OCR Further Mechanics 2021 June — Question 4 12 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2021
SessionJune
Marks12
TopicCentre of Mass 2
TypeLamina with hole removed
DifficultyChallenging +1.3 This is a multi-part Further Maths mechanics question requiring symmetry arguments, a standard but algebraically involved proof of a centre of mass formula, and application to a composite lamina problem with equilibrium. While it involves several steps and careful coordinate geometry, the techniques are standard for FM mechanics (composite bodies, moments principle). The proof in part (b) is bookwork-level for this specification, and parts (c)-(d) follow established methods without requiring novel insight.
Spec6.04b Find centre of mass: using symmetry6.04d Integration: for centre of mass of laminas/solids

4 Fig. 4.1 shows a uniform lamina in the shape of a sector of a circle of radius \(r\) and angle \(2 \theta\) where \(\theta\) is in radians. The sector consists of a triangle \(O A B\) and a segment bounded by the chord \(A B\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8859baf3-f8e8-4fbf-b54f-34f550b02c26-03_358_545_543_255} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
\end{figure}
  1. Explain why the centre of mass of the segment lies on the radius through the midpoint of \(A B\).
  2. Show that the distance of the centre of mass of the segment from \(O\) is \(\frac { 2 r \sin ^ { 3 } \theta } { 3 ( \theta - \sin \theta \cos \theta ) }\). A uniform circular lamina of radius 5 units is placed with its centre at the origin, \(O\), of an \(x - y\) coordinate system. A component for a machine is made by removing and discarding a segment from the lamina. The radius of the circle from which the segment is formed is 3 units and the centre of this circle is \(O\). The centre of the straight edge of the segment has coordinates \(( 0,2 )\) and this edge is perpendicular to the \(y\)-axis (see Fig. 4.2). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8859baf3-f8e8-4fbf-b54f-34f550b02c26-03_748_743_1594_251} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
    \end{figure}
  3. Find the \(y\)-coordinate of the centre of mass of the component, giving your answer correct to 3 significant figures. \(C\) is the point on the component with coordinates \(( 0,5 )\). The component is now placed horizontally and supported only at \(O\). A particle of mass \(m \mathrm {~kg}\) is placed on the component at \(C\) and the component and particle are in equilibrium.
  4. Find the mass of the component in terms of \(m\). Total Marks for Question Set 3: 40

4 Fig. 4.1 shows a uniform lamina in the shape of a sector of a circle of radius $r$ and angle $2 \theta$ where $\theta$ is in radians. The sector consists of a triangle $O A B$ and a segment bounded by the chord $A B$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8859baf3-f8e8-4fbf-b54f-34f550b02c26-03_358_545_543_255}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Explain why the centre of mass of the segment lies on the radius through the midpoint of $A B$.
\item Show that the distance of the centre of mass of the segment from $O$ is $\frac { 2 r \sin ^ { 3 } \theta } { 3 ( \theta - \sin \theta \cos \theta ) }$.

A uniform circular lamina of radius 5 units is placed with its centre at the origin, $O$, of an $x - y$ coordinate system. A component for a machine is made by removing and discarding a segment from the lamina. The radius of the circle from which the segment is formed is 3 units and the centre of this circle is $O$. The centre of the straight edge of the segment has coordinates $( 0,2 )$ and this edge is perpendicular to the $y$-axis (see Fig. 4.2).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8859baf3-f8e8-4fbf-b54f-34f550b02c26-03_748_743_1594_251}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
\item Find the $y$-coordinate of the centre of mass of the component, giving your answer correct to 3 significant figures.\\
$C$ is the point on the component with coordinates $( 0,5 )$. The component is now placed horizontally and supported only at $O$. A particle of mass $m \mathrm {~kg}$ is placed on the component at $C$ and the component and particle are in equilibrium.
\item Find the mass of the component in terms of $m$.

Total Marks for Question Set 3: 40
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2021 Q4 [12]}}