3d rigid body equations of motion
Euler’s second law, when applied to rotational motion, states that the sum of external moments (torques) about a point is equal to the rate of change of angular momentum about that point:
\sum \mathbf{M}_O = \frac{\mathrm{d}\mathbf{L}_O}{\mathrm{d}t}
A particularly useful case arises when considering the mass center (C) of the rigid body. The equation simplifies to:
\sum \mathbf{M}_C = \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}
For an arbitrary point P in the body, I need to account for the motion of the mass center relative to P:
\sum \mathbf{M}_P = \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t} + \mathbf{r}_{PC} \times m\mathbf{a}_C
where \mathbf{r}_{PC} is the position vector from point P to the mass center C, and \mathbf{a}_C is the acceleration of the mass center C.
It is often convenient to use a body-fixed reference frame 2, with axes \mathbf{i}_2, \mathbf{j}_2, \mathbf{k}_2, welded to the body. This choice ensures that the moments and products of inertia remain constant over time. To compute the derivative of the angular momentum, I use the transport theorem:
\frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}\bigg|_1 = \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}\bigg|_2 + \boldsymbol{\omega}_{2/1} \times \mathbf{L}_C
The angular momentum is related to the inertia matrix and angular velocity:
\mathbf{L}_C = \mathbf{I}_C \boldsymbol{\omega}
In the body-fixed frame, the moment of inertia is constant, so:
\frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}\bigg|_2 = \mathbf{I}_C \dot{\boldsymbol{\omega}}
Combining these expressions, I arrive at Euler’s equations.
When I choose the principal axes as my reference frame, the product of inertia terms vanish, significantly simplifying the equations:
\begin{aligned} \sum M_{C_x} = & I_{xx}^C \dot{\omega}_x - \left(I_{yy}^C - I_{zz}^C\right) \omega_y \omega_z \\ \sum M_{C_y} = & I_{yy}^C \dot{\omega}_y - \left(I_{zz}^C - I_{xx}^C\right) \omega_z \omega_x\\ \sum M_{C_z} = & I_{zz}^C \dot{\omega}_z - \left(I_{xx}^C - I_{yy}^C\right) \omega_x \omega_y \end{aligned}
Sometimes, it is useful to introduce an intermediate reference frame 3 where the angular momentum or external moments can be expressed conveniently. The equations then take the form:
\sum \mathbf{M}_C = \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}\bigg|_3 = \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}\bigg|_2 + \boldsymbol{\omega}_{3/2} \times \mathbf{L}_C
This approach can be helpful when dealing with certain symmetries or other problem-specific characteristics.
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