| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina in equilibrium with applied force |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question with routine techniques: part (a) is a bookwork proof using volumes of revolution, part (b)(i) requires standard integration for lamina centroid, and part (b)(ii) involves straightforward equilibrium equations with moments. All techniques are textbook exercises requiring no novel insight, making it slightly easier than average for Further Maths M3. |
| 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 | Mark | Guidance |
| \(\int \pi x y^2 \, dx = \int_0^a \pi x(a^2 - x^2) \, dx\) | M1 | Limits not required |
| \(= \pi\left[\frac{1}{2}a^2x^2 - \frac{1}{4}x^4\right]_0^a\) | A1 | For \(\frac{1}{2}a^2x^2 - \frac{1}{4}x^4\) |
| \(= \frac{1}{4}\pi a^4\) | A1 | |
| \(\bar{x} = \dfrac{\frac{1}{4}\pi a^4}{\frac{2}{3}\pi a^3}\) | M1 | |
| \(= \frac{3}{8}a\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area is \(\int_1^4 (2 - \sqrt{x}) \, dx\) | M1 | Limits not required |
| \(= \left[2x - \frac{2}{3}x^{3/2}\right]_1^4 \left(= \frac{4}{3}\right)\) | A1 | For \(2x - \frac{2}{3}x^{\frac{3}{2}}\) |
| \(\int xy \, dx = \int_1^4 x(2 - \sqrt{x}) \, dx\) | M1 | Limits not required |
| \(= \left[x^2 - \frac{2}{5}x^{5/2}\right]_1^4 \left(= \frac{13}{5}\right)\) | A1 | For \(x^2 - \frac{2}{5}x^{\frac{5}{2}}\) |
| \(\bar{x} = \dfrac{\frac{13}{5}}{\frac{4}{3}} = \dfrac{39}{20} = 1.95\) | A1 | |
| \(\int \frac{1}{2}y^2 \, dx = \int_1^4 \frac{1}{2}(2-\sqrt{x})^2 \, dx\) | M1 | \(\int(2-\sqrt{x})^2 \, dx\) or \(\int\left((2-y)^2 - 1\right)y \, dy\) |
| \(= \left[2x - \frac{4}{3}x^{3/2} + \frac{1}{4}x^2\right]_1^4 \left(= \frac{5}{12}\right)\) | A2 | For \(2x - \frac{4}{3}x^{\frac{3}{2}} + \frac{1}{4}x^2\) or \(\frac{3}{2}y^2 - \frac{4}{3}y^3 + \frac{1}{4}y^4\); Give A1 for two terms correct, or all correct with \(\frac{1}{2}\) omitted |
| \(\bar{y} = \dfrac{\frac{5}{12}}{\frac{4}{3}} = \dfrac{5}{16} = 0.3125\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Taking moments about A: \(T_C \times 3 - W \times 0.95 = 0\) | M1, A1 | Moments equation (no force omitted); Any correct moments equation (May involve both \(T_A\) and \(T_C\)); Accept \(Wg\) or \(W = \frac{4}{3}, \frac{4}{3}g\) |
| \(T_A + T_C = W\) | M1 | Resolving vertically (or a second moments equation) |
| \(T_A = \frac{41}{60}W\), \(T_C = \frac{19}{60}W\) | A1 | Accept \(0.68W\), \(0.32W\) |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \pi x y^2 \, dx = \int_0^a \pi x(a^2 - x^2) \, dx$ | M1 | Limits not required |
| $= \pi\left[\frac{1}{2}a^2x^2 - \frac{1}{4}x^4\right]_0^a$ | A1 | For $\frac{1}{2}a^2x^2 - \frac{1}{4}x^4$ |
| $= \frac{1}{4}\pi a^4$ | A1 | |
| $\bar{x} = \dfrac{\frac{1}{4}\pi a^4}{\frac{2}{3}\pi a^3}$ | M1 | |
| $= \frac{3}{8}a$ | E1 | |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area is $\int_1^4 (2 - \sqrt{x}) \, dx$ | M1 | Limits not required |
| $= \left[2x - \frac{2}{3}x^{3/2}\right]_1^4 \left(= \frac{4}{3}\right)$ | A1 | For $2x - \frac{2}{3}x^{\frac{3}{2}}$ |
| $\int xy \, dx = \int_1^4 x(2 - \sqrt{x}) \, dx$ | M1 | Limits not required |
| $= \left[x^2 - \frac{2}{5}x^{5/2}\right]_1^4 \left(= \frac{13}{5}\right)$ | A1 | For $x^2 - \frac{2}{5}x^{\frac{5}{2}}$ |
| $\bar{x} = \dfrac{\frac{13}{5}}{\frac{4}{3}} = \dfrac{39}{20} = 1.95$ | A1 | |
| $\int \frac{1}{2}y^2 \, dx = \int_1^4 \frac{1}{2}(2-\sqrt{x})^2 \, dx$ | M1 | $\int(2-\sqrt{x})^2 \, dx$ or $\int\left((2-y)^2 - 1\right)y \, dy$ |
| $= \left[2x - \frac{4}{3}x^{3/2} + \frac{1}{4}x^2\right]_1^4 \left(= \frac{5}{12}\right)$ | A2 | For $2x - \frac{4}{3}x^{\frac{3}{2}} + \frac{1}{4}x^2$ or $\frac{3}{2}y^2 - \frac{4}{3}y^3 + \frac{1}{4}y^4$; Give A1 for two terms correct, or all correct with $\frac{1}{2}$ omitted |
| $\bar{y} = \dfrac{\frac{5}{12}}{\frac{4}{3}} = \dfrac{5}{16} = 0.3125$ | A1 | |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Taking moments about A: $T_C \times 3 - W \times 0.95 = 0$ | M1, A1 | Moments equation (no force omitted); Any correct moments equation (May involve both $T_A$ and $T_C$); Accept $Wg$ or $W = \frac{4}{3}, \frac{4}{3}g$ |
| $T_A + T_C = W$ | M1 | Resolving vertically (or a second moments equation) |
| $T_A = \frac{41}{60}W$, $T_C = \frac{19}{60}W$ | A1 | Accept $0.68W$, $0.32W$ |
---2
\begin{enumerate}[label=(\alph*)]
\item A uniform solid hemisphere of volume $\frac { 2 } { 3 } \pi a ^ { 3 }$ is formed by rotating the region bounded by the $x$-axis, the $y$-axis and the curve $y = \sqrt { a ^ { 2 } - x ^ { 2 } }$ for $0 \leqslant x \leqslant a$, through $2 \pi$ radians about the $x$-axis.
Show that the $x$-coordinate of the centre of mass of the hemisphere is $\frac { 3 } { 8 } a$.
\item A uniform lamina is bounded by the $x$-axis, the line $x = 1$, and the curve $y = 2 - \sqrt { x }$ for $1 \leqslant x \leqslant 4$. Its corners are $\mathrm { A } ( 1,1 ) , \mathrm { B } ( 1,0 )$ and $\mathrm { C } ( 4,0 )$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the centre of mass of the lamina.
The lamina is suspended with AB vertical and BC horizontal by light vertical strings attached to A and C , as shown in Fig. 2. The weight of the lamina is $W$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{023afdfb-21b6-40fe-9a09-e6769667ee7b-2_346_684_1672_772}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
\item Find the tensions in the two strings in terms of $W$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2010 Q2 [18]}}