Problem source

A smooth horizontal plane rotates with constant angular velocity $\Omega$ about a fixed vertical axis through a fixed point $O$ of the plane. The point $A$ is fixed in the plane and $OA=a.$ A particle $P$ lies on the plane and is joined to $A$ by a light rod of length $b(>a)$ freely pivoted at $A$. Initially $OAP$ is a straight line and $P$ is moving with speed $(a+2\sqrt{ab})\Omega$ perpendicular to $OP$ in the same sense as $\Omega.$ At time $t,$ $AP$ makes an angle $\phi$ with $OA$ produced. Obtain an expression for the component of the acceleration of $P$ perpendicular to $AP$ in terms of $\dfrac{\mathrm{d}^{2}\phi}{\mathrm{d}t^{2}},\phi,a,b$ and $\Omega.$
Hence find $\dfrac{\mathrm{d}\phi}{\mathrm{d}t}$, in terms of $\phi,a,b$ and $\Omega,$ and show that $P$ never crosses $OA.$

Solution source

Set up coordinate axes so that at time $t$ $OA$ is the $x$-axis, and all rotations are counter-clockwise.

Then if $OA = \mathbf{a}$, $AP = \mathbf{x}$ and $OP = \mathbf{p}$ we have:

\begin{align*}
\mathbf{a} &= \binom{a}{0} \\
\dot{\mathbf{a}} &= \binom{0}{-a \Omega} \\
\ddot{\mathbf{a}} &= \binom{-a \Omega^2}{0} \\
\\
\mathbf{x} &= \binom{b \cos \phi }{b \sin \phi } \\
\dot{\mathbf{x}} &= b \dot{\phi} \binom{-\sin \phi}{\cos \phi} \\
\ddot{\mathbf{x}} &= \binom{-b \ddot{\phi} \sin \phi-b \dot{\phi}^2 \cos \phi }{b \ddot{\phi} \cos \phi-b \dot{\phi}^2 \sin \phi} \\
\\
\ddot{\mathbf{p}} &= \binom{-a \Omega^2 +-b \ddot{\phi} \sin \phi-b \dot{\phi}^2 \cos \phi }{b \ddot{\phi} \cos \phi-b \dot{\phi}^2 \sin \phi} 
\end{align*}

We can take a dot product with $\mathbf{n} = \binom{-\sin \phi}{\cos \phi}$ to obtain the component perpendicular to $AP$, which is:

\begin{align*}
&& \binom{-\sin \phi}{\cos \phi} \cdot \ddot{x} &= a \Omega^2 \sin \phi + b \ddot{\phi}
\end{align*}

Noticing that this component must be $0$ (since the only force acting on $P$ is the rod), this must be equal to zero.

\begin{align*}
&& 0 &= a \Omega^2 \sin \phi + b \ddot{\phi} \\
\Rightarrow && 0 &=  a \Omega^2 \dot{\phi} \sin \phi + b\dot{\phi} \ddot{\phi} \\
\Rightarrow && C &= -a \Omega^2 \cos \phi + \tfrac12 b \dot{\phi}^2 
\end{align*}

Noticing that the initial conditions are $\phi = 0$ and $\dot{\phi} = 2\sqrt{\frac{a}{b}} \Omega$,  so

\begin{align*}
&& C &= -a \Omega^2+ \tfrac12 b \left ( 2\sqrt{\frac{a}{b}} \Omega \right)^2 \\
&&&= -a \Omega^2 + 2a \Omega^2 \\
&&&= a \Omega^2\\
\Rightarrow && \dot{\phi} &=\sqrt{\frac{2}{b} \left ( a \Omega^2 + a \Omega^2 \cos \phi \right)} \\
&&&= \Omega \sqrt{\frac{2a}{b}} \sqrt{1+ \cos \phi} \\
&&& =  \Omega \sqrt{\frac{2a}{b}}\sqrt{2} \cos \tfrac{\phi}{2} \\
\Rightarrow && \int \sec \tfrac{\phi}{2} \d \phi &= 2 \Omega \sqrt{\frac{a}{b}}t \\
\Rightarrow && \tfrac12 \ln | \sec \tfrac{\phi}{2}+\tan \tfrac{\phi}{2} | &= 2 \Omega \sqrt{\frac{a}{b}}t + C \\
t = 0, \phi = 0: && C = 0 \\
\Rightarrow && \sec \tfrac{\phi}{2}+\tan \tfrac{\phi}{2} &= e^{4 \Omega \sqrt{\frac{a}{b}}t}
\end{align*}

Since when $t > 0$ this is positive and larger than $1$ we cannot have $\phi = 0$ and since it remains below infinite $\phi$ cannot reach $\pi$. Therefore it cannot cross $OA$