Problem
Consider in Figure 1 three identical cubes \({\cal A}\), \({\cal B}\), \({\cal C}\) of side length \(l\), interconnected along particular edges. Cube \({\cal B}\) is hinged along axis \((A,\boldsymbol{\hat{a}}_2)\) relative to cube \({\cal A}\), while cube \({\cal C}\) is hinged along axis \((B,\boldsymbol{\hat{b}}_1)\) relative to cube \({\cal C}\). The orientations of the cubes are defined by the two (oriented) angles \(\psi\) and \(\theta\) as shown.
1. Sketch the rotation diagrams between the bases of \({\cal A}\), \({\cal B}\) and \({\cal C}\).
2. Find the kinematic screw \(\{{\cal V}_{{\cal B}/{\cal A}}\}\) resolved at \(A\) and the kinematic screw \(\{{\cal V}_{{\cal C}/{\cal B}}\}\) resolved at \(B\). Deduce the expression of kinematic screw \(\{{\cal V}_{{\cal C}/{\cal A}}\}\).
3. Then the velocity of point \(C\) relative to \({\cal A}\).
Formulas:
1. Time-Derivative Formula. Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), and an arbitrary vector function \(\boldsymbol{U}(t)\), \[ \Big(\frac{d\boldsymbol{U}}{dt} \Big)_{\cal E}=\Big(\frac{d\boldsymbol{U}}{dt} \Big)_{\cal F}+ \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{U} \]
2. Triple product formula: \[ \boldsymbol{A}\times (\boldsymbol{B}\times \boldsymbol{C})= (\boldsymbol{A}\cdot \boldsymbol{C}) \boldsymbol{B}- (\boldsymbol{A}\cdot \boldsymbol{B}) \boldsymbol{C}$ \]
3. Rigid Body Kinematics: Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), for any two points \(P\) and \(Q\), and at any given time, \[ \boldsymbol{v}_{Q \in {\cal F}/{\cal E}}= \boldsymbol{v}_{P \in {\cal F}/{\cal E}}+ \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ} \] \[ \boldsymbol{a}_{Q \in {\cal F}/{\cal E}}= \boldsymbol{a}_{P \in {\cal F}/{\cal E}}+ \boldsymbol{\alpha}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ} + \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times (\boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ}) \]
4. Change of Referential Formulas. Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), for any particle \(P\), and at any given time, \[ \boldsymbol{v}_{P/{\cal E}} = \boldsymbol{v}_{P/{\cal F}} + \boldsymbol{v}_{P\in {\cal F}/ {\cal E}} \] \[ \boldsymbol{a}_{P/{\cal E}} = \boldsymbol{a}_{P/{\cal F}} + \boldsymbol{a}_{P\in {\cal F}/ {\cal E}} + 2 \boldsymbol{\omega}_{{\cal F}/{\cal E}} \times \boldsymbol{v}_{P/{\cal F}} \]
5. Loop formulas: Given 3 rigid bodies \({\cal A}\), \({\cal B}\), \({\cal C}\), and an arbitrary point \(P\), \[ \boldsymbol{\omega}_{{\cal A}/{\cal B}}+ \boldsymbol{\omega}_{{\cal B}/{\cal C}}+ \boldsymbol{\omega}_{{\cal C}/{\cal A}} = \boldsymbol{0} \] \[ \boldsymbol{v}_{P\in {\cal A}/ {\cal B}}+ \boldsymbol{v}_{P\in {\cal B}/ {\cal C}}+ \boldsymbol{v}_{P\in {\cal C}/ {\cal A}}=\boldsymbol{0} \]
Solution
1. First we sketch the rotation diagrams between the bases of \({\cal A}\), \({\cal B}\) and \({\cal C}\) (this must be compatible with the data shown in Figure 1. See Figure 1S. The unit vectors (not shown on the original sketch) are all defined so that the basis \((\boldsymbol{\hat{a}}_1, \boldsymbol{\hat{a}}_2 ,\boldsymbol{\hat{a}}_3)\) \((\boldsymbol{\hat{b}}_1, \boldsymbol{\hat{b}}_2 ,\boldsymbol{\hat{b}}_3)\) and \((\boldsymbol{\hat{c}}_1, \boldsymbol{\hat{c}}_2 ,\boldsymbol{\hat{c}}_3)\) are right-handed.
Note: we have \(\boldsymbol{\hat{b}}_2= \boldsymbol{\hat{a}}_2\) and \(\boldsymbol{\hat{c}}_1 = \boldsymbol{\hat{b}}_1\).
2. The joints \({\cal A}/{\cal B}\) and \({\cal C}/{\cal B}\) are pivots, about axes \((A,\boldsymbol{\hat{a}}_1)\) and \((B,\boldsymbol{\hat{b}}_1)\), respectively. We then obtain the expressions of the kinematic screws \(\{{\cal V}_{{\cal B}/{\cal A}}\}\) and \(\{{\cal V}_{{\cal C}/{\cal B}}\}\) \[ \{{\cal V}_{{\cal B}/{\cal A}}\}= \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{a}}_2 \\ \boldsymbol{0}\end{Bmatrix}_{A} \qquad \{{\cal V}_{{\cal C}/{\cal B}}\}= \begin{Bmatrix} - \dot{\theta}\boldsymbol{\hat{b}}_1 \\ \boldsymbol{0}\end{Bmatrix}_{B} \] We can then obtain \(\{{\cal V}_{{\cal C}/{\cal B}}\}\) resolved at \(B\): \[ \{{\cal V}_{{\cal C}/{\cal A}}\}= \{{\cal V}_{{\cal C}/{\cal B}}\} + \{{\cal V}_{{\cal B}/{\cal A}}\}= \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{a}}_2- \dot{\theta}\boldsymbol{\hat{b}}_1 \\ \dot{\psi}\boldsymbol{\hat{a}}_2 \times \boldsymbol{r}_{AB} \end{Bmatrix}_{B} =\begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{a}}_2- \dot{\theta}\boldsymbol{\hat{b}}_1 \\ \dot{\psi}\boldsymbol{\hat{a}}_2 \times \boldsymbol{r}_{AB} \end{Bmatrix}_{B} =\begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{a}}_2- \dot{\theta}\boldsymbol{\hat{b}}_1 \\ \ell\dot{\psi}(\boldsymbol{\hat{b}}_1 - \boldsymbol{\hat{b}}_3) \end{Bmatrix}_{B} \] where we used \(\boldsymbol{r}_{AB}= \ell (\boldsymbol{\hat{b}}_1 +\boldsymbol{\hat{b}}_2 +\boldsymbol{\hat{b}}_3)\).
3. To find the velocity of point \(C\) relative to \({\cal A}\), we use the kinematic screw \(\{{\cal V}_{{\cal C}/{\cal A}}\}\) and \(\boldsymbol{r}_{BC}=\ell (\boldsymbol{\hat{c}}_3 -\boldsymbol{\hat{c}}_2- \boldsymbol{\hat{c}}_1)\): \[ \boldsymbol{v}_{C/{\cal A}}= \boldsymbol{v}_{B/{\cal A}}+ \boldsymbol{\omega}_{{\cal C}/{\cal A}}\times \boldsymbol{r}_{BC} = \ell\dot{\psi}(\boldsymbol{\hat{b}}_1 - \boldsymbol{\hat{b}}_3)+ (\dot{\psi}\boldsymbol{\hat{b}}_2- \dot{\theta}\boldsymbol{\hat{b}}_1)\times \ell (\boldsymbol{\hat{c}}_3 -\boldsymbol{\hat{c}}_2- \boldsymbol{\hat{c}}_1) \] After evaluating the cross-products, we obtain the expression \[ \boldsymbol{v}_{C/{\cal A}} =\ell\dot{\psi}(\boldsymbol{\hat{b}}_1 - \boldsymbol{\hat{b}}_3)+\ell\dot{\psi}(\cos\theta+\sin\theta) \boldsymbol{\hat{b}}_1 + l\dot{\psi}\boldsymbol{\hat{b}}_3 + \ell\dot{\theta}(\boldsymbol{\hat{c}}_2 +\boldsymbol{\hat{c}}_3) \] After simplifications, we obtain \[ \boxed{ \boldsymbol{v}_{C/{\cal A}}= \ell\dot{\psi}(1+\cos\theta+\sin\theta) \boldsymbol{\hat{b}}_1 + \ell\dot{\theta}(\boldsymbol{\hat{c}}_2 +\boldsymbol{\hat{c}}_3) } \]
© 2024 R. Valéry Roy.