| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with straight rod/wire components only |
| Difficulty | Standard +0.3 This is a systematic centre of mass calculation requiring careful bookkeeping of multiple components (lamina + 6 rods) in both 2D and 3D configurations. While it involves several steps and coordinate systems, the techniques are entirely standard for M2: finding individual centres of mass, using the formula $\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$, and applying equilibrium conditions. The 3D extension in part (ii) and the suspension angle in part (iii) add length but not conceptual difficulty beyond typical M2 questions. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(28\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = 16\begin{pmatrix}2\\2\end{pmatrix} + 2\begin{pmatrix}5\\0\end{pmatrix} + 2\begin{pmatrix}6\\1\end{pmatrix} + 2\begin{pmatrix}5\\2\end{pmatrix} + 2\begin{pmatrix}0\\5\end{pmatrix} + 2\begin{pmatrix}1\\6\end{pmatrix} + 2\begin{pmatrix}2\\5\end{pmatrix}\) | M1 | Complete method |
| B1 | Total mass correct | |
| B1 | 3 c.m. correct (or 4 \(x\)- or \(y\)-values correct) | |
| \(\bar{x} = 2.5\) | A1 | |
| \(\bar{y} = 2.5\) | A1 | [Allow A0 A1 if only error is in total mass]. [If \(\bar{x} = \bar{y}\) claimed by symmetry and only one component worked, replace final A1, A1 by B1 explicit claim of symmetry, A1 for the 2.5] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\bar{x} = \bar{y}\) | B1 | Or by direct calculation |
| \(28\bar{x} = 16 \times 2 + 6 \times 4 + 2 \times 0 + 2 \times 1 + 2 \times 2\) | M1 | Dealing with 'folded' parts for \(\bar{x}\) or for \(\bar{z}\) |
| \(\bar{x} = \frac{31}{14}\) (2.21428...) | A1 | At least 3 terms correct for \(\bar{x}\) |
| A1 | ||
| \(\bar{z} = \frac{8 \times (-1) + 4 \times (-2)}{28} = -\frac{4}{7}\) (-0.57142...) | A1 | All terms correct allowing sign errors |
| A1 | ||
| Distance is \(\sqrt{\left(\frac{31}{14}\right)^2 + \left(\frac{31}{14}\right)^2 + \left(\frac{4}{7}\right)^2}\) | M1 | Use of Pythagoras in 3D on their c.m. |
| \(= 3.18318...\) so \(3.18\) m (3 s.f.) | F1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| [Diagram with A above, \(\alpha\) shown, 3.18318... and 4/7 lengths, C at base] | M1 | c.m. clearly directly below A |
| B1 | Diagram showing \(\alpha\) and known lengths (or equivalent). FT their values. Award if final answer follows their values | |
| \(\sin\alpha = \frac{4}{7} / 3.18318...\) | M1 | Appropriate expression for \(\alpha\). FT their values |
| so \(\alpha = 10.3415...\) so \(10.3°\) (3 s.f.) | A1 | cao |
# Question 3:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $28\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = 16\begin{pmatrix}2\\2\end{pmatrix} + 2\begin{pmatrix}5\\0\end{pmatrix} + 2\begin{pmatrix}6\\1\end{pmatrix} + 2\begin{pmatrix}5\\2\end{pmatrix} + 2\begin{pmatrix}0\\5\end{pmatrix} + 2\begin{pmatrix}1\\6\end{pmatrix} + 2\begin{pmatrix}2\\5\end{pmatrix}$ | M1 | Complete method |
| | B1 | Total mass correct |
| | B1 | 3 c.m. correct (or 4 $x$- or $y$-values correct) |
| $\bar{x} = 2.5$ | A1 | |
| $\bar{y} = 2.5$ | A1 | [Allow A0 A1 if only error is in total mass]. [If $\bar{x} = \bar{y}$ claimed by symmetry and only one component worked, replace final A1, A1 by B1 explicit claim of symmetry, A1 for the 2.5] |
## Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\bar{x} = \bar{y}$ | B1 | Or by direct calculation |
| $28\bar{x} = 16 \times 2 + 6 \times 4 + 2 \times 0 + 2 \times 1 + 2 \times 2$ | M1 | Dealing with 'folded' parts for $\bar{x}$ or for $\bar{z}$ |
| $\bar{x} = \frac{31}{14}$ (2.21428...) | A1 | At least 3 terms correct for $\bar{x}$ |
| | A1 | |
| $\bar{z} = \frac{8 \times (-1) + 4 \times (-2)}{28} = -\frac{4}{7}$ (-0.57142...) | A1 | All terms correct allowing sign errors |
| | A1 | |
| Distance is $\sqrt{\left(\frac{31}{14}\right)^2 + \left(\frac{31}{14}\right)^2 + \left(\frac{4}{7}\right)^2}$ | M1 | Use of Pythagoras in 3D on **their** c.m. |
| $= 3.18318...$ so $3.18$ m (3 s.f.) | F1 | |
## Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| [Diagram with A above, $\alpha$ shown, 3.18318... and 4/7 lengths, C at base] | M1 | c.m. clearly directly below A |
| | B1 | Diagram showing $\alpha$ and known lengths (or equivalent). FT their values. Award if final answer follows **their** values |
| $\sin\alpha = \frac{4}{7} / 3.18318...$ | M1 | Appropriate expression for $\alpha$. FT **their** values |
| so $\alpha = 10.3415...$ so $10.3°$ (3 s.f.) | A1 | cao |
---3 Fig. 3.1 shows an object made up as follows. ABCD is a uniform lamina of mass $16 \mathrm {~kg} . \mathrm { BE } , \mathrm { EF }$, FG, HI, IJ and JD are each uniform rods of mass 2 kg . ABCD, BEFG and HIJD are squares lying in the same plane. The dimensions in metres are shown in the figure.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{43d5bbfb-8726-4bcd-a73d-01728d532e98-4_627_648_429_735}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
(i) Find the coordinates of the centre of mass of the object, referred to the axes shown in Fig.3.1.
The rods are now re-positioned so that BEFG and HIJD are perpendicular to the lamina, as shown in Fig. 3.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{43d5bbfb-8726-4bcd-a73d-01728d532e98-4_442_666_1510_722}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
(ii) Find the $x$-, $y$-and $z$-coordinates of the centre of mass of the object, referred to the axes shown in Fig. 3.2. Calculate the distance of the centre of mass from A .
The object is now freely suspended from A and hangs in equilibrium with AC at $\alpha ^ { \circ }$ to the vertical.\\
(iii) Calculate $\alpha$.
\hfill \mbox{\textit{OCR MEI M2 2005 Q3 [17]}}