# Application of a pendulum

1. Oct 19, 2006

### DocZaius

This is a theoretical question:

Say you bore a hole through earth, suck out all its air, and drop an object in one side. Let's say it takes 42 minutes for the object to come out the other side and come to complete stop (only to then drop back down, of course)

Now let's say you drop an object twice as heavy, still 42 minutes right?

Ok. Now say you bore a straight hole from Los Angeles to New York in this planet, put a completely frictionless track through that hole (I know, I know, bear with me) and stick a train on it in Los Angeles, letting gravity accelerate it. Will it arrive in New York in 42 minutes? Will it arrive from any point to any other point in 42 minutes?

If so, we've got a pendulum there, right?

Again, please approach this theoretically, oh and of course earth isn't spinning and is a perfect sphere..

What do you think, is this true?

Last edited: Oct 19, 2006
2. Oct 19, 2006

I started to say "no", but I'm going to have to think about it longer. Where did you get the "42 minutes"? It should be easy to calculate the time. The gravitational force, as you move through the earth, is proprotional to $r$ so the acceleration is also proportional to r. Knowing that the acceleration is g at r= R, the radius of the earth, we get $$\frac{d^2x}{dt^2}= -\frac{gr}{R}[/itex] and can solve for the motion. Although we are moving in a straight line from Los Angeles to New York, the force in that direction changes in much the same way the force along the arc of a pendulum does. You might well be correct. I'll think about it. 3. Oct 19, 2006 ### tim_lou well, you will have to resolve the force vectors into components. [tex]\frac{d^2x}{dt^2}=\frac{-g\sqrt{x^2+d^2}}{R}*\frac{x}{\sqrt{x^2+d^2}}$$
where d is the distance between the mid-point of the track to the center of the earth.

at the end, the differiential equation is the same, the period should be the same.

4. Oct 19, 2006

### Integral

Staff Emeritus
Such a "train" could be a high speed transoceanic or continental transit system for a low energy cost. HOWEVER, construction expenses would astronomical and just think of the engineering difficulties. It would be interesting to compute just how deep you would have to dig to get, say a 8 hr trip from NY to LA.

Yet another interesting question would be to calculate the "ideal" path would it be Parabolic???

5. Oct 19, 2006

### quinn

The real problem here is that you need a frictionless track. Not going to happen.

Another problem would be derailing due to corolis forces if one didn't have the train take the proper "twisted" path. I am also not entirely sure how the corolis forces would affect the motion/period of the train if one crossed lataitudes...

All in all the biggest obstactle would be to dig a tunnel that long,... how long did it take the "Chunnel" to be built" Let's say 10 years... the Chunnel is 26 miles long,.. let's say 30 miles,.. and from LA to NYC is about 3000 miles,... this tunnel is going to take roughly 100 years?...

6. Oct 20, 2006

### DocZaius

This is a completely theoretical question which has to do with the application of the pendulum's fundamentals into a sphere/tunnel experiment. I just wanted to see some good discussion on this problem.

HallsofIvy, I got 42 minutes from my own calculations, but here's a page that goes through them:
http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html

From all the thought I put into it, a tunnel from any point to any point on that sphere will give the same period of oscillation. I would love to be shown to be wrong though.

The easiest way for me to think about the change from a hole through the center to any tunnel through earth, is that even though the distance to travel is shorter, the vector which actually does work is much reduced, since the majority of gravity is pulling perpendicular to you.

Oh and again to be completely clear, I am not proposing this as an actual possibility of course, this is just theory.

7. Oct 20, 2006

### Staff: Mentor

But tim_lou already showed that you are right! Since the effective "spring constant" is the same regardless of where the hole is dug, the period of oscillation is the same.