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Re: Spreadsheet Analysis of Rotating Stick



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