On Kinematic Description of the Motion of a Rigid Body

Abstract

A system of ordinary differential equations has been derived for a vector of finite rotation corresponding to Euler’s theorem: the vector of finite rotation is directed along the axis of finite rotation of a solid, and its length is equal to the angle of plane rotation around this axis. The system of equations is explicitly resolved with respect to the time derivative of the rotation vector components. The right part of the system depends on the rotation vector and the angular velocity vector in the principle axes. The obtained system of equations is shown to be equivalent to the system of equations for quaternions. The coordinates of the orts of the principle axes of a rigid body in fixed axes are expressed in terms of finite rotation angles and the components of angular velocity using simple analytical formulas.

APPENDIX

1.1 A1 DERIVATION OF A FORMULA FOR ANGULAR VELOCITY IN PRINCIPAL AXES

According to Eq. (4.1), the matrix $A(\boldsymbol\omega )$ is equal to a product of two matrices

$$\begin{gathered} \frac{d}{{dt}}C(\boldsymbol\varphi ) = (A({\mathbf{e}})\cos \theta + A{{({\mathbf{e}})}^{2}}\sin \theta )\dot {\theta } + A({\mathbf{\dot {e}}})\sin \theta + (A({\mathbf{\dot {e}}})A({\mathbf{e}}) + A({\mathbf{e}})A({\mathbf{\dot {e}}}))(1 - \cos \theta ), \\ C( - \boldsymbol\varphi ) = E - A({\mathbf{e}})\sin \theta + A{{({\mathbf{e}})}^{2}}(1 - \cos \theta ). \\ \end{gathered} $$

It can be represented in the form

$$\begin{gathered} A(\boldsymbol\omega ) = \left( {\frac{d}{{dt}}C(\boldsymbol\varphi )} \right)C( - \boldsymbol\varphi ) = \dot {\theta }F + \sin \theta {{F}_{1}} + {{\sin }^{2}}\theta {{F}_{2}} \\ \, + (1 - \cos \theta ){{F}_{3}} + {{(1 - \cos \theta )}^{2}}{{F}_{4}} + \sin \theta (1 - \cos \theta ){{F}_{5}}. \\ \end{gathered} $$

(A1.1)

Using equalities (2.2) and

$$\begin{gathered} {\mathbf{e}} \cdot {\mathbf{\dot {e}}} = 0,\quad A({\mathbf{e}})A({\mathbf{\dot {e}}})A({\mathbf{e}}){\mathbf{R}} = {\mathbf{e}} \times ({\mathbf{\dot {e}}} \times ({\mathbf{e}} \times {\mathbf{R}})) = {\mathbf{e}} \times {\mathbf{e}}\,({\mathbf{\dot {e}}} \cdot {\mathbf{R}}) = 0, \\ A({\mathbf{e}})A({\mathbf{\dot {e}}})A({\mathbf{e}})A({\mathbf{e}}){\mathbf{R}} = {\mathbf{e}} \times ({\mathbf{\dot {e}}} \times ({\mathbf{e}} \times ({\mathbf{e}} \times {\mathbf{R}}))) = {\mathbf{e}} \times ({\mathbf{\dot {e}}} \times ({\mathbf{e}}({\mathbf{e}} \cdot {\mathbf{R}}) - {\mathbf{R}})) \\ \, = {\mathbf{\dot {e}}}\,({\mathbf{e}} \cdot {\mathbf{R}}) - {\mathbf{\dot {e}}}\,({\mathbf{e}} \cdot {\mathbf{R}}) = 0, \\ \end{gathered} $$

we sequentially find

$$\begin{gathered} F = A({\mathbf{e}}),\quad {{F}_{1}} = A({\mathbf{\dot {e}}}),\quad {{F}_{2}} = - A({\mathbf{\dot {e}}})A({\mathbf{e}}),\quad {{F}_{3}} = A({\mathbf{\dot {e}}})A({\mathbf{e}}) + A({\mathbf{e}})A({\mathbf{\dot {e}}}), \\ {{F}_{4}} = \left( {A({\mathbf{\dot {e}}})A({\mathbf{e}}) + A({\mathbf{e}})A({\mathbf{\dot {e}}})} \right)A{{({\mathbf{e}})}^{2}} = - A({\mathbf{\dot {e}}})A({\mathbf{e}}) + A({\mathbf{e}})A({\mathbf{\dot {e}}})A{{({\mathbf{e}})}^{2}} = - A({\mathbf{\dot {e}}})A({\mathbf{e}}), \\ {{F}_{5}} = A({\mathbf{\dot {e}}})A{{({\mathbf{e}})}^{2}} - A({\mathbf{\dot {e}}})A{{({\mathbf{e}})}^{2}} - A({\mathbf{e}})A({\mathbf{\dot {e}}})A({\mathbf{e}}) = 0. \\ \end{gathered} $$

Substituting the obtained formulas into (A1.1), we arrive at

$$A(\boldsymbol\omega ) = \dot {\theta }A({\mathbf{e}}) + \sin\theta A({\mathbf{\dot {e}}}) + (1 - \cos \theta )(A({\mathbf{e}})A({\mathbf{\dot {e}}}) - A({\mathbf{\dot {e}}})A({\mathbf{e}})).$$

The identity

$$(A({\mathbf{e}})A({\mathbf{\dot {e}}}) - A({\mathbf{\dot {e}}})A({\mathbf{e}})){\mathbf{R}} = {\mathbf{e}} \times ({\mathbf{\dot {e}}} \times {\mathbf{R}}) - {\mathbf{\dot {e}}} \times ({\mathbf{e}} \times {\mathbf{R}}) = ({\mathbf{e}} \times {\mathbf{\dot {e}}}) \times {\mathbf{R}}$$

yields that $A({\mathbf{e}})A({\mathbf{\dot {e}}}) - A({\mathbf{\dot {e}}})A({\mathbf{e}}) = ({\mathbf{e}} \times {\mathbf{\dot {e}}})$ and

$$A(\boldsymbol\omega ) = A(\dot {\theta }{\mathbf{e}} + {\text{sin}}\theta {\mathbf{\dot {e}}} + (1 - {\text{cos}}\theta ){\mathbf{e}} \times {\mathbf{\dot {e}}}),$$

(A1.2)

which proves Eq. (4.4).

1.2 A2 DERIVATION OF AN EQUATION FOR ANGULAR VELOCITY IN MOVING AXES

Euler’s law for the velocity distribution in rigid body axes (RCS) has the form

$$\begin{gathered} {{{\mathbf{v}}}_{*}} = C( - \boldsymbol\varphi ){\mathbf{v}} = C( - \boldsymbol\varphi )A(\boldsymbol\omega ){\mathbf{R}} = C( - \boldsymbol\varphi )\frac{d}{{dt}}C(\boldsymbol\varphi )C( - \boldsymbol\varphi ){\mathbf{R}} \\ \, = C( - \boldsymbol\varphi )\frac{d}{{dt}}C(\boldsymbol\varphi )C( - \boldsymbol\varphi )C(\boldsymbol\varphi ){{{\mathbf{R}}}_{*}} \\ \Rightarrow {{{\mathbf{v}}}_{*}} = A({{\boldsymbol\omega }_{*}}){{{\mathbf{R}}}_{*}},\quad A({{\boldsymbol\omega }_{*}}) = C( - \boldsymbol\varphi )\left( {\frac{d}{{dt}}C(\boldsymbol\varphi )} \right). \\ \end{gathered} $$

(A2.1)

Identity (4.2) can be used to represent Eq. (A1.2) as

$$ - A(\boldsymbol\omega ) = C(\boldsymbol\varphi )\left( {\frac{d}{{dt}}C( - \boldsymbol\varphi )} \right) = A\left( { - \dot {\theta }\,{\mathbf{e}} - \sin\theta \,{\mathbf{\dot {e}}} - (1 - \cos \theta )\,{\mathbf{e}} \times {\mathbf{\dot {e}}})} \right).$$

Changing in this formula the sign of $\boldsymbol\varphi$ to the opposite and, respectively, the sign of ${\mathbf{e}}$ in the right-hand side, we obtain a formula for antisymmetric matrix (A2.1):

$$A({{\boldsymbol\omega }_{*}}) = C( - \boldsymbol\varphi )\left( {\frac{d}{{dt}}C(\boldsymbol\varphi )} \right) = A\left( {\dot {\theta }\,{\mathbf{e}} + sin\theta \,{\mathbf{\dot {e}}} - (1 - \cos \theta ){\mathbf{e}} \times {\mathbf{\dot {e}}}} \right),$$

which yields Eq. (4.5) for rotational velocity in moving axes, which was to be proved.