This problem is solved in two ways, using different body coordinates system, showing that the final answer is the same.
Two coordinates systems are used. The first one has its origin at point \(O\) and rotates along with the long bar. This is the one shown above with \(xyz\) coordinates. The unit vectors for this coordinates system are \(\overset{\rightarrow }{e}_{x},\overset{\rightarrow }{e}_{y},\overset{\rightarrow }{e}_{z}\). This coordinates system is rotating relative to inertial frame with angular velocity \(\omega _{1}\overset{\rightarrow }{e}_{y}\). The second coordinate system is centered at point \(C\) and rotates with the disk \(D\) (it can be imagined to be painted on disk \(D\) to make it more clear that it moves with the disk).
The second coordinates system (the one on the disk) will use unit vectors \(\overset{\rightarrow }{i},\overset{\rightarrow }{j},\overset{\rightarrow }{k}\). It rotates with angular velocity \(\omega _{2}\overset{\rightarrow }{k}\) relative to the first one. The following diagram illustrates this relation.
\(\overset{\rightarrow }{k}\) and \(\overset{\rightarrow }{e}_{z}\) are always pointing the same direction for all time. But only at the snap shot shown in the problem diagram that \(\overset{\rightarrow }{e}_{x}=\overset{\rightarrow }{i}\) and \(\overset{\rightarrow }{e}_{y}=\overset{\rightarrow }{j}\). So this problem will be solved at the snapshot time.
Given the above, a vector that represents the position of the center of the disk \(D\) relative the the first coordinate system is shown in this diagram
It is important to see that \(\overset{\rightarrow }{R}\) is rotating and not fixed in inertial frame. It is fixed in length, but it is attached to the first coordinates system, and not to the inertial frame, hence it rotates with first coordinate system and hence will have an \(\overset{\centerdot }{\overset{\rightarrow }{R}}_{r}\) term show up in the equations below due to this.
\(\overset{\rightarrow }{\rho }\) is vector that represents the position of point \(Q\) on the disk. It goes from \(C\) to \(Q\).
From point of view of the second coordinates system, the ant (point \(Q\)) appears to move in straight line, since an observer standing on the disk is rotating with the same angular velocity as the ant as it moves away from the origin of the disk.
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}
Now \(\overset{\centerdot }{\overset{\rightarrow }{r}}=\left ( \overset{\centerdot }{\overset{\rightarrow }{r}}\right ) _{r}+\left ( \overset{\rightarrow }{\omega }\times \overset{\rightarrow }{r}\right ) \) is applied to Eq.(1) above.\begin{equation} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q}=\overset{\centerdot }{\overset{\rightarrow }{R}}_{r}+\left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) +\overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \overset{\rightarrow }{\rho }\right ) \tag{2} \end{equation}
But \(\overset{\centerdot }{\overset{\rightarrow }{R}}_{r}\) since it does not change in length. Hence \begin{equation} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q}=\left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) +\overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \overset{\rightarrow }{\rho }\right ) \tag{2A} \end{equation} and taking derivatives again gives\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}\times \overset{\rightarrow }{R}\right ) +\left ( \overset{\rightarrow }{\omega }_{1}\times \left ( \overset{\centerdot }{\overset{\rightarrow }{R}}_{r}+\left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) \right ) \right ) \\ & +\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}+\overset{\centerdot }{\overset{\rightarrow }{\omega }}_{2}\right ) \times \overset{\rightarrow }{\rho }\\ & +\left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \left ( \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \overset{\rightarrow }{\rho }\right ) \right ) \end{align*}
In the above equation, since \(\overset{\rightarrow }{R}\) does not change in length, hence all its time derivatives are zero, and the above simplifies to\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}\times \overset{\rightarrow }{R}\right ) +\left ( \overset{\rightarrow }{\omega }_{1}\times \left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) \right ) +\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}\\ & +\left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}+\overset{\centerdot }{\overset{\rightarrow }{\omega }}_{2}\right ) \times \overset{\rightarrow }{\rho }\\ & +\left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \left ( \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \overset{\rightarrow }{\rho }\right ) \right ) \end{align*}
Or
\begin{align} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}\times \overset{\rightarrow }{R}\right ) +\left ( \overset{\rightarrow }{\omega }_{1}\times \left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) \right ) +\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}\tag{3A}\\ & +\overset{\rightarrow }{\omega }_{2}\times \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }_{1}\times \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}\times \overset{\rightarrow }{\rho }+\overset{\centerdot }{\overset{\rightarrow }{\omega }}_{2}\times \overset{\rightarrow }{\rho }\nonumber \\ & +\left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \left ( \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \overset{\rightarrow }{\omega }_{2}\times \overset{\rightarrow }{\rho }+\overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{\rho }\right ) \right ) \nonumber \end{align}
Eq. (2A) and (3A) above give the answers needed. The rest is just writing down each of the above vectors in component terms. Snapshot time is used as was described above.
At the snapshot time, \[ \overset{\rightarrow }{\omega }_{1}=\omega _{1}\overset{\rightarrow }{j}\] and \[ \overset{\rightarrow }{\omega }_{2}=\omega _{2}\overset{\rightarrow }{k}\]
And\[ \overset{\rightarrow }{\rho }=\rho \overset{\rightarrow }{i}\]
The relative velocity of \(\overset{\rightarrow }{\rho }\) is given by\[ \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}=\overset{\centerdot }{\rho }\overset{\rightarrow }{i}\]
And the relative acceleration of \(\overset{\rightarrow }{\rho }\) is given by\[ \overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}=\overset{\centerdot \centerdot }{\rho }\overset{\rightarrow }{i}\]
and, at the snapshot time, \[ \overset{\rightarrow }{R}=L_{1}\overset{\rightarrow }{i}+L_{2}\overset{\rightarrow }{k}-L_{3}\overset{\rightarrow }{j}\]
All terms in Eq. (2A) are now known. Hence\begin{align} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) +\overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \overset{\rightarrow }{\rho }\right ) \nonumber \\ & =\left ( \omega _{1}\overset{\rightarrow }{j}\right ) \times \left ( L_{1}\overset{\rightarrow }{i}+L_{2}\overset{\rightarrow }{k}-L_{3}\overset{\rightarrow }{j}\right ) +\left ( \overset{\centerdot }{\rho }\overset{\rightarrow }{i}\right ) +\left ( \left ( \omega _{2}\overset{\rightarrow }{k}+\omega _{1}\overset{\rightarrow }{j}\right ) \times \rho \overset{\rightarrow }{i}\right ) \nonumber \\ & =-\overset{\rightarrow }{k}\left ( \omega _{1}L_{1}\right ) +\overset{\rightarrow }{i}\left ( \omega _{1}L_{2}\right ) +\overset{\centerdot }{\rho }\overset{\rightarrow }{i}+\overset{\rightarrow }{j}\omega _{2}\rho -\omega _{1}\rho \overset{\rightarrow }{k}\nonumber \\ & =\left ( \omega _{1}L_{2}+\overset{\centerdot }{\rho }\right ) \overset{\rightarrow }{i}+\omega _{2}\rho \overset{\rightarrow }{j}-\omega _{1}\left ( L_{1}+\rho \right ) \overset{\rightarrow }{k}\tag{4} \end{align}
At the instance shown \(\rho =\frac{0.75}{2}=\allowbreak 0.375\,\) and \(\overset{\centerdot }{\rho }\left ( t\right ) =1.5\) 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 Eq. (3) becomes\begin{align*} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \left ( 1.2\right ) \left ( 0.7\right ) +1.5\right ) \overset{\rightarrow }{i}+\left ( 0.5\right ) \left ( 0.375\right ) \overset{\rightarrow }{j}-\left ( 1.2\right ) \left ( 2.5+0.375\,\right ) \overset{\rightarrow }{k}\\ & =2.34\overset{\rightarrow }{i}+0.1875\overset{\rightarrow }{j}-3.\,\allowbreak 45\overset{\rightarrow }{k} \end{align*}
Therefore\begin{align*} \left \vert \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q}\right \vert & =\sqrt{2.34^{2}+0.1875^{2}+3.\,\allowbreak 45^{2}}\\ & =4.1729\qquad ft/\sec \end{align*}
From Eq.(3A) becomes
\begin{align} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}\times \overset{\rightarrow }{R}\right ) +\left ( \overset{\rightarrow }{\omega }_{1}\times \left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) \right ) +\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}\tag{3A}\\ & +\overset{\rightarrow }{\omega }_{2}\times \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\overset{\rightarrow }{\omega }_{1}\times \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}\times \overset{\rightarrow }{\rho }+\overset{\centerdot }{\overset{\rightarrow }{\omega }}_{2}\times \overset{\rightarrow }{\rho }\nonumber \\ & +\left ( \overset{\rightarrow }{\omega }_{2}+\overset{\rightarrow }{\omega }_{1}\right ) \times \left ( \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+\left ( \overset{\rightarrow }{\omega }_{2}\times \overset{\rightarrow }{\rho }+\overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{\rho }\right ) \right ) \nonumber \end{align}
Hence
\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\centerdot }{\omega }_{1}\overset{\rightarrow }{j}\times \left ( L_{1}\overset{\rightarrow }{i}+L_{2}\overset{\rightarrow }{k}-L_{3}\overset{\rightarrow }{j}\right ) \right ) +\left ( \omega _{1}\overset{\rightarrow }{j}\times \left ( \omega _{1}\overset{\rightarrow }{j}\times \left ( L_{1}\overset{\rightarrow }{i}+L_{2}\overset{\rightarrow }{k}-L_{3}\overset{\rightarrow }{j}\right ) \right ) \right ) \\ & +\overset{\centerdot \centerdot }{\rho }\overset{\rightarrow }{i}\\ & +\left ( \omega _{2}\overset{\rightarrow }{k}\times \overset{\centerdot }{\rho }\overset{\rightarrow }{i}\right ) \\ & +\left ( \omega _{1}\overset{\rightarrow }{j}\times \overset{\centerdot }{\rho }\overset{\rightarrow }{i}\right ) \\ & +\left ( \overset{\centerdot }{\omega }_{1}\overset{\rightarrow }{j}\times \rho \overset{\rightarrow }{i}\right ) \\ & +\left ( \overset{\centerdot }{\omega }_{2}\overset{\rightarrow }{k}\times \rho \overset{\rightarrow }{i}\right ) \\ & +\left ( \omega _{2}\overset{\rightarrow }{k}+\omega _{1}\overset{\rightarrow }{j}\right ) \times \left ( \overset{\centerdot }{\rho }\overset{\rightarrow }{i}+\left ( \omega _{2}\overset{\rightarrow }{k}\times \rho \overset{\rightarrow }{i}+\omega _{1}\overset{\rightarrow }{j}\times \rho \overset{\rightarrow }{i}\right ) \right ) \end{align*}
Hence\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =-\overset{\centerdot }{\omega }_{1}L_{1}\overset{\rightarrow }{k}+\overset{\centerdot }{\omega }_{1}L_{2}\overset{\rightarrow }{i}-\omega _{1}^{2}L_{1}\overset{\rightarrow }{i}-\omega _{1}^{2}L_{2}\overset{\rightarrow }{k}\\ & +\overset{\centerdot \centerdot }{\rho }\left ( t\right ) \overset{\rightarrow }{i}\\ & +\omega _{2}\overset{\centerdot }{\rho }\left ( t\right ) \overset{\rightarrow }{j}\\ & -\omega _{1}\overset{\centerdot }{\rho }\overset{\rightarrow }{k}\\ & +\overset{\centerdot }{\omega }_{1}\rho \overset{\rightarrow }{k}\\ & +\overset{\centerdot }{\omega }_{2}\rho \overset{\rightarrow }{j}\\ & +\left ( \omega _{2}\overset{\rightarrow }{k}+\omega _{1}\overset{\rightarrow }{j}\right ) \times \left ( \overset{\centerdot }{\rho }\overset{\rightarrow }{i}+\omega _{2}\rho \overset{\rightarrow }{j}-\omega _{1}\rho \overset{\rightarrow }{k}\right ) \end{align*}
Or
\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =-\overset{\centerdot }{\omega }_{1}L_{1}\overset{\rightarrow }{k}+\overset{\centerdot }{\omega }_{1}L_{2}\overset{\rightarrow }{i}-\omega _{1}^{2}L_{1}\overset{\rightarrow }{i}-\omega _{1}^{2}L_{2}\overset{\rightarrow }{k}\\ & +\overset{\centerdot \centerdot }{\rho }\left ( t\right ) \overset{\rightarrow }{i}\\ & +\omega _{2}\overset{\centerdot }{\rho }\left ( t\right ) \overset{\rightarrow }{j}\\ & -\omega _{1}\overset{\centerdot }{\rho }\overset{\rightarrow }{k}\\ & +\overset{\centerdot }{\omega }_{1}\rho \overset{\rightarrow }{k}\\ & +\overset{\centerdot }{\omega }_{2}\rho \overset{\rightarrow }{j}\\ & +\left ( \omega _{2}\overset{\centerdot }{\rho }\overset{\rightarrow }{j}-\omega _{2}^{2}\rho \overset{\rightarrow }{i}\right ) +\left ( -\omega _{1}\overset{\centerdot }{\rho }\overset{\rightarrow }{k}-\omega _{1}^{2}\rho \overset{\rightarrow }{i}\right ) \end{align*}
Collecting terms\begin{equation} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q}=\overset{\rightarrow }{i}\left ( \overset{\centerdot }{\omega }_{1}L_{2}-\omega _{1}^{2}\left ( L_{1}+\rho \right ) -\omega _{2}^{2}\rho +\overset{\centerdot \centerdot }{\rho }\right ) +\overset{\rightarrow }{j}\left ( 2\omega _{2}\overset{\centerdot }{\rho }+\overset{\centerdot }{\omega }_{2}\rho \right ) -\overset{\rightarrow }{k}\left ( \omega _{1}^{2}L_{2}+\overset{\centerdot }{\omega }_{1}\left ( L_{1}+\rho \right ) +2\omega _{1}\overset{\centerdot }{\rho }\right ) \tag{5} \end{equation}
At the instance shown \(\rho \left ( t\right ) =\frac{0.75}{2}=0.375\,\) and \(\overset{\centerdot }{\rho }\left ( t\right ) =1.5\) ft/sec,\(\overset{\centerdot \centerdot }{\rho }\left ( t\right ) =0.8\) 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, \(\overset{\centerdot }{\omega }_{2}=0.25\) rad/sec\(^{2}\),\(\overset{\centerdot }{\omega }_{1}=0.6\) rad/sec\(^{2}\), hence the above becomes\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\overset{\rightarrow }{i}\left ( \left ( 0.6\right ) 0.7-\left ( 1.2\right ) ^{2}\left ( 2.5+0.375\right ) -0.5^{2}\left ( 0.375\right ) +0.8\right ) \\ & +\overset{\rightarrow }{j}\left ( 2\left ( 0.5\right ) 1.5+\left ( 0.25\right ) 0.375\,\right ) \\ & -\overset{\rightarrow }{k}\left ( 1.2^{2}\left ( 0.7\right ) +0.6\left ( 2.5+0.375\right ) +2\left ( 1.2\right ) 1.5\right ) \end{align*}
Therefore\[ \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q}=-3.0138\overset{\rightarrow }{i}+1.5938\overset{\rightarrow }{j}-6.333\overset{\rightarrow }{k}\]
Therefore \begin{align*} \left \vert \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q}\right \vert & =\sqrt{3.0138^{2}+1.5938^{2}+6.333^{2}}\\ & =7.192\,4\text{ ft/sec}^{2} \end{align*}
In this case, the local body coordinates \(\overset{\rightarrow }{i},\overset{\rightarrow }{j},\overset{\rightarrow }{k}\). \(\ \)is attached to the bar labeled \(L_{3}\) and hence does not rotate with the disk, as shown in this diagram
The main difference between this set up and the first case, is that now \(\overset{\rightarrow }{k}\) and \(\overset{\rightarrow }{e}_{z}\) are still pointing the same direction for all time but now also \(\overset{\rightarrow }{e}_{x}\)and\(\overset{\rightarrow }{i}\) are always pointing in same direction, as well as \(\overset{\rightarrow }{e}_{y}\)and \(\overset{\rightarrow }{j}\). And now body frame \(\overset{\rightarrow }{i},\overset{\rightarrow }{j},\overset{\rightarrow }{k}\) does not rotate relative to frame \(\overset{\rightarrow }{e}_{x},\overset{\rightarrow }{e}_{y},\overset{\rightarrow }{e}_{z}\). The two frames are actually fixed to each others, and only difference is that the origin of one is displaced from the other by the vector \(\overset{\rightarrow }{R}\).
Using the same equations (2A) and (3A), the only difference is in writing down the components of the vectors.\begin{align} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) +\overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}+ \overbrace{\left ( \overset{\rightarrow }{\omega }_{2}\times \overset{\rightarrow }{\rho }\right ) }^{\text{This term is zero now}} \tag{2A}\\ \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}\times \overset{\rightarrow }{R}\right ) +\left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) +\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r}+\overbrace{2\left ( \overset{\rightarrow }{\omega }_{2}\times \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}\right ) +\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{2}\times \overset{\rightarrow }{\rho }\right ) +\overset{\rightarrow }{\omega }_{2}\times \overset{\rightarrow }{\omega }_{2}\times \overset{\rightarrow }{\rho }}^{\text{This term is zero in this case}} \tag{3A} \end{align}
Since now frame \(\overset{\rightarrow }{i},\overset{\rightarrow }{j},\overset{\rightarrow }{k}\) does not rotate relative to the frame \(\overset{\rightarrow }{e}_{x},\overset{\rightarrow }{e}_{y},\overset{\rightarrow }{e}_{z}\), then the above simplifies to\begin{align} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) +\overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}\tag{2AA}\\ \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\centerdot }{\overset{\rightarrow }{\omega }}_{1}\times \overset{\rightarrow }{R}\right ) +\left ( \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{\omega }_{1}\times \overset{\rightarrow }{R}\right ) +\overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r} \tag{3AA} \end{align}
Now the vector \(\overset{\rightarrow }{\rho }\) is\[ \overset{\rightarrow }{\rho }=\left ( \rho \cos \theta \right ) \overset{\rightarrow }{i}+\left ( \rho \sin \theta \right ) \overset{\rightarrow }{j}+0\overset{\rightarrow }{k}\]
The relative velocity of \(\overset{\rightarrow }{\rho }\) is now given by\[ \overset{\centerdot }{\overset{\rightarrow }{\rho }}_{r}=\left ( \overset{\centerdot }{\rho }\cos \theta -\rho \overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{i}+\left ( \overset{\centerdot }{\rho }\sin \theta +\rho \overset{\centerdot }{\theta }\cos \theta \right ) \overset{\rightarrow }{j}+0\overset{\rightarrow }{k}\]
And the relative acceleration of \(\overset{\rightarrow }{\rho }\) is given by\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{\rho }}_{r} & =\left ( \left ( \overset{\centerdot \centerdot }{\rho }\cos \theta -\overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\sin \theta \right ) -\left ( \overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\sin \theta +\rho \overset{\centerdot \centerdot }{\theta }\sin \theta +\rho \overset{\centerdot }{\theta }^{2}\cos \theta \right ) \right ) \overset{\rightarrow }{i}\\ & +\left ( \left ( \overset{\centerdot \centerdot }{\rho }\sin \theta +\overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\cos \theta \right ) +\left ( \overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\cos \theta +\rho \overset{\centerdot \centerdot }{\theta }\cos \theta -\rho \overset{\centerdot }{\theta }^{2}\sin \theta \right ) \right ) \overset{\rightarrow }{j}\\ & +0\overset{\rightarrow }{k} \end{align*}
All remaining vectors are the same as the first case. In particular\[ \overset{\rightarrow }{R}=L_{1}\overset{\rightarrow }{i}+L_{2}\overset{\rightarrow }{k}-L_{3}\overset{\rightarrow }{j}\]
However, this vector is now valid for all time, and not only at the snapshot. Hence Eq. (2AA) now can be written down as\begin{align*} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \omega _{1}\overset{\rightarrow }{j}\right ) \times \left ( L_{1}\overset{\rightarrow }{i}+L_{2}\overset{\rightarrow }{k}-L_{3}\overset{\rightarrow }{j}\right ) +\left ( \overset{\centerdot }{\rho }\cos \theta -\rho \overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{i}+\left ( \overset{\centerdot }{\rho }\sin \theta +\rho \overset{\centerdot }{\theta }\cos \theta \right ) \overset{\rightarrow }{j}\\ & =-\overset{\rightarrow }{k}\left ( \omega _{1}L_{1}\right ) +\overset{\rightarrow }{i}\left ( \omega _{1}L_{2}\right ) +\left ( \overset{\centerdot }{\rho }\cos \theta -\rho \overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{i}+\left ( \overset{\centerdot }{\rho }\sin \theta +\rho \overset{\centerdot }{\theta }\cos \theta \right ) \overset{\rightarrow }{j}\\ & =\left ( \omega _{1}L_{2}+\overset{\centerdot }{\rho }\cos \theta -\rho \overset{\centerdot }{\theta }\sin \theta \right ) \overset{\rightarrow }{i}+\left ( \overset{\centerdot }{\rho }\sin \theta +\rho \overset{\centerdot }{\theta }\cos \theta \right ) \overset{\rightarrow }{j}-\omega _{1}L_{1}\overset{\rightarrow }{k} \end{align*}
Since \(\overset{\centerdot }{\theta }=\omega _{2}\) then\begin{equation} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q}=\left ( \omega _{1}L_{2}+\overset{\centerdot }{\rho }\cos \theta -\rho \omega _{2}\sin \theta \right ) \overset{\rightarrow }{i}+\left ( \overset{\centerdot }{\rho }\sin \theta +\rho \omega _{2}\cos \theta \right ) \overset{\rightarrow }{j}-\omega _{1}L_{1}\overset{\rightarrow }{k} \tag{6} \end{equation}
Now, at the snapshot time, \(\theta =0^{0}\), hence the above simplifies to\begin{equation} \overset{\centerdot }{\overset{\rightarrow }{r}}_{Q}=\left ( \omega _{1}L_{2}+\overset{\centerdot }{\rho }\right ) \overset{\rightarrow }{i}+\omega _{2}\rho \overset{\rightarrow }{j}-\omega _{1}L_{1}\overset{\rightarrow }{k} \tag{6A} \end{equation}
Comparing the above Eq. (6A) to Eq. (4) found in the first case, it is seen to be the same, as expected. The difference is that Eq. (6) is valid for all time, while Eq. (4) is valid at the snapshot only. Now the acceleration will be found from Eq. (3AA)\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\left ( \overset{\centerdot }{\omega }_{1}\overset{\rightarrow }{j}\times \left ( L_{1}\overset{\rightarrow }{i}+L_{2}\overset{\rightarrow }{k}-L_{3}\overset{\rightarrow }{j}\right ) \right ) +\left ( \omega _{1}\overset{\rightarrow }{j}\times \left ( \omega _{1}\overset{\rightarrow }{j}\times \left ( L_{1}\overset{\rightarrow }{i}+L_{2}\overset{\rightarrow }{k}-L_{3}\overset{\rightarrow }{j}\right ) \right ) \right ) \\ & +\left ( \left ( \overset{\centerdot \centerdot }{\rho }\cos \theta -\overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\sin \theta \right ) -\left ( \overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\sin \theta +\rho \overset{\centerdot \centerdot }{\theta }\sin \theta +\rho \overset{\centerdot }{\theta }^{2}\cos \theta \right ) \right ) \overset{\rightarrow }{i}\\ & +\left ( \left ( \overset{\centerdot \centerdot }{\rho }\sin \theta +\overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\cos \theta \right ) +\left ( \overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\cos \theta +\rho \overset{\centerdot \centerdot }{\theta }\cos \theta -\rho \overset{\centerdot }{\theta }^{2}\sin \theta \right ) \right ) \overset{\rightarrow }{j} \end{align*}
Hence\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =-\overset{\centerdot }{\omega }_{1}L_{1}\overset{\rightarrow }{k}+\overset{\centerdot }{\omega }_{1}L_{2}\overset{\rightarrow }{i}-\omega _{1}\omega _{1}L_{1}\overset{\rightarrow }{i}-\omega _{1}\omega _{1}L_{2}\overset{\rightarrow }{k}\\ & +\left ( \left ( \overset{\centerdot \centerdot }{\rho }\cos \theta -\overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\sin \theta \right ) -\left ( \overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\sin \theta +\rho \overset{\centerdot \centerdot }{\theta }\sin \theta +\rho \overset{\centerdot }{\theta }^{2}\cos \theta \right ) \right ) \overset{\rightarrow }{i}\\ & +\left ( \left ( \overset{\centerdot \centerdot }{\rho }\sin \theta +\overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\cos \theta \right ) +\left ( \overset{\centerdot }{\rho }\overset{\centerdot }{\theta }\cos \theta +\rho \overset{\centerdot \centerdot }{\theta }\cos \theta -\rho \overset{\centerdot }{\theta }^{2}\sin \theta \right ) \right ) \overset{\rightarrow }{j} \end{align*}
But \(\overset{\centerdot }{\theta }=\omega _{2}\) and \(\overset{\centerdot }{\theta }^{2}=\omega _{2}^{2}\) and \(\overset{\centerdot \centerdot }{\theta }=\overset{\centerdot }{\omega }_{2}\)hence the above becomes\begin{align*} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =-\overset{\centerdot }{\omega }_{1}L_{1}\overset{\rightarrow }{k}+\overset{\centerdot }{\omega }_{1}L_{2}\overset{\rightarrow }{i}-\omega _{1}\omega _{1}L_{1}\overset{\rightarrow }{i}-\omega _{1}\omega _{1}L_{2}\overset{\rightarrow }{k}\\ & +\left ( \left ( \overset{\centerdot \centerdot }{\rho }\cos \theta -\overset{\centerdot }{\rho }\omega _{2}\sin \theta \right ) -\left ( \overset{\centerdot }{\rho }\omega _{2}\sin \theta +\rho \overset{\centerdot }{\omega }_{2}\sin \theta +\rho \omega _{2}^{2}\cos \theta \right ) \right ) \overset{\rightarrow }{i}\\ & +\left ( \left ( \overset{\centerdot \centerdot }{\rho }\sin \theta +\overset{\centerdot }{\rho }\omega _{2}\cos \theta \right ) +\left ( \overset{\centerdot }{\rho }\omega _{2}\cos \theta +\rho \overset{\centerdot }{\omega }_{2}\cos \theta -\rho \omega _{2}^{2}\sin \theta \right ) \right ) \overset{\rightarrow }{j} \end{align*}
Collecting terms\begin{align} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q} & =\overset{\rightarrow }{i}\left ( \overset{\centerdot }{\omega }_{1}L_{2}-\omega _{1}^{2}L_{1}+\overset{\centerdot \centerdot }{\rho }\cos \theta -2\overset{\centerdot }{\rho }\omega _{2}\sin \theta -\rho \overset{\centerdot }{\omega }_{2}\sin \theta -\rho \omega _{2}^{2}\cos \theta \right ) \nonumber \\ & +\overset{\rightarrow }{j}\left ( \overset{\centerdot \centerdot }{\rho }\sin \theta +2\overset{\centerdot }{\rho }\omega _{2}\cos \theta +\rho \overset{\centerdot }{\omega }_{2}\cos \theta -\rho \omega _{2}^{2}\sin \theta \right ) \nonumber \\ & +\overset{\rightarrow }{k}\left ( -\overset{\centerdot }{\omega }_{1}L_{1}-\omega _{1}^{2}L_{2}\right ) \tag{7} \end{align}
Now, at snapshot, where \(\theta =0^{0}\), the above simplifies to\begin{equation} \overset{\centerdot \centerdot }{\overset{\rightarrow }{r}}_{Q}=\overset{\rightarrow }{i}\left ( \overset{\centerdot }{\omega }_{1}L_{2}-\omega _{1}^{2}L_{1}+\overset{\centerdot \centerdot }{\rho }-\rho \omega _{2}^{2}\right ) +\overset{\rightarrow }{j}\left ( 2\overset{\centerdot }{\rho }\omega _{2}+\rho \overset{\centerdot }{\omega }_{2}\right ) +\overset{\rightarrow }{k}\left ( -\overset{\centerdot }{\omega }_{1}L_{1}-\omega _{1}^{2}L_{2}\right ) \tag{7A} \end{equation}
Comparing the Eq. (7A) above to Eq. (5) found in the first case, it is seen they are the same. The difference is that Eq. (7) now can be used for all time, while Eq. (5) was valid only at the snapshot.