| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| 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 to find centroids of a lamina and solid of revolution, followed by equilibrium analysis on an inclined plane. While it involves multiple parts and careful integration, the techniques are routine for M3 students: standard centroid formulas, volume of revolution, and applying equilibrium conditions with geometry. The calculations are somewhat lengthy but conceptually straightforward with no novel problem-solving required. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \int_0^9 (3 - \sqrt{x})\,dx\) | M1 | |
| \(= \left[3x - \frac{2}{3}x^{\frac{3}{2}}\right]_0^9\ (= 9)\) | A1 | For \(3x - \frac{2}{3}x^{\frac{3}{2}}\) |
| \(A\bar{x} = \int xy\,dx = \int_0^9 x(3 - \sqrt{x})\,dx\) | M1 | For \(\int xy\,dx\) |
| \(= \left[\frac{3}{2}x^2 - \frac{2}{5}x^{\frac{5}{2}}\right]_0^9\ (= 24.3)\) | A1 | For \(\frac{3}{2}x^2 - \frac{2}{5}x^{\frac{5}{2}}\) |
| \(\bar{x} = \frac{24.3}{9} = 2.7\) | A1 | |
| \(A\bar{y} = \int \frac{1}{2}y^2\,dx = \int_0^9 \frac{1}{2}(3 - \sqrt{x})^2\,dx\) | M1 | For \(\int \ldots y^2\,dx\); or \(\int_0^3 (3-y)^2 y\,dy\) |
| Expanding (three terms) and integrating | M1 | Allow one error |
| \(= \left[\frac{9}{2}x - 2x^{\frac{3}{2}} + \frac{1}{4}x^2\right]_0^9\ (= 6.75)\) | A1 | For \(\frac{9}{2}x - 2x^{\frac{3}{2}} + \frac{1}{4}x^2\); or \(\frac{9}{2}y^2 - 2y^3 + \frac{1}{4}y^4\) |
| \(\bar{y} = \frac{6.75}{9} = 0.75\) | A1 [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = \int_2^5 \pi(25 - x^2)\,dx\) | M1 | For \(\int\ldots(25 - x^2)\,dx\) |
| \(= \pi\left[25x - \frac{1}{3}x^3\right]_2^5\ (= 36\pi)\) | A1 | For \(25x - \frac{1}{3}x^3\) |
| \(V\bar{x} = \int \pi xy^2\,dx = \int_2^5 \pi x(25 - x^2)\,dx\) | M1 | For \(\int xy^2\,dx\) |
| \(= \pi\left[\frac{25}{2}x^2 - \frac{1}{4}x^4\right]_2^5\ \left(= \frac{441\pi}{4}\right)\) | A1 | For \(\frac{25}{2}x^2 - \frac{1}{4}x^4\) |
| \(\bar{x} = \frac{\frac{441\pi}{4}}{36\pi} = \frac{49}{16}\ (= 3.0625)\) | A1 [5] | Accept \(3.1\) from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Diagram showing CG position] | M1 | CG is vertical (may be implied); lenient if CG drawn, needs to be quite accurate if CG not drawn |
| Using triangle OGC or equivalent | M1 | |
| \(\frac{\sin\theta}{5} = \frac{\sin 25°}{\bar{x}}\) | M1 | FT is \(\sin^{-1}\left(\frac{2.113}{\bar{x}}\right)\); provided \(2.113 < \bar{x} < 5\) |
| \(\theta = 43.6°\) | A1 [4] | Accept \(43°\) or \(44°\) from correct work |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \int_0^9 (3 - \sqrt{x})\,dx$ | M1 | |
| $= \left[3x - \frac{2}{3}x^{\frac{3}{2}}\right]_0^9\ (= 9)$ | A1 | For $3x - \frac{2}{3}x^{\frac{3}{2}}$ |
| $A\bar{x} = \int xy\,dx = \int_0^9 x(3 - \sqrt{x})\,dx$ | M1 | For $\int xy\,dx$ |
| $= \left[\frac{3}{2}x^2 - \frac{2}{5}x^{\frac{5}{2}}\right]_0^9\ (= 24.3)$ | A1 | For $\frac{3}{2}x^2 - \frac{2}{5}x^{\frac{5}{2}}$ |
| $\bar{x} = \frac{24.3}{9} = 2.7$ | A1 | |
| $A\bar{y} = \int \frac{1}{2}y^2\,dx = \int_0^9 \frac{1}{2}(3 - \sqrt{x})^2\,dx$ | M1 | For $\int \ldots y^2\,dx$; or $\int_0^3 (3-y)^2 y\,dy$ |
| Expanding (three terms) and integrating | M1 | Allow one error |
| $= \left[\frac{9}{2}x - 2x^{\frac{3}{2}} + \frac{1}{4}x^2\right]_0^9\ (= 6.75)$ | A1 | For $\frac{9}{2}x - 2x^{\frac{3}{2}} + \frac{1}{4}x^2$; or $\frac{9}{2}y^2 - 2y^3 + \frac{1}{4}y^4$ |
| $\bar{y} = \frac{6.75}{9} = 0.75$ | A1 [9] | |
---
## Question 4(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \int_2^5 \pi(25 - x^2)\,dx$ | M1 | For $\int\ldots(25 - x^2)\,dx$ |
| $= \pi\left[25x - \frac{1}{3}x^3\right]_2^5\ (= 36\pi)$ | A1 | For $25x - \frac{1}{3}x^3$ |
| $V\bar{x} = \int \pi xy^2\,dx = \int_2^5 \pi x(25 - x^2)\,dx$ | M1 | For $\int xy^2\,dx$ |
| $= \pi\left[\frac{25}{2}x^2 - \frac{1}{4}x^4\right]_2^5\ \left(= \frac{441\pi}{4}\right)$ | A1 | For $\frac{25}{2}x^2 - \frac{1}{4}x^4$ |
| $\bar{x} = \frac{\frac{441\pi}{4}}{36\pi} = \frac{49}{16}\ (= 3.0625)$ | A1 [5] | Accept $3.1$ from correct working |
---
## Question 4(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Diagram showing CG position] | M1 | CG is vertical (may be implied); lenient if CG drawn, needs to be quite accurate if CG not drawn |
| Using triangle OGC or equivalent | M1 | |
| $\frac{\sin\theta}{5} = \frac{\sin 25°}{\bar{x}}$ | M1 | FT is $\sin^{-1}\left(\frac{2.113}{\bar{x}}\right)$; provided $2.113 < \bar{x} < 5$ |
| $\theta = 43.6°$ | A1 [4] | Accept $43°$ or $44°$ from correct work |4
\begin{enumerate}[label=(\alph*)]
\item A uniform lamina occupies the region bounded by the $x$-axis, the $y$-axis and the curve $y = 3 - \sqrt { x }$ for $0 \leqslant x \leqslant 9$. Find the coordinates of the centre of mass of this lamina.
\item Fig. 4.1 shows the region bounded by the line $x = 2$ and the part of the circle $y ^ { 2 } = 25 - x ^ { 2 }$ for which $2 \leqslant x \leqslant 5$. This region is rotated about the $x$-axis to form a uniform solid of revolution $S$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86dd0c01-970d-4b67-9a6c-5df276a4a2be-5_675_659_479_705}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find the $x$-coordinate of the centre of mass of $S$.
The solid $S$ rests in equilibrium with its curved surface in contact with a rough plane inclined at $25 ^ { \circ }$ to the horizontal. Fig. 4.2 shows a vertical section containing AB , which is a diameter and also a line of greatest slope of the flat surface of $S$. This section also contains XY, which is a line of greatest slope of the plane.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86dd0c01-970d-4b67-9a6c-5df276a4a2be-5_494_560_1615_749}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
\item Find the angle $\theta$ that AB makes with the horizontal.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2012 Q4 [18]}}