OCR Further Mechanics 2020 November — Question 7

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2020
SessionNovember
TopicCentre of Mass 2

7 Fig. 7.1 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]{831ba5da-df19-43bb-b163-02bbddb4e2b8-6_364_556_342_246} \captionsetup{labelformat=empty} \caption{Fig. 7.1}
\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. 7.2). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{831ba5da-df19-43bb-b163-02bbddb4e2b8-6_766_757_1400_244} \captionsetup{labelformat=empty} \caption{Fig. 7.2}
    \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 \(m\).