Direction of rotation in Feynman's Wobbling Plate

Question

For cube size 0.1,1,1, Mass=100; Initial angular velocity=ω(0)=[1;0.25;0.5]; we get

Moment of inertia tensor=I=[16.67,0,0;0,8.42,0;0,0,8.42]; Angular momentum magnitude=|L|=17.32;

If body axes=x’,y’,z’;

θ(t)=θ(0)=acos(I1ω1/|L|)=0.27; [rad; x’ with respect to L]

ϕ(t)=t*|L|/I3=2.05*t; [rad; x’ rotation about L]

Ψ(t)=tω1(I3-I1)/I3=-0.97t; [rad; z’ rotation about x’]

1)What’s the meaning of the negative(-0.97), is it mean that z’ has left-handed rotation about x’?

2)x’ preccess about L every 3.06s; and z’ rotate about x’ every -6.41s; But if we use the simulation in the link and draw z’, we see that after 6s z’ complete more than 1 circle about x’, why? It should do so only after 6.41s doesn’t it?

simulation link: http://www.ialms.net/sim/3d-rigid-body-simulation/

Answer 1

The Theory behind this animation

Euler Equation:

$$I\vec{\dot{\omega}}+\vec{\omega}\times (I\,\vec{\omega})=0$$

with :

$$I=\left[ \begin {array}{ccc} T_{{x}}&0&0\\ 0&T_{{y}}&0 \\ 0&0&T_{{y}}\end {array} \right]$$

$$\vec\omega=\left[ \begin {array}{c} \omega_{{x}}\\ \omega_{{y} }\\ \omega_{{z}}\end {array} \right] $$

with the initial conditions $~\omega_x(0)=\omega_{x0}~,~\omega_y(0)=\omega_{y0}~,~\omega_z(0)=\omega_{z0}~$ you can obtain the solution for $~\vec{\omega}(\tau)$

$$\omega_x(\tau)=\omega_{x0}$$ $$\omega_y(\tau)=-\omega_{{{ z0}}}\sin \left( {\frac {\omega_{{{ x0}}} \left( -T_ {{y}}+T_{{x}} \right) \tau}{T_{{y}}}} \right) +\omega_{{{ y0}}}\cos \left( {\frac {\omega_{{{ x0}}} \left( -T_{{y}}+T_{{x}} \right) \tau}{T_{{y}}}} \right) $$ $$\omega_z(\tau)=\omega_{{{ z0}}}\cos \left( {\frac {\omega_{{{ x0}}} \left( -T_{ {y}}+T_{{x}} \right) \tau}{T_{{y}}}} \right) +\omega_{{{ y0}}}\sin \left( {\frac {\omega_{{{ x0}}} \left( -T_{{y}}+T_{{x}} \right) \tau}{T_{{y}}}} \right) $$

Rotation Matrix

The rotation matrix between the body and inertial system is:

$$S= \left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \varphi \right) &-\sin \left( \varphi \right) \\ 0 &\sin \left( \varphi \right) &\cos \left( \varphi \right) \end {array} \right] \,\left[ \begin {array}{ccc} \cos \left( \vartheta \right) &0&\sin \left( \vartheta \right) \\ 0&1&0 \\ -\sin \left( \vartheta \right) &0&\cos \left( \vartheta \right) \end {array} \right] \, \left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( \psi \right) &0\\ \sin \left( \psi \right) &\cos \left( \psi \right) &0\\ 0&0&1\end {array} \right] $$

from here you get the equation:

$$\left[ \begin {array}{c} \dot\varphi \\ \dot\vartheta \\ \dot\psi \end {array} \right] = \left[ \begin {array}{ccc} {\frac {\cos \left( \psi \right) }{\cos \left( \vartheta \right) }}&-{\frac {\sin \left( \psi \right) }{\cos \left( \vartheta \right) }}&0\\ \sin \left( \psi \right) &\cos \left( \psi \right) &0\\ -{\frac { \sin \left( \vartheta \right) \cos \left( \psi \right) }{\cos \left( \vartheta \right) }}&{\frac {\sin \left( \vartheta \right) \sin \left( \psi \right) }{\cos \left( \vartheta \right) }}&1\end {array} \right] \,\left[ \begin {array}{c} \omega_{{x}}\\ \omega_{{y} }\\ \omega_{{z}}\end {array} \right] $$

you don't have analytical solution for this ODE, you have to solve it numerical after that you obtain $~\varphi(\tau)~,\vartheta(\tau)~,\psi(\tau)~$ so you can animate the rigid body with the rotation matrix $~S=S(\tau)$