If I rotate a coin with dimensions: 10X10X1 about 1 of the big axes(10) in space, where there is no torque, will the rotation will be stable just like a frisbee or a football regular rotations are? please prove mathematically? my proof tells me that this is not stable, am I wrong, where? my proof: (assume coin is spheroid):
For other readers, a relevant reference for the equations used above is the Wikipedia page on the tennis racket theorem.
Your proof demonstrates that, to first order, the stability is marginal. This is what you showed by demonstrating that (to first order) $d \omega_x' = d \omega_z' = 0$. That is, it is on the border of stable and unstable. Any initial perturbation will neither grow nor shrink but will stay the same size.
At the end you show that $d \omega_y'$ is given by a constant that has the same sign as $d \omega_z$. I could not find an error in the derivation after a few minutes, so although I am not completely sure it is correct, it probably is. The conclusion is probably still incorrect, though. A perturbation that grows linearly in time might still be called marginally stable by most physicists. It is still quite different than the usual stable/unstable cases, where the perturbation will either grow or shrink exponentially in time.
with the "tennis racket theorem" you get this differential equation:
$\ddot{\omega}_y=-k\,\omega_y$
so the solution is: (with the initial condition $\omega_y(0)=A\,,\quad D(\omega)_y(0)=0$ )
$\omega_y(t)=A\cos(\sqrt{k}) $
if $k \ge 0$ you get a stable solution.
if $k < 0$ the solution is unstable ( $\omega_y(t)=A\,\cosh(\sqrt{k}\,t)$
with:
$k={\frac {{\Omega_{{x}}}^{2} \left( -I{\it yy}+{\it Ixx} \right) \left( {\it Ixx}-{\it Izz} \right) }{{\it Iyy}\,{\it Izz}}} $
your case:
$Ixx=Iyy=\frac{M*(b^2+s^2)}{5}$
$Izz=\frac{2\,M\,b^2}{5}$
$\rightarrow$
$k=0$ stable system!
Due to its disc-like shape, an ideal coin is symmetric and it has an infinite number of major axes (the axes aligned with the diameters of the coin). So technically, you have an infinite number of equilibrium points for the dynamics of the angular velocity. None of these equilibrium points is stable.
In the body-fixed coordinate system, centered at the center of the coin, with axes $x$ and $y$ along two of the major axes of the coin, and the $z-$axis perpendicular to the disc of the coin, the inertia tensor has zero off-diagonal components and $I_1 = I_2 = \frac{m}{12}(3r^2 + h^2)$ and $I_3 = \frac{m}{2}r^2$ where $r = 10h$ is the radius of the coin and $h$ is its thickness. Thus $I_3 > I_2 = I_1$
The equations of motion (Euler's equations) are \begin{align} &I_1\frac{d\omega_1}{dt} = (I_1 - I_3)\, \omega_2\, \omega_3\\ &I_1\frac{d\omega_2}{dt} = (I_3 - I_1)\, \omega_3\, \omega_1\\ &I_3\frac{d\omega_3}{dt} = (I_1 - I_1)\, \omega_1\, \omega_2 = 0 \end{align} so if we set $k = \frac{I_3 - I_1}{I_1}$ the equations of motion become \begin{align} &\frac{d\omega_1}{dt} = -\, k\, \omega_2\, \omega_3\\ &\frac{d\omega_2}{dt} = k\, \omega_3\, \omega_1\\ &\frac{d\omega_3}{dt} = 0 \end{align} The third equation yields $\omega_3(t) = \omega_3^0$ for all $t$, so the equations of motion reduce to \begin{align} &\frac{d\omega_1}{dt} = -\, (k\, \omega_3^0)\, \omega_2\\ &\frac{d\omega_2}{dt} = (k\, \omega_3^0)\, \omega_1\\ &\omega_3 = \omega_3^0 \end{align} The two by two linear system \begin{align} &\frac{d\omega_1}{dt} = -\, (k\, \omega_3^0)\, \omega_2\\ &\frac{d\omega_2}{dt} = (k\, \omega_3^0)\, \omega_1\\ \end{align} has the solutions \begin{align} &\omega_1 = \omega_1^0\, \cos\Big((k\omega_3^0)\, t\Big) - \omega_2^0\, \sin\Big((k\omega_3^0)\, t\Big)\\ &\omega_2 = \omega_1^0\, \sin\Big((k\omega_3^0)\, t\Big) + \omega_2^0\, \cos\Big((k\omega_3^0)\, t\Big)\\ \end{align} and the solution to the original system, describing the precession of the angular velocity, is \begin{align} &\omega_1 = \omega_1^0\, \cos\Big((k\omega_3^0)\, t\Big) - \omega_2^0\, \sin\Big((k\omega_3^0)\, t\Big)\\ &\omega_2 = \omega_1^0\, \sin\Big((k\omega_3^0)\, t\Big) + \omega_2^0\, \cos\Big((k\omega_3^0)\, t\Big)\\ &\omega_3 = \omega_3^0 \end{align} If you want to understand the time-evolution of the angular velocity of the coin nearby a major axis, you can simply consider the $x-$axis, due to coin's symmetry. Then, start with an initial angular velocity $(\omega_1^0, \omega_2^0, \omega_3^0)$ with $\omega_2^0 = 0$ and $\omega_3^0 =\varepsilon$ a very small number. Then the time-evolution of the angular velocity is \begin{align} &\omega_1 = \omega_1^0\, \cos\big(k\varepsilon\, t\big)\\ &\omega_2 = \omega_1^0\, \sin\big(k\varepsilon\, t\big)\\ &\omega_3 = \varepsilon \end{align} which shows that staring from initial angular velocity $(\omega_1^0, 0, \varepsilon)$ after time $t = \frac{\pi}{2\,k\varepsilon}$ the angular velocity will be $(0, \omega_1^0, \varepsilon)$ and then at time $t = \frac{\pi}{k\varepsilon}$ it will be $(-\omega_1^0, 0, \varepsilon)$ which is almost diametrically opposing the initial angular velocity. This means that very slowly, the axis of rotation aligned with the angular velocity will be drifting away from the $x-$axis. Hence, the latter is unstable. However, if you toss a coin, you would not see this clearly, unless you are in space, because the drift of the axis is very slow and the coin is going to land before any significant drift has occurred.
You can also see this behavior geometrically, using Poincot's ellipsoids: the one coming from the conservation of angular momentum $$\big(I_1\omega_1\big)^2 + \big(I_1\omega_2\big)^2 + \big(I_3\omega_3\big)^2 = c_1$$ and the other coming from the conservation of energy $$I_1\big(\omega_1\big)^2 +I_1 \big(\omega_2\big)^2 + I_3\big(\omega_3\big)^2 = c_2$$ The trajectory of the angular velocity is along one of the intersection curves of the two ellipsoids. In the case of the symmetric coin, the ellipsoids are rotational and have $z-$axes aligned, so the trajectory is a circle.
Edit: To make things a bit easier to visualize, it is a good idea to look at the dynamics of the angular momentum $\vec{l} = (l_1, l_2, l_3)$ in the body-fixed frame and compare it to its representation $\vec{L} = (L_1, L_2, L_3)$ in the world frame. There is a linear transformation between the angular momentum and the angular velocity in terms of the inertia tensor, i.e. \begin{align*} &l_1 = I_1\, \omega_1\\ &l_2 = I_1\, \omega_2\\ &l_3 = I_3\, \omega_3\\ \end{align*} so the dynamics of the angular velocity and the angular momentum are equivalent (isomorphic), i.e. whatever happens to one of them, the same thing happens to the other. Then the Euler's equations of motion become \begin{align} &\frac{d l_1}{dt} = \frac{(I_1 - I_3)}{I_1I_3}\, l_2\, l_3\\ &\frac{d l_2}{dt} = \frac{(I_3 - I_1)}{I_1I_3}\, l_3\, l_1\\ &\frac{d l_3}{dt} = \frac{(I_1 - I_1)}{I_1^2}\, \omega_1\, \omega_2 = 0 \end{align} so if we set this time $m = \frac{I_3 - I_1}{I_1I_1}$ the equations of motion become \begin{align} &\frac{d l_1}{dt} = -\, m\, l_2\, l_3\\ &\frac{d l_2}{dt} = m\, l_3\, l_1\\ &\frac{d l_3}{dt} = 0 \end{align} The third equation yields $l_3(t) = l_3^0$ for all $t$, so the equations of motion reduce to \begin{align} &\frac{d l_1}{dt} = -\, (m\, l_3^0)\, l_2\\ &\frac{d l_2}{dt} = (m\, l_3^0)\, l_1\\ &l_3 = l_3^0 \end{align} As before, the two by two linear system \begin{align} &\frac{d l_1}{dt} = -\, (m\, l_3^0)\, l_2\\ &\frac{d l_2}{dt} = (m\, l_3^0)\, l_1\\ \end{align} has the solutions \begin{align} &l_1 = l_1^0\, \cos\Big((m l_3^0)\, t\Big) - l_2^0\, \sin\Big((m l_3^0)\, t\Big)\\ &l_2 = l_1^0\, \sin\Big((m l_3^0)\, t\Big) + \omega_2^0\, \cos\Big((m l_3^0)\, t\Big)\\ \end{align} and the solution to the original system, describing the precession of the angular momentum, is \begin{align} &l_1 = l_1^0\, \cos\Big((m l_3^0)\, t\Big) - l_2^0\, \sin\Big((m l_3^0)\, t\Big)\\ &l_2 = \omega_1^0\, \sin\Big((m l_3^0)\, t\Big) + \omega_2^0\, \cos\Big((m l_3^0)\, t\Big)\\ &l_3 = l_3^0 \end{align}
Just like with the angular velocity, let us explore the stability of the $x-$axis of the body-fixed system. Observe, the $x-$axis is aligned with a diameter of the coin (this is not the $z-$axis which is goes through the center of the coin and is perpendicular to the coin!). We are in the body fixed frame. Let us start again from an angular momentum $(l_1, 0, \delta)$ where $\delta$ is a very small number. Then the time-evolution of the angular velocity is \begin{align} &l_1 = l_1^0\, \cos\big(m\delta\, t\big)\\ &l_2 = l_1^0\, \sin\big(m\delta \, t\big)\\ &l_3 = \delta \end{align} Denote $|\vec{l}| = \sqrt{l_1^2 + l_2^2 + l_3^2} = \sqrt{(l_1^0)^2 + \delta^2}$ and let us look at the change of distance between the angular momentum $\vec{l}$ and the $x-$axis $\vec{x} = ( \sqrt{(l_1^0)^2 + \delta^2}, 0, 0)$: \begin{align} \text{dist}\big(\vec{l}(t), \vec{x}\big)^2 &= |\vec{l}(t) - \vec{x}|^2={(l_1 - \sqrt{(l_1^0)^2 + \delta^2})^2 + l_2^2 + l_3^2}\\ &= {(l_1^0\, \cos\big(m\delta\, t\big) - p_0)^2 + (l_1^0\, \sin\big(m\delta \, t\big))^2 + \delta^2}\\ &= 2(l_1^0)^2 + 2\delta^2 - 2l_1^0\sqrt{(l_1^0)^2 + \delta^2}\, \cos\big(m\delta\, t\big) \end{align} This distance squared can be interpreted as a Lyapunov function and it changes from very closed to zero to a fairly large number. Since distance is the same in any Cartesian coordinate frame, this distance is the same in the body-fixed frame and in the world frame. Simply, in the body-fixed frame the $x-$axis $\vec{x}$ is fixed while the angular momentum $\vec{l}$ changes with time, while in the world coordinate frame the angular momentum $\vec{L}$ is fixed and the body-fixed axis $\vec{x}$ changes with time. In the body fixed frame the angular momentum $\vec{l}$ precesses along a circle around the body-fixed $z-$axis, while in the world frame the body-fixed $z-$axis precesses along a circle around the fixed angular momentum $\vec{L}$. Do not confuse the $x-$ and the $z-$axis. Their behaviors in the world system are different.
