Integrate Acceleration to Find Time?

[Trapezoid]

This question is from my calculus assignment but I apologize if it belongs on the a physics board regardless.

Homework Statement

A particle of mass m is attracted towards a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If a particle is released from rest at a distance L from 0, how long will it take to reach 0?

2. The attempt at a solution
$$F = ma = \frac{k}{d}$$
where d is the distance from 0 and k is a proportionality constant. Therefore, $$a = \frac{d^2x}{dt^2} = \frac{k}{md}$$
I know that integration is required but I don't see how to use it to find time. Could somebody give me a tip as to how to proceed?

Thanks,
Trapezoid

 

[Robert1986]

Well, imagine that the question asked you for an equation that gave the position of the particle as a function of time. E.g. $s(t) = a_0 + a_1t + a_2t^2$. You know how to do this, right? Now, once you have the equation, just put the distance traveled on the LHS and solve for $t$.

 

[RoshanBBQ]

This question is from my calculus assignment but I apologize if it belongs on the a physics board regardless.

Homework Statement

A particle of mass m is attracted towards a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If a particle is released from rest at a distance L from 0, how long will it take to reach 0?

2. The attempt at a solution
$$F = ma = \frac{k}{d}$$
where d is the distance from 0 and k is a proportionality constant. Therefore, $$a = \frac{d^2x}{dt^2} = \frac{k}{md}$$
I know that integration is required but I don't see how to use it to find time. Could somebody give me a tip as to how to proceed?

Thanks,
Trapezoid

$$x\frac{d^2x}{dt^2} = \frac{k}{m}$$
d is x. Is that what you were forgetting?

 

[HallsofIvy]

You say "d is the distance from 0 and k is a proportionality constant" but you don't say what "x" is! I think you mean x to be also the distance from 0 at time t so your differential equation is really
$$\frac{d^2x}{dt^2}-= \frac{k}{mx}$$

You will have to solve the differential equation for x, then set it equal to 0 and solve for t.

Now, that is not easy to integrate directly but you can use a technique called "quadrature".
Let v= dx/dt so that $d^2x/dt^2= dv/dt$. By the chain rule, $dv/dt= (dv/dx)(dx/dt)= v dx/dt$. So your equation is
$$v\frac{dv}{dx}= \frac{k}{mx}$$
a separable equation.

$$v dv= \frac{k}{m}\frac{dx}{x}$$
Integrating,
$$(1/2)v^2= \frac{k}{m}ln(|x|)+ C$$
Since the object was released "distance L" (x= L) at rest (v= 0) we have
$$(1/2)0^2= \frac{k}{m}ln(L)+ C$$
so
$$C= -\frac{k}{m}ln(L)$$
and the equation for v becomes
$$\frac{1}{2}v^2= \frac{k}{m}ln(x)- \frac{k}{m}ln(L)= \frac{k}{m}ln\left(\frac{x}{L}\right)$$
(I am assuming that L is positive and as x changes from L to 0, it will be positive so we don't need the absolute value.)
$$v= \frac{dx}{dt}= \sqrt{\frac{2k}{m}ln\left(\frac{x}{L}\right)}$$

Integrate that, solve for x as a function of t, set it equal to 0, and solve for t.

 

[Trapezoid]

Thanks for all the help everybody. I'll take a closer look to see if I can't find the answer from here.

 

[RoshanBBQ]

I thought it would just be
$$x\frac{d^2x}{dt^2} = \frac{k}{m}$$
$$\int \int x dxdx = \frac{k}{m} \int \int dtdt$$
$$\frac{x^3}{6}=\frac{k}{m}\frac{t^2}{2}+c$$
$$x^3=3\frac{k}{m}t^2+c$$
Here is a nice junction to find c (assuming x = 0 is at the point and x = L is the initial condition, L > 0):
$$L^3 = 0 + c \rightarrow c = L^3$$
then
$$x^3=3\frac{k}{m}t^2+L^3$$
$$x(t)=\left ( 3\frac{k}{m}t^2+L^3 \right )^{1/3}$$

You then solve for the time for which x(t) = 0. Note, k must be < 0 for such a solution.

 

[Trapezoid]

Thanks HallsofIvy and RoshanBBQ,

Unless I've misunderstood, the method outlined by HallsofIvy leaves me with the integral $\int \frac{dx}{\sqrt{\ln(\frac{x}{L})}}$ which I cannot even begin to solve. With regards to RoshanBBQ's solution, it seems odd to me that k would be negative since constants in examples done in class are always specified to be positive. That being said, a negative k would imply that the force acts in the negative direction, which seems consistent with the problem. Is my thinking here correct?

Trapezoid

 

[RoshanBBQ]

Thanks HallsofIvy and RoshanBBQ,

Unless I've misunderstood, the method outlined by HallsofIvy leaves me with the integral $\int \frac{dx}{\sqrt{\ln(\frac{x}{L})}}$ which I cannot even begin to solve. With regards to RoshanBBQ's solution, it seems odd to me that k would be negative since constants in examples done in class are always specified to be positive. That being said, a negative k would imply that the force acts in the negative direction, which seems consistent with the problem. Is my thinking here correct?

Trapezoid

It just depends on how you set up the problem.I'm not entirely comfortable with the answer I found, though. Consider when t -> infinity. You would expect x -> 0. My solution has x become the 3rd root of an increasingly negative number...

 

[Trapezoid]

Hi all,

Sorry to bump but we got the solutions to the assignment so I thought I'd post some more guidance in case anybody else is having trouble with a similar question.

The method outlined by HallsofIvy is correct. Without giving too much away, to solve the last integral, substitute $u = \frac{L}{x}$ and use the gamma function.