In spherical coordinates the position vector and velocity vector of a car is given by\begin{align*} \vec{r} & =R\vec{e}_{r}+\theta \vec{e}_{\theta }+\phi \vec{e}_{\phi }\\ \vec{v} & =\dot{R}\vec{e}_{r}+R\dot{\phi }\vec{e}_{\phi }+R\dot{\theta }\sin \phi \vec{e}_{\theta } \end{align*}
Since \(R\) is constant, then \(\dot{R}=0\). It is also given that \(\dot{\theta }=\omega .\) The above becomes\begin{equation} \vec{v}=R\dot{\phi }\vec{e}_{\phi }+R\omega \sin \phi \vec{e}_{\theta } \tag{1} \end{equation}
Given that\begin{align} R\sin \left ( \frac{\pi }{2}-\phi \right ) & =\frac{h}{2}\left ( 1-2\cos 2\theta \right ) \nonumber \\ \cos \left ( \phi \right ) & =\frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \tag{2} \end{align}
And taking derivative w.r.t. \begin{align} -\dot{\phi }\sin \phi & = \frac{2h}{R}\dot{\theta }\sin \left ( 2\theta \right ) \nonumber \\ \dot{\phi } & =\frac{-2h\dot{\theta }\sin \left ( 2\theta \right ) }{R\sin \phi } \tag{3} \end{align}
Substituting the above in Eq. (1) gives\[ \vec{v}=\frac{-2h\dot{\theta }\sin \left ( 2\theta \right ) }{\sin \phi }\vec{e}_{\phi }+R\omega \sin \phi \vec{e}_{\theta }\]
But from Eq. (2) \[ \phi =\arccos \left ( \frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \right ) \] hence \(\vec{v}\) becomes\[ \vec{v}=\frac{-2h\dot{\theta }\sin \left ( 2\theta \right ) }{\sin \left ( \arccos \left ( \frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \right ) \right ) }\vec{e}_{\phi }+R\omega \sin \left ( \arccos \left ( \frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \right ) \right ) \vec{e}_{\theta }\]
When \(\theta =\frac{\pi }{4},\sin \left ( 2\theta \right ) =1\) and \(\cos 2\theta =0\) the above becomes \[ \vec{v} = -\frac{2h\omega }{\sin \left ( \arccos \left ( \frac{h}{2R}\right ) \right ) }\vec{e}_{\phi }+R\omega \sin \left ( \arccos \left ( \frac{h}{2R}\right ) \right ) \vec{e}_{\theta }\] But \[ \sin \left ( \arccos \left ( x\right ) \right ) =\sqrt{1-x^{2}}\] hence the above becomes\[ \vec{v} = -\frac{2h\omega }{\sqrt{1-\frac{h^{2}}{4R^{2}}}}\vec{e}_{\phi }+R\omega \sqrt{1-\frac{h^{2}}{4R^{2}}}\vec{e}_{\theta }\] Therefore, the \(\vec{e}_{\phi }\) component is \(\frac{-2h\omega }{\sqrt{1-\frac{h^{2}}{4R^{2}}}}\) and the \(\vec{e}_{\theta }\) is \(R\omega \sqrt{1-\frac{h^{2}}{4R^{2}}}\) and the \(\vec{e}_{r}\) component is zero.
The \(\theta \) component of the acceleration is given from eq. (1.30) in the class handout book as\[ R\ddot{\theta }\sin \phi +2\dot{R}\dot{\theta }\sin \phi +2R\dot{\phi }\dot{\theta }\cos \phi \] Since \(\dot{R}=0\) and \(\ddot{\theta }=0\) (angular velocity is constant and the length of the swing arm is also constant) the above expression reduces to\[ 2R\dot{\phi }\omega \cos \phi \] From Eq. (3) in part(a), using \(\dot{\phi }=\frac{-2h\omega \sin \left ( 2\theta \right ) }{R\sin \phi }\) and \(\phi =\arccos \left ( \frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \right ) ,\) the above simplifies to\begin{align*} & 2R\left ( \frac{-2h\omega \sin \left ( 2\theta \right ) }{R\sin \left ( \arccos \left ( \frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \right ) \right ) }\right ) \omega \cos \left ( \arccos \left ( \frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \right ) \right ) \\ & 2R\left ( \frac{-2h\omega \sin \left ( 2\theta \right ) }{R\sin \left ( \arccos \left ( \frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \right ) \right ) }\right ) \omega \left ( \frac{h}{2R}\left ( 1-2\cos 2\theta \right ) \right ) \end{align*}
When \(\theta =\frac{\pi }{4},\sin \left ( 2\theta \right ) =1\) and \(\cos 2\theta =0,\) the \(\theta \) component of the acceleration becomes\begin{align*} & \left ( \frac{-4h\omega }{\sin \left ( \arccos \left ( \frac{h}{2R}\right ) \right ) }\right ) \frac{\omega h}{2R}\\ & \left ( \frac{-4h\omega }{\sqrt{1-\left ( \frac{h}{2R}\right ) ^{2}}}\right ) \frac{\omega h}{2R}\\ & \frac{-2}{R}\frac{h^{2}\omega ^{2}}{\sqrt{1-\frac{h^{2}}{4R^{2}}}} \end{align*}
Let \(L\) be the side length of the cone (\(2\) ft. in current diagram) and \(\vec{\omega }_{c}\) the angular velocity vector of the cone around its own axes, and \(r\) the cone base radius. Let \(\vec{\omega }_{total}\) be the angular velocity of cone w.r.t. the rigid frame \(XYZ\) (inertial frame), Hence vector additions gives\begin{equation} \vec{\omega }_{total}=\vec{\omega }_{c}+\vec{\omega }_{z} \tag{1} \end{equation}
No slipping implies \[ L\omega _{z}=r\omega _{c}\] Or\[ \omega _{c}=\frac{L}{r}\omega _{z}\] Hence Eq. (1) becomes\[ \vec{\omega }_{total}=\left ( 1+\frac{L}{r}\right ) \omega _{z}\vec{k}\]
Since \(\frac{r}{L}=\tan 20^{0},\) then \(r=L\tan 20^{0}\) and the above simplifies to\begin{align*} \vec{\omega }_{total} & =\left ( 1+\frac{1}{\tan 20^{0}}\right ) \left ( 4\right ) \vec{k}\\ & =14.989\vec{k} \end{align*}
The total angular acceleration of the cone is\begin{align*} \overrightarrow{\dot{\omega }}_{total} & =\frac{d}{dt}\left ( \left ( 1+\frac{L}{r}\right ) \omega _{z}\vec{k}\right ) \\ & =\left ( 1+\frac{1}{\tan 20^{0}}\right ) \dot{\omega }_{z}\vec{k} \end{align*}
But \(\dot{\omega }_{z}=3\) rad/sec\(^{2}\) hence\[ \overrightarrow{\dot{\omega }}_{total}=11.241\vec{k}\]
The position vector \(\vec{r}\) can be written as\[ \vec{r}=R\cos \phi \vec{i}+R\sin \phi \vec{j}+z\vec{k}\]
Taking derivatives w.r.t in the inertial frame, and since the unit vectors \(\vec{i},\vec{j},\vec{k}\) do not change in this frame, the following result is obtained\[ \vec{v}=\dot{R}\cos \phi \vec{i}-R\dot{\phi }\sin \phi \vec{i}+\dot{R}\sin \phi \vec{j}+R\dot{\phi }\cos \phi \vec{j}+\dot{z}\vec{k}\]
Since \(R\) do not change with time, the above simplifies to\[ \vec{v}=-R\dot{\phi }\sin \phi \vec{i}+R\dot{\phi }\cos \phi \vec{j}+\dot{z}\vec{k}\]
and the acceleration vector is\begin{align*} \vec{a} & =-\dot{R}\dot{\phi }\sin \phi \vec{i}-R\ddot{\phi }\sin \phi \vec{i}-R\dot{\phi }\dot{\phi }\cos \phi \vec{i}\\ & +\dot{R}\dot{\phi }\cos \phi \vec{j}+R\ddot{\phi }\cos \phi \vec{j}-R\dot{\phi }\dot{\phi }\sin \phi \vec{j}\\ & +\ddot{z}\vec{k} \end{align*}
Since \(R\) do not change with time, the above simplifies to\begin{align*} \vec{a} & =-R\ddot{\phi }\sin \phi \vec{i}-R\dot{\phi }\dot{\phi }\cos \phi \vec{i}\\ & +R\ddot{\phi }\cos \phi \vec{j}-R\dot{\phi }\dot{\phi }\sin \phi \vec{j}\\ & +\ddot{z}\vec{k} \end{align*}
Since \(\phi =2t\), then \(\dot{\phi }=2,\ddot{\phi }=0\) and \(z=t^{2},\) \(\dot{z}=2t,\ddot{z}=2.\) Substituting these values in the above two expressions for velocity and acceleration gives\begin{align*} \vec{v} & =-3\sin \left ( 2t\right ) \vec{i}+3\cos \left ( 2t\right ) \vec{j}+2t\vec{k}\\ \vec{a} & =-\left ( 1.5\right ) 4\cos \left ( 2t\right ) \vec{i}-\left ( 1.5\right ) 4\sin \left ( 2t\right ) \vec{j}+2\vec{k}\\ & =-6\cos \left ( 2t\right ) \vec{i}-6\sin \left ( 2t\right ) \vec{j}+2\vec{k} \end{align*}
At \(t=0.25\) second, \begin{align*} \vec{v} & =-3\sin \left ( 0.5\right ) \vec{i}+3\cos \left ( 0.5\right ) \vec{j}+0.5\vec{k}\\ & =-1.438\,\vec{i}+2.633\vec{j}+0.5\allowbreak k\qquad \lbrack ft/s]\\ \vec{a} & =-6\cos \left ( 0.5\right ) \vec{i}-6\sin \left ( 0.5\right ) \vec{j}+2\vec{k}\\ & =-5.266\vec{i}-2.877\allowbreak \vec{j}+2\vec{k}\qquad \lbrack ft/s^{2}] \end{align*}
\begin{align*} \vec{e}_{t} & =\frac{\vec{v}}{\left \vert \vec{v}\right \vert }\\ & =\frac{-1.438\,\vec{i}+2.633\vec{j}+0.5\allowbreak k}{\sqrt{1.438^{2}+2.633^{2}+0.5^{2}}}\\ & =\frac{-1.438\,\allowbreak \vec{i}+2.633\vec{j}+0.5\allowbreak k}{3.0414}\\ & =-0.473\allowbreak \vec{i}+0.866\,\vec{j}+0.164\vec{k} \end{align*}
and\[ \vec{a}_{t}=\left ( \vec{a}\cdot \vec{e}_{t}\right ) \vec{e}_{t}\]
But\begin{align*} \left ( \vec{a}\cdot \vec{e}_{t}\right ) & =\left ( -5.266\vec{i}-2.877\allowbreak \vec{j}+2\vec{k}\right ) \cdot \left ( -0.473\allowbreak \vec{i}+0.866\,\vec{j}+0.164\vec{k}\right ) \\ & =0.329\, \end{align*}
Hence\begin{align*} \vec{a}_{t} & =0.329\,\left ( -0.473\allowbreak \vec{i}+0.866\,\vec{j}+0.164\vec{k}\right ) \\ & =-0.155\vec{i}+0.285\vec{j}+0.054\vec{k}\qquad \lbrack ft/s^{2}] \end{align*}
Since \(\vec{a}=\vec{a}_{t}+\vec{a}_{n}\) then \begin{align*} \vec{a}_{n} & =\vec{a}-\vec{a}_{t}\\ & =\left ( -5.266\vec{i}-2.877\allowbreak \vec{j}+2\vec{k}\right ) -\left ( -0.155\vec{i}+0.284\vec{j}+0.054\vec{k}\right ) \\ & =-5.110\vec{i}-3.161\vec{j}+1.946\vec{k}\qquad \left [ ft/s^{2}\right ] \end{align*}
Hence \begin{align*} \vec{e}_{n} & =\frac{\vec{a}_{n}}{\left \vert \vec{a}_{n}\right \vert }=\frac{-5.110\vec{i}-3.161\vec{j}+1.946\vec{k}}{\sqrt{5.110^{2}+3.161^{2}\,+1.946^{2}}}\\ & =-0.809\vec{i}-0.5\,\vec{j}+0.308\,\vec{k} \end{align*}
Hence\begin{align*} \vec{e}_{b} & =\vec{e}_{t}\times \vec{e}_{n}\\ & =\left ( -0.473\allowbreak \vec{i}+0.866\,\vec{j}+0.164\vec{k}\right ) \times \left ( -0.809\vec{i}-0.5\,\vec{j}+0.308\,\vec{k}\right ) \\ & =\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ -0.473 & 0.866 & 0.164\\ -0.809 & -0.5 & 0.308 \end{vmatrix} \\ & =\vec{i}\left ( 0.866\times 0.308+0.164\times 0.5\right ) -\vec{j}\left ( -0.473\times 0.308+0.164\times 0.809\right ) +\vec{k}\left ( 0.473\times 0.5+0.866\times 0.809\right ) \\ & =0.349\vec{i}-0.0127\,\vec{j}+0.937\,\vec{k} \end{align*}
The speed and acceleration was found above as\begin{align*} \vec{v} & =-1.438\,\vec{i}+2.633\vec{j}+0.5\allowbreak \vec{k}\qquad \left [ ft/s\right ] \\ \vec{a} & =-5.266\vec{i}-2.877\allowbreak \vec{j}+2\vec{k}\qquad \left [ ft/s^{2}\right ] \end{align*}
since\begin{align*} \left \vert \vec{a}_{n}\right \vert & =\frac{\dot{s}^{2}}{\rho }\\ & =\frac{\left \vert \vec{v}\right \vert ^{2}}{\rho } \end{align*}
Hence \[ \rho =\frac{\left \vert \vec{v}\right \vert ^{2}}{\left \vert \vec{a}_{n}\right \vert }=\frac{\left \vert -1.438\,\vec{i}+2.633\vec{j}+0.5\allowbreak k\right \vert ^{2}}{\left \vert -5.110\vec{i}-3.161\vec{j}+1.946\vec{k}\right \vert }=\frac{1.438\,^{2}+2.633^{2}+0.5^{2}}{\sqrt{5.110^{2}+3.161^{2}+1.946^{2}}}=1.465\qquad \left [ ft\right ] \]
But \[ k=\frac{1}{\rho }=\frac{1}{1.465}=0.683 \]
\begin{align*} \dot{\theta } & =\frac{\dot{s}}{\rho }\\ & =\frac{\left \vert \vec{v}\right \vert }{\rho } \end{align*}
At \(t=0.25\) sec.\begin{align*} \dot{\theta } & =\frac{\left \vert -1.438\,\vec{i}+2.633\vec{j}+0.5\vec{k}\right \vert }{1.465}\\ & =\frac{\sqrt{1.438\,^{2}+2.633^{2}+0.5^{2}}}{1.465}\\ & =2.076\qquad \left [ rad/\sec \right ] \, \end{align*}
\begin{align*} \vec{v}\times \vec{a} & =\vec{v}\times \left ( \vec{a}_{t}+\vec{a}_{n}\right ) \\ & =\left ( \vec{v}\times \vec{a}_{t}\right ) +\left ( \vec{v}\times \vec{a}_{n}\right ) \\ & =\left ( \left \vert \vec{v}\right \vert \vec{e}_{t}\times \left \vert \vec{a}_{t}\right \vert \vec{e}_{t}\right ) +\left ( \left \vert \vec{v}\right \vert \vec{e}_{t}\times \left \vert \vec{a}_{n}\right \vert \vec{e}_{n}\right ) \\ & =\left \vert \vec{v}\right \vert \left \vert \vec{a}_{t}\right \vert \left ( \vec{e}_{t}\times \vec{e}_{t}\right ) +\left \vert \vec{v}\right \vert \left \vert \vec{a}_{n}\right \vert \left ( \vec{e}_{t}\times \vec{e}_{n}\right ) \end{align*}
But \(\vec{e}_{t}\times \vec{e}_{t}=0\) and \(\vec{e}_{t}\times \vec{e}_{n}\) using the right-hand rule is \(\vec{e}_{b}\) hence\[ \vec{v}\times \vec{a}=\left \vert \vec{v}\right \vert \left \vert \vec{a}_{n}\right \vert \vec{e}_{b}\]
This is a vector parallel to \(\vec{e}_{b}\) of magnitude \(\left \vert \vec{v}\right \vert \left \vert \vec{a}_{n}\right \vert \)
There is local frame of reference attached to the inner gimbal as shown in the following diagram
Given these, the angular velocity vector of the fly wheel can be written as (in terms of local coordinates system)
\begin{equation} \vec{\omega }_{wheel} = \omega _{1} \vec{e}_{x} - \dot{\theta } \vec{e}_{y}+\omega _{2}\vec{e}_{z} \tag{1} \end{equation}
Hence, taking derivatives
\[{\dot{\vec{\omega }}_{wheel}}=\dot{\omega }_{1}\vec{e}_{x}+{\omega }_{1}\dot{\vec{e}}_{x}-\ddot{\theta }\vec{e}_{y}-\dot{\theta }\dot{\vec{e}}_{y}+\dot{\omega }_{2}\vec{e}_{z}+\omega _{2}\dot{\vec{e}}_{z}\]
But \(\dot{\omega }_{2}=0\,\) then
\begin{equation}{\dot{\vec{\omega }}_{wheel}}=\dot{\omega }_{1}\vec{e}_{x}+{\omega }_{1}\dot{\vec{e}}_{x}-\ddot{\theta }\vec{e}_{y}-\dot{\theta }\dot{\vec{e}}_{y}+\omega _{2}\dot{\vec{e}}_{z}\tag{2} \end{equation}
But\begin{align*} \dot{\vec{e}}_{x} & =\omega _{e_{x}}\times \vec{e}_{x}\\ & =\left ( -\dot{\theta }\vec{j}+\omega _{1}\vec{i}\right ) \times \vec{e}_{x}\\ & =\left ( -\dot{\theta }\vec{j}\times \vec{e}_{x}\right ) +\left ( \omega _{1}\vec{i}\times \vec{e}_{x}\right ) \\ & =\dot{\theta }\vec{e}_{z}-\sin \theta \vec{e}_{y} \end{align*}
and
\begin{align*} \dot{\vec{e}}_{y} & =\omega _{e_{y}}\times \vec{e}_{y}\\ & =\omega _{1}\vec{i}\times \vec{e}_{y}\\ & =\omega _{1}\vec{e}_{z} \end{align*}
and
\begin{align*} \dot{\vec{e}}_{z} & =\omega _{e_{z}}\times \vec{e}_{z}\\ & =\left ( -\dot{\theta }\vec{j}+\omega _{1}\vec{i}\right ) \times \vec{e}_{y}\\ & =\left ( -\dot{\theta }\vec{j}\times \vec{e}_{y}\right ) +\left ( \omega _{1}\vec{i}\times \vec{e}_{y}\right ) \\ & =-\dot{\theta }\vec{e}_{x}\sin \omega _{1}t+\omega _{1}\vec{e}_{z} \end{align*}
Assuming \(t=0\) is when the instance taken, the above becomes (we are not given time)
\[ \dot{\vec{e}}_{z}=\omega _{1}\vec{e}_{z}\]
Hence Eq. (2) becomes
\begin{align}{\dot{\vec{\omega }}_{wheel}} & =\dot{\omega }_{1}\vec{e}_{x}+{\omega }_{1}\dot{\vec{e}}_{x}-\ddot{\theta }\vec{e}_{y}-\dot{\theta }\dot{\vec{e}}_{y}+\omega _{2}\dot{\vec{e}}_{z}\nonumber \\ & =\dot{\omega }_{1}\vec{e}_{x}+{\omega }_{1}\left ( \dot{\theta }\vec{e}_{z}-\sin \theta \vec{e}_{y}\right ) -\ddot{\theta }\vec{e}_{y}-\dot{\theta }\left ( \omega _{1}\vec{e}_{z}\right ) +\omega _{2}\left ( \omega _{1}\vec{e}_{z}\right ) \nonumber \\ & =\dot{\omega }_{1}\vec{e}_{x}-\vec{e}_{y}\left ( \ddot{\theta }+{\omega }_{1}\sin \theta \right ) +\omega _{2}\omega _{1}\vec{e}_{z}\tag{3} \end{align}
Since \(\omega _{2}=6000\) rev/min or \(\frac{6000\left ( 2\pi \right ) }{60}=200\pi \) rad/sec, \(\omega _{1}=10\ \)rad/\(\sec ,\dot{\omega }_{1}=100\ \)rad/sec\(^{2}\),\(\dot{\theta }=6\) rad/sec,\(\ddot{\theta }=-90\) rad/sec\(^{2}\), \(\theta =120^{0},\)then Eq. (1) becomes
\begin{align*} \vec{\omega }_{wheel} & =10\vec{e}_{x}-6\vec{e}_{y}+200\pi \vec{e}_{z}\\ \left \vert \vec{\omega }_{wheel}\right \vert & =\sqrt{10^{2}+6^{2}+\left ( 200\pi \right ) ^{2}}\\ & =628.\,\allowbreak 43\text{ rad/sec} \end{align*}
and Eq. (3) becomes
\begin{align*}{\dot{\vec{\omega }}_{wheel}} &{=}100\vec{e}_{x}-\vec{e}_{y}\left ( -90+{10}\sin 120^{0}\right ) +\vec{e}_{z}\left ( 2000\pi \right ) \\ & =100\vec{e}_{x}-\vec{e}_{y}\left ( -90+{10}\frac{\sqrt{3}}{2}\right ) +\vec{e}_{z}\left ( 2000\pi \right ) \\ & =100\vec{e}_{x}+81.34\vec{e}_{y}+6283.2\vec{e}_{z} \end{align*}
Hence
\begin{align*} \left \vert{\dot{\vec{\omega }}_{wheel}}\right \vert & =\sqrt{100^{2}+81.34^{2}+6283.2^{2}}\\ & =6284.\,\allowbreak 5\text{ rad/sec}^{2} \end{align*}