| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring: (i) recall of the standard result for a quarter-circle lamina (4r/3π), (ii) straightforward composite body calculation using moments, and (iii) basic equilibrium geometry. All techniques are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(d = (2x 6 \sin\pi/4)/3\pi/4\) | M1 A1 | must be correct formula with rads |
| \(d = 3.61\) | A1 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(d \cos 45° = "2.55"\) | B1 | may be implied |
| \(5\bar{x} = 3 \times 3 + 2 \times "2.55"\) | M1 A1 | moments must not have areas |
| \(\bar{x} = 2.82\) | A1 | |
| \(5\bar{y} = 3 \times 6 + 2 \times (12 + "2.55")\) | M1 A1 | 2kg/3kg misread (swap) gives (2.73,11.13) 0 = 21.7° |
| \(\bar{y} = 9.42\) | A1 | (MR – 2) (max 7 for (ii) + (iii)) |
| 7 | SR -1 for \(\bar{x}, \bar{y}\) swap |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = 2.82/8.58\) | M1 | M0 for their \(\bar{x}/\bar{y}\) |
| \(\theta = 18.2°\) | A1 2 | their \(\bar{x}/(18 - \bar{y})\) |
### Part (i)
$d = (2x 6 \sin\pi/4)/3\pi/4$ | M1 A1 | must be correct formula with rads
$d = 3.61$ | A1 2 |
### Part (ii)
$d \cos 45° = "2.55"$ | B1 | may be implied
$5\bar{x} = 3 \times 3 + 2 \times "2.55"$ | M1 A1 | moments must not have areas
$\bar{x} = 2.82$ | A1 |
$5\bar{y} = 3 \times 6 + 2 \times (12 + "2.55")$ | M1 A1 | 2kg/3kg misread (swap) gives (2.73,11.13) 0 = 21.7°
$\bar{y} = 9.42$ | A1 | (MR – 2) (max 7 for (ii) + (iii))
| | 7 | SR -1 for $\bar{x}, \bar{y}$ swap
### Part (iii)
$\tan\theta = 2.82/8.58$ | M1 | M0 for their $\bar{x}/\bar{y}$
$\theta = 18.2°$ | A1 2 | their $\bar{x}/(18 - \bar{y})$ | 11(i)
Fig. 1
Fig. 1 shows a uniform lamina $B C D$ in the shape of a quarter circle of radius 6 cm . Show that the distance of the centre of mass of the lamina from $B$ is 3.60 cm , correct to 3 significant figures.
A uniform rectangular lamina $A B D E$ has dimensions $A B = 12 \mathrm {~cm}$ and $A E = 6 \mathrm {~cm}$. A single plane object is formed by attaching the rectangular lamina to the lamina $B C D$ along $B D$ (see Fig. 2). The mass of $A B D E$ is 3 kg and the mass of $B C D$ is 2 kg .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e85c2bf4-21a8-4d9a-93c5-d5679b2a8233-3_959_447_1123_849}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
(ii) Taking $x$ - and $y$-axes along $A E$ and $A B$ respectively, find the coordinates of the centre of mass of the object.
The object is freely suspended at $C$ and rests in equilibrium.\\
(iii) Calculate the angle that $A C$ makes with the vertical.
\hfill \mbox{\textit{OCR M2 2009 Q5 [11]}}