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If $\vec{\omega} =\vec{ \omega}(\phi,\theta,\dot{\phi},\dot{\theta})$ then $$ \vec{\alpha} = \frac{\partial \vec{\omega}}{\partial \phi} \dot{\phi} + \frac{\partial \vec{\omega}}{\partial \dot{\phi}} \ddot{\phi} + \frac{\partial \vec{\omega}}{\partial \theta} \dot{\theta} + \frac{\partial \vec{\omega}}{\partial \dot{\theta}} \ddot{\theta}$$ Note also that ...


The inertia ellipsoid is computed from an integral about an axis - in other words you rotate the object. This will "smooth out" any symmetries and typically increase the symmetry. Sorry this is a "early morning" intuitive explanation - maybe someone else will give you a more formal answer.


Angular displacement is an example of what's generally called a pseudovector. This is a quantity that is similar to a regular vector, except for the fact that it behaves differently under improper rotations such as reflections (it gains an additional sign flip). Any quantity which is the cross-product of two polar vectors will generally be a pseudovector. ...


It's obvious that the kinetic energy must be compute with respect to the inertial frame. Using Lagrangian method for your problem is very similar to applying this method in order to derive the equations of motion of a gyroscope that can be found anywhere (for example here). The only difference between your case and a gyroscope is that in your case, the ...


Imagine what happens when $\Delta \phi$ keeps increasing to make a full rotation of $360^ \circ$. Then the angle of $P_2$ increases by $360^ \circ$ so that $P_2$ comes back to $P_1$. Also we know that after the full rotation $\vec{v}_1$ must be equal to $\vec{v}_2$ again. Since $\vec{v}_2$ is going around in a circle at the same time $P_2$ does, its angle ...


The perpendicularity of the velocities and radii, plus the fact that both triangle are iscoceles, guarantees that the triangles are similar. In similar triangles, all corresponding angles are equal.

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