| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Solid on inclined plane - toppling |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics question combining centre of mass calculations with equilibrium analysis. Part (a) requires routine integration for a 2D lamina. Part (b)(i) uses the standard Pappus formula for solids of revolution (given as 'show that'), and part (b)(ii) applies standard equilibrium conditions (resolving forces and taking moments) to find friction coefficient. While it requires multiple techniques and careful setup, all methods are textbook-standard for M3 with no novel insight required. |
| Spec | 6.04c 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 |
| Area \(= \int_0^2 x^3\, dx = \left[\frac{x^4}{4}\right]_0^2 = 4\) | M1 A1 | |
| \(\bar{x} = \frac{1}{4}\int_0^2 x \cdot x^3\, dx = \frac{1}{4}\int_0^2 x^4\, dx = \frac{1}{4}\left[\frac{x^5}{5}\right]_0^2 = \frac{1}{4}\cdot\frac{32}{5} = \frac{8}{5} = 1.6\) | M1 A1 | |
| \(\bar{y} = \frac{1}{4}\int_0^2 \frac{1}{2}(x^3)^2\, dx = \frac{1}{8}\int_0^2 x^6\, dx = \frac{1}{8}\left[\frac{x^7}{7}\right]_0^2 = \frac{1}{8}\cdot\frac{128}{7} = \frac{16}{7} \approx 2.29\) | M1 A1 M1 A1 | |
| Centre of mass: \((1.6,\ \frac{16}{7})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Volume \(= \pi\int_1^2 (4-x^2)\, dx = \pi\left[4x - \frac{x^3}{3}\right]_1^2 = \pi\left[(8-\frac{8}{3})-(4-\frac{1}{3})\right] = \pi \cdot \frac{11}{3} - \pi\) | M1 A1 | |
| \(= \pi\left(\frac{16}{3} - \frac{11}{3}\right) = \frac{5\pi}{3}\)... evaluates to \(\frac{11\pi}{3} - \frac{11\pi}{3}\)... | Check arithmetic carefully | |
| \(\bar{x}\cdot V = \pi\int_1^2 x(4-x^2)\, dx = \pi\left[2x^2 - \frac{x^4}{4}\right]_1^2 = \pi\left[(8-4)-(2-\frac{1}{4})\right] = \pi \cdot \frac{9}{4}\)... | M1 A1 | |
| \(\bar{x} = \frac{9\pi/4}{V}\); shown to give \(\bar{x} = 1.35\) | DM1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Flat face is at \(x=1\); CoM is at \(x=1.35\), so distance from flat face \(= 1.35 - 1 = 0.35\ \text{m}\) | B1 | |
| Radius at flat face \(= \sqrt{4-1} = \sqrt{3}\) | B1 | |
| Taking moments about base contact point: \(T \times \sqrt{3} = W \times 0.35\) (string horizontal, weight vertical) | M1 | |
| Friction \(F = T\), Normal \(N = W\) | ||
| \(\mu \geq \frac{T}{N} = \frac{W \times 0.35}{W \times \sqrt{3}} = \frac{0.35}{\sqrt{3}} \approx 0.202\) | A1 |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $= \int_0^2 x^3\, dx = \left[\frac{x^4}{4}\right]_0^2 = 4$ | M1 A1 | |
| $\bar{x} = \frac{1}{4}\int_0^2 x \cdot x^3\, dx = \frac{1}{4}\int_0^2 x^4\, dx = \frac{1}{4}\left[\frac{x^5}{5}\right]_0^2 = \frac{1}{4}\cdot\frac{32}{5} = \frac{8}{5} = 1.6$ | M1 A1 | |
| $\bar{y} = \frac{1}{4}\int_0^2 \frac{1}{2}(x^3)^2\, dx = \frac{1}{8}\int_0^2 x^6\, dx = \frac{1}{8}\left[\frac{x^7}{7}\right]_0^2 = \frac{1}{8}\cdot\frac{128}{7} = \frac{16}{7} \approx 2.29$ | M1 A1 M1 A1 | |
| Centre of mass: $(1.6,\ \frac{16}{7})$ | | |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Volume $= \pi\int_1^2 (4-x^2)\, dx = \pi\left[4x - \frac{x^3}{3}\right]_1^2 = \pi\left[(8-\frac{8}{3})-(4-\frac{1}{3})\right] = \pi \cdot \frac{11}{3} - \pi$ | M1 A1 | |
| $= \pi\left(\frac{16}{3} - \frac{11}{3}\right) = \frac{5\pi}{3}$... evaluates to $\frac{11\pi}{3} - \frac{11\pi}{3}$... | | Check arithmetic carefully |
| $\bar{x}\cdot V = \pi\int_1^2 x(4-x^2)\, dx = \pi\left[2x^2 - \frac{x^4}{4}\right]_1^2 = \pi\left[(8-4)-(2-\frac{1}{4})\right] = \pi \cdot \frac{9}{4}$... | M1 A1 | |
| $\bar{x} = \frac{9\pi/4}{V}$; shown to give $\bar{x} = 1.35$ | DM1 A1 | |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Flat face is at $x=1$; CoM is at $x=1.35$, so distance from flat face $= 1.35 - 1 = 0.35\ \text{m}$ | B1 | |
| Radius at flat face $= \sqrt{4-1} = \sqrt{3}$ | B1 | |
| Taking moments about base contact point: $T \times \sqrt{3} = W \times 0.35$ (string horizontal, weight vertical) | M1 | |
| Friction $F = T$, Normal $N = W$ | | |
| $\mu \geq \frac{T}{N} = \frac{W \times 0.35}{W \times \sqrt{3}} = \frac{0.35}{\sqrt{3}} \approx 0.202$ | A1 | |4
\begin{enumerate}[label=(\alph*)]
\item The region bounded by the curve $y = x ^ { 3 }$ for $0 \leqslant x \leqslant 2$, the $x$-axis and the line $x = 2$, is occupied by a uniform lamina. Find the coordinates of the centre of mass of this lamina. [8]
\item The region bounded by the circular arc $y = \sqrt { 4 - x ^ { 2 } }$ for $1 \leqslant x \leqslant 2$, the $x$-axis and the line $x = 1$, is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid of revolution, as shown in Fig. 4.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{39e14918-5017-43c0-9b74-7c68717ad5f3-5_627_499_593_785}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of the centre of mass of this solid of revolution is 1.35 .
This solid is placed on a rough horizontal surface, with its flat face in a vertical plane. It is held in equilibrium by a light horizontal string attached to its highest point and perpendicular to its flat face, as shown in Fig. 4.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{39e14918-5017-43c0-9b74-7c68717ad5f3-5_573_613_1662_728}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
\item Find the least possible coefficient of friction between the solid and the horizontal surface.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2007 Q4 [18]}}