One rotating frame is used. The rotating coordinate system is attached to the rotating bar shown above with axis \(xyz\) with its origin at point \(O\). The vector \(\boldsymbol{\rho }\) goes from point \(O\) to point \(Q\) as shown in this diagram
\(\boldsymbol{\rho }\) is vector that represents the position of point \(Q\) on the disk relative to the rotating frame. Let current distance of point \(Q\) from center of disk be \(r\left ( t\right ) \) and angle be \(\theta \left ( t\right ) \) where \(\dot{\theta }\left ( t\right ) =\omega _{2}\) as shown in this diagram
The position of \(Q\) as seen in inertial frame is therefore\begin{equation} \overset{\rightarrow }{r}_{Q}=\overset{\rightarrow }{R}+\overset{\rightarrow }{\rho } \tag{1} \end{equation}
But \(\overset{\rightarrow }{R}=0\) here. And \[ \boldsymbol{\rho }=\left ( L_{1}+r\cos \theta \right ) \mathbf{i}+\left ( -L_{3}+r\sin \theta \right ) \mathbf{j}+L_{2}\mathbf{k}\]
Hence the total velocity is \begin{equation} \mathbf{V}_{Q}=\dot{\boldsymbol{\rho }}_{r}+\left ( \boldsymbol{\omega }_{1}\times \overset{\rightarrow }{\rho }\right ) \tag{2} \end{equation}
Where \begin{align*} \dot{\boldsymbol{\rho }}_{r} & =\left ( \dot{r}\cos \theta -r\dot{\theta }\sin \theta \right ) \mathbf{i}+\left ( \dot{r}\sin \theta +r\dot{\theta }\cos \theta \right ) \mathbf{j}\\ & =\left ( \dot{r}\cos \theta -r\omega _{2}\sin \theta \right ) \mathbf{i}+\left ( \dot{r}\sin \theta +r\omega _{2}\cos \theta \right ) \mathbf{j} \end{align*}
and\begin{align*} \boldsymbol{\omega }_{1}\times \boldsymbol{\rho } & =\omega _{1}\mathbf{j}\times \left ( \left ( L_{1}+r\cos \theta \right ) \mathbf{i}+\left ( -L_{3}+r\sin \theta \right ) \mathbf{j}+L_{2}\mathbf{k}\right ) \\ & =-\omega _{1}\left ( L_{1}+r\cos \theta \right ) \overset{\rightarrow }{k}+\omega _{1}L_{2}\mathbf{i} \end{align*}
Hence Eq. (2) becomes\begin{align} \mathbf{V}_{Q} & =\left ( \dot{r}\cos \theta -r\omega _{2}\sin \theta \right ) \mathbf{i}+\left ( \dot{r}\sin \theta +r\omega _{2}\cos \theta \right ) \mathbf{j}-\omega _{1}\left ( L_{1}+r\cos \theta \right ) \mathbf{k}+\omega _{1}L_{2}\mathbf{i}\nonumber \\ & =\left ( \dot{r}\cos \theta -r\omega _{2}\sin \theta +\omega _{1}L_{2}\right ) \mathbf{i}+\left ( \dot{r}\sin \theta +r\omega _{2}\cos \theta \right ) \mathbf{j}-\omega _{1}\left ( L_{1}+r\cos \theta \right ) \mathbf{k} \tag{3} \end{align}
At the snapshot, \(\theta =0\) and \(\dot{\theta }=\omega _{2}\) and \(r=\frac{0.75}{2}=0.375\) ft, and \(\dot{r}=1.5\) ft/sec,\(L_{1}=2.5,L_{2}=0.7,L_{3}=1.4,\omega _{2}=0.5\) rad/sec, \(\omega _{1}=1.2\) rad/sec, Hence the above becomes\[ \mathbf{V}_{Q}=\left ( \dot{r}+\omega _{1}L_{2}\right ) \mathbf{i}+r\omega _{2}\mathbf{j}-\omega _{1}\left ( L_{1}+r\right ) \mathbf{k}\]
Now it is evaluated using the numerical values given\begin{align*} \mathbf{V}_{Q} & =\left ( 1.5+1.2\left ( 0.7\right ) \right ) \mathbf{i}+0.375\left ( 0.5\right ) \mathbf{j}-1.2\left ( 2.5+0.375\right ) \mathbf{k}\\ & =2.34\mathbf{i}+0.187\,5\mathbf{j}-3.45\mathbf{k} \end{align*}
Hence \begin{align*} \left \vert \mathbf{V}_{Q}\right \vert & =\sqrt{2.34^{2}+0.1875^{2}+3.45^{2}}\\ & =4.172\,9\qquad ft/sec \end{align*}
To find absolute acceleration, the derivative of Eq. (2) is\begin{align} \mathbf{a}_{Q} & =\frac{d}{dt}\left ( \dot{\boldsymbol{\rho }}_{r}+\left ( \boldsymbol{\omega }_{1}\times \boldsymbol{\rho }\right ) \right ) \nonumber \\ & =\ddot{\boldsymbol{\rho }}_{r}+\left ( \boldsymbol{\omega }_{1}\times \dot{\boldsymbol{\rho }}_{r}\right ) + \left ( \dot{\boldsymbol{\omega }}_{1}\times \boldsymbol{\rho }\right ) +\left ( \boldsymbol{\omega }_{1}\times \left ( \dot{\boldsymbol{\rho }}_{r}+\left ( \boldsymbol{\omega }_{1}\times \boldsymbol{\rho }\right ) \right ) \right ) \nonumber \\ & =\ddot{\boldsymbol{\rho }}_{r} +2\left (\boldsymbol{\omega }_{1}\times \dot{\boldsymbol{\rho }}_{r}\right ) +\left ( \dot{\boldsymbol{\omega }}_{1}\times \boldsymbol{\rho }\right ) +\left ( \boldsymbol{\omega }_{1}\times \left ( \boldsymbol{\omega }_{1}\times \boldsymbol{\rho }\right ) \right ) \tag{5} \end{align}
Each term in the above is now found\begin{align*} \ddot{\boldsymbol{\rho }}_{r} & =\frac{d}{dt}\left [ \left ( \dot{r}\cos \theta -r\dot{\theta }\sin \theta \right ) \mathbf{i}+\left ( \dot{r}\sin \theta +r\dot{\theta }\cos \theta \right ) \mathbf{j}\right ] \\ & =\left ( \left ( \ddot{r}\cos \theta -\dot{r}\dot{\theta }\sin \theta \right ) -\left ( \dot{r}\dot{\theta }\sin \theta +r\ddot{\theta }\sin \theta +r\dot{\theta }^{2}\cos \theta \right ) \right ) \mathbf{i}\\ & +\left ( \left ( \ddot{r}\sin \theta +\dot{r}\dot{\theta }\cos \theta \right ) +\left ( \dot{r}\dot{\theta }\cos \theta +r\ddot{\theta }\cos \theta -r\dot{\theta }^{2}\sin \theta \right ) \right ) \mathbf{j} \end{align*}
Hence\[ \ddot{\boldsymbol{\rho }}_{r}=\left ( \ddot{r}\cos \theta -2\dot{r}\dot{\theta }\sin \theta -r\ddot{\theta }\sin \theta -r\dot{\theta }^{2}\cos \theta \right ) \mathbf{i}+\left ( \ddot{r}\sin \theta +2\dot{r}\dot{\theta }\cos \theta +r\ddot{\theta }\cos \theta -r\dot{\theta }^{2}\sin \theta \right ) \mathbf{j}\]
And \begin{align*} \boldsymbol{\omega }_{1}\times \dot{\boldsymbol{\rho }}_{r} & =\omega _{1}\mathbf{j}\times \left ( \left ( \dot{r}\cos \theta -r\dot{\theta }\sin \theta \right ) \mathbf{i}+\left ( \dot{r}\sin \theta +r\dot{\theta }\cos \theta \right ) \mathbf{j}\right ) \\ & =-\omega _{1}\left ( \dot{r}\cos \theta -r\dot{\theta }\sin \theta \right ) \mathbf{k} \end{align*}
And \begin{align*} \left ( \dot{\boldsymbol{\omega }}_{1}\times \boldsymbol{\rho }\right ) & =\left ( \dot{\omega }_{1}\mathbf{j}\times \left ( \left ( L_{1}+r\cos \theta \right ) \mathbf{i}+\left ( -L_{3}+r\sin \theta \right ) \mathbf{j}+L_{2}\mathbf{k}\right ) \right ) \\ & =-\dot{\omega }_{1}\left ( L_{1}+r\cos \theta \right ) \mathbf{k}+\left ( \dot{\omega }_{1}L_{2}\right ) \mathbf{i} \end{align*}
And finally\begin{align*} \boldsymbol{\omega }_{1}\times \left ( \overset{\rightarrow }{\omega }_{1}\times \boldsymbol{\rho }\right ) & =\omega _{1}\mathbf{j}\times \left ( \omega _{1}\mathbf{j}\times \left ( \left ( L_{1}+r\cos \theta \right ) \overset{\rightarrow }{i}+\left ( -L_{3}+r\sin \theta \right ) \mathbf{j}+L_{2}\mathbf{k}\right ) \right ) \\ & =\omega _{1}\mathbf{j}\times \left ( -\omega _{1}\left ( L_{1}+r\cos \theta \right ) \mathbf{k}+\omega _{1}L_{2}\mathbf{i}\right ) \\ & =-\omega _{1}^{2}\left ( L_{1}+r\cos \theta \right ) \overset{\rightarrow }{i}-\omega _{1}^{2}L_{2}\mathbf{k} \end{align*}
Now all terms in Eq. (5) are known. Hence Eq. (5) becomes\begin{align} \mathbf{a}_{Q} & =\ddot{\boldsymbol{\rho }}_{r}+2\left ( \boldsymbol{\omega }_{1}\times \dot{\boldsymbol{\rho }}_{r}\right ) +\left ( \dot{\boldsymbol{\omega }}_{1}\times \boldsymbol{\rho }\right ) +\left ( \overset{\rightarrow }{\omega }_{1}\times \left ( \boldsymbol{\omega }_{1}\times \boldsymbol{\rho }\right ) \right ) \nonumber \\ & =\left ( \ddot{r}\cos \theta -2\dot{r}\dot{\theta }\sin \theta -r\ddot{\theta }\sin \theta -r\dot{\theta }^{2}\cos \theta \right ) \mathbf{i}+\left ( \ddot{r}\sin \theta +2\dot{r}\dot{\theta }\cos \theta +r\ddot{\theta }\cos \theta -r\dot{\theta }^{2}\sin \theta \right ) \mathbf{j}\nonumber \\ & +2\left ( -\omega _{1}\left ( \dot{r}\cos \theta -r\dot{\theta }\sin \theta \right ) \mathbf{k}\right ) \nonumber \\ & +\left ( -\dot{\omega }_{1}\left ( L_{1}+r\cos \theta \right ) \mathbf{k}+\left ( \dot{\omega }_{1}L_{2}\right ) \mathbf{i}\right ) \nonumber \\ & +\left ( -\omega _{1}^{2}\left ( L_{1}+r\cos \theta \right ) \mathbf{i}-\omega _{1}^{2}L_{2}\mathbf{k}\right ) \tag{6} \end{align}
At snapshot time, \(\theta =0\), and the above simplifies to (noting that \(\dot{\theta }=\omega _{2}\) and \(\ddot{\theta }=\dot{\omega }_{2}\)\begin{align*} \mathbf{a}_{Q} & =\left ( \ddot{r}-r\omega _{2}^{2}\right ) \mathbf{i}+\left ( 2\dot{r}\omega _{2}+r\dot{\omega }_{2}\right ) \mathbf{j}-2\omega _{1}\dot{r}\mathbf{k}-\dot{\omega }_{1}\left ( L_{1}+r\right ) \mathbf{k}+\dot{\omega }_{1}L_{2}\mathbf{i}-\omega _{1}^{2}\left ( L_{1}+r\right ) \mathbf{i}-\omega _{1}^{2}L_{2}\mathbf{k}\\ & =\left ( \ddot{r}-r\omega _{2}^{2}+\dot{\omega }_{1}L_{2}-\omega _{1}^{2}\left ( L_{1}+r\right ) \right ) \mathbf{i}+\left ( 2\dot{r}\omega _{2}+r\dot{\omega }_{2}\right ) \overset{\rightarrow }{j}-\left ( 2\omega _{1}\dot{r}+\dot{\omega }_{1}\left ( L_{1}+r\right ) +\omega _{1}^{2}L_{2}\right ) \mathbf{k} \end{align*}
At the instance shown \(r=\frac{0.75}{2}=0.375\,\) and \(\dot{r}=1.5\) ft/sec,\(\ddot{r}=0.8\) ft/sec\(^{2}\),\(L_{1}=2.5,L_{2}=0.7,L_{3}=1.4,\omega _{2}=0.5\) rad/sec, \(\omega _{1}=1.2\) rad/sec, \(\dot{\omega }_{2}=0.25\) rad/sec\(^{2}\),\(\dot{\omega }_{1}=0.6\) rad/sec\(^{2}\), hence the above becomes\begin{align*} \mathbf{a}_{Q} & =\left ( 0.8-0.375\left ( 0.5^{2}\right ) +0.6\left ( 0.7\right ) -1.2^{2}\left ( 2.5+0.375\right ) \right ) \mathbf{i}\\ & +\left ( 2\left ( 1.5\right ) 0.5+\left ( 0.375\right ) 0.25\right ) \mathbf{j}\\ & -\left ( 2\left ( 1.2\right ) 1.5+0.6\left ( 2.5+0.375\right ) +1.2^{2}\left ( 0.7\right ) \right ) \mathbf{k} \end{align*}
Therefore\begin{equation} \mathbf{a}_{Q}=-3.013\,8\mathbf{i}+1.5938\mathbf{j}-6.333\mathbf{k} \tag{7} \end{equation}
Hence \begin{align*} \left \vert \mathbf{a}_{Q}\right \vert & =\sqrt{3.0138^{2}+1.5938^{2}+6.333^{2}}\\ & =7.192\,4\text{ ft/sec}^{2} \end{align*}
The total angular velocity \(\boldsymbol{\omega }_{G}\) of the disk \(G\) using body coordinates \(\left \{ \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\right \} \) is\begin{align} \boldsymbol{\omega }_{G} & =\dot{\psi }\mathbf{e}_{3}+\dot{\theta }\mathbf{e}_{2}+\dot{\phi }\cos \theta \mathbf{e}_{3}\nonumber \\ & =\dot{\theta }\mathbf{e}_{2}+\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \overset{\rightarrow }{e}_{3} \tag{1} \end{align}
To find the acceleration, the rate of change of the above vector is taken. When taking rate of change of each unit vector \(\mathbf{e}\) the following will be used\[ \dot{\mathbf{e}}=\overset{\rightarrow }{\omega }_{e}\times \mathbf{e}\]
Where \(\boldsymbol{\omega }_{e}\) is the angular rate that the unit vector \(\mathbf{e}\) rotates relative to the inertial frame. Hence Eq. (1) becomes\begin{align} \dot{\boldsymbol{\omega }}_{G} & =\frac{d}{dt}\left ( \dot{\theta }\mathbf{e}_{2}\right ) +\frac{d}{dt}\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \mathbf{e}_{3}\nonumber \\ & =\ddot{\theta }\mathbf{e}_{2}+\dot{\theta }\left ( \boldsymbol{\omega }_{e_{2}}\times \mathbf{e}_{2}\right ) +\left ( \ddot{\phi }\cos \theta -\dot{\phi }\dot{\theta }\sin \theta +\ddot{\psi }\right ) \mathbf{e}_{3}+\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \left ( \overset{\rightarrow }{\omega }_{e_{3}}\times \mathbf{e}_{3}\right ) \tag{2} \end{align}
What is left is to find \(\boldsymbol{\omega }_{e_{2}}\times \mathbf{e}_{2}\) and \(\boldsymbol{\omega }_{e_{3}}\times \mathbf{e}_{3}\).\begin{align*} \boldsymbol{\omega }_{e_{2}}\times \mathbf{e}_{2} & =\left ( \dot{\theta }\mathbf{e}_{2}+\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \overset{\rightarrow }{e}_{3}\right ) \times \mathbf{e}_{2}\\ & =-\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \mathbf{e}_{1} \end{align*}
And\begin{align*} \boldsymbol{\omega }_{e_{3}}\times \mathbf{e}_{3} & =\left ( \dot{\theta }\mathbf{e}_{2}+\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \overset{\rightarrow }{e}_{3}\right ) \times \mathbf{e}_{3}\\ & =\dot{\theta }\mathbf{e}_{1} \end{align*}
Hence Eq. (2) becomes\begin{align*} \dot{\boldsymbol{\omega }}_{G} & =\ddot{\theta }\mathbf{e}_{2}+\dot{\theta }\left ( -\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \mathbf{e}_{1}\right ) +\left ( \ddot{\phi }\cos \theta -\dot{\phi }\dot{\theta }\sin \theta +\ddot{\psi }\right ) \mathbf{e}_{3}+\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \left ( \dot{\theta }\mathbf{e}_{1}\right ) \\ & =\mathbf{e}_{1}\left ( -\dot{\theta }\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) +\dot{\theta }\left ( \dot{\phi }\cos \theta +\dot{\psi }\right ) \right ) +\ddot{\theta }\mathbf{e}_{2}+\left ( \ddot{\phi }\cos \theta -\dot{\phi }\dot{\theta }\sin \theta +\ddot{\psi }\right ) \mathbf{e}_{3}\\ & =\ddot{\theta }\mathbf{e}_{2}+\left ( \ddot{\phi }\cos \theta -\dot{\phi }\dot{\theta }\sin \theta +\ddot{\psi }\right ) \mathbf{e}_{3} \end{align*}