| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with hemisphere and cylinder/cone |
| Difficulty | Standard +0.8 This is a challenging M2 centre of mass problem requiring: (i) combining standard formulae for hemisphere and cone centres of mass with geometric reasoning involving tan α, then algebraic manipulation to reach a specific form; (ii) applying the toppling condition (centre of mass must be within base of support) which requires geometric analysis and solving a trigonometric inequality. The multi-step nature, algebraic complexity, and need to connect geometry with mechanics principles places this above average difficulty, though it follows a recognizable M2 pattern. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(h = r\tan\alpha\) | B1 | Seen anywhere and in any form |
| \(x\left(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\right) = \frac{1}{3}\pi r^2 h \times \frac{h}{4} - \frac{2}{3}\pi r^3 \times \frac{3}{8}r\) | M1 | Table of values idea |
| (working to result) | A1 | |
| \(x = \dfrac{r(\tan^2\alpha - 3)}{8 + 4\tan\alpha}\) | A1 | AG; www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x < 0\) | B1 | May be implied |
| Solve \(\tan^2\alpha - 3 < 0\) | M1 | Condone = |
| \(\alpha < 60\) | A1 | Condone \(\leq\) throughout |
| [3] | SC Use of \(=\) or \(>\) throughout. Max B0 M1 A0 |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $h = r\tan\alpha$ | B1 | Seen anywhere and in any form |
| $x\left(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\right) = \frac{1}{3}\pi r^2 h \times \frac{h}{4} - \frac{2}{3}\pi r^3 \times \frac{3}{8}r$ | M1 | Table of values idea |
| (working to result) | A1 | |
| $x = \dfrac{r(\tan^2\alpha - 3)}{8 + 4\tan\alpha}$ | A1 | AG; www |
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## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x < 0$ | B1 | May be implied |
| Solve $\tan^2\alpha - 3 < 0$ | M1 | Condone = |
| $\alpha < 60$ | A1 | Condone $\leq$ throughout |
| | [3] | SC Use of $=$ or $>$ throughout. Max B0 M1 A0 |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5addd79d-d502-455c-936f-27005483164e-2_655_334_440_861}
\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 $x = \frac { r \left( \tan ^ { 2 } \alpha - 3 \right) } { 8 + 4 \tan \alpha }$.
The toy is placed on a horizontal surface with the hemisphere in contact with the surface. The toy is released from rest from the position in which the common plane circular face is vertical (see Fig. 2).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5addd79d-d502-455c-936f-27005483164e-2_193_670_1827_699}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
(ii) Find the set of values of $\alpha$ such that the toy moves to the upright position.
\hfill \mbox{\textit{OCR M2 2012 Q2 [7]}}