We finally arrive at the top, namely the Euler equations for rigid bodies. I mentioned the inertia tensor in previous section, but I have decided to eschew from it since there are a multitude of good resources on it. What matters to us at this point is that it is a matrix and not only that but a symmetric matrix. Given that this is the case, we can always diagonalize it so that it simply has entries on the diagonal. These entries we shall call \( I_x, I_y, I_z \). These are the so called principal axes. Now if we jump into a co-ordinate system which coincide with these axes then the angular momentum L = \( \left( I_x\omega_x, I_y \omega_y , I_z \omega_z \right) \). The angular moment as observed in the inertial reference is $$ \dot{L_{i.f}} = \dot{L_{r.f}}+ \Omega \times L = \Gamma $$ where i.f is "inertial reference frame" , r.f is "rotating reference frame" and gamma is the external torque. If we plug in what L is we arrive at the blessed equations
\begin{align}
\Gamma_x & = \dot{L_x} - (I_y - I_z) \omega_z \omega_y\\
\Gamma_y &= \dot{L_y}- (I_z - I_x) \omega_y \omega_z\\
\Gamma_z &= \dot{L_z}-(I_x - I_y) \omega_x \omega_ y \\
\end{align}
Notice that the time derivate does not affect I, that is the purpose of using these principal axes. Now supposing external torque is zero, then the angular moment L should be constant. Here is the proof:
1) If external torque is zero, angular momentum is constant
Multiply with ith equation above by \( I_i\omega_i\) and add all three equations
\begin{align}
I_x^2 \omega_x \dot{\omega_x} + I_y^2 \omega_y \dot{\omega_y}+ I_z^2 \omega_z \dot{\omega_z} &= \left(I_x(I_y - I_z) + (I_z - I_x)I_y + (I_x - I_y)I_z \right) \omega_z \omega_y\omega_x \\
I_x^2 \omega_x \dot{\omega_x} + I_y^2 \omega_y \dot{\omega_y}+ I_z^2 \omega_z \dot{\omega_z} &= 0\\
L_x\dot{L_x}+ L_y\dot{L_y}+L_z\dot{L_z} &= 0\\
\frac{d}{dt}\frac{\left( L \cdot L\right)}{2} &= 0\\
\frac{d}{dt}\frac{L^2}{2} &= 0\\
\dot{L} &= 0 \hspace{20 mm} QED
\end{align}
2) If we have a lamina rotating freely(no torques) about a point O of the lamina. The component of \( \omega \) in the plane is constant.
Again since we have no torque we start from Euler's equations with no torque, but we use a slight of hand which should not look like one since we have seen it before, namely use the perpendicular axis theorem. This gets rid of the \( I_x \) and \(I_y\) but we are getting ahead of ourselves:
\begin{align}
I_x \dot{\omega_x} &= (I_y - I_z) \omega_y \omega_z\\
I_y \dot{\omega_y} &= (I_z - I_x) \omega_z \omega_x\\
I_z \dot{\omega_z} &= (I_x - I_y) \omega_y \omega_x\\
\end{align}
Use perpendicular axis theorem
\begin{align}
\frac{I_z}{2} \dot{\omega_x} &= (\frac{I_z}{2} - I_z) \omega_y \omega_z\\
\frac{I_z}{2} \dot{\omega_y} &= (I_z - \frac{I_z}{2}) \omega_z \omega_x\\
I_z \dot{\omega_z}& = (\frac{I_z - I_z}{2}) \omega_y \omega_x\\
\end{align}
The new set of equations
\begin{align}
\dot{\omega_x}& = - \omega_y \omega_z\\
\dot{\omega_y} &= \omega_z \omega_x\\
\end{align}
Multiply the top equation above by \(\omega_x\) and the bottom by \( \omega_y\) and then add the two to get \(\omega_x \dot{\omega_x}+ \omega_y\dot{\omega_y} = 0 \). But this is merely \(\frac{d}{dt}\left( \omega_x^2 + \omega_y^2\right) \)= 0. Voila
\begin{align}
\Gamma_x & = \dot{L_x} - (I_y - I_z) \omega_z \omega_y\\
\Gamma_y &= \dot{L_y}- (I_z - I_x) \omega_y \omega_z\\
\Gamma_z &= \dot{L_z}-(I_x - I_y) \omega_x \omega_ y \\
\end{align}
Notice that the time derivate does not affect I, that is the purpose of using these principal axes. Now supposing external torque is zero, then the angular moment L should be constant. Here is the proof:
1) If external torque is zero, angular momentum is constant
Multiply with ith equation above by \( I_i\omega_i\) and add all three equations
\begin{align}
I_x^2 \omega_x \dot{\omega_x} + I_y^2 \omega_y \dot{\omega_y}+ I_z^2 \omega_z \dot{\omega_z} &= \left(I_x(I_y - I_z) + (I_z - I_x)I_y + (I_x - I_y)I_z \right) \omega_z \omega_y\omega_x \\
I_x^2 \omega_x \dot{\omega_x} + I_y^2 \omega_y \dot{\omega_y}+ I_z^2 \omega_z \dot{\omega_z} &= 0\\
L_x\dot{L_x}+ L_y\dot{L_y}+L_z\dot{L_z} &= 0\\
\frac{d}{dt}\frac{\left( L \cdot L\right)}{2} &= 0\\
\frac{d}{dt}\frac{L^2}{2} &= 0\\
\dot{L} &= 0 \hspace{20 mm} QED
\end{align}
2) If we have a lamina rotating freely(no torques) about a point O of the lamina. The component of \( \omega \) in the plane is constant.
Again since we have no torque we start from Euler's equations with no torque, but we use a slight of hand which should not look like one since we have seen it before, namely use the perpendicular axis theorem. This gets rid of the \( I_x \) and \(I_y\) but we are getting ahead of ourselves:
\begin{align}
I_x \dot{\omega_x} &= (I_y - I_z) \omega_y \omega_z\\
I_y \dot{\omega_y} &= (I_z - I_x) \omega_z \omega_x\\
I_z \dot{\omega_z} &= (I_x - I_y) \omega_y \omega_x\\
\end{align}
Use perpendicular axis theorem
\begin{align}
\frac{I_z}{2} \dot{\omega_x} &= (\frac{I_z}{2} - I_z) \omega_y \omega_z\\
\frac{I_z}{2} \dot{\omega_y} &= (I_z - \frac{I_z}{2}) \omega_z \omega_x\\
I_z \dot{\omega_z}& = (\frac{I_z - I_z}{2}) \omega_y \omega_x\\
\end{align}
The new set of equations
\begin{align}
\dot{\omega_x}& = - \omega_y \omega_z\\
\dot{\omega_y} &= \omega_z \omega_x\\
\end{align}
Multiply the top equation above by \(\omega_x\) and the bottom by \( \omega_y\) and then add the two to get \(\omega_x \dot{\omega_x}+ \omega_y\dot{\omega_y} = 0 \). But this is merely \(\frac{d}{dt}\left( \omega_x^2 + \omega_y^2\right) \)= 0. Voila
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