Wednesday, July 13, 2016

Nothing To Do...With An Egg

Actually the parametric is just,

\(x=1-sin(\cfrac{\pi}{2}-s)*cos(s)\)

\(y=sin(\cfrac{\pi}{2}-s)\)

\(0\le s\le\pi/2\)

Since, \(sin(\cfrac{\pi}{2}-x)=cos(x)\)

\(x=1-y^2\)

\(y=\pm\sqrt{1-x}\)

A plot of (1-x)^(1/2) and -(1-x)^(1/2) gives,


which looks more like a wave front than egg.  And a blimp,


a bunch of pile up waves.  This however suggests that there is something wrong with the expression,

\(x_g=rcos(\theta_1)=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)cos(\theta_1)\)

from the post "Egg Shaped Egg" dated 12 Jul 2106.

\(x_g\) here, starts with a maximum value, when \(r=0\) and \(\theta_1=0\) and end up with \(x_g=0\) when \(\theta_1=\pi/2\).

If we considers the effects of \(F_{D}=F_{drag}cos(\theta)\).  If the particle is at terminal speed, \(F.c=constant\), \(F\) is a constant, then \(F_{D}\) is fully compensated by the internal forces of particle, after it deformed.  If the particle deforms in the same way in all directions, the length along the line of action of the force \(F_{drag}sin(\theta)\) is \(r\), so the length along \(F_D\) is,

\(\cfrac{F_{drag}sin(\theta)}{r}=\cfrac{F_{drag}cos(\theta)}{x_g}\)

\(x_g=\cfrac{rcos(\theta)}{sin(\theta)}\)

But \(r=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)\)

\(x_g=\cfrac{|F_{drag}|}{\delta m.\omega^2}cos(\theta)\)

which is still a circle!  What happened?  The statement " If the particle deforms in the same way in all directions..." changes the particle back into a circle/sphere.  If we instead allow the changes along the surface of particle to dictate the change along \(x_g\), where like the point at the tip of the particle, an elemental mass on the surface if moved by a centripetal force tangentially by an infinitesimal distance.


The displace/distortion tangential to the surface due to \(F_{D}=F_{drag}cos(\theta)\) is,

\(\cfrac{F_{drag}sin(\theta)}{r}=\cfrac{F_{drag}cos(\theta)}{x_t}\)

\(x_t=\cfrac{rcos(\theta)}{sin(\theta)}\)

the projection of this length along the horizontal axis is,

\(x_g=x_tcos(\pi/2-\theta)=\cfrac{rcos(\theta)}{sin(\theta)}.sin(\theta)\)

Since,  \(r=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)\)

So,

\(x_g=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)cos(\theta)\)

which is the same as before.  And the egg remains.