| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Challenging +1.2 This is a standard Further Maths M3 centre of mass question requiring solid of revolution integration (part a) and lamina centre of mass with equilibrium (part b). While it involves multiple integration setups and the hyperbola adds mild algebraic complexity, these are routine techniques for this module with no novel problem-solving required. The equilibrium angle calculation in (b)(ii) is straightforward trigonometry once the centre of mass is found. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Volume is \(\int\pi y^2\,dx = \int_k^{4k}\pi(x^2-k^2)\,dx\) | M1 | For \(\int(x^2-k^2)\,dx\) |
| \(= \pi\left[\frac{1}{3}x^3 - k^2x\right]_k^{4k}\) \((=18\pi k^3)\) | A1 | For \(\frac{1}{3}x^3 - k^2x\) |
| \(\int\pi xy^2\,dx = \int_k^{4k}\pi(x^3-k^2x)\,dx\) | M1 | For \(\int xy^2\,dx\) |
| \(= \pi\left[\frac{1}{4}x^4 - \frac{1}{2}k^2x^2\right]_k^{4k}\) \((=\frac{225\pi k^4}{4})\) | A1A1 | For \(\frac{1}{4}x^4\) and \(-\frac{1}{2}k^2x^2\) |
| \(\bar{x} = \frac{\frac{225}{4}\pi k^4}{18\pi k^3}\) | M1 | Dependent on previous M1M1 |
| \(= \frac{25k}{8} = 3.125k\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area is \(\int_0^{2a}\frac{x^3}{a^2}\,dx\) | M1 | For \(\int\frac{x^3}{a^2}\,dx\) |
| \(= \left[\frac{x^4}{4a^2}\right]_0^{2a}\) \((=4a^2)\) | A1 | For \(\frac{x^4}{4a^2}\) |
| \(\int xy\,dx = \int_0^{2a}\frac{x^4}{a^2}\,dx\) | M1 | For \(\int xy\,dx\) |
| \(= \left[\frac{x^5}{5a^2}\right]_0^{2a}\) \((=\frac{32a^3}{5})\) | A1 | For \(\frac{x^5}{5a^2}\) |
| \(\bar{x} = \frac{\frac{32}{5}a^3}{4a^2} = \frac{8a}{5} = 1.6a\) | A1 | |
| \(\int\frac{1}{2}y^2\,dx = \int_0^{2a}\frac{x^6}{2a^4}\,dx\) | M1 | For \(\int y^2\,dx\) or \(\int(2a-x)y\,dy\) |
| \(= \left[\frac{x^7}{14a^4}\right]_0^{2a}\) \((=\frac{64a^3}{7})\) | A1 | For \(\frac{x^7}{14a^4}\) or \(ay^2 - \frac{3}{7}a^{\frac{2}{3}}y^{\frac{7}{3}}\) |
| \(\bar{y} = \frac{\frac{64}{7}a^3}{4a^2} = \frac{16a}{7}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Centre of mass is vertically below A | M1 | May be implied |
| \(\tan\theta = \frac{2a-\bar{x}}{8a-\bar{y}} = \frac{\frac{2}{5}a}{\frac{40}{7}a}\) \((=0.07)\) | M1 | Condone reciprocal |
| Angle is \(4.00°\) (3 sf) | A1 |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Volume is $\int\pi y^2\,dx = \int_k^{4k}\pi(x^2-k^2)\,dx$ | M1 | For $\int(x^2-k^2)\,dx$ |
| $= \pi\left[\frac{1}{3}x^3 - k^2x\right]_k^{4k}$ $(=18\pi k^3)$ | A1 | For $\frac{1}{3}x^3 - k^2x$ |
| $\int\pi xy^2\,dx = \int_k^{4k}\pi(x^3-k^2x)\,dx$ | M1 | For $\int xy^2\,dx$ |
| $= \pi\left[\frac{1}{4}x^4 - \frac{1}{2}k^2x^2\right]_k^{4k}$ $(=\frac{225\pi k^4}{4})$ | A1A1 | For $\frac{1}{4}x^4$ and $-\frac{1}{2}k^2x^2$ |
| $\bar{x} = \frac{\frac{225}{4}\pi k^4}{18\pi k^3}$ | M1 | Dependent on previous M1M1 |
| $= \frac{25k}{8} = 3.125k$ | A1 | |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Area is $\int_0^{2a}\frac{x^3}{a^2}\,dx$ | M1 | For $\int\frac{x^3}{a^2}\,dx$ |
| $= \left[\frac{x^4}{4a^2}\right]_0^{2a}$ $(=4a^2)$ | A1 | For $\frac{x^4}{4a^2}$ |
| $\int xy\,dx = \int_0^{2a}\frac{x^4}{a^2}\,dx$ | M1 | For $\int xy\,dx$ |
| $= \left[\frac{x^5}{5a^2}\right]_0^{2a}$ $(=\frac{32a^3}{5})$ | A1 | For $\frac{x^5}{5a^2}$ |
| $\bar{x} = \frac{\frac{32}{5}a^3}{4a^2} = \frac{8a}{5} = 1.6a$ | A1 | |
| $\int\frac{1}{2}y^2\,dx = \int_0^{2a}\frac{x^6}{2a^4}\,dx$ | M1 | For $\int y^2\,dx$ or $\int(2a-x)y\,dy$ |
| $= \left[\frac{x^7}{14a^4}\right]_0^{2a}$ $(=\frac{64a^3}{7})$ | A1 | For $\frac{x^7}{14a^4}$ or $ay^2 - \frac{3}{7}a^{\frac{2}{3}}y^{\frac{7}{3}}$ |
| $\bar{y} = \frac{\frac{64}{7}a^3}{4a^2} = \frac{16a}{7}$ | A1 | |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Centre of mass is vertically below A | M1 | May be implied |
| $\tan\theta = \frac{2a-\bar{x}}{8a-\bar{y}} = \frac{\frac{2}{5}a}{\frac{40}{7}a}$ $(=0.07)$ | M1 | Condone reciprocal |
| Angle is $4.00°$ (3 sf) | A1 | |4
\begin{enumerate}[label=(\alph*)]
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2dd5719-bef3-45f2-afd2-c481e6a4b129-5_705_501_260_863}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
The region $R$, shown in Fig. 4.1, is bounded by the curve $x ^ { 2 } - y ^ { 2 } = k ^ { 2 }$ for $k \leqslant x \leqslant 4 k$ and the line $x = 4 k$, where $k$ is a positive constant. Find the $x$-coordinate of the centre of mass of the uniform solid of revolution formed when $R$ is rotated about the $x$-axis.
\item A uniform lamina occupies the region bounded by the curve $y = \frac { x ^ { 3 } } { a ^ { 2 } }$ for $0 \leqslant x \leqslant 2 a$, the $x$-axis and the line $x = 2 a$, where $a$ is a positive constant. The vertices of the lamina are $\mathrm { O } ( 0,0 ) , \mathrm { A } ( 2 a , 8 a )$ and $\mathrm { B } ( 2 a , 0 )$, as shown in Fig. 4.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2dd5719-bef3-45f2-afd2-c481e6a4b129-5_714_509_1546_858}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the centre of mass of the lamina.
\item The lamina is freely suspended from the point A and hangs in equilibrium. Find the angle that AB makes with the vertical.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2011 Q4 [18]}}