Re: Vertical fall. A paradox?

[LUDWIK KOWALSKI <KOWALSKIL@alpha.montclair.edu>]

John's answer and my question, are quoted below the dashed line. I am not 
fully satisfied with the answer. Let me add a description of a simple demo
to remove the ambiguity of the term "twisted blades".

1) Take a cork from a champain bottle and turn its wider side down. Drop
   it and it falls down along the vertical axis. There is no rotation.
2) With a sharp knife make a "meridional cut" in the cork and insert an
   index cart into it. Drop the cork and see it falling without rotation.
   The fins are parallel to the axis, the "fin angle" of John is zero.
3) Cut the card (only that part which is above the cork) along the axis.
   Bend each half so that its plane is no longer parallel to the corks axis.
   The fin angle is 30 to 45 degrees. Looking from the side you would see 
   the card's profile as a letter Y. Something as below. Drop this "rigid"
                           object and abserve its spinning. It is clear
        \       /          that the terminal angular velocity is rapidly  
         \     /           established when the "fins" are large. It can
          \   /            be measured with a camcorder. If I had time
           \ /             I would check John's formula against the real
            I              data. The "mean radius" of the fins can
          **I**            easilly be calculated for the flat rectangular
          **I**            card sections. 
          *****            What I would like to know now is the reason for
          *****            saying that "the fins should slice down through
         *******           the air without a net torque." The argument that
        *********          "this is necessary to conserve energy" would not 
        *********          satisfy me in this context.
                                                          Ludwik Kowalski

P.S. The left blade above is closer to me than the right blade; do not 
forget that the blades were part of a single index card. The direction
of rotation is counter-clock-wise when looking from above.
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On Sun, 9 Nov 1997, LUDWIK KOWALSKI wrote:

> Now suppose the object is a  rigid "bomb" whose fins are twisted. Its 
> rotational kinetic energy (spinning about the vertical axis) also increases 
> at the expense of potential energy. Will the angular velocity continue to 
> increase after the terminal linear velocity is reached? What is the  maximum 
> possible rotational energy? We know that KErot can not exceed (m*g*h-KEtt), 
> where KEtt is the terminal translational KE. In fact, KErot must be smaller 
> than (m*g*h-KEtt) because part of the initial potential energy is thermalized.
> What mechanism is responsible the "terminal angular velocity"? Why should the 
> net torque be zero at some rate of rotation? Constant terminal torque? Yes. 
> Constant terminal omega?  (???).

Ludwik,

At terminal conditions the fins should slice through the air without any
net torque one way or the other.  I would expect a relationship something
like the following to hold:

  (terminal angular speed) x ("mean radius" of the fins)
  ------------------------------------------------------ = tan(fin angle)
                  (terminal linear speed)

where the fin angle is the angle the fins make with the vertical.  Slower
rotation than this would cause air resistance to both brake the fall and
increase the rotation; faster rotation would cause the air to both drive
the device downward and brake the rotation. 
                                                   John