Variable acceleration problem.

In summary: But it is not the solution for the problem. I wish I had my calculus text. It's not on my bookshelf, it's on the other side of the country.I think I've had enough for tonight. I don't know if I'm going to get much more done. I've never seen an integral like that.Thanks for your help Tom.
  • #1
frankR
91
0
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
 
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  • #2
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
 
Last edited:
  • #3
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.
 
Last edited:
  • #4
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
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
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
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
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
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
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
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
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.
 

1. What is variable acceleration?

Variable acceleration refers to the change in an object's velocity over time. This means that the object's speed and/or direction are changing at a varying rate.

2. How do you calculate variable acceleration?

To calculate variable acceleration, you need to know the initial velocity, final velocity, and the time it took for the change in velocity to occur. The equation for variable acceleration is (final velocity - initial velocity) / time.

3. What is the difference between constant and variable acceleration?

Constant acceleration occurs when an object's velocity changes at a steady rate, while variable acceleration occurs when the rate of change in velocity is not constant.

4. What are some real-world examples of variable acceleration?

Some examples of variable acceleration include a car accelerating and decelerating in traffic, a rollercoaster speeding up and slowing down on different parts of the track, and a swinging pendulum.

5. How does variable acceleration affect an object's motion?

Variable acceleration can affect an object's motion by changing its speed and/or direction. The object may speed up, slow down, or change direction depending on the rate of change in acceleration.

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