Géométrie des masses, cinétique et dynamique des solides indéformables

Vector integral condition defining the center of mass $G$ of a continuous body

$\int<em>V \mathbf{GM}\,dm = \mathbf{0}$, which is equivalent to $\mathbf{OG} = \frac{1}{m}\int</em>V \mathbf{OM}\,dm$.

Effect of a geometric or material symmetry element on the location of the center of mass $G$

If a system possesses an element of symmetry (such as a symmetry plane, axis, or center), the center of mass $G$ must lie within that symmetry element.

Integral formula for the moment of inertia $I_\Delta$ of a continuous body about an axis $\Delta$

$I<em>\Delta = \int</em>V r^2\,dm$, where $r$ represents the perpendicular distance from the mass element $dm$ to the axis $\Delta$.

Integral expressions for the Cartesian moments of inertia $A, B, C$ of a solid

$A = \int (y^2 + z^2)\,dm$, $B = \int (x^2 + z^2)\,dm$, and $C = \int (x^2 + y^2)\,dm$.

Integral expressions for the products of inertia $D, E, F$ of a solid

$D = \int y z\,dm$, $E = \int z x\,dm$, and $F = \int x y\,dm$.

Definition of the inertia operator $\mathbf{I}_O(\mathbf{u})$ of a solid at point $O$ applied to a vector $\mathbf{u}$

$$\mathbf{I}<em>O(\mathbf{u}) = \int</em>S \mathbf{OM} \wedge (\mathbf{u} \wedge \mathbf{OM})\,dm$$

Formula for the moment of inertia $I_\Delta$ about an axis directed by unit vector $\mathbf{u}$ via the inertia operator

$$I<em>\Delta = \mathbf{u} \cdot \mathbf{I}</em>O(\mathbf{u})$$

Huygens' theorem relating the inertia matrix at point $O$ to that at the center of mass $G$

$\mathbf{I}<em>{O,R} = \mathbf{I}</em>{G,R} + \mathbf{I}<em>{O,R}(m, G)$, where $\mathbf{I}</em>{O,R}(m, G)$ is the inertia matrix of the total mass $m$ concentrated at $G$.

Significance of the principal axes of inertia at a point $O$

They are the orthogonal eigenvectors of the inertia operator $\mathbf{I}_O$. In this basis, the inertia matrix is diagonal and all products of inertia are zero.

The two defining elements of the kinetic torsor of a system at point $O$

The resultant linear momentum $\mathbf{P} = m\mathbf{V}<em>G$ and the angular momentum $\mathbf{\sigma}</em>O = \int_S \mathbf{OM} \wedge \mathbf{V}(M)\,dm$.

Transport formula for the angular momentum $\mathbf{\sigma}$ between two points $A$ and $B$

$$\mathbf{\sigma}<em>B = \mathbf{\sigma}</em>A + m\mathbf{V}_G \wedge \mathbf{AB}$$

Angular momentum of a rigid solid $S$ at its center of mass $G$ in terms of angular velocity $\mathbf{\Omega}$

$$\mathbf{\sigma}<em>G(S/R</em>0) = \mathbf{I}<em>G(\mathbf{\Omega}(S/R</em>0))$$

General relation between dynamic moment $\mathbf{\delta}<em>A$ and angular momentum $\mathbf{\sigma}</em>A$ at an arbitrary point $A$

$$\mathbf{\delta}<em>A = \frac{d\mathbf{\sigma}</em>A}{dt} + \mathbf{V}<em>A \wedge (m\mathbf{V}</em>G)$$

Three conditions where the dynamic moment $\mathbf{\delta}<em>A$ simplifies to $\frac{d\mathbf{\sigma}</em>A}{dt}$

1. Point $A$ is fixed in the reference frame ($\mathbf{V}<em>A = \mathbf{0}$). 2. Point $A$ is the center of mass $G$. 3. Velocity vectors $\mathbf{V}</em>A$ and $\mathbf{V}_G$ are parallel.

Koenig's theorem for the kinetic energy $T$ of a rigid solid

$T = \frac{1}{2}m\mathbf{V}<em>G^2 + \frac{1}{2}\mathbf{\Omega} \cdot \mathbf{I}</em>G(\mathbf{\Omega})$, combining translational kinetic energy of $G$ and rotational kinetic energy about $G$.

Kinetic energy formula for a rigid solid rotating about a fixed axis $Oz_0$

$T = \frac{1}{2}I<em>{zz}\omega^2$, where $I</em>{zz}$ is the moment of inertia about $Oz_0$ and $\omega$ is the angular speed.