\includegraphics{figure_4}

As shown in the diagram, $AB$ is a long thin rod which is fixed vertically with $A$ above $B$. One end of a light inextensible string of length $1$ m is attached to $A$ and the other end is attached to a particle $P$ of mass $m_1$ kg. One end of another light inextensible string of length $1$ m is also attached to $P$. Its other end is attached to a small smooth ring $R$, of mass $m_2$ kg, which is free to move on $AB$.

Initially, $P$ moves in a horizontal circle of radius $0.6$ m with constant angular velocity $\omega$ rad s$^{-1}$. The magnitude of the tension in string $AP$ is denoted by $T_1$ N while that in string $PR$ is denoted by $T_2$ N.

\begin{enumerate}[label=(\alph*)]
\item By considering forces on $R$, express $T_2$ in terms of $m_2$. [2]

\item Show that
\begin{enumerate}[label=(\roman*)]
\item $T_1 = \frac{4g}{5}(m_1 + m_2)$. [2]

\item $\omega^2 = \frac{4g(m_1 + 2m_2)}{4m_1}$. [3]
\end{enumerate}

\item Deduce that, in the case where $m_1$ is much bigger than $m_2$, $\omega \approx 3.5$. [2]
\end{enumerate}

In a different case, where $m_1 = 2.5$ and $m_2 = 2.8$, $P$ slows down. Eventually the system comes to rest with $P$ and $R$ hanging in equilibrium.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the total energy lost by $P$ and $R$ as the angular velocity of $P$ changes from the initial value of $\omega$ rad s$^{-1}$ to zero. [5]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS ASFM Mechanics 2021 Q4 [14]}}