SPS SPS FM Mechanics 2024 January — Question 4 10 marks

Exam BoardSPS
ModuleSPS FM Mechanics (SPS FM Mechanics)
Year2024
SessionJanuary
Marks10
TopicCentre of Mass 1
TypeSuspended lamina equilibrium angle
DifficultyStandard +0.8 This is a multi-part centre of mass problem requiring composite lamina calculations, algebraic manipulation with given hints, verification that the centre of mass lies on the perimeter (involving the golden ratio), and finally finding an equilibrium angle. While the algebraic result is provided and the problem is structured with guidance, it requires careful coordinate geometry, understanding of composite bodies, and geometric reasoning about equilibrium. The golden ratio appearance and perimeter condition add conceptual depth beyond routine centre of mass exercises, placing it moderately above average difficulty.
Spec6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids
4. A uniform heavy lamina occupies the region shaded in Fig. 3. This region is formed by removing a square of side 1 unit from a square of side \(a\) units (where \(a > 1\) ). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5f9a87c6-2255-4178-ab04-441bb0cc4ce0-08_558_594_299_699} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure} Relative to the axes shown in Fig. 3, the centre of mass of the lamina is at \(( \bar { x } , \bar { y } )\).
  1. Show that \(\bar { x } = \bar { y } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }\).
    [0pt] [You may need to use the result \(\frac { a ^ { 3 } - 1 } { 2 \left( a ^ { 2 } - 1 \right) } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }\).]
  2. Show that the centre of mass of the lamina lies on its perimeter if \(a = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )\).
  3. With the value of \(a = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )\) the lamina is suspended from A and hangs in equilibrium. Find the angle that the line OA makes with the vertical.
    [0pt] [Question 4 Continued] \section*{5.}
4.

A uniform heavy lamina occupies the region shaded in Fig. 3. This region is formed by removing a square of side 1 unit from a square of side $a$ units (where $a > 1$ ).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5f9a87c6-2255-4178-ab04-441bb0cc4ce0-08_558_594_299_699}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

Relative to the axes shown in Fig. 3, the centre of mass of the lamina is at $( \bar { x } , \bar { y } )$.\\
(i) Show that $\bar { x } = \bar { y } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }$.\\[0pt]
[You may need to use the result $\frac { a ^ { 3 } - 1 } { 2 \left( a ^ { 2 } - 1 \right) } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }$.]\\
(ii) Show that the centre of mass of the lamina lies on its perimeter if $a = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )$.\\
(iii) With the value of $a = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )$ the lamina is suspended from A and hangs in equilibrium. Find the angle that the line OA makes with the vertical.\\[0pt]
[Question 4 Continued]

\section*{5.}

\hfill \mbox{\textit{SPS SPS FM Mechanics 2024 Q4 [10]}}
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