| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 10 |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of composite lamina |
| Difficulty | Standard +0.8 This is a multi-part Further Maths mechanics question requiring integration to find centre of mass (part ii), then composite body analysis with equilibrium conditions involving trigonometry (part iii). While systematic, it demands multiple techniques (integration, composite CoM formula, equilibrium geometry) and careful coordinate work across several steps, placing it moderately above average difficulty. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| 1 | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(a \int_0^{a^2} x(2 - x) dx\) | M1 | Attempt to integrate to find area; Ignore wrong/missing limits for M mark |
| \(\frac{4}{3}a\) | A1 | BC |
| \(\frac{4}{3}\bar{a}y = a^2 \int_0^{2} \frac{1}{2}x^2(2-x)^2 dx\) | M1 | Use of \(\bar{A}y = \frac{1}{2}\int y^2 dx\); Ignore wrong/missing limits for M mark |
| \(= \frac{8}{15}a^2\) | A1 | BC; Integral correctly evaluated |
| \(\bar{y} = \frac{15a^2}{4a} = \frac{5}{4}a\) | A1 [5] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(3M\bar{Y} = 2M \times \frac{1}{2}a + M \times (a + \frac{2}{3}a)\) | M1 | Moments about base; or other horizontal axis |
| \(\bar{Y} = \frac{4}{3}a\) | A1 | oe (eg \(\bar{Y} = -\frac{1}{4}a\) wrt original x-axis) |
| \(\tan 20° = \frac{1}{\frac{4}{3}a}\) | M1 | Use of CoM position to find a |
| So value of a is 3.43 (3 sf) | A1 [4] |
## (i)
1 | B1 [1] |
## (ii)
$a \int_0^{a^2} x(2 - x) dx$ | M1 | Attempt to integrate to find area; Ignore wrong/missing limits for M mark
$\frac{4}{3}a$ | A1 | BC
$\frac{4}{3}\bar{a}y = a^2 \int_0^{2} \frac{1}{2}x^2(2-x)^2 dx$ | M1 | Use of $\bar{A}y = \frac{1}{2}\int y^2 dx$; Ignore wrong/missing limits for M mark
$= \frac{8}{15}a^2$ | A1 | BC; Integral correctly evaluated
$\bar{y} = \frac{15a^2}{4a} = \frac{5}{4}a$ | A1 [5] | AG
## (iii)
$3M\bar{Y} = 2M \times \frac{1}{2}a + M \times (a + \frac{2}{3}a)$ | M1 | Moments about base; or other horizontal axis
$\bar{Y} = \frac{4}{3}a$ | A1 | oe (eg $\bar{Y} = -\frac{1}{4}a$ wrt original x-axis)
$\tan 20° = \frac{1}{\frac{4}{3}a}$ | M1 | Use of CoM position to find a
So value of a is 3.43 (3 sf) | A1 [4] |
---2 The region bounded by the $x$-axis and the curve $y = a x ( 2 - x )$, where $a$ is a constant, is occupied by a uniform lamina $L _ { 1 }$ (see Fig. 1). Units on the axes are metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a8c9d007-e67f-4637-9e74-630ba9a91442-2_385_349_906_849}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
(i) Write down the value of the $x$-coordinate of the centre of mass of $L _ { 1 }$.\\
(ii) Show that the $y$-coordinate of the centre of mass of $L _ { 1 }$ is $\frac { 2 } { 5 } a$.
The mass of $L _ { 1 }$ is $M \mathrm {~kg}$. A uniform rectangular lamina of width 2 m and height $a \mathrm {~m}$ is made from a different material from that of $L _ { 1 }$ and has a mass of $2 M \mathrm {~kg}$. A new lamina, $L _ { 2 }$, is formed by joining the straight edge of $L _ { 1 }$ to an edge of the rectangular lamina of length 2 m (see Fig. 2).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a8c9d007-e67f-4637-9e74-630ba9a91442-2_547_273_1772_890}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
$L _ { 2 }$ is freely suspended from one of its right-angled corners and hangs in equilibrium with its edge of length 2 m making an angle of $20 ^ { \circ }$ with the horizontal.\\
(iii) Find the value of $a$, giving your answer correct to 3 significant figures.
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q2 [10]}}