\(\omega^2=\cfrac{F_{drag}}{\delta m}\cfrac{sin(\theta)}{r}\)
where \({F_{drag}}\) is a force not the distance from the origin to the elemental mass, \(\delta m\) on the surface of the particle. The particle will deform such that it rotates as a whole at a constant angular velocity.
\(\omega^2=A.{F_{drag}}\)
If we define that when the particle spins at \(\omega_o\) , it will present an \(S_n=38.5\), and that \(\omega\) is proportional to \(S_n\),
\(\cfrac{\omega}{\omega_o}=\cfrac{S_n}{38.5}\)
And we consider that the particle starts at \(S_n=39\) when \(\omega=0\), spinning decreases \(S_n\) and at \(\omega=\omega_o\), \(S_n=38.5\), we have,
\(S_n=39-0.5\cfrac{\omega}{\omega_o}\)
\(S_n=39-\cfrac{0.5}{\omega_o}\sqrt{AF_{drag}}\)
as \(F_{drag}\propto v^2\)
\(S_n=39-\cfrac{0.5S}{\omega_o}v\)
where \(S\) is a constant.
where \(\omega\le\omega_o\). And we define \(\omega_o\) to be the the maximum angular velocity attained by the particle as \(B\) oscillates, assuming that the particle does not continue to increase in spin when \(S_n=38.5\).
Are we ready to make \(B\) spin? It is Bu...ood as in wood.
