Solution
A single rotating coordinates system (body fixed) was used with its origin at the center of disk and rotates with the disk as shown below
The absolute velocity and absolute acceleration of the particle can now be found as follows\[ \mathbf{v}=\mathbf{\dot{R}}+\boldsymbol{\dot{\rho }}_{r}+\left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \] But \(\mathbf{\dot{R}}=0\) since the center of the C.S. does not move relative to the center of the disk. \(\boldsymbol{\omega }=-\omega \mathbf{k}\), \(\boldsymbol{\rho }=\rho \mathbf{i}\) and \(\boldsymbol{\dot{\rho }}_{r}=\dot{\rho }_{r}\mathbf{i}\), therefore\[ \mathbf{v}=\dot{\rho }_{r}\mathbf{i}+\left ( -\omega \mathbf{k}\times \rho \mathbf{i}\right ) =\dot{\rho }_{r}\mathbf{i}-\omega \rho \mathbf{j}\] The absolute acceleration is\begin{align*} \mathbf{a} & =\mathbf{\ddot{R}}+\boldsymbol{\ddot{\rho }}_{r}+\left ( \boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }\right ) +\boldsymbol{\omega }\times \left ( \boldsymbol{\dot{\rho }}_{r}+\left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \right ) \\ & =\mathbf{\ddot{R}}+\boldsymbol{\ddot{\rho }}_{r}+\left ( \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\right ) +\left ( \boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }\right ) +\left ( \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\right ) +\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \\ & =\mathbf{\ddot{R}}+\boldsymbol{\ddot{\rho }}_{r}+2\left ( \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\right ) +\left ( \boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }\right ) +\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \end{align*}
But \(\mathbf{\ddot{R}}=0\), \(\boldsymbol{\omega }=-\omega \mathbf{k}\), \(\boldsymbol{\rho }=\rho \mathbf{i}\), \(\boldsymbol{\dot{\rho }}_{r}=\dot{\rho }_{r}\mathbf{i}\), \(\boldsymbol{\ddot{\rho }}_{r}=\ddot{\rho }\mathbf{i}\) and \(\boldsymbol{\dot{\omega }}=-\dot{\omega }k=0\) since \(\dot{\omega }=0\), therefore\begin{align*} \mathbf{a} & =\ddot{\rho }_{r}\mathbf{i}+2\left ( -\omega \mathbf{k}\times \dot{\rho }_{r}\mathbf{i}\right ) +\left ( -\omega \mathbf{k}\right ) \times \left ( -\omega \mathbf{k}\times \rho \mathbf{i}\right ) \\ & =\ddot{\rho }_{r}\mathbf{i}-2\omega \dot{\rho }_{r}\mathbf{j}+\left ( -\omega \mathbf{k}\right ) \times \left ( -\omega \rho _{r}\mathbf{j}\right ) \\ & =-\omega ^{2}\rho _{r}\mathbf{i+}\ddot{\rho }_{r}\mathbf{i}-2\omega \dot{\rho }_{r}\mathbf{j}\\ & =\left ( -\omega ^{2}\rho _{r}+\ddot{\rho }_{r}\right ) \mathbf{i}-2\omega \dot{\rho }_{r}\mathbf{j} \end{align*}
The particular has acceleration in the \(x\) and \(y\) directions. To find how long it takes to travel to the edge, the equation of motion in the \(x\) direction is first found.
Using Newton’s first law in the \(x\) direction, the total external forces acting in the \(x\) direction is zero. Hence \(\mathbf{f}_{x}=m\mathbf{a}_{x}\) gives\begin{align*} m\left ( -\omega ^{2}\rho _{r}+\ddot{\rho }_{r}\right ) & =0\\ \ddot{\rho }_{r}-\omega ^{2}\rho _{r} & =0 \end{align*}
This is a second order ODE. It is constant coefficients. The roots of the characteristic equation can be used for the solution. The roots are \(\lambda =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{\pm \sqrt{4\omega ^{2}}}{2}=\pm \omega \), hence the general solution is given by\[ \rho _{r}=Ae^{\omega t}+Be^{-\omega t}\] The constants \(A,B\) are found from initial conditions. When \(t=0\), \(\rho _{r}=b\), hence \begin{equation} b=A+B \tag{1} \end{equation} Taking derivative of the general solution gives \[ \dot{\rho }_{r}=\omega Ae^{\omega t}-\omega Be^{-\omega t}\] But when \(t=0\), \(\dot{\rho }_{r}\left ( 0\right ) =0\) hence\begin{align} 0 & =\omega A-\omega B\nonumber \\ 0 & =A-B \tag{2} \end{align}
From Eqs (1) and (2) the values of \(A,B\) are found to be\[ A=B=\frac{b}{2}\] The general solution becomes\begin{align*} \rho _{r}\left ( t\right ) & =\frac{b}{2}e^{\omega t}+\frac{b}{2}e^{-\omega t}\\ \rho _{r}\left ( t\right ) & =b\cosh \left ( \omega t\right ) \end{align*}
Solving for time \(t\) when \(\rho _{r}\left ( t\right ) =R\) results in\begin{align*} R & =b\cosh \left ( \omega t\right ) \\ t & =\frac{1}{\omega }\operatorname{arccosh}\left ( \frac{R}{b}\right ) \end{align*}
Here is a plot showing the time it takes to reach the edge for \(\omega =1\) rad/sec and \(R=1\), as \(b\) is changed from \(10^{-3}\) (very close to the origin) to \(1\) (the edge). Clearly when \(b=R\) the time is zero, and when \(b=\frac{R}{2}\) the time is found to be \(\operatorname{arccosh}\left ( 2\right ) =1.31\) sec.
The above shows that the time to reach the edge is not linear with the distance, but it is almost linear between \(20\%\) and \(80\%\) of the distance to the edge.
Solution
The first step is to find the angular velocity vector \(\boldsymbol{\omega }\) of the body C.S. in terms of Euler rates.
Using the above diagram the velocity vector \(\boldsymbol{\omega }\) can be written as (Eq. 1.99, page 85, class notes book).\begin{equation} \begin{Bmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end{Bmatrix} =\begin{bmatrix} \sin \theta \sin \phi & \cos \phi & 0\\ \sin \theta \cos \phi & -\sin \phi & 0\\ \cos \theta & 0 & 1 \end{bmatrix}\begin{Bmatrix} \dot{\psi }\\ \dot{\theta }\\ \dot{\phi }\end{Bmatrix} \tag{1} \end{equation} Therefore, in vector form the above becomes\begin{equation} \boldsymbol{\omega }=\mathbf{i}\left ( \sin \theta \sin \phi \dot{\psi }+\cos \phi \dot{\theta }\right ) +\mathbf{j}\left ( \sin \theta \cos \phi \dot{\psi }-\sin \phi \dot{\theta }\right ) +\mathbf{k}\left ( \cos \theta \dot{\psi }+\dot{\phi }\right ) \tag{2} \end{equation} The position vector of the \(p\) is \(\boldsymbol{\rho }\) given as (in the equation below, \(r\) represents the radius of the satellite, which is shown in the diagram as \(R\). It was replaced by by small \(r\) so not to confuse this letter with the standard vector \(\mathbf{R}\) that is commonly used in the main equations below).\[ \boldsymbol{\rho }=x\mathbf{i}+\left ( r+\xi \right ) \mathbf{j}+z\mathbf{k}\] Since \(r\) is constant then the relative velocity of \(p\) is\[ \boldsymbol{\dot{\rho }}_{r}=\dot{x}\mathbf{i}+\dot{\xi }\mathbf{j}+\dot{z}\mathbf{k}\]
The absolute velocity of \(P\) is\[ \mathbf{v}=\mathbf{\dot{R}}+\boldsymbol{\dot{\rho }}_{r}+\left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \] \(\boldsymbol{\omega }\times \boldsymbol{\rho }\) is now calculated\begin{align*} \boldsymbol{\omega }\times \boldsymbol{\rho } & =\mathbf{i}\left ( -z\sin \theta \cos \phi \dot{\psi }-z\sin \phi \dot{\theta }-\left ( r+\xi \right ) \left ( \cos \theta \dot{\psi }+\dot{\phi }\right ) \right ) \\ & +\mathbf{j}\left ( z\sin \theta \sin \phi \dot{\psi }+z\cos \phi \dot{\theta }+x\cos \theta \dot{\psi }+x\dot{\phi }\right ) \\ & +\mathbf{k}\left ( \left ( r+\xi \right ) \sin \theta \sin \phi \dot{\psi }+\left ( r+\xi \right ) \cos \phi \dot{\theta }-x\sin \theta \cos \phi \dot{\psi }+x\sin \phi \dot{\theta }\right ) \end{align*}
Collecting terms, the absolute velocity is simplified to\begin{align*} \mathbf{v} & =\mathbf{i}\left ( v_{X}+\dot{x}-z\sin \theta \cos \phi \dot{\psi }-z\sin \phi \dot{\theta }-\left ( r+\xi \right ) \left ( \cos \theta \dot{\psi }+\dot{\phi }\right ) \right ) \\ & +\mathbf{j}\left ( v_{Y}+\dot{x}+z\sin \theta \sin \phi \dot{\psi }+z\cos \phi \dot{\theta }+x\cos \theta \dot{\psi }+x\dot{\phi }\right ) \\ & +\mathbf{k}\left ( v_{Z}+\dot{z}+\left ( r+\xi \right ) \sin \theta \sin \phi \dot{\psi }+\left ( r+\xi \right ) \cos \phi \dot{\theta }-x\sin \theta \cos \phi \dot{\psi }+x\sin \phi \dot{\theta }\right ) \end{align*}
The absolute acceleration of \(P\) is\begin{equation} \mathbf{a}=\mathbf{\ddot{R}}+\boldsymbol{\ddot{\rho }}_{r}+2\left ( \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\right ) +\left ( \boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }\right ) +\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \tag{3} \end{equation} \(\mathbf{\ddot{R}}\) is given in the problem as \(\left ( a_{X}\mathbf{i}+a_{Y}\mathbf{j}+a_{Z}\mathbf{k}\right ) \) and \(\boldsymbol{\ddot{\rho }}_{r}=\ddot{x}\mathbf{i}+\ddot{\xi }\mathbf{j}+\ddot{z}\mathbf{k}\). The remaining term to calculate is \(\boldsymbol{\dot{\omega }}\)
Taking derivative w.r.t time of Eq. (2) above results in\begin{align*} \boldsymbol{\dot{\omega }} & =\mathbf{i}\left ( \frac{d}{dt}\left ( \sin \theta \sin \phi \dot{\psi }+\cos \phi \dot{\theta }\right ) \right ) +\mathbf{j}\left ( \frac{d}{dt}\left ( \sin \theta \cos \phi \dot{\psi }-\sin \phi \dot{\theta }\right ) \right ) +\mathbf{k}\left ( \frac{d}{dt}\left ( \cos \theta \dot{\psi }+\dot{\phi }\right ) \right ) \\ & \\ & =\mathbf{i}\left ( \dot{\theta }\dot{\psi }\cos \theta \sin \phi +\dot{\psi }\dot{\phi }\sin \theta \cos \phi +\ddot{\psi }\sin \theta \sin \phi -\dot{\phi }\dot{\theta }\sin \phi +\ddot{\theta }\cos \phi \right ) \\ & +\mathbf{j}\left ( -\ddot{\theta }\sin \phi -\dot{\phi }\dot{\theta }\cos \phi +\ddot{\psi }\sin \theta \cos \phi +\dot{\psi }\dot{\theta }\cos \theta \cos \phi -\dot{\psi }\dot{\phi }\sin \theta \sin \phi \right ) \\ & +\mathbf{k}\left ( \ddot{\psi }\cos \theta -\dot{\psi }\dot{\theta }\sin \theta +\ddot{\phi }\right ) \end{align*}
Since the angular accelerations are all constant, all terms above with second time derivatives can be set to zero. Hence \(\boldsymbol{\dot{\omega }}\) simplifies to\begin{align*} \boldsymbol{\dot{\omega }} & =\mathbf{i}\left ( -\dot{\phi }\dot{\theta }\sin \phi +\dot{\theta }\dot{\psi }\cos \theta \sin \phi +\dot{\psi }\dot{\phi }\sin \theta \cos \phi \right ) \\ & +\mathbf{j}\left ( -\dot{\phi }\dot{\theta }\cos \phi +\dot{\psi }\dot{\theta }\cos \theta \cos \phi -\dot{\psi }\dot{\phi }\sin \theta \sin \phi \right ) \\ & +\mathbf{k}\left ( -\dot{\psi }\dot{\theta }\sin \theta \right ) \end{align*}
Now Eq. (3) can be evaluated. Each term is first evaluated. \(\boldsymbol{\omega }\times \boldsymbol{\rho }\) was found in part (1). \(\boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\) is similar to \(\boldsymbol{\omega }\times \boldsymbol{\rho }\), but \(\boldsymbol{\rho }\) is changed to \(\boldsymbol{\dot{\rho }}\). The derivation of \(\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \) is too complicated to do by hand and was done on the computer. Here is the final result of each component of \(\mathbf{a}\) in as \(\{a_{x},a_{y},a_{z}\}\).
This is the result of evaluating Eq. (3)
\[ a_x= a_X+2 \xi \theta ' \sin (\theta ) \psi ' \cos ^2(\phi )-\xi \left (\theta '\right )^2 \sin (\phi ) \cos (\phi )-2 \cos (\theta ) \xi ' \psi '+\frac{1}{2} \xi \sin ^2(\theta ) \left (\psi '\right )^2 \sin (2 \phi )-2 \xi ' \phi '+2 r \theta ' \sin (\theta ) \psi ' \cos ^2(\phi )-r \left (\theta '\right )^2 \sin (\phi ) \cos (\phi )+\frac{1}{2} r \sin ^2(\theta ) \left (\psi '\right )^2 \sin (2 \phi )+x''+2 x \theta ' \sin (\theta ) \psi ' \sin (\phi ) \cos (\phi )-x \left (\theta '\right )^2 \sin ^2(\phi )-\frac{1}{4} x \cos (2 \theta ) \left (\psi '\right )^2-2 x \cos (\theta ) \psi ' \phi '-\frac{1}{2} x \sin ^2(\theta ) \left (\psi '\right )^2 \cos (2 \phi )-\frac{3}{4} x \left (\psi '\right )^2-x \left (\phi '\right )^2-2 \theta ' z' \sin (\phi )+2 \sin (\theta ) \psi ' z' \cos (\phi )+2 z \theta ' \cos (\theta ) \psi ' \cos (\phi )+\frac{1}{2} z \sin (2 \theta ) \left (\psi '\right )^2 \sin (\phi ) \]
\[ a_y =a_Y-2 \xi \theta ' \sin (\theta ) \psi ' \sin (\phi ) \cos (\phi )-\xi \left (\theta '\right )^2 \cos ^2(\phi )-\frac{1}{4} \xi \cos (2 \theta ) \left (\psi '\right )^2-2 \xi \cos (\theta ) \psi ' \phi '+\frac{1}{2} \xi \sin ^2(\theta ) \left (\psi '\right )^2 \cos (2 \phi ) + \xi ''-\frac{3}{4} \xi \left (\psi '\right )^2 - \xi \left (\phi '\right )^2-2 r \theta ' \sin (\theta ) \psi ' \sin (\phi ) \cos (\phi ) - r \left (\theta '\right )^2 \cos ^2(\phi )-\frac{1}{4} r \cos (2 \theta ) \left (\psi '\right )^2-2 r \cos (\theta ) \psi ' \phi '+\frac{1}{2} r \sin ^2(\theta ) \left (\psi '\right )^2 \cos (2 \phi )-\frac{3}{4} r \left (\psi '\right )^2-r \left (\phi '\right )^2+2 \cos (\theta ) x' \psi '+2 x' \phi '-2 x \theta ' \sin (\theta ) \psi ' \sin ^2(\phi )-x \left (\theta '\right )^2 \sin (\phi ) \cos (\phi )+\frac{1}{2} x \sin ^2(\theta ) \left (\psi '\right )^2 \sin (2 \phi )-2 \theta ' z' \cos (\phi )-2 \sin (\theta ) \psi ' z' \sin (\phi )-2 z \theta ' \cos (\theta ) \psi ' \sin (\phi )+\frac{1}{2} z \sin (2 \theta ) \left (\psi '\right )^2 \cos (\phi ) \]
\[ a_z =a_Z+2 \theta ' \xi ' \cos (\phi )-2 \xi \theta ' \phi ' \sin (\phi )+2 \sin (\theta ) \xi ' \psi ' \sin (\phi )+2 \xi \sin (\theta ) \psi ' \phi ' \cos (\phi )+\frac{1}{2} \xi \sin (2 \theta ) \left (\psi '\right )^2 \cos (\phi )-2 r \theta ' \phi ' \sin (\phi )+2 r \sin (\theta ) \psi ' \phi ' \cos (\phi )+\frac{1}{2} r \sin (2 \theta ) \left (\psi '\right )^2 \cos (\phi )+2 \theta ' x' \sin (\phi )-2 \sin (\theta ) x' \psi ' \cos (\phi )+2 x \theta ' \phi ' \cos (\phi )+2 x \sin (\theta ) \psi ' \phi ' \sin (\phi )+\frac{1}{2} x \sin (2 \theta ) \left (\psi '\right )^2 \sin (\phi )+z''-z \left (\theta '\right )^2-z \sin ^2(\theta ) \left (\psi '\right )^2 \]