x^2
m*x^2*d^2x=-k*dt^2
m[inte][inte]x^2dxdx=-k[inte][inte]dtdt

After solving the double integrals and pluging in the constants I get.

t = sqrt[-1/(6k)*m(16b^4)]

I'm going to be really embarrased if my calculus is wrong.

So am I doing wrong. I still have no idea where a pi comes from!

Thanks

Frank
 

Answers and Replies

  • #2
Tom Mattson
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Originally posted by frankR
m*d^2x/dt^2=-k/x^2
m*x^2*d^2x=-k*dt^2
Here's your problem. d2x/dt2 cannot be separated as dxdx/dtdt.

You'll have much better luck by noting the following:

a=dv/dt
a=(dv/dx)(dx/dt)
a=(dv/dx)(v)

or

a=v(dv/dx)

Now, try that with your force:

-k/x2=mv(dv/dx)
(-k/x2)dx=mvdv

You can get x in terms of v, and then simply note that v=dx/dt and integrate again to find the time. Give it a shot and let me know if you need more help.

edit: fixed an omission
 
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  • #3
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I get -b*sqrt(2)/(3*sqrt(k/(b*m)) = sqrt(2*b^3*m/9k)

I still don't see where a pi could come from.
 
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  • #4
Tom Mattson
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Originally posted by frankR
I get -b*sqrt(2)/(3*sqrt(k/(b*m)) = sqrt(2*b^3*m/9k)

I still don't see where a pi could come from.
Can you show your work? That will help a lot!

Here's how I started:

-(k/x2)dx=mvdv

(sorry, I accidentally left the 'm' out in my last post, but I have fixed it)

-k[inte]bx(1/x2)dx=m[inte]0vvdv
(k/x)-(k/b)=(1/2)mv2

Are you with me so far? If so, can you show me how you did the second integral?
 
  • #5
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Originally posted by Tom

(k/x)-(k/b)=(1/2)mv2
Except for up to there. I made an algebriac mistake.

k/x - k/b != k/(x-b)

Let me start over. Give me a couple minutes.

Thanks for your help so far.
 
  • #6
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Okay cool.

The integral ins't too prety so I let me Ti-89 solve it.

I get t = -Sin([oo])b^3/2*sqrt(2m)[pi]/4sqrt(k)

Which is = [pi]sqrt[Sin([oo])*b^3m/8k]

What the heck does Sin([oo]) mean?

Thanks. I just need to slow down. Stop making dumb mistakes!

I need to stop forgetting that chain rule property.

Thanks a lot!

Frank
 
  • #7
Tom Mattson
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Originally posted by frankR
What the heck does Sin([oo]) mean?
Uh-oh.

It sin([oo]) is indeterminate, because it keeps oscillating. That means you can't assign a value to it. Me smells a mistake here.
 
  • #8
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Originally posted by Tom
Uh-oh.

It sin([oo]) is indeterminate, because it keeps oscillating.
I'm it's the software in my calculator.

I get a headache looking at this integral:

sqrt[m/2][inte]dx/sqrt[k/x-k/b]

I'm sure it's not as hard as it looks. I can drag up my old Calculus text, or look for a tabulated solution, to solve it another way.
 
  • #9
Tom Mattson
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I would try to work it into a form that matches an integral in a table. My guess is that it will turn out to be an inverse trig function, which is how you will get a π out of it.
 
  • #10
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Originally posted by Tom
I would try to work it into a form that matches an integral in a table. My guess is that it will turn out to be an inverse trig function, which is how you will get a π out of it.
I've been playing with it for half an hour, I'm not getting anywhere. It's beyond my mathematical abilities. Everything I do I still get an indeterminate solution.

If b>0, which it is, then I believe we get a solution.
 
  • #11
Tom Mattson
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Originally posted by frankR
I've been playing with it for half an hour, I'm not getting anywhere. It's beyond my mathematical abilities. Everything I do I still get an indeterminate solution.
Without knowing what you've done, I really can't say much on this.

If b>0, which it is, then I believe we get a solution.
Physically, you should get a solution, because this force is the same as a point mass or a point charge. Mathematically however, there seems to be a problem, because you encounter an integral like this:

[inte]b0((x/(b-x))1/2dx

whose integrand diverges at the lower boundary. Still, there should be a way to handle this improper integral.
 
  • #12
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Originally posted by Tom
[inte]b0((x/(b-x))1/2dx
That's the form I found.

If x is under the following condition:

0<=x<b

The solution is fine.
 

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