Pendulum project update

Bert Freudenberg bert
Mon May 5 08:53:40 PDT 2003


Am Donnerstag, 24.04.03 um 15:50 Uhr schrieb Phil Firsenbaum:

> So, I've done a little reading (Physics Made Simple, etc,) and most 
> recently consulted with a physics professor who happens to be in the 
> family. As a result, I'm fairly (if not completely) convinced that 
> creating a realistic pendulum is well beyond what 5th graders could 
> do. I'm not even sure if I'll ever be able to complete the project 
> myself.
> I have learned a lot about pendulums, gravity, etc.
> I realize now that my pendulum does not reflect reality because it 
> moves at a steady rate through it's swing when, in fact, a pendulum 
> accelerates in its downward motion and decelerates in its upward 
> motion.
> In order to simulate reality I need to be able to change the rate of 
> acceleration of the heading of my pendulum. I now have a formula that 
> would accomplish this, however, it includes a square root function. Is 
> that possible in the etoy environment?

I guess you are a little bit blinded by all the math. It's really 
simple - the acceleration of the pendulum depends on its position. If 
it is vertical, you have zero acceleration (because the force of 
gravity is straight down and does not cause the pendulum to swing). If 
it is horizontal, you have the maximum acceleration, again because the 
force points straight down, but now this is exactly the direction to 
make the pendulum rotate.

This trivially maps to an etoy (just increase the speed by the 
acceleration value), the only obstacle is to get the acceleration 
depending on the current angle. This is what the "weighing angles" 
discussion was all about. Either you do this (best for 5th graders I 
guess), or you "measure" it, like in the pendulum project I sent last 
week. It works fine without any trigonometry or square roots: You 
basically just take the _horizontal_ extent of a line that represents 
your pendulum. If the line is vertical, its horizontal extent is zero. 
If the line is horizontal, its horizontal extent is maximal. You still 
need a sign for the force, which you can get by checking the extent 
relative to the line's reference point.

HTH

-- Bert




More information about the Squeakland mailing list