# Tough Math problem, above my head, any help much appreciated.

1. Jan 16, 2012

### Trevormbarker

1. The problem statement, all variables and given/known data
Essesntially me and my friend for our physics ISP (which is like a final project in ontario) are building this gravitational torsional balance: http://www.fourmilab.ch/gravitation/foobar/

Our set up is very similar, with our main balance arm being 30cm long , 5cm wide and 5cm high. The weights we are using are 50 ounces each (one on each side of the arm) and we are putting a 50kg weight 90 degress between each wight on the arm, 5cm off the radius.
were trying to impress our physics teacher by showing him the theoretical value for how long the wieghts should take to attract gravitationally.
2. Relevant equations

Both of our math skills are limited to differntial calculus and the little intergral ive tried to teach myself over the internet, i know that since acceleration is always changing it will involve integration of the acceleration function, and im guessing this displacemnt too, anyways i know this problem is exceptioanlly tough so any help would be much appreciated.

3. The attempt at a solution
Honestly dont know where to start. Wrote f=Gmm(1/r^2) then subsituted expressions in for r and got stuck.

2. Jan 17, 2012

### lanedance

To solve completely you would need to solve a differential equation, that of a damped harmonic oscillator, which should be quite google-able .

But we can make some approximations, so lets assume you have a mass at rest M

You bring the test mass m close to it, which generates a force f=GMm(1/r^2)

This will cause the mass to accelerate f=Ma, rearranging for a gives
a= Gm/r^2

As soon as the mass begins to move toward the test mass, it will experience a spring force (and the damping force due to the water brake, but lets neglect that for now)
F=-kx

so the acceleration is
a = Gm/r^2-kx/M

And it will move until the forces (spring and grav.) are balanced which is equivalent to when the acceleration is zero. Knowing the accelartion as a function of distance, you should be able to solve for the time it takes to get to the equilibrium point.

As the damping term is not included, solving the above equation you would actually find that it oscillates about that zero point - however you could still use it to get a feel for the time it takes to get there

have a look at the following wiki entry
http://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator

in particular the follwing diagram, which can be thought of as position vs time
http://en.wikipedia.org/wiki/File:Damping_1.svg

The blue is the undamped motion. The red is what is called critically damped. You can see that ~3-4times after the blue first reaches the equilibirum point, the critically damped curve is very close to the equilibrium point

3. Jan 17, 2012

### Trevormbarker

Yeah we actually did the experiment yesterday, and it worked pretty well, and it did pretty much what you just predidicted. Thanks alot for the help! I'll see if we can work through that.
but also since their are 2 masses would the problem not have to take that into consideration, im guessing that the time isnt just half of what it is if their was only one mass.