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Re: [Phys-L] no spin decay



*If* one wanted to model *both* the high Reynolds number quadratic velocity dissipation from the drag (involving the moving of the mass of air in the way of the moving spinner lobes out of their way as they move) *and* model the low Reynolds number viscous shear dissipation in the wet friction of the bearings and shearing of the viscous air near the slowly moving lobes in one single overall model, then the simplest way to do it is to consider the deceleration to be a simple quadratic function of the speed with a nonzero linear term but a zero constant term, i.e.

- dv/dt = k*v^2 + g*v

where k in the characteristic inverse length for the quadratic drag in the high speed regime, and g is the characteristic exponential decay rate for the linear decay parameter regime. It is fairly easy to solve the ODE for this problem since it boils down to a doable integral. The result (for initial speed v_0 @ time 0) is

v = v(t) = v_0/((1+p)*exp(g*t) - p)

where the dimensionless parameter p is defined by

p == k*v_0/g

If p << 1 then the motion is dominated by the linear dissipation term and the motion approaches the simple exponential decay with amplitude v_0 and decay rate g.

If p is moderately valued and the long time limit g*t >> 1 is investigated the behavior is *still* an exponential decay with decay rate g, but with a modified amplitude of v_0/(1+p).

If p >> 1 and the time is short compared to 1/g, i.e. g*t << 1, then the quadratic drag term dominates and the decay is characteristic of a quadratically dissipative decay, i.e. 1/v approaches 1/v_0 + k*t in its behavior.

Note the above equation for the motion can be rearranged into the form

ln(v_0/v + p) = ln(1+p) + g*t

So a simple curve fitting strategy (involving only least squares line fits to supposedly line-like data rather than a full multi-parameter nonlinear curve fit) would be to plot ln(v_0/v + p) vs t for a range of trial values for p and then keep the graph that gives the straightest line for the data. The slope of the straightest line would be g and the optimal trial parameter p could be checked against the intercept of the fitted line to see if it is indeed ln(1+p). If the match to the intercept with the optimally straight line p value is a good match then the fit is good overall and the p value used would give the parameter k as

k = p*g/v_0.

But it would probably be best to self-consistently and iteratively home in on the optimal p value by some sort of (fixed point) iterative scheme whereby one takes the intercept of the previous trial least squares fitted line and uses that to get the next value of p to try in the ordinate of the next fitted least squares fitted graph and iterate until the optimal p is found.


Brian W wrote:

No access to Physics Teacher, but the URL to the spinner demo
was interesting

https://www.youtube.com/watch?v=G7hh42AgBlQ&feature=youtu.be

Grabbing the screen plot here

<http://s880.photobucket.com/user/betwys/media/ScreenHunter_355%20Sep.%2017%2021.31.jpg.html?sort=3&o=0>

Allows x,y estimates to be extracted via Graph Grabber (a free
download) leading to a fit to this function:

rate = 13.7123447*exp( -0.0891256151* time^(2/3) )

The plot of this fit given here:
<http://s880.photobucket.com/user/betwys/media/rotation.jpg.html?sort=3&o=0>

Notice the amusing precision of the model from such rough and
ready data!

Brian W

This fit is admittedly pretty good for a simple 2 parameter fit over the time range shown. But unfortunately it can't asymptotically have the correct behavior in the long time limit. All fits to such a family of functions will not decay as fast as the actual data do at long times. We know at long times the behavior has to approach a usual exponential decay, but the family of fitting curves has the constant exponent, 2/3, of the time, t, in the exponential. If you look carefully at the fit you will notice that the data begin to fall below the fit at large times simply because the fitting function simply can't decay fast enough there.

If one works backwards from Brian's family of fitting functions to the corresponding ODE they solve, i.e. find the dissipative decay model ODE for

v = A*exp(-B*t^(2/3)),

the DE one ends up with is

dv/dt = -(2/3)*(B^(3/2))*v/sqrt(ln(A) - ln(v)) .

Needless to say such a functional form for the dissipative force as a function of the spinning velocity is quite peculiar, to say the least.

David Bowman