OCR MEI M4 2006 June — Question 2 12 marks

Exam BoardOCR MEI
ModuleM4 (Mechanics 4)
Year2006
SessionJune
Marks12
PaperDownload PDF ↗
TopicMoments
TypePotential energy with inextensible strings or gravity only
DifficultyChallenging +1.2 This is a multi-step mechanics problem requiring geometric analysis, potential energy formulation, and stability investigation. While it involves several techniques (trigonometry, differentiation, second derivative test), the geometric setup is clearly guided ('show that'), and the equilibrium/stability analysis follows standard M4 procedures. It's moderately harder than average due to the multi-part nature and requiring careful geometric reasoning, but doesn't demand exceptional insight.
Spec3.04b Equilibrium: zero resultant moment and force6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces
2 A rigid circular hoop of radius \(a\) is fixed in a vertical plane. At the highest point of the hoop there is a small smooth pulley, P. A light inextensible string AB of length \(\frac { 5 } { 2 } a\) is passed over the pulley. A particle of mass \(m\) is attached to the string at \(\mathrm { B } . \mathrm { PB }\) is vertical and angle \(\mathrm { APB } = \theta\). A small smooth ring of mass \(m\) is threaded onto the hoop and attached to the string at A . This situation is shown in Fig. 2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c97056a9-4156-4ecd-a80e-1a82c81ab824-2_568_549_1306_758} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. Show that \(\mathrm { PB } = \frac { 5 } { 2 } a - 2 a \cos \theta\) and hence show that the potential energy of the system relative to P is \(V = - m g a \left( 2 \cos ^ { 2 } \theta - 2 \cos \theta + \frac { 5 } { 2 } \right)\).
  2. Hence find the positions of equilibrium and investigate their stability.
2 A rigid circular hoop of radius $a$ is fixed in a vertical plane. At the highest point of the hoop there is a small smooth pulley, P. A light inextensible string AB of length $\frac { 5 } { 2 } a$ is passed over the pulley.

A particle of mass $m$ is attached to the string at $\mathrm { B } . \mathrm { PB }$ is vertical and angle $\mathrm { APB } = \theta$. A small smooth ring of mass $m$ is threaded onto the hoop and attached to the string at A . This situation is shown in Fig. 2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{c97056a9-4156-4ecd-a80e-1a82c81ab824-2_568_549_1306_758}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

(i) Show that $\mathrm { PB } = \frac { 5 } { 2 } a - 2 a \cos \theta$ and hence show that the potential energy of the system relative to P is $V = - m g a \left( 2 \cos ^ { 2 } \theta - 2 \cos \theta + \frac { 5 } { 2 } \right)$.\\
(ii) Hence find the positions of equilibrium and investigate their stability.

\hfill \mbox{\textit{OCR MEI M4 2006 Q2 [12]}}
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Q1 12 Q2 12 Q3 24 Q4 24