Re: [Phys-L] no spin decay

[John Denker <jsd@av8n.com>]

On 09/17/2017 01:38 PM, David Bowman wrote:

> if one wants to verify or test for an exponential decay the time
> honored way is to plot the logarithm of the speed (or frequency) vs.
> time (or equivalently, make a semi-log-scaled graph).  If the graph
> is adequately represented as a straight line then the decay is
> exponential, and the intercept gives the logarithm of the initial
> speed/frequency, while the negative of the slope gives the decay
> rate.

Yes.

There's a good reason for that.  The semi-log paper is a good
match to the expected exponential.

> OTOH, if one wants to test for or verify a quadratic velocity
> dissipative decay, the time honored way is to plot the *reciprocal*
> of the speed (or frequency) vs time.  If the dissipative forces are
> quadratic in the velocity the plot will again be a straight line with
> the intercept giving the reciprocal of the initial speed and the
> slope giving the quadratic decay constant.

Yes.

There's a good reason for that, too.

I was disappointed in the video at this point:
  https://www.youtube.com/watch?v=G7hh42AgBlQ&t=471

where it said that the nonlinear differential equation was
"really hard to solve".  Gimme a break.  It is /easier/ to
solve than the exponential-decay equation.  No transcendental
functions required.  The equation of interest is:

       dv/dt = - Cd v^2

where Cd is the coefficient of drag.

You should be able to solve that almost by inspection.  The
first guess, that it is a power-law function of t, is not
exactly correct, but it's very close.  To cut to the chase:
substitute u := 1/v and find the equation of motion for u,
starting from du/dt.

Now you know why plotting 1/v versus time is the Right Thing
... assuming that Cd is constant (or nearly so) over the range
of speeds of interest.

==============

OTOH you can plot the data as-is, and compare to theory
directly:
  https://www.av8n.com/physics/img48/fidget.png

The yellow curve is my hyperbola.
The orange-ish points are the original data, as they appear
in the youtube image.

The fit is pretty good in the early going.  It's verrrry much
better than the fit to the exponential.  We can very confidently
reject the exponential hypothesis.

At the end, the decay is slightly faster than the simple
model would predict ... which is unsurprising, because we
know the coefficient of drag increases when the Reynolds
number gets to be smaller than 1000 or so.

============

Using the correct model, your ability to increase the spin
time by whacking it harder is more constrained than it would
be in the exponential model.

If you extrapolate the yellow curve back in time, you get
to /infinite/ velocity at a finite (and not very large)
past time.