| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| 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 Further Maths M3 centre of mass question requiring integration for a solid of revolution (using Pappus or volume integrals), followed by a straightforward toppling condition (vertical line through COM must pass through pivot edge). The integration is routine for this level, and the toppling geometry is a standard application. More challenging than basic A-level due to the solid of revolution setup, but follows well-established M3 techniques without requiring novel insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = \int \pi x^2\, dy = \int_0^8 \pi(4 - \frac{1}{2}y)\, dy\) | M1 | \(\pi\) may be omitted throughout; limits not required for M marks |
| \(= \pi\left[4y - \frac{1}{4}y^2\right]_0^8 = 16\pi\) | A1 | |
| \(V\bar{y} = \int \pi y x^2\, dy\) | M1 | |
| \(= \int_0^8 \pi y(4 - \frac{1}{2}y)\, dy\) | A1 | |
| \(= \pi\left[2y^2 - \frac{1}{6}y^3\right]_0^8 = \frac{128}{3}\pi\) | A1 | |
| \(\bar{y} = \frac{\frac{128}{3}\pi}{16\pi}\) | M1 | Dependent on M1M1 |
| \(= \frac{8}{3}\ (\approx 2.67)\) | A1 | 7 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CM is vertically above lower corner | M1 | |
| \(\tan\theta = \frac{2}{\bar{y}} = \frac{2}{\frac{8}{3}} = \left(\frac{3}{4}\right)\) | M1, A1 | Trig in triangle including \(\theta\); dependent on previous M1; correct expression for \(\tan\theta\) or \(\tan(90-\theta)\). *Notes: \(\tan\theta = \frac{2}{\text{cand's }\bar{y}}\) implies M1M1A1; \(\tan\theta = \frac{\text{cand's }\bar{y}}{2}\) implies M1M1; \(\tan\theta = \frac{1}{\text{cand's }\bar{y}}\) without further evidence is M0M0* |
| \(\theta = 36.9°\) (0.6435 rad) | A1 | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = \int_{-2}^{2}(8 - 2x^2)\, dx\) | M1 | *or* \((2)\int_0^8 \sqrt{4 - \frac{1}{2}y}\, dy\) |
| \(= \left[8x - \frac{2}{3}x^3\right]_{-2}^{2} = \frac{64}{3}\) | A1 | |
| \(A\bar{y} = \int_{-2}^{2}\frac{1}{2}(8-2x^2)^2\, dx\) | M1 | *or* \((2)\int_0^8 y\sqrt{4 - \frac{1}{2}y}\, dy\) *(M0 if \(\frac{1}{2}\) is omitted)* |
| \(= \left[32x - \frac{16}{3}x^3 + \frac{2}{5}x^5\right]_{-2}^{2}\) | M1 | For \(32x - \frac{16}{3}x^3 + \frac{2}{5}x^5\); allow one error |
| \(= \frac{1024}{15}\) | A1 | |
| \(\bar{y} = \frac{\frac{1024}{15}}{\frac{64}{3}}\) | M1 | Dependent on first two M1s |
| \(= \frac{16}{5} = 3.2\) | A1 | 7 marks total |
# Question 4(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \int \pi x^2\, dy = \int_0^8 \pi(4 - \frac{1}{2}y)\, dy$ | M1 | $\pi$ may be omitted throughout; limits not required for M marks |
| $= \pi\left[4y - \frac{1}{4}y^2\right]_0^8 = 16\pi$ | A1 | |
| $V\bar{y} = \int \pi y x^2\, dy$ | M1 | |
| $= \int_0^8 \pi y(4 - \frac{1}{2}y)\, dy$ | A1 | |
| $= \pi\left[2y^2 - \frac{1}{6}y^3\right]_0^8 = \frac{128}{3}\pi$ | A1 | |
| $\bar{y} = \frac{\frac{128}{3}\pi}{16\pi}$ | M1 | Dependent on M1M1 |
| $= \frac{8}{3}\ (\approx 2.67)$ | A1 | **7 marks total** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CM is vertically above lower corner | M1 | |
| $\tan\theta = \frac{2}{\bar{y}} = \frac{2}{\frac{8}{3}} = \left(\frac{3}{4}\right)$ | M1, A1 | Trig in triangle including $\theta$; dependent on previous M1; correct expression for $\tan\theta$ or $\tan(90-\theta)$. *Notes: $\tan\theta = \frac{2}{\text{cand's }\bar{y}}$ implies M1M1A1; $\tan\theta = \frac{\text{cand's }\bar{y}}{2}$ implies M1M1; $\tan\theta = \frac{1}{\text{cand's }\bar{y}}$ without further evidence is M0M0* |
| $\theta = 36.9°$ (0.6435 rad) | A1 | **4 marks total** |
---
# Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = \int_{-2}^{2}(8 - 2x^2)\, dx$ | M1 | *or* $(2)\int_0^8 \sqrt{4 - \frac{1}{2}y}\, dy$ |
| $= \left[8x - \frac{2}{3}x^3\right]_{-2}^{2} = \frac{64}{3}$ | A1 | |
| $A\bar{y} = \int_{-2}^{2}\frac{1}{2}(8-2x^2)^2\, dx$ | M1 | *or* $(2)\int_0^8 y\sqrt{4 - \frac{1}{2}y}\, dy$ *(M0 if $\frac{1}{2}$ is omitted)* |
| $= \left[32x - \frac{16}{3}x^3 + \frac{2}{5}x^5\right]_{-2}^{2}$ | M1 | For $32x - \frac{16}{3}x^3 + \frac{2}{5}x^5$; allow one error |
| $= \frac{1024}{15}$ | A1 | |
| $\bar{y} = \frac{\frac{1024}{15}}{\frac{64}{3}}$ | M1 | Dependent on first two M1s |
| $= \frac{16}{5} = 3.2$ | A1 | **7 marks total** |4
\begin{enumerate}[label=(\alph*)]
\item A uniform solid of revolution is obtained by rotating through $2 \pi$ radians about the $y$-axis the region bounded by the curve $y = 8 - 2 x ^ { 2 }$ for $0 \leqslant x \leqslant 2$, the $x$-axis and the $y$-axis.
\begin{enumerate}[label=(\roman*)]
\item Find the $y$-coordinate of the centre of mass of this solid.
The solid is now placed on a rough plane inclined at an angle $\theta$ to the horizontal. It rests in equilibrium with its circular face in contact with the plane as shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2a4afead-e772-4d86-bc8d-86ffa5bca507-4_511_568_616_831}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
\item Given that the solid is on the point of toppling, find $\theta$.
\end{enumerate}\item Find the $y$-coordinate of the centre of mass of a uniform lamina in the shape of the region bounded by the curve $y = 8 - 2 x ^ { 2 }$ for $- 2 \leqslant x \leqslant 2$, and the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2008 Q4 [18]}}