Re: Spreadsheet Analysis of Rotating Stick

[Robert Carlson <Raacc@AOL.COM>]

Brian writes:

> If you model a meter stick pivoted at one end, starting from the upper
>   vertical, I would have thought you could set the initial conditions
>  as angular velocity = a small starting rate
>  angular position = a small displacement to avoid the quasi stable upright.
>
>  ! initial conditions
>  ao = Acos(q0)
>  w0 = 1E-6
>  q0 = 1E-6
>
>  !real time model, iterate from here.
>  a1 = Acos(q0)
>  w1 = wo+a1(dt)
>  q1 = qo+w1(dt)
>
>  I seem to recall that with dt set small enough the approximations
>  built into this very simple approach are not onerous. But you have
>  a spread sheet ready to go. So I expect you can soon tell if this
>  has sacrificed accuracy.   It is true that in particularly crucial
>  numerical integrations, some form of predictor/corrector or other
>  elaboration like Newton-Raphson is preferred for the
>  summastion/integration.
>
>  This expression...
>  >q(i) = 4t^2Acos(q(i-2))+2q(i-2)-q(i-4)
>  (q(i) depends on q(i-2) and q(i-4))
>
>  ...was evocative of a sample digital signal processing scheme
>  for filtering in the frequency domain. If it discards i-1 and i-3
>  there is a curve fitting loss, it seems to me. I wish I were better
>  equiped to discuss this in more detail.

Brian,

I agree, but I didn't want to iterate, I wanted to use a simple model, shoot,
and see how close it would come.  If one uses a five point derivative, rather
than a three point derivative, this increases the number of initial
conditions.

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?

The small angle approximation gives 2.01 s which is of course not correct
here.  In Tipler, an approximation for T is given as:

T = To(1+ 0.25sin(0.5B)^2+0.1406sin(0.5B)^4+ . . .)

Where To = 2.01 s, and B is the release angle measured from the vertical of
the bob at its lowest point.

For B = 90 degrees this gives T = 2.33 s.  However, this only includes terms
through sin(0.5B) to the fourth power.

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