Solution
Two rotating coordinates systems are used as shown in this diagram
The origin of CS 2 is at point \(O\) and attached to capsule itself. CS 1 origin is at top of column and attached to column.
Motion in 1 (CS 1 is the reference frame now)\[ \mathbf{V}_{p/1}=\mathbf{\dot{R}}_{o/1}+\boldsymbol{\omega }_{2/1} \times \boldsymbol{\rho }_{2p}+\boldsymbol{\dot{\rho }}_{2p,r} \] The above is the velocity of point \(P\) as seen in C.S. \(1\). The vector \(\boldsymbol{\rho }_{2p}\) goes from the origin of C.S. \(2\) to \(P\). And the \(\mathbf{\dot{R}}_{o/1}\) is the velocity of origin of C.S. \(2\) as seen in C.S. 1. and \(\boldsymbol{\dot{\rho }}_{2p,r}\) is the velocity of \(P\) relative to C.S. 2. Therefore\begin{align*} \boldsymbol{\rho }_{2p} & =-3\mathbf{k+}2\mathbf{i}\\ \boldsymbol{\dot{\rho }}_{2p,r} & =5\mathbf{k}\\ \mathbf{\dot{R}}_{o/1} & =0\\ \boldsymbol{\omega }_{2/1} & =\omega _{3}\mathbf{j+}\omega _{2}\mathbf{\mathbf{i}=}6\mathbf{j}+5\mathbf{i} \end{align*}
Therefore\begin{align*} \mathbf{V}_{p/1} & =\left ( 6\mathbf{j}+5\mathbf{i}\right ) \times \left ( -3\mathbf{k+}2\mathbf{i}\right ) +5\mathbf{k}\\ & =-18\mathbf{i+}15\mathbf{j}-7\mathbf{k} \end{align*}
\[ \mathbf{V}_{p}=\mathbf{\dot{R}}_{1}+\boldsymbol{\omega }_{1}\times \boldsymbol{\rho }_{1p}+\boldsymbol{\dot{\rho }}_{1p,r}\] The above is the absolute velocity of point \(P\). The vector \(\boldsymbol{\rho }_{1p}\) goes from the origin of C.S. \(1\) to \(P\). And the \(\mathbf{\dot{R}}_{1}\) is the absolute velocity of origin of C.S. \(1\) and \(\boldsymbol{\dot{\rho }}_{1p,r}\) is the velocity of \(P\) relative to C.S. 1 which we found above as \(\mathbf{V}_{p/1}\). The only quantity we need to find now is \(\boldsymbol{\rho }_{1p}\). At the instance shown it is simply\[ \boldsymbol{\rho }_{1p}=12\mathbf{i}-3\mathbf{k}\] But the above is only valid at this instance. Now we can find the absolute velocity\begin{align*} \boldsymbol{\dot{\rho }}_{1p,r} & =-18\mathbf{i+}15\mathbf{j}-7\mathbf{k}\\ \mathbf{\dot{R}}_{1} & =0\\ \boldsymbol{\omega }_{1} & =\omega _{1}\mathbf{k=}4\mathbf{k} \end{align*}
Therefore\begin{align*} \mathbf{V}_{p} & =4\mathbf{k}\times \left ( 12\mathbf{i}-3\mathbf{k}\right ) +\left ( -18\mathbf{i+}15\mathbf{j}-7\mathbf{k}\right ) \\ & =-18\mathbf{i+}63\mathbf{j}-7\mathbf{k} \end{align*}
Hence \(\left \vert \mathbf{V}_{p}\right \vert =\sqrt{18^{2}+63^{2}+7^{2}}=65.894\) ft/sec.
Motion in 1 (CS 1 is the reference frame now)\begin{equation} \mathbf{a}_{p/1}=\mathbf{\ddot{R}}_{o/1}+2\left ( \boldsymbol{\omega }_{2/1}\times \boldsymbol{\dot{\rho }}_{2p,r}\right ) +\left ( \boldsymbol{\dot{\omega }}_{2/1}\times \boldsymbol{\rho }_{2p}\right ) +\boldsymbol{\omega }_{2/1}\times \left ( \boldsymbol{\omega }_{2/1}\times \boldsymbol{\rho }_{2p}\right ) +\boldsymbol{\ddot{\rho }}_{2p,r} \tag{1} \end{equation} The above is the acceleration of point \(P\) as seen in C.S. \(1\). \begin{align*} \boldsymbol{\rho }_{2p} & =-3\mathbf{k+}2\mathbf{i}\\ \boldsymbol{\dot{\rho }}_{2p,r} & =5\mathbf{k}\\ \mathbf{\ddot{R}}_{o/1} & =0\\ \boldsymbol{\omega }_{2/1} & =\omega _{3}\mathbf{j+}\omega _{2}\mathbf{\mathbf{i}=}6\mathbf{j}+5\mathbf{i}\\ \boldsymbol{\dot{\omega }}_{2/1} & =\dot{\omega }_{3}\mathbf{j}+\left ( \omega _{2}\mathbf{\mathbf{i}}\times \omega _{3}\mathbf{j}\right ) +\dot{\omega }_{2}\mathbf{i}+\left ( \omega _{3}\mathbf{j}\times \omega _{2}\mathbf{\mathbf{i}}\right ) \\ & =2\mathbf{j}+\left ( 5\mathbf{\mathbf{i}}\times 6\mathbf{j}\right ) +0\mathbf{i}+\left ( 6\mathbf{j}\times 5\mathbf{\mathbf{i}}\right ) \\ & =2\mathbf{j}+30\mathbf{k-}30\mathbf{k}\\ & =2\mathbf{j} \end{align*}
To find \(\boldsymbol{\ddot{\rho }}_{2p,r}\), which is acceleration of point \(p\) relative to CS 2, we look at each angular acceleration on its own. Due to \(\omega _{2}\) , using this diagram
So the point \(p\) appears to move is the opposite direction with tangential acceleration \(\left ( -3\dot{\omega }_{2}\right ) \mathbf{j}\) and normal acceleration \(3\omega _{2}^{2}\mathbf{k}\). Now looking at effect due to \(\omega _{3}\) as seen in this diagram
So the point \(p\) appears to move is the opposite direction with tangential acceleration \(-\left ( \sqrt{13}\dot{\omega }_{3}\right ) \sin \theta \mathbf{i-}\left ( \sqrt{13}\dot{\omega }_{3}\right ) \cos \theta \mathbf{k}\) and normal acceleration \(-\left ( \sqrt{13}\omega _{3}^{2}\right ) \cos \theta \mathbf{i+}\left ( \sqrt{13}\omega _{3}^{2}\right ) \sin \theta \mathbf{k}\) Where \(\theta =\tan ^{-1}\left ( \frac{3}{2}\right ) \), hence \(\cos \theta =\frac{2}{\sqrt{13}}\) and \(\sin \theta =\frac{3}{\sqrt{13}}\)therefore\begin{align*} \boldsymbol{\ddot{\rho }}_{2p,r} & =-a_{r}\mathbf{k+}\overset{\text{due to }\omega _{2}}{\overbrace{\left ( -3\dot{\omega }_{2}\right ) \mathbf{j+}3\omega _{2}^{2}\mathbf{k}}}\\ & \mathbf{+}\overset{\text{due to }\omega _{3}}{\overbrace{-\left ( \sqrt{13}\dot{\omega }_{3}\right ) \sin \theta \mathbf{i-}\left ( \sqrt{13}\dot{\omega }_{3}\right ) \cos \theta \mathbf{k}-\left ( \sqrt{13}\omega _{3}^{2}\right ) \cos \theta \mathbf{i+}\left ( \sqrt{13}\omega _{3}^{2}\right ) \sin \theta \mathbf{k}}} \end{align*}
or (note \(\dot{\omega }_{3}\) is negative, since it is shown in diagram as moving in clockwise circular arrow)\begin{align*} \boldsymbol{\ddot{\rho }}_{2p,r} & =-32\mathbf{k+}\overset{\text{due to }\omega _{2}}{\overbrace{3\left ( 25\right ) \mathbf{k}}}-\left ( \sqrt{13}\left ( -2\right ) \right ) \frac{3}{\sqrt{13}}\mathbf{i-}\left ( \sqrt{13}\left ( -2\right ) \right ) \frac{2}{\sqrt{13}}\mathbf{k}-\left ( \sqrt{13}36\right ) \frac{2}{\sqrt{13}}\mathbf{i+}\left ( \sqrt{13}36\right ) \frac{3}{\sqrt{13}}\mathbf{k}\\ & =-32\mathbf{k+}75\mathbf{k}-6\mathbf{i-}4\mathbf{k-}72\mathbf{i+}108\mathbf{k}\\ & =-76\mathbf{i+}147\mathbf{k} \end{align*}
Therefore from Eq. (1)\begin{align*} \mathbf{a}_{p/1} & =\mathbf{\ddot{R}}_{o/1}+2\left ( \boldsymbol{\omega }_{2/1}\times \boldsymbol{\dot{\rho }}_{2p,r}\right ) +\left ( \boldsymbol{\dot{\omega }}_{2/1}\times \boldsymbol{\rho }_{2p}\right ) +\boldsymbol{\omega }_{2/1}\times \left ( \boldsymbol{\omega }_{2/1}\times \boldsymbol{\rho }_{2p}\right ) +\boldsymbol{\ddot{\rho }}_{2p,r}\\ \mathbf{a}_{p/1} & =0+2\left ( \left ( 6\mathbf{j}+5\mathbf{i}\right ) \times 5\mathbf{k}\right ) +\left ( 2\mathbf{j}\times \left ( -3\mathbf{k+}2\mathbf{i}\right ) \right ) +\left ( 6\mathbf{j}+5\mathbf{i}\right ) \times \left ( \left ( 6\mathbf{j}+5\mathbf{i}\right ) \times \left ( -3\mathbf{k+}2\mathbf{i}\right ) \right ) +\left ( -76\mathbf{i+}147\mathbf{k}\right ) \\ \mathbf{a}_{p/1} & =-94\mathbf{i+}10\mathbf{j+}326\mathbf{k} \end{align*}
\begin{equation} \mathbf{a}_{p}=\mathbf{\ddot{R}}_{1}+2\left ( \boldsymbol{\omega }_{1}\times \boldsymbol{\dot{\rho }}_{1p,r}\right ) +\left ( \boldsymbol{\dot{\omega }}_{1}\times \boldsymbol{\rho }_{1p}\right ) +\boldsymbol{\omega }_{1}\times \left ( \boldsymbol{\omega }_{1}\times \boldsymbol{\rho }_{1p}\right ) +\boldsymbol{\ddot{\rho }}_{1p,r}\tag{2} \end{equation} The above is the absolute acceleration of point \(P\). At the instance shown\begin{align*} \boldsymbol{\rho }_{1p} & =12\mathbf{i}-3\mathbf{k}\\ \boldsymbol{\dot{\rho }}_{1p,r} & =-18\mathbf{i+}15\mathbf{j}-7\mathbf{k}\\ \mathbf{\ddot{R}}_{1} & =0\\ \boldsymbol{\omega }_{1} & =\omega _{1}\mathbf{k=}4\mathbf{k}\\ \boldsymbol{\dot{\omega }}_{1} & =\dot{\omega }_{1}\mathbf{k=}3\mathbf{k} \end{align*}
and \(\boldsymbol{\ddot{\rho }}_{1p,r}\) we found above which is \(\mathbf{a}_{p/1}\), hence Eq. (2) becomes\begin{align*} \mathbf{a}_{p} & =2\left ( 4\mathbf{k}\times \left ( -18\mathbf{i+}15\mathbf{j}-7\mathbf{k}\right ) \right ) +\left ( 3\mathbf{k}\times \left ( 12\mathbf{i}-3\mathbf{k}\right ) \right ) +4\mathbf{k}\times \left ( 4\mathbf{k}\times \left ( 12\mathbf{i}-3\mathbf{k}\right ) \right ) +\left ( -94\mathbf{i+}10\mathbf{j+}326\mathbf{k}\right ) \\ & =-406\mathbf{i-}98\mathbf{j+}326\mathbf{k} \end{align*}
Therefore\begin{align*} \left \vert \mathbf{a}_{p}\right \vert & =\sqrt{406^{2}+98^{2}+326^{2}}\\ & =529.\,\allowbreak 83\text{ ft/sec}^{2} \end{align*}
Two rotating CS are used as shown in this diagram
The origin of CS 2 and CS 1 are both at the same point is at point \(O\)
Motion in first CS (first CS is the reference frame now)\[ \mathbf{V}_{Q/1}=\mathbf{\dot{R}}_{2/1}+\boldsymbol{\omega }_{2/1}\times \boldsymbol{\rho }_{2Q}+\boldsymbol{\dot{\rho }}_{2Q,r}\] The above is the velocity of point \(Q\) as seen in first C.S. The vector \(\boldsymbol{\rho }_{2p}\) goes from the origin of second C.S. to \(Q\). And the \(\mathbf{\dot{R}}_{2/1}\) is the velocity of origin of second C.S. as seen in first C.S. and \(\boldsymbol{\dot{\rho }}_{2Q,r}\) is the velocity of \(Q\) relative to second C.S. Therefore\begin{align*} \boldsymbol{\rho }_{2Q} & =6\mathbf{i+j}\\ \boldsymbol{\dot{\rho }}_{2p,r} & =\left ( 1\times \omega _{1}\right ) \mathbf{k=}10\mathbf{k}\\ \mathbf{\dot{R}}_{2/1} & =0\\ \boldsymbol{\omega }_{2/1} & =\omega _{2}\mathbf{k=k} \end{align*}
Therefore\begin{align*} \mathbf{V}_{Q/1} & =\mathbf{k}\times \left ( 6\mathbf{i+j}\right ) +10\mathbf{k}\\ & =-\mathbf{i+}6\mathbf{j}+10\mathbf{k} \end{align*}
Motion in inertial frame (ground) \[ \mathbf{V}_{Q}=\mathbf{\dot{R}}_{1}+\boldsymbol{\omega }_{first}\times \boldsymbol{\rho }_{1Q}+\boldsymbol{\dot{\rho }}_{1Q,r}\] The above is the absolute velocity of point \(Q\). The vector \(\boldsymbol{\rho }_{1Q}\) goes from the origin of first C.S.to \(Q\). And the \(\mathbf{\dot{R}}_{1}\) is the absolute velocity of origin of first C.S. and \(\boldsymbol{\dot{\rho }}_{1Q,r}\) is the velocity of \(Q\) relative to first C.S. which we found above as \(\mathbf{V}_{Q/1}\). The only quantity we need to find now is \(\boldsymbol{\rho }_{1Q}\). At the instance shown it is simply\[ \boldsymbol{\rho }_{1p}=6\mathbf{i+j}\] But the above is only valid at this instance. Now we can find the absolute velocity\begin{align*} \boldsymbol{\dot{\rho }}_{1Q,r} & =-\mathbf{i+}6\mathbf{j}+10\mathbf{k}\\ \mathbf{\dot{R}}_{1} & =0\\ \boldsymbol{\omega }_{first} & =\omega _{3}\mathbf{j=}2\mathbf{j} \end{align*}
Therefore\begin{align*} \mathbf{V}_{Q} & =2\mathbf{j}\times \left ( 6\mathbf{i+j}\right ) +\left ( -\mathbf{i+}6\mathbf{j}+10\mathbf{k}\right ) \\ & =-\mathbf{i+}6\mathbf{j}-2\mathbf{k} \end{align*}
Hence \(\left \vert \mathbf{V}_{Q}\right \vert =\sqrt{1^{2}+6^{2}+2^{2}}=\allowbreak 6.403\) ft/sec.
Motion in 1 (first CS is the reference frame now)\begin{equation} \mathbf{a}_{Q/1}=\mathbf{\ddot{R}}_{2/1}+2\left ( \boldsymbol{\omega }_{2/1}\times \boldsymbol{\dot{\rho }}_{2Q,r}\right ) +\left ( \boldsymbol{\dot{\omega }}_{2/1}\times \boldsymbol{\rho }_{2Q}\right ) +\boldsymbol{\omega }_{2/1}\times \left ( \boldsymbol{\omega }_{2/1}\times \boldsymbol{\rho }_{2Q}\right ) +\boldsymbol{\ddot{\rho }}_{2Q,r}\tag{1} \end{equation} The above is the acceleration of point \(Q\) as seen in first C.S. \begin{align*} \boldsymbol{\rho }_{2Q} & =6\mathbf{i+j}\\ \boldsymbol{\dot{\rho }}_{2Q,r} & =10\mathbf{k}\\ \mathbf{\ddot{R}}_{2/1} & =0\\ \boldsymbol{\omega }_{2/1} & =\mathbf{k}\\ \boldsymbol{\dot{\omega }}_{2/1} & =\alpha _{2}\mathbf{k}+\left ( 0\times \mathbf{k}\right ) =3\mathbf{k} \end{align*}
To find \(\boldsymbol{\ddot{\rho }}_{2Q,r}\), which is acceleration of point \(Q\) relative to second CS we look at this diagram
Hence
\[ \boldsymbol{\ddot{\rho }}_{2Q,r}=\mathbf{-}\omega _{1}^{2}\mathbf{j=}-100\mathbf{j}\] Therefore from Eq. (1)\begin{align*} \mathbf{a}_{Q/1} & =\mathbf{\ddot{R}}_{2/1}+2\left ( \boldsymbol{\omega }_{2/1}\times \boldsymbol{\dot{\rho }}_{2Q,r}\right ) +\left ( \boldsymbol{\dot{\omega }}_{2/1}\times \boldsymbol{\rho }_{2Q}\right ) +\boldsymbol{\omega }_{2/1}\times \left ( \boldsymbol{\omega }_{2/1}\times \boldsymbol{\rho }_{2Q}\right ) +\boldsymbol{\ddot{\rho }}_{2Q,r}\\ \mathbf{a}_{p/1} & =0+2\left ( \mathbf{k}\times 10\mathbf{k}\right ) +\left ( 3\mathbf{k}\times \left ( 6\mathbf{i+j}\right ) \right ) +\mathbf{k}\times \left ( \mathbf{k}\times \left ( 6\mathbf{i+j}\right ) \right ) -100\mathbf{j}\\ \mathbf{a}_{p/1} & =-9\mathbf{i-}83\mathbf{j} \end{align*}
Motion in inertial frame (ground) \begin{equation} \mathbf{a}_{Q}=\mathbf{\ddot{R}}_{1}+2\left ( \boldsymbol{\omega }_{first}\times \boldsymbol{\dot{\rho }}_{1Q,r}\right ) +\left ( \boldsymbol{\dot{\omega }}_{first}\times \boldsymbol{\rho }_{1Q}\right ) +\boldsymbol{\omega }_{first}\times \left ( \boldsymbol{\omega }_{first}\times \boldsymbol{\rho }_{1Q}\right ) +\boldsymbol{\ddot{\rho }}_{1Q,r}\tag{2} \end{equation} The above is the absolute acceleration of point \(Q\). At the instance shown \begin{align*} \boldsymbol{\rho }_{1Q} & =6\mathbf{i+j}\\ \boldsymbol{\dot{\rho }}_{1Q,r} & =\boldsymbol{\dot{\rho }}_{2Q,r}=10\mathbf{k}\\ \mathbf{\ddot{R}}_{1} & =0\\ \boldsymbol{\omega }_{first} & =\omega _{3}\mathbf{j=}2\mathbf{j}\\ \boldsymbol{\dot{\omega }}_{1} & =-\alpha _{3}\mathbf{j+}\left ( 0\times \omega _{3}\mathbf{j}\right ) =-4\mathbf{j} \end{align*} and \(\boldsymbol{\ddot{\rho }}_{1p,r}\) we found above which is \(\mathbf{a}_{p/1}\), hence Eq. (2) becomes\begin{align*} \mathbf{a}_{p} & =2\left ( 2\mathbf{j}\times 10\mathbf{k}\right ) +\left ( -4\mathbf{j}\times \left ( 6\mathbf{i+j}\right ) \right ) +2\mathbf{j}\times \left ( 2\mathbf{j}\times \left ( 6\mathbf{i+j}\right ) \right ) +\left ( -9\mathbf{i-}83\mathbf{j}\right ) \\ & =7\mathbf{i-}83\mathbf{j+}24\mathbf{k} \end{align*}
Therefore\begin{align*} \left \vert \mathbf{a}_{p}\right \vert & =\sqrt{7^{2}+83^{2}+24^{2}}\\ & =86.683\text{ ft/sec}^{2} \end{align*}
