| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina with hole removed |
| Difficulty | Standard +0.8 This is a multi-part centre of mass question requiring integration proofs for standard results (semicircle, cone), then applying composite body techniques with specific numerical values, and finally using equilibrium conditions. While the individual techniques are standard M3 content, the combination of proof, calculation, and equilibrium application across multiple parts with careful bookkeeping makes this moderately challenging but still within typical Further Maths scope. |
| 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/Working | Marks | Guidance |
| \(\int\frac{1}{2}y^2\,dx = \int_{-a}^{a}\frac{1}{2}(a^2-x^2)\,dx\) | M1 | For integral of \((a^2-x^2)\) |
| \(= \left[\frac{1}{2}(a^2x - \frac{1}{3}x^3)\right]_{-a}^{a} = \frac{2}{3}a^3\) | A1 | |
| \(\bar{y} = \frac{\frac{2}{3}a^3}{\frac{1}{2}\pi a^2}\) | M1 | Dependent on previous M1 |
| \(= \frac{4a}{3\pi}\) | E1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V = \int\pi y^2\,dx = \int_0^h\pi(mx)^2\,dx\) | M1 | For integral of \(x^2\), or use of \(V=\frac{1}{3}\pi r^2 h\) and \(r=mh\) |
| \(= \left[\frac{1}{3}\pi m^2x^3\right]_0^h = \frac{1}{3}\pi m^2h^3\) | A1 | |
| \(\int\pi xy^2\,dx = \int_0^h\pi x(mx)^2\,dx\) | M1 | For integral of \(x^3\) |
| \(= \left[\frac{1}{4}\pi m^2x^4\right]_0^h = \frac{1}{4}\pi m^2h^4\) | A1 | |
| \(\bar{x} = \frac{\frac{1}{4}\pi m^2h^4}{\frac{1}{3}\pi m^2h^3} = \frac{3}{4}h\) | M1 | Dependent on M1 for integral of \(x^3\) |
| E1 | ||
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m_1 = \frac{1}{3}\pi\times0.7^2\times2.4\rho = \frac{1}{3}\pi\rho\times1.176\) | ||
| \(VG_1 = 1.8\) | B1 | For \(m_1\) and \(m_2\) (or volumes) |
| \(m_2 = \frac{1}{3}\pi\times0.4^2\times1.1\rho = \frac{1}{3}\pi\rho\times0.176\) | ||
| \(VG_2 = 1.3 + \frac{3}{4}\times1.1 = 2.125\) | B1 | or \(\frac{1}{4}\times1.1\) from base |
| \((m_1-m_2)(VG) + m_2(VG_2) = m_1(VG_1)\) | M1 | Attempt formula for composite body |
| \((VG) + 0.176\times2.125 = 1.176\times1.8\) | F1 | |
| Distance (VG) is 1.74 m | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| VQG is a right-angle | M1 | |
| \(VQ = VG\cos\theta\) where \(\tan\theta = \frac{0.7}{2.4}\) \((\theta=16.26°)\) | M1 | |
| \(VQ = 1.7428\times\frac{24}{25}\) | ||
| \(= 1.67\) m | A1 | ft is \(VG\times0.96\) |
| Total: 3 |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\frac{1}{2}y^2\,dx = \int_{-a}^{a}\frac{1}{2}(a^2-x^2)\,dx$ | M1 | For integral of $(a^2-x^2)$ |
| $= \left[\frac{1}{2}(a^2x - \frac{1}{3}x^3)\right]_{-a}^{a} = \frac{2}{3}a^3$ | A1 | |
| $\bar{y} = \frac{\frac{2}{3}a^3}{\frac{1}{2}\pi a^2}$ | M1 | Dependent on previous M1 |
| $= \frac{4a}{3\pi}$ | E1 | |
| **Total: 4** | | |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = \int\pi y^2\,dx = \int_0^h\pi(mx)^2\,dx$ | M1 | For integral of $x^2$, or use of $V=\frac{1}{3}\pi r^2 h$ and $r=mh$ |
| $= \left[\frac{1}{3}\pi m^2x^3\right]_0^h = \frac{1}{3}\pi m^2h^3$ | A1 | |
| $\int\pi xy^2\,dx = \int_0^h\pi x(mx)^2\,dx$ | M1 | For integral of $x^3$ |
| $= \left[\frac{1}{4}\pi m^2x^4\right]_0^h = \frac{1}{4}\pi m^2h^4$ | A1 | |
| $\bar{x} = \frac{\frac{1}{4}\pi m^2h^4}{\frac{1}{3}\pi m^2h^3} = \frac{3}{4}h$ | M1 | Dependent on M1 for integral of $x^3$ |
| | E1 | |
| **Total: 6** | | |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m_1 = \frac{1}{3}\pi\times0.7^2\times2.4\rho = \frac{1}{3}\pi\rho\times1.176$ | | |
| $VG_1 = 1.8$ | B1 | For $m_1$ and $m_2$ (or volumes) |
| $m_2 = \frac{1}{3}\pi\times0.4^2\times1.1\rho = \frac{1}{3}\pi\rho\times0.176$ | | |
| $VG_2 = 1.3 + \frac{3}{4}\times1.1 = 2.125$ | B1 | or $\frac{1}{4}\times1.1$ from base |
| $(m_1-m_2)(VG) + m_2(VG_2) = m_1(VG_1)$ | M1 | Attempt formula for composite body |
| $(VG) + 0.176\times2.125 = 1.176\times1.8$ | F1 | |
| Distance (VG) is 1.74 m | A1 | |
| **Total: 5** | | |
## Part (b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| VQG is a right-angle | M1 | |
| $VQ = VG\cos\theta$ where $\tan\theta = \frac{0.7}{2.4}$ $(\theta=16.26°)$ | M1 | |
| $VQ = 1.7428\times\frac{24}{25}$ | | |
| $= 1.67$ m | A1 | ft is $VG\times0.96$ |
| **Total: 3** | | |4
\begin{enumerate}[label=(\alph*)]
\item The region bounded by the $x$-axis and the semicircle $y = \sqrt { a ^ { 2 } - x ^ { 2 } }$ for $- a \leqslant x \leqslant a$ is occupied by a uniform lamina with area $\frac { 1 } { 2 } \pi a ^ { 2 }$. Show by integration that the $y$-coordinate of the centre of mass of this lamina is $\frac { 4 a } { 3 \pi }$.
\item A uniform solid cone is formed by rotating the region between the $x$-axis and the line $y = m x$, for $0 \leqslant x \leqslant h$, through $2 \pi$ radians about the $x$-axis.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of the centre of mass of this cone is $\frac { 3 } { 4 } h$.\\[0pt]
[You may use the formula $\frac { 1 } { 3 } \pi r ^ { 2 } h$ for the volume of a cone.]\\
From such a uniform solid cone with radius 0.7 m and height 2.4 m , a cone of material is removed. The cone removed has radius 0.4 m and height 1.1 m ; the centre of its base coincides with the centre of the base of the original cone, and its axis of symmetry is also the axis of symmetry of the original cone. Fig. 4 shows the resulting object; the vertex of the original cone is V, and A is a point on the circumference of its base.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b8573ee2-771c-4a93-88d9-346a9da94494-5_716_1228_1027_497}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
\item Find the distance of the centre of mass of this object from V .
This object is suspended by a string attached to a point Q on the line VA, and hangs in equilibrium with VA horizontal.
\item Find the distance VQ.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2009 Q4 [18]}}