\(|F^{*}_{c}|=|F_{drag}|sin(\theta)=\delta m.r.\omega^2\)
where \(\delta m\) is an elemental mass on the surface of the egg. \(r\) is the radial distance from the axis along the direction of travel and \(\theta\) the angle at the \(C.G\) between \(F_{drag}\) and the rotational axis.
\(r=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)\)
This is not a polar plot, \(r\) is the distance from the horizontal rotational axis.
\(y_g=r\)
The distance along the rotational axis of \(\delta m\) from the \(C.G\) is given by,
\(x_g=rcos(\theta_1)=\cfrac{|F_{drag}|}{\delta m.\omega^2}sin(\theta)cos(\theta_1)\)
This may be wrong Pls refer to "Nothing To Do...With An Egg" dated 13 Jul 2016
We plot the parametric pair,
\(x=1-sin(\cfrac{\pi}{2}-s)*cos(s)\) and
\(y=sin(\cfrac{\pi}{2}-s)\)
\(0\le s\le\pi/2\)
Implemented in the plotting software, \(s\) is measured with reference to the \(y=0\) axis. In the intended graph, \(\theta\) starts at \(s=\pi/2\) at the front tip and ends with \(s=0\) at the half circle defined by an axis that divides the particle into front and back. \(\theta_1\) is always measured with reference to the \(x=0\) axis and start from \(0\) to \(\pi\) and varies linearly with \(s\). So, \(\theta\rightarrow (\pi/2-s)\) and \(\theta_1\rightarrow s\). Up to this point, the front tip will be at the origin where \(s=0\), to invert the plot \(x\rightarrow 1-x\).
As spin, \(\omega\), increases, the front tip sharpens. As drag force, \(|F_{drag}|\), increases, the front tip flattens. The ratio,
\(R_D=\cfrac{|F_{drag}|}{\omega^2}\)
a drag-spin ratio determines the shape of the particle.
The front tip is always flat as \(\delta m\) moves tangentially away under the action of \(F_c\) a centripetal force. The front top does not sharpen with a discontinuity in gradient.
What happen to the back of the particle?
