Escape Velocity and Particle Position

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Homework Help Overview

The discussion revolves around a particle of mass m moving in one dimension with a given potential energy function. The original poster seeks to determine the position function x(t) for the particle when it starts at the origin with an initial velocity equal to the escape velocity.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive the position function but encounters difficulties with integration, particularly with logarithmic terms. Some participants question the clarity of the original poster's attempts and request more detailed explanations of the work shown. Others suggest considering conserved quantities to aid in finding expressions for velocity and position.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. The original poster has identified a potential method involving energy conservation but has not yet fully articulated the steps taken or the resulting expressions. There is a collaborative effort to clarify the original poster's approach and to guide them toward a solution.

Contextual Notes

There are indications of challenges with formatting and expressing mathematical equations correctly in the forum, which may affect the clarity of communication. The original poster has acknowledged these issues while attempting to convey their reasoning.

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Homework Statement


A particle of mass m moving in one dimension has potential energy [itex]U(x)=U_0[2(\frac{x}{a})^2-(\frac{x}{a})^4][/itex], where [itex]U_0[/itex] and a are positive constants.

What is the function x(t) for the position of the particle when it starts at the origin with a velocity in the positive direction equal to the escape velocity?

Homework Equations


The escape velocity is [itex]$\sqrt{\frac{2U_0}{m}}$[/itex] which escapes at x=a.

The Attempt at a Solution


I was able to determine the function for the force easily, but I ran into many problems with logs when trying to integrate it twice and ended up with logarithms that could not be paired up so that I could actually solve for x.

Having a fair amount of trouble getting the latex code to work on these forums properly... Looks like it's because of Chrome.
 
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What exactly is the question?
 
Sorry, I had been having so much trouble when trying to get it done with Chrome (restarting the post multiple times) that I forgot to include that part. It's there now.
 
Could you show us what you tried and got? It's hard to see where the problem is if all you do is give a vague description of what you tried.
 
[itex]$(\frac{1}{a^2})(x \ln (x)-x-\frac{1}{2}(x\ln(x^2-a^2)+a\ln(x^2-a^2)-2x))+C$[/itex] all over ma^2 I believe. It's difficult getting this stuff to show up correctly on the forums :s But basically something that I can't really continue with.
 
Cider said:
[itex]$(\frac{1}{a^2})(x \ln (x)-x-\frac{1}{2}(x\ln(x^2-a^2)+a\ln(x^2-a^2)-2x))+C$[/itex] all over ma^2 I believe. It's difficult getting this stuff to show up correctly on the forums :s But basically something that I can't really continue with.

What is this expression supposed to represent (position, velocity, force, the numer of apples in a carton of width [itex]x[/itex])? And how did you get it?

Posting an expression without any work shown or description of what the expression is and where it came from doesn't help us to see where you are going wrong.

In any case, can you think of a conserved quantity that will help you get an expression for velocity? Can you determine position as a function of time from that expression?
 
I figured out what I was doing wrong. The correct method was the second I tried, but I had represented something in a way that prevented me from seeing the ease with which I could have solved it. Basically the process would use E=1/2 mv^2 + U(r), which I can simplify so that I get a first order differential equation that requires some simple partial fractions to integrate. Thanks though.
 

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