Question 13.3: A homogeneous three-axial ellipsoid with the moments of iner...
A homogeneous three-axial ellipsoid with the moments of inertia Θ_{1},Θ_{2},Θ_{3} rotates with the angular velocity \dot{φ} about the principal axis of inertia 3. The axis 3 rotates with \dot{ϑ} about the axis \overline{AB}. The axis \overline{AB} passes through the center of gravity and is perpendicular to 3. Find the kinetic energy.
We decompose the angular velocity ω into its components along the principal axes of inertia:
ω = (ω_{1},ω_{2},ω_{3}), where ω_{1} = \dot{ϑ} cos \varphi, ω_{2} =−\dot{ϑ} sin \varphi, ω_{3} = \dot{\varphi}.
The kinetic energy is then
T = \frac{1}{2} \sum\limits_{i}{Θ_{i}ω^{2}_{i}}= \frac{1}{2} (Θ_{1} cos^{2} \varphi +Θ_{2} sin^{2} \varphi)\dot{ϑ}^{2} + \frac{1}{2} \dot{\varphi}^{2}.
The ellipsoid shall now be symmetric, Θ_{1} = Θ_{2}; the axis \overline{AB} is tilted from the third axis by the angle α. For the total angular velocity, we have
ω= \dot{\varphi} e_{3} + \dot{ϑ} e_{AB}.We decompose the unit vector e_{AB} along the axis \overline{AB} with respect to the principal axes
e_{AB} = e_{3} · cos α +(cos \varphi e_{1} −sin \varphi e_{2}) sin α.Thus, the components of ω along the directions of the principal axes are
ω_{1} = sin α cos \varphi \dot{ϑ},ω_{2} =−sin α sin \varphi \dot{ϑ},
ω_{3} = \dot{\varphi} + cos α \dot{ϑ}.
Hence, the kinetic energy reads
T = \frac{1}{2} Θ_{1} sin^{2} α \dot{ϑ}^{2} + \frac{1}{2}Θ_{3}(\dot{\varphi} + \dot{ϑ} cos α)^{2}.For α = 90°, we obtain for the first case a rotation ellipsoid.
