Mass under a force

whitey · #1 $Old$ August 25th, 2009, 11:59 AM

Not sure if this one should go here.

Basically the first two bits are easy

$V(x)=18/x^2 -9/x$

With the equilibrium =4

The second part proving the bounded region isn't too bad either we can use conservation of energy

E=T+V quite easily

showing total energy =-1J and then

creating a quadtratic
-1= $V(x)=18/x^2 -9/x$

$-x^2 = 18-9x$

$0= x^2-9x+18$

$0=(x-6)(x-3)$

thus reaches 0 velocity between the points 3 and 6

The third part confuses me. I don't want an answer. Just an idea where I am heading.

I hope this is in the right place

how can we work out time when given x co-ordinates?

running-gag · #2 $Old$ August 25th, 2009, 01:04 PM

Hi

As you said the conservation of energy can be expressed as
$\frac{18}{x^2}-\frac{9}{x}+\frac12\:v^2 = -1$

This gives
$\frac12\:v^2 = -\frac{18}{x^2}+\frac{9}{x}-1 = -\frac{x^2-9x+18}{x^2} = -\frac{(x-3)(x-6)}{x^2}$

and finally $v^2 = -\frac{2(x-3)(x-6)}{x^2}$

Since $v^2$ is positive we have $3 \leq x \leq 6$ and since $v = \frac{dx}{dt}$

$T = \int_{x=3}^{x=6} \frac{x}{\sqrt{-2(x-3)(x-6)}}\:dx$

Haytham · #3 $Old$ August 25th, 2009, 01:06 PM

$dt=\frac {dx}{v}$

edit: too late

whitey · #4 $Old$ August 25th, 2009, 01:31 PM

Thankyou to both of you