Re: Spreadsheet Analysis of Rotating Stick

[Robert A Cohen <bbq@ESU.EDU>]

Just for fun, I decided to compare the time indicated below (someone
else mentioned that a better estimate was 2.369) with the time
assuming the average speed of the object was equal to
    alpha * sqrt(2gh)
where sqrt(2gh) is the maximum speed of the object (KE = mgh).

The total time would be the total distance (pi*h) divided by the average
speed, or
              pi*sqrt(h/2g) / alpha

Since the total time is 2.369, I solved for alpha (using h = 1) and got
alpha = 0.29954.  Is it just a coincidence that it comes out to be very
close to 0.3 or is there some way to show that alpha is 0.3 for some
reason?  I didn't do the analysis for other values of h.  Perhaps someone
can do that for me.

Just curious.

----------------------------------------------------------
| Robert Cohen               Department of Physics       |
|                            East Stroudsburg University |
| bbq@esu.edu                East Stroudsburg, PA  18301 |
| http://www.esu.edu/~bbq/   (570) 422-3428              |
----------------------------------------------------------

On Sun, 28 Nov 1999, Robert Carlson wrote:

> [snip]
>
> Here is a question for anyone on the list.  What is the period of a point
> mass attached to a string of length one meter, released from rest from the
> horizontal?
>
> [snip]
>
> The spreadsheet gives T = 2.37 s.
>
> Symon goes through the solution that develops the power series, but only
> takes it to sin(0.5B) to the second power.  I'm lazy, and wonder what the
> exact solution is.  With limited resources at home, and trying to follow the
> Vikings game, what is the next term in the power series for T, and what is
> the predicted period?
>
> Bob Carlson