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fiona1218 · Sep 9, 2008, 03:25 AM

A shaft is drilled from brisbane to perth along the chord connecting these two towns. A trolley moves along the shaft only using gravity for its motion. There is no resistance to the motion. You are investigate the result of motion.
You need to find:
1. acceleration
2. max and min velocity
3. time of travel

I have find this motion is simple harmonic motion, but I am stuck on finding the time and max velocity. I asked my teacher to check my answer. He said it was wrong, but he cannot tell why. Help me out!!

ebaines · Sep 9, 2008, 06:13 AM

This problem is very similar to a pendulum, but is complicated by the fact that the force of gravity varies along the length of the tunnel as its depth below the surface of the earth varies. If you ignore that fact, then the force on the trolley acting in the direction of the tunnel is

$ F = mg \cdot sin \theta $
For small angles:
$ sin \theta \approx \theta $
so
$ F = mg \theta $
The equation of motion is therefore:
$ F = ma = -mg \theta $
where
$ a = R \ddot {\theta} $
and R is the radius of the earth.

So
$ \ddot {\theta} = -\frac gR \theta $
The solution is of the form
$ \theta = A sin ( \sqrt {\frac gR } t ) $
where A = magnitude of oscillation = the angle of half the length of the tunnel = arcsin(L/2R) which is approximately L/2R. Likewise, $\theta \approx x/R$ . So restating in terms of x:
$ x = \frac L2 sin( \sqrt {\frac g R} t $

The velocity is the derivative of this, and the acceleration is the second derivative. Can you take it from here?

fiona1218 · Sep 9, 2008, 06:25 AM

we are using the formulae of a=g sin wt
we integrate acceleration to find velocity and then s (displacement)
then we can find acceleration is related to the displacement and we say it is SHM.

this is sort of way my teacher want us to follow. But I find a problem to reach the right answer this this way...

and by the way, what is the "L" in your answer?

thank you for your answer. I really appreciate that.

ebaines · Sep 9, 2008, 06:45 AM

L is the length of the tunnel.

One problem with starting with a = g sin(wt) is that this says that the max acceleration is g, but it's not. To accelerate at g would mean the trolley is in free fall, like falling down a well, but in this problem the tunnel is not vertical. So the equation to start with is a = Cgsin(wt), where C is a constant. Using the argument I gave earlier, you can see that C = L/2R.

Integrate to find velocity:

v = -(Cg/w) cos(wt)

and integrate again to find displacement:

x = -(Cg/w^2) sin (wt)

You know the max displacement is L/2. So now you can solve for w. Hope this helps.

fiona1218 · Sep 9, 2008, 07:17 AM

Thanks mate!