6. A uniform solid right circular cone has base radius \(r\) and height \(h\).
- Use algebraic integration to show that the distance of the centre of mass of the cone from its vertex is \(\frac { 3 } { 4 } h\).
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[You may assume that the volume of a cone of base radius \(r\) and height \(h\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\) ]
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2273ca38-1e16-44ab-ae84-f4c576cbb8f9-20_394_716_632_621}
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\caption{Figure 3}
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A solid \(S\) is formed by joining a uniform right circular solid cone of mass \(5 m\) to a uniform solid hemisphere, of radius \(r\) and mass \(k m\) where \(k < 20\). The cone has base radius \(r\) and height \(6 r\). The plane face of the cone coincides with the plane face of the hemisphere. The centre of the plane face of the cone is \(O\) and the point \(A\) is on the circular edge of this plane face, as shown in Figure 3. - Find the distance from \(O\) to the centre of mass of \(S\).
The solid is suspended from \(A\) and hangs freely in equilibrium. The angle between the axis of the cone and the horizontal is \(30 ^ { \circ }\).
- Find, to the nearest whole number, the value of \(k\).