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Analyzing the angular acceleration vector of a moving rigid body

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Abstract

This paper presents a novel and systematic approach for obtaining the angular acceleration vector of a moving rigid body. The novelty of the proposed method lies in the particular form of writing the pose of the moving rigid body, as well as in the procedure to compute its time derivatives. The derivation process goes directly to the very foundations of rotational motion and exploits the phenomenological connection between orientation, angular velocity, angular acceleration, and spatial motion of a rigid body. Hence, as a remarkable result, a symbolic expression for the angular acceleration vector arises naturally without the need to solve the inverse acceleration problem. The novel and general expression of the angular acceleration vector involves relationships between the position, velocity, and acceleration vectors of three non-collinear points of the body, which can be easily understood and physically interpreted without particular knowledge of specialized techniques or advanced mathematical tools. Due to its vector nature, the expression for the angular acceleration vector proposed in this paper is relatively simple, as well as, it is very robust against computational singularities. Two fully detailed case studies demonstrate the robustness of the proposed angular acceleration vector compared with other expressions appearing in the literature.

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Notes

  1. It should be noted that Eq. (8) can be equivalently written as \({\textbf{r}}_{P/O} = {\textbf{p}}_1 + {\textbf{R}} \, {\textbf{r}}\), where the \((3 \times 3)\) matrix \({\textbf{R}} \equiv [{{\textbf{u}}} \,\, {{\textbf{v}}} \,\, {{\textbf{w}}}]\) describes the orientation of the rigid body with respect to reference frame XYZ, and \({\textbf{r}} \equiv (u, v, w)^T\).

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Acknowledgements

The authors are very thankful to the National Council of Humanities, Science and Technology of México (CONAHCYT) for supporting this research through SNI (National System of Researchers) fellowships.

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Authors and Affiliations

  1. Departamento de Ingeniería Mecánica, Universidad de Guanajuato, DICIS, 36885, Salamanca, Guanajuato, Mexico

    J. Jesús Cervantes-Sánchez, José M. Rico-Martínez, Mario A. García-Murillo & Marco A. Gómez-Álvarez

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  1. J. Jesús Cervantes-Sánchez
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  2. José M. Rico-Martínez
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  3. Mario A. García-Murillo
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  4. Marco A. Gómez-Álvarez
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Correspondence to J. Jesús Cervantes-Sánchez.

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Appendix 1

Appendix 1

The objective of this appendix is to present a detailed derivation of the nine vector terms \(\dot{{\textbf{n}}}_1, \dot{{\textbf{n}}}_2, \cdots \dot{{\textbf{n}}}_9\) involved into Eq. (30). To this end, the general idea is to include only \({\textbf{u}}\), \(\dot{{\textbf{u}}}\), \(\ddot{{\textbf{u}}}\), \({\textbf{m}}\), \(\dot{{\textbf{m}}}\), and \(\ddot{{\textbf{m}}}\), since these vectors are directly related to the position, the velocity, and the acceleration of the three non-collinear points of the moving rigid body under analysis.

  1. (1)

    Computation of the first term, \(\dot{{\textbf{n}}}_1\).

The first term, namely, \(\dot{{\textbf{n}}}_1\), has been previously defined in Eq. (30), and it may be handled as follows:

$$\begin{aligned}{} & {} \dot{{\textbf{n}}}_1 \equiv (\ddot{{\textbf{v}}} \! \cdot \! {{\textbf{w}}}) {\textbf{u}} = \{ (k_1 \ddot{{\textbf{m}}} - k_2 \ddot{{\textbf{u}}}) \cdot ({\textbf{u}} \times {\textbf{v}}) \} {\textbf{u}} \nonumber \\{} & {} \quad = k_1 \{ \ddot{{\textbf{m}}} \cdot ({\textbf{u}} \times {\textbf{v}}) \} {\textbf{u}} - k_2 \{ \ddot{{\textbf{u}}} \cdot ({\textbf{u}} \times {\textbf{v}}) \} {\textbf{u}}. \end{aligned}$$
(89)

By using vector product identities, we have that:

$$\begin{aligned}{} & {} ({\textbf{u}} \!\times \! {\textbf{v}}) \!\times \! ({\textbf{u}} \!\times \! \ddot{{\textbf{m}}}) = \{ \ddot{{\textbf{m}}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{v}}) \} {\textbf{u}} - \{ {\textbf{u}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{v}}) \} \ddot{{\textbf{m}}} \nonumber \\{} & {} \quad = \{ \ddot{{\textbf{m}}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{v}}) \} {\textbf{u}}, \,\,\, {\textbf{u}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{v}}) = 0. \end{aligned}$$
(90)
$$\begin{aligned}{} & {} ({\textbf{u}} \!\times \! {\textbf{v}}) \!\times \! ({\textbf{u}} \!\times \! \ddot{{\textbf{u}}}) = \{ \ddot{{\textbf{u}}} \cdot ({\textbf{u}} \!\times \! {\textbf{v}}) \} {\textbf{u}} - \{ {\textbf{u}} \cdot ({\textbf{u}} \!\times \! {\textbf{v}}) \} \ddot{{\textbf{u}}}\nonumber \\{} & {} \quad = \{ \ddot{{\textbf{u}}} \cdot ({\textbf{u}} \!\times \! {\textbf{v}}) \} {\textbf{u}}, \,\,\, {\textbf{u}} \cdot ({\textbf{u}} \!\times \! {\textbf{v}}) = 0. \end{aligned}$$
(91)

where

$$\begin{aligned} {\textbf{u}} \!\times \! {\textbf{v}} = {\textbf{u}} \!\times \! ( k_1 {\textbf{m}} - k_2 {\textbf{u}}) = k_1 ({\textbf{u}} \!\times \! {\textbf{m}}) - k_2 ({\textbf{u}} \!\times \! {\textbf{u}}) = k_1 ({\textbf{u}} \!\times \! {\textbf{m}}), \,\,\, {\textbf{u}} \!\times \! {\textbf{u}} = {\textbf{0}}. \end{aligned}$$
(92)

In this way, according to the rules of cross vector products, Eq. (89) may be written as follows:

$$\begin{aligned} \dot{{\textbf{n}}}_1 = k_1^2 \, (\ddot{{\textbf{m}}} \!\times \! {\textbf{u}}) \!\times \! ({\textbf{u}} \!\times \! {{\textbf{m}}}) - k_1 k_2 \, (\ddot{{\textbf{u}}} \!\times \! {\textbf{u}}) \!\times \! ({\textbf{u}} \!\times \! {{\textbf{m}}}). \end{aligned}$$
(93)
  1. (2)

    Computation of the second term, \(\dot{{\textbf{n}}}_2\).

The algebraic handling of the second term, namely, \(\dot{{\textbf{n}}}_2\), is described below:

$$\begin{aligned} \dot{{\textbf{n}}}_2 \equiv (\dot{{\textbf{v}}} \! \cdot \! \dot{{{\textbf{w}}}}) {\textbf{u}} = \{ (k_1 \dot{{\textbf{m}}} - k_2 \dot{{\textbf{u}}} ) \cdot (\dot{{\textbf{u}}} \times {\textbf{v}} + {\textbf{u}} \times \dot{{\textbf{v}}}) \} {\textbf{u}} \end{aligned}$$
(94)

where

$$\begin{aligned}{} & {} \dot{{\textbf{u}}} \times {\textbf{v}} = \dot{{\textbf{u}}} \times ( k_1 {\textbf{m}} - k_2 {\textbf{u}}) = k_1 (\dot{{\textbf{u}}} \times {\textbf{m}}) - k_2 (\dot{{\textbf{u}}} \times {\textbf{u}}) \end{aligned}$$
(95)
$$\begin{aligned}{} & {} {\textbf{u}} \times \dot{{\textbf{v}}} = {{\textbf{u}}} \times ( k_1 \dot{{\textbf{m}}} - k_2 \dot{{\textbf{u}}}) = k_1 ({{\textbf{u}}} \times \dot{{\textbf{m}}}) - k_2 ({{\textbf{u}}} \times \dot{{\textbf{u}}}) \end{aligned}$$
(96)

Thus, vector \(\dot{{\textbf{n}}}_2\) becomes:

$$\begin{aligned} \dot{{\textbf{n}}}_2 = k_1^2 \, \{ \dot{{\textbf{m}}} \cdot (\dot{{\textbf{u}}} \times {\textbf{m}} ) \} {\textbf{u}} - k_1 k_2 \, \{ \dot{{\textbf{u}}} \cdot ( {{\textbf{u}}} \times \dot{{\textbf{m}}} ) \} {\textbf{u}} \end{aligned}$$
(97)

Next, by resorting to the following vector product identity, it is found that:

$$\begin{aligned}{} & {} (\dot{{\textbf{u}}} \!\times \! {\textbf{u}}) \!\times \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \!=\! \{ \dot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \} {\textbf{u}} \!-\! \{ {\textbf{u}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \} \dot{{\textbf{u}}} \nonumber \\{} & {} \quad \!=\! \{ \dot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \} {\textbf{u}}, \,\, {\textbf{u}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \!=\! 0. \end{aligned}$$
(98)

Therefore, it is finally obtained that:

$$\begin{aligned} \dot{{\textbf{n}}}_2 = k_1^2 \, \{ \dot{{\textbf{m}}} \cdot (\dot{{\textbf{u}}} \times {\textbf{m}} ) \} {\textbf{u}} - k_1 k_2 \, (\dot{{\textbf{u}}} \!\times \! {\textbf{u}}) \!\times \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}). \end{aligned}$$
(99)
  1. (3)

    Computation of the third term, \(\dot{{\textbf{n}}}_3\).

This section shows the computation of the third term, namely, \(\dot{{\textbf{n}}}_3\). The process begins with the following expression:

$$\begin{aligned} \dot{{\textbf{n}}}_3 \equiv (\dot{{\textbf{v}}} \! \cdot \! {{\textbf{w}}}) \dot{{\textbf{u}}} = \{ (k_1 \dot{{\textbf{m}}} - k_2 \dot{{\textbf{u}}} ) \cdot ({\textbf{u}} \times {\textbf{v}}) \} \dot{{\textbf{u}}} \end{aligned}$$
(100)

where

$$\begin{aligned} {\textbf{u}} \!\times \! {\textbf{v}} = {\textbf{u}} \!\times \! ( k_1 {\textbf{m}} - k_2 {\textbf{u}}) = k_1 ({\textbf{u}} \!\times \! {\textbf{m}}) - k_2 ({\textbf{u}} \!\times \! {\textbf{u}}) = k_1 ({\textbf{u}} \!\times \! {\textbf{m}}), \,\,\, {\textbf{u}} \!\times \! {\textbf{u}} = {\textbf{0}}. \end{aligned}$$
(101)

Then, the third term can be expressed as:

$$\begin{aligned} \dot{{\textbf{n}}}_3 = k_1^2 \, \{ \dot{{\textbf{m}}} \cdot ({{\textbf{u}}} \times {\textbf{m}} ) \} \dot{{\textbf{u}}} - k_1 k_2 \, \{ \dot{{\textbf{u}}} \cdot ({{\textbf{u}}} \times {\textbf{m}} ) \} \dot{{\textbf{u}}}. \end{aligned}$$
(102)
  1. (4)

    Computation of the fourth term, \(\dot{{\textbf{n}}}_4\).

The algebraic manipulation of the fourth term is as follows:

$$\begin{aligned}{} & {} \dot{{\textbf{n}}}_4 \equiv (\ddot{{\textbf{u}}} \! \cdot \! {\textbf{w}}) {\textbf{v}} = \{ \ddot{{\textbf{u}}} \cdot ({\textbf{u}} \times {\textbf{v}}) \} {\textbf{v}} = \{ \ddot{{\textbf{u}}} \cdot ( k_1 {\textbf{u}} \times {\textbf{m}}) \} ( k_1 {\textbf{m}} - k_2 {\textbf{u}}) \nonumber \\{} & {} \dot{{\textbf{n}}}_4 = k_1^2 \, \{ \ddot{{\textbf{u}}} \cdot ({{\textbf{u}}} \times {\textbf{m}} ) \} {{\textbf{m}}} - k_1 k_2 \, \{ \ddot{{\textbf{u}}} \cdot ({{\textbf{u}}} \times {\textbf{m}} ) \} {{\textbf{u}}}. \end{aligned}$$
(103)

Recalling the following vector product identities:

$$\begin{aligned}{} & {} \!\!\!\!\!\! (\ddot{{\textbf{u}}} \!\times \! {\textbf{m}}) \!\times \! ({\textbf{u}} \!\times \! {{\textbf{m}}}) \!=\! \{ \ddot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{m}}) \} {\textbf{m}} \!-\! \{ {\textbf{m}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{m}}) \} \ddot{{\textbf{u}}} \!=\! \{ \ddot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{m}}) \} {\textbf{m}}, \,\, {\textbf{m}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{m}}) \!=\! 0. \end{aligned}$$
(104)
$$\begin{aligned}{} & {} \!\!\!\!\!\! (\ddot{{\textbf{u}}} \!\times \! {\textbf{u}}) \!\times \! ({\textbf{u}} \!\times \! {{\textbf{m}}}) \!=\! \{ \ddot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{m}}) \} {\textbf{u}} \!-\! \{ {\textbf{u}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{m}}) \} \ddot{{\textbf{u}}} \!=\! \{ \ddot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{m}}) \} {\textbf{u}}, \,\, {\textbf{u}} \!\cdot \! ({\textbf{u}} \!\times \! {\textbf{m}}) \!=\! 0. \end{aligned}$$
(105)

Having laid the necessary groundwork, we get the following result:

$$\begin{aligned} \dot{{\textbf{n}}}_4 = k_1^2 \, (\ddot{{\textbf{u}}} \!\times \! {\textbf{m}}) \!\times \! ({\textbf{u}} \!\times \! {{\textbf{m}}}) - k_1 k_2 \, (\ddot{{\textbf{u}}} \!\times \! {\textbf{u}}) \!\times \! ({\textbf{u}} \!\times \! {{\textbf{m}}}). \end{aligned}$$
(106)
  1. (5)

    Computation of the fifth term, \(\dot{{\textbf{n}}}_5\).

The fifth term can be formulated in a way that yields a convenient vector expression, which starts with the relation:

$$\begin{aligned}{} & {} \dot{{\textbf{n}}}_5 \equiv (\dot{{\textbf{u}}} \! \cdot \! \dot{{\textbf{w}}}){\textbf{v}} = \{ \dot{{\textbf{u}}} \cdot (\dot{{\textbf{u}}} \times {\textbf{v}} + {\textbf{u}} \times \dot{{\textbf{v}}}) \} {\textbf{v}} = \{ \dot{{\textbf{u}}} \cdot (\dot{{\textbf{u}}} \times {\textbf{v}}) + \dot{{\textbf{u}}} \cdot ({\textbf{u}} \times \dot{{\textbf{v}}}) \} {\textbf{v}} \nonumber \\{} & {} \dot{{\textbf{n}}}_5 = \{ \dot{{\textbf{u}}} \cdot ({\textbf{u}} \times \dot{{\textbf{v}}}) \} {\textbf{v}}, \,\,\, \dot{{\textbf{u}}} \cdot (\dot{{\textbf{u}}} \times {\textbf{v}}) = 0. \end{aligned}$$
(107)

where

$$\begin{aligned}{} & {} \!\!\!\!\!\!\!\! {\textbf{u}} \times \dot{{\textbf{v}}} = {\textbf{u}} \times (k_1 \dot{{\textbf{m}}} - k_2 \dot{{\textbf{u}}} ) = k_1 ({\textbf{u}} \times \dot{{\textbf{m}}}) - k_2 ({\textbf{u}} \times \dot{{\textbf{u}}}) \end{aligned}$$
(108)
$$\begin{aligned}{} & {} \!\!\!\!\!\!\!\! \dot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{v}}}) \!=\! k_1 \{ \dot{{\textbf{u}}} \!\cdot \! ({{\textbf{u}}} \!\times \! \dot{{\textbf{m}}} ) \} \!-\! k_2 \{ \dot{{\textbf{u}}} \!\cdot \! ({{\textbf{u}}} \!\times \! \dot{{\textbf{u}}} ) \} \!=\! k_1 \{ \dot{{\textbf{u}}} \!\cdot \! ({{\textbf{u}}} \!\times \! \dot{{\textbf{m}}} ) \}, \,\, \dot{{\textbf{u}}} \!\cdot \! ({{\textbf{u}}} \!\times \! \dot{{\textbf{u}}} ) \!=\! 0. \end{aligned}$$
(109)

Then Eq. (107) becomes:

$$\begin{aligned} \dot{{\textbf{n}}}_5 = k_1 \, \{ \dot{{\textbf{u}}} \cdot ({{\textbf{u}}} \!\times \! \dot{{\textbf{m}}} ) \} ( k_1 {\textbf{m}} - k_2 {\textbf{u}} ) = k_1^2 \, \{ \dot{{\textbf{u}}} \cdot ({{\textbf{u}}} \times \dot{{\textbf{m}}} ) \} {{\textbf{m}}} - k_1 k_2 \, \{ \dot{{\textbf{u}}} \cdot ({{\textbf{u}}} \times \dot{{\textbf{m}}} ) \} {{\textbf{u}}}. \end{aligned}$$
(110)

To complete the reduction process, we now use the well-known vector product identity:

$$\begin{aligned} (\dot{{\textbf{u}}} \!\times \! {\textbf{u}}) \!\times \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \!=\! \{ \dot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \} {\textbf{u}} - \{ {\textbf{u}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \} \dot{{\textbf{u}}} \!=\! \{ \dot{{\textbf{u}}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) \} {\textbf{u}}, \,\, {\textbf{u}} \!\cdot \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}) = 0. \end{aligned}$$
(111)

Thus, the sought expression is, therefore:

$$\begin{aligned} \dot{{\textbf{n}}}_5 = k_1^2 \, \{ \dot{{\textbf{u}}} \cdot ({{\textbf{u}}} \times \dot{{\textbf{m}}} ) \} {{\textbf{m}}} - k_1 k_2 \, (\dot{{\textbf{u}}} \!\times \! {\textbf{u}}) \!\times \! ({\textbf{u}} \!\times \! \dot{{\textbf{m}}}). \end{aligned}$$
(112)
  1. (6)

    Computation of the sixth term, \(\dot{{\textbf{n}}}_6\).

In this section, we examine another means of expressing the so-called sixth term. To this end, in the first instance, we have the following equation:

$$\begin{aligned}{} & {} \dot{{\textbf{n}}}_6 \equiv (\dot{{\textbf{u}}} \! \cdot \! {\textbf{w}}) \dot{{\textbf{v}}} = \{ \dot{{\textbf{u}}} \! \cdot \! ({\textbf{u}} \times {\textbf{v}}) \} \dot{{\textbf{v}}} = \{ \dot{{\textbf{u}}} \! \cdot \! ( k_1 {\textbf{u}} \times {{\textbf{m}}}) \} \dot{{\textbf{v}}} \nonumber \\{} & {} \quad = \{ \dot{{\textbf{u}}} \! \cdot \! ( k_1 {\textbf{u}} \times {{\textbf{m}}}) \} (k_1 \dot{{\textbf{m}}} - k_2 \dot{{\textbf{u}}} ) \end{aligned}$$
(113)

and, after some vector algebra, we obtain the final result given by:

$$\begin{aligned} \dot{{\textbf{n}}}_6 = k_1^2 \, \{ \dot{{\textbf{u}}} \cdot ({{\textbf{u}}} \times {\textbf{m}} ) \} \dot{{\textbf{m}}} - k_1 k_2 \, \{ \dot{{\textbf{u}}} \cdot ({{\textbf{u}}} \times {\textbf{m}} ) \} \dot{{\textbf{u}}}. \end{aligned}$$
(114)
  1. (7)

    Computation of the seventh term, \(\dot{{\textbf{n}}}_7\).

In this section, we analyze the mathematical form associated with the seventh term described in Eq. (30), namely:

$$\begin{aligned}{} & {} \!\!\!\!\!\!\!\! \dot{{\textbf{n}}}_7 \equiv (\ddot{{\textbf{u}}} \! \cdot \! {{\textbf{v}}}) {\textbf{w}} = \{ \ddot{{\textbf{u}}} \! \cdot \! ( k_1 {\textbf{m}} - k_2 {\textbf{u}} ) \} {\textbf{w}} = \{ \ddot{{\textbf{u}}} \! \cdot \! ( k_1 {\textbf{m}} - k_2 {\textbf{u}} ) \} ({\textbf{u}} \times {\textbf{v}}) \nonumber \\{} & {} \!\!\!\!\!\!\!\! \dot{{\textbf{n}}}_7 \!=\! \{ \ddot{{\textbf{u}}} \! \cdot \! ( k_1 {\textbf{m}} - k_2 {\textbf{u}} ) \} (k_1 {\textbf{u}} \times {{\textbf{m}}} ) \!=\! k_1^2 \, ( \ddot{{\textbf{u}}} \cdot {{\textbf{m}}} ) ( {\textbf{u}} \times {{\textbf{m}}} ) \!-\! k_1 k_2 \, ( \ddot{{\textbf{u}}} \cdot {{\textbf{u}}} ) ( {\textbf{u}} \times {{\textbf{m}}} ) \end{aligned}$$
(115)

The first term of the above equation may be conveniently transformed by the following vector identity:

$$\begin{aligned} \ddot{{\textbf{u}}} \times \{ ( {\textbf{u}} \times {{\textbf{m}}} ) \times {{\textbf{m}}} \} = ( \ddot{{\textbf{u}}} \cdot {{\textbf{m}}} ) ( {\textbf{u}} \times {{\textbf{m}}} ) - \{\ddot{{\textbf{u}}} \cdot ( {\textbf{u}} \times {{\textbf{m}}} ) \} {\textbf{m}} \end{aligned}$$
(116)

Thus, the seventh term is given by:

$$\begin{aligned} \dot{{\textbf{n}}}_7 = k_1^2 \, \ddot{{\textbf{u}}} \times \{ ( {\textbf{u}} \times {{\textbf{m}}} ) \times {{\textbf{m}}} \} + k_1^2 \, \{\ddot{{\textbf{u}}} \cdot ( {\textbf{u}} \times {{\textbf{m}}} ) \} {\textbf{m}} - k_1 k_2 \, ( \ddot{{\textbf{u}}} \cdot {{\textbf{u}}} ) ( {\textbf{u}} \times {{\textbf{m}}} ). \end{aligned}$$
(117)
  1. (8)

    Computation of the eighth term, \(\dot{{\textbf{n}}}_8\).

We now consider an alternative derivation of the formula for the eighth term that was previously defined. The procedure is as follows:

$$\begin{aligned}{} & {} \!\!\!\!\!\!\!\! \dot{{\textbf{n}}}_8 \equiv (\dot{{\textbf{u}}} \! \cdot \! \dot{{{\textbf{v}}}}) {\textbf{w}} = \{ \dot{{\textbf{u}}} \! \cdot \! (k_1 \dot{{\textbf{m}}} - k_2 \dot{{\textbf{u}}} ) \} {\textbf{w}} = \{ \dot{{\textbf{u}}} \! \cdot \! (k_1 \dot{{\textbf{m}}} - k_2 \dot{{\textbf{u}}} ) \} ({\textbf{u}} \times {\textbf{v}}) \nonumber \\{} & {} \!\!\!\!\!\!\!\! \dot{{\textbf{n}}}_8 \!=\! \{ \dot{{\textbf{u}}} \! \cdot \! (k_1 \dot{{\textbf{m}}} \!-\! k_2 \dot{{\textbf{u}}} ) \} (k_1 {\textbf{u}} \!\times \! {{\textbf{m}}}) \!=\! k_1^2 ( \dot{{\textbf{u}}} \cdot \dot{{\textbf{m}}} ) ( {\textbf{u}} \times {{\textbf{m}}} ) \!-\! k_1 k_2 ( \dot{{\textbf{u}}} \!\cdot \! \dot{{\textbf{u}}} ) ( {\textbf{u}} \!\times \! {{\textbf{m}}} ) \end{aligned}$$
(118)

To get a more convenient form of the first term of the above equation, we resort to the following vector identity:

$$\begin{aligned} \dot{{\textbf{u}}} \times \{ ( {\textbf{u}} \times {{\textbf{m}}} ) \times \dot{{\textbf{m}}} \} = ( \dot{{\textbf{u}}} \cdot \dot{{\textbf{m}}} ) ( {\textbf{u}} \times {{\textbf{m}}} ) - \{\dot{{\textbf{u}}} \cdot ( {\textbf{u}} \times {{\textbf{m}}} ) \} \dot{{\textbf{m}}} \end{aligned}$$
(119)

Therefore, the eighth term gets the following form:

$$\begin{aligned} \dot{{\textbf{n}}}_8 = k_1^2 \, \dot{{\textbf{u}}} \times \{ ( {\textbf{u}} \times {{\textbf{m}}} ) \times \dot{{\textbf{m}}} \} + k_1^2 \, \{\dot{{\textbf{u}}} \cdot ( {\textbf{u}} \times {{\textbf{m}}} ) \} \dot{{\textbf{m}}} - k_1 k_2 \, ( \dot{{\textbf{u}}} \cdot \dot{{\textbf{u}}} ) ( {\textbf{u}} \times {{\textbf{m}}} ). \end{aligned}$$
(120)
  1. (9)

    Computation of the ninth term, \(\dot{{\textbf{n}}}_9\).

This section is dedicated to finding an alternative formula for the ninth term defined in Eq. (30). This term can be expressed as:

$$\begin{aligned} \dot{{\textbf{n}}}_9 \equiv (\dot{{\textbf{u}}} \! \cdot \! {{\textbf{v}}}) \dot{{\textbf{w}}} = \{ \dot{{\textbf{u}}} \! \cdot \! ( k_1 {\textbf{m}} - k_2 {\textbf{u}} ) \} \dot{{\textbf{w}}} = \{ k_1 ( \dot{{\textbf{u}}} \! \cdot {\textbf{m}} ) - k_2 ( \dot{{\textbf{u}}} \! \cdot {\textbf{u}} ) \} (\dot{{\textbf{u}}} \times {\textbf{v}} + {\textbf{u}} \times \dot{{\textbf{v}}}) \end{aligned}$$
(121)

where

$$\begin{aligned}{} & {} \dot{{\textbf{u}}} \times {\textbf{v}} = \dot{{\textbf{u}}} \times ( k_1 {\textbf{m}} - k_2 {\textbf{u}}) = k_1 (\dot{{\textbf{u}}} \times {\textbf{m}}) - k_2 (\dot{{\textbf{u}}} \times {\textbf{u}}) \end{aligned}$$
(122)
$$\begin{aligned}{} & {} {\textbf{u}} \times \dot{{\textbf{v}}} = {{\textbf{u}}} \times ( k_1 \dot{{\textbf{m}}} - k_2 \dot{{\textbf{u}}}) = k_1 ({{\textbf{u}}} \times \dot{{\textbf{m}}}) - k_2 ({{\textbf{u}}} \times \dot{{\textbf{u}}}) \end{aligned}$$
(123)
$$\begin{aligned}{} & {} \dot{{\textbf{u}}} \! \cdot {\textbf{u}} = 0, \quad \dot{{\textbf{u}}} \times {\textbf{u}} = -{{\textbf{u}}} \times \dot{{\textbf{u}}}. \end{aligned}$$
(124)

From these relationships, we have that:

$$\begin{aligned} \dot{{\textbf{n}}}_9 = k_1^2 \, ( \dot{{\textbf{u}}} \cdot {\textbf{m}} ) ( \dot{{\textbf{u}}} \times {{\textbf{m}}} ) + k_1^2 \, ( \dot{{\textbf{u}}} \cdot {\textbf{m}} ) ( {{\textbf{u}}} \times \dot{{\textbf{m}}} ). \end{aligned}$$
(125)

The algebraic process continues using the following two vector identities:

$$\begin{aligned}{} & {} \!\!\!\!\!\!\!\!\!\!\! \dot{{\textbf{u}}} \!\times \! \{ ( \dot{{\textbf{u}}} \!\times \! {{\textbf{m}}} ) \!\times \! {{\textbf{m}}} \} \!=\! ( \dot{{\textbf{u}}} \!\cdot \! {{\textbf{m}}} ) ( \dot{{\textbf{u}}} \!\times \! {{\textbf{m}}} ) \!-\! \{\dot{{\textbf{u}}} \!\cdot \! ( \dot{{\textbf{u}}} \!\times \! {{\textbf{m}}} ) \} {{\textbf{m}}} \nonumber \\{} & {} \quad \!=\! ( \dot{{\textbf{u}}} \!\cdot \! {{\textbf{m}}} ) ( \dot{{\textbf{u}}} \!\times \! {{\textbf{m}}} ), \, \dot{{\textbf{u}}} \!\cdot \! ( \dot{{\textbf{u}}} \!\times \! {{\textbf{m}}} ) \!=\! 0. \end{aligned}$$
(126)
$$\begin{aligned}{} & {} \!\!\!\!\!\!\!\!\!\!\! \dot{{\textbf{u}}} \!\times \! \{ ( {{\textbf{u}}} \!\times \! \dot{{\textbf{m}}} ) \!\times \! {{\textbf{m}}} \} \!=\! ( \dot{{\textbf{u}}} \!\cdot \! {{\textbf{m}}} ) ( {{\textbf{u}}} \!\times \! \dot{{\textbf{m}}} ) \!-\! \{\dot{{\textbf{u}}} \!\cdot \! ( {{\textbf{u}}} \!\times \! \dot{{\textbf{m}}} ) \} {{\textbf{m}}}. \end{aligned}$$
(127)

By using the foregoing identities it is obtained the final result given by:

$$\begin{aligned} \dot{{\textbf{n}}}_9 = k_1^2 \, \dot{{\textbf{u}}} \!\times \! \{ ( \dot{{\textbf{u}}} \!\times \! {{\textbf{m}}} ) \!\times \! {{\textbf{m}}} \} + k_1^2 \, \dot{{\textbf{u}}} \!\times \! \{ ( {{\textbf{u}}} \!\times \! \dot{{\textbf{m}}} ) \!\times \! {{\textbf{m}}} \} + k_1^2 \, \{\dot{{\textbf{u}}} \!\cdot \! ( {{\textbf{u}}} \!\times \! \dot{{\textbf{m}}} ) \} {{\textbf{m}}}. \end{aligned}$$
(128)

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Cervantes-Sánchez, J.J., Rico-Martínez, J.M., García-Murillo, M.A. et al. Analyzing the angular acceleration vector of a moving rigid body. Meccanica 59, 89–106 (2024). https://doi.org/10.1007/s11012-023-01741-7

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