SPS SPS FM 2020 May — Question 10 4 marks

Exam BoardSPS
ModuleSPS FM (SPS FM)
Year2020
SessionMay
Marks4
TopicCentre of Mass 1
TypeComposite solid with hemisphere and cylinder/cone
DifficultyStandard +0.8 This is a standard composite centre of mass problem requiring multiple steps: finding volumes of hemisphere and cone, locating individual centres of mass, and applying the composite formula. The trigonometric manipulation to reach the given form adds moderate algebraic complexity, but the overall approach is methodical and follows established techniques for Further Maths Mechanics.
Spec6.04d Integration: for centre of mass of laminas/solids
10. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ab2949b2-11f2-4682-ab0c-25ecee2d665a-5_643_325_388_822} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} A child's toy is a uniform solid consisting of a hemisphere of radius \(r \mathrm {~cm}\) joined to a cone of base radius \(r \mathrm {~cm}\). The curved surface of the cone makes an angle \(\alpha\) with its base. The two shapes are joined at the plane faces with their circumferences coinciding (see Fig. 1). The distance of the centre of mass of the toy above the common circular plane face is \(x \mathrm {~cm}\).
[0pt] [The volume of a sphere is \(\frac { 4 } { 3 } \pi r ^ { 3 }\) and the volume of a cone is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
  1. Show that \(x = \frac { r \left( \tan ^ { 2 } \alpha - 3 \right) } { 8 + 4 \tan \alpha }\).
10.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{ab2949b2-11f2-4682-ab0c-25ecee2d665a-5_643_325_388_822}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

A child's toy is a uniform solid consisting of a hemisphere of radius $r \mathrm {~cm}$ joined to a cone of base radius $r \mathrm {~cm}$. The curved surface of the cone makes an angle $\alpha$ with its base. The two shapes are joined at the plane faces with their circumferences coinciding (see Fig. 1). The distance of the centre of mass of the toy above the common circular plane face is $x \mathrm {~cm}$.\\[0pt]
[The volume of a sphere is $\frac { 4 } { 3 } \pi r ^ { 3 }$ and the volume of a cone is $\frac { 1 } { 3 } \pi r ^ { 2 } h$.]\\
(i) Show that $x = \frac { r \left( \tan ^ { 2 } \alpha - 3 \right) } { 8 + 4 \tan \alpha }$.\\

\hfill \mbox{\textit{SPS SPS FM 2020 Q10 [4]}}
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