Problem 5.5
To find the kinematic screw of \({\cal A}\) relative to \({\cal E}\) (at that instant), we need to find \(\boldsymbol{\omega}_{{\cal A}/{\cal E}} =\boldsymbol{\omega}\). We must solve equation \[ \boldsymbol{\omega}\times \boldsymbol{r}_{BC}= \boldsymbol{v}_C - \boldsymbol{v}_B \] or (using basis \((\boldsymbol{\hat{a}}_1,\boldsymbol{\hat{a}}_2,\boldsymbol{\hat{a}}_3)\)) \[ \boldsymbol{\omega}\times a \boldsymbol{\hat{a}}_3 = v(- \boldsymbol{\hat{a}}_1- \boldsymbol{\hat{a}}_2 ) \Longrightarrow \boldsymbol{\hat{a}}_3\times(\boldsymbol{\omega}\times a \boldsymbol{\hat{a}}_3) = v\boldsymbol{\hat{a}}_3\times(- \boldsymbol{\hat{a}}_1- \boldsymbol{\hat{a}}_2 ) \] This gives only two components for \(\boldsymbol{\omega}\): \[ \boldsymbol{\omega}= \omega_3 \boldsymbol{\hat{a}}_3 + \frac{v}{a} (- \boldsymbol{\hat{a}}_2 + \boldsymbol{\hat{a}}_1) \] To find remaining component \(\omega_3\) on \(\boldsymbol{\hat{a}}_3\), we use equation \(\boldsymbol{\omega}\times \boldsymbol{r}_{AB} = \boldsymbol{v}_B -\boldsymbol{v}_A\) \[ \boldsymbol{\hat{a}}_3 \times(\boldsymbol{\omega}\times \boldsymbol{r}_{AB}) = \boldsymbol{\hat{a}}_3 \times(\boldsymbol{v}_A -\boldsymbol{v}_B) \] which gives (noting that \(\boldsymbol{\hat{a}}_3 \cdot \boldsymbol{r}_{AB} = 0\)) \[ - \omega_3 \boldsymbol{r}_{AB} = v \boldsymbol{\hat{a}}_3 \times(- 2 \boldsymbol{\hat{a}}_1 + 2 \boldsymbol{\hat{a}}_2 +2 \boldsymbol{\hat{a}}_3) = 2 v (-\boldsymbol{\hat{a}}_2 - \boldsymbol{\hat{a}}_2)\Longrightarrow \omega_3 = \frac{2v}{a} \] We conclude that \(\boxed{\boldsymbol{\omega}= \frac{v}{a} (\boldsymbol{\hat{a}}_1 - \boldsymbol{\hat{a}}_2+ 2\boldsymbol{\hat{a}}_3)}\) and that \[ \{ {\cal V}_{{\cal A}/ {\cal E}} \} = \left\{ \begin{array}{cc} \displaystyle\frac{v}{a} (\boldsymbol{\hat{a}}_1 - \boldsymbol{\hat{a}}_2+ 2\boldsymbol{\hat{a}}_3) \\ \\ v (2 \boldsymbol{\hat{a}}_1 + \boldsymbol{\hat{a}}_2 -3 \boldsymbol{\hat{a}}_3) \end{array} \right\}_A \] To find instantaneous screw axis \(\Delta\), we seek the points \(Q\) s.t. \(\boldsymbol{v}_Q = p \boldsymbol{\omega}\): \[ \boldsymbol{\omega}\times\boldsymbol{r}_{AQ} = p\boldsymbol{\omega}- \boldsymbol{v}_A \qquad{(1)} \] This equation necessarily implies that \(\boldsymbol{\omega}\cdot (p\boldsymbol{\omega}-\boldsymbol{v}_A) = 0\): this gives pitch \(p\) \[ \boxed{ p = \frac{\boldsymbol{\omega}\cdot\boldsymbol{v}_A}{\omega^2} = -5a/6 } \] Then, we take the cross-product of (1) with \(\boldsymbol{\omega}\): \[ \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r}_{AQ}) =\boldsymbol{\omega}\times( p\boldsymbol{\omega}- \boldsymbol{v}_A) \Longrightarrow \lambda\boldsymbol{\omega}- \boldsymbol{\omega}^2\boldsymbol{r}_{AQ} = - \boldsymbol{\omega}\times\boldsymbol{v}_A \] where \(\lambda\) is unknown. Setting \(\lambda=0\) gives a particular point on \(\Delta\). After evaluating the cross-product \(\boldsymbol{\omega}\times\boldsymbol{v}_A\), we find that \(\Delta\) is defined as the set of points \(Q\) which satisfy \[ \boldsymbol{r}_{AQ} = \mu (\boldsymbol{\hat{a}}_1 - \boldsymbol{\hat{a}}_2+ 2\boldsymbol{\hat{a}}_3) +\frac{a}{6} ( \boldsymbol{\hat{a}}_1 +7 \boldsymbol{\hat{a}}_2 +3 \boldsymbol{\hat{a}}_3) \qquad (\mu \in \mathbb{R}) \] We can now write the kinematic screw as the sum \[ \{ {\cal V}_{{\cal A}/ {\cal E}} \} = \left\{ \begin{array}{cc} \boldsymbol{0}\\ \\ \displaystyle -\frac{5v}{6} (\boldsymbol{\hat{a}}_1 - \boldsymbol{\hat{a}}_2+ 2\boldsymbol{\hat{a}}_3) \end{array} \right\} + \left\{ \begin{array}{cc} \displaystyle\frac{v}{a} (\boldsymbol{\hat{a}}_1 - \boldsymbol{\hat{a}}_2+ 2\boldsymbol{\hat{a}}_3) \\ \\ \boldsymbol{0} \end{array} \right\}_{Q^*} \] where \(Q^*\) is the point defined by \(\boldsymbol{r}_{AQ^*} = (a/6) ( \boldsymbol{\hat{a}}_1 +7 \boldsymbol{\hat{a}}_2 +3 \boldsymbol{\hat{a}}_3)\). This expresses the fact that body \({\cal A}\) is in helical motion about \(\Delta\) at the specified instant.