| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving composite shapes (arc + straight line). Part (i) applies a given formula, part (ii) uses standard weighted mean calculations with arc length πr and straight line length. Part (iii) involves basic equilibrium with the centre of mass directly below the pivot. All techniques are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| Semi-circle: \(r=8\), \(\theta = \frac{\pi}{2}\) | M1 | Correct identification of parameters |
| \(CG = \frac{8\sin(\pi/2)}{\pi/2} = \frac{8}{\pi/2} = \frac{16}{\pi}\) | A1 | |
| Centre C is at \((0,0)\); G is \(\frac{16}{\pi}\) to the left, so \(x = -\frac{16}{\pi}\); \(y\)-coordinate of centre \(= 8\) | A1 | Confirm both coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Length of semi-arc \(= \pi \times 8 = 8\pi\) | B1 | |
| Length of OB \(= 72\) | B1 | |
| \(\bar{x}\): taking moments about \(y\)-axis | M1 | Correct method for composite |
| \(8\pi \times \left(-\frac{16}{\pi}\right) + 72 \times 36 = (8\pi + 72)\bar{x}\) | A1 | Correct moment equation for \(x\) |
| \(\bar{x} = \frac{-128 + 2592}{8\pi + 72} = \frac{2464}{8\pi + 72} \approx 25.37\) | A1 | |
| \(\bar{y}\): \(8\pi \times 8 + 72 \times 0 = (8\pi + 72)\bar{y}\) | M1 | |
| \(\bar{y} = \frac{64\pi}{8\pi + 72} \approx 2.07\) | A1 | Both values confirmed to 2 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Diagram: G shown vertically below A, with horizontal and vertical distances marked | B1 | |
| Horizontal distance of G from A \(= 25.37 - 0 = 25.37\) (but A is at \(x=0\); OB along \(x\)-axis, A at top of arc) | M1 | Correct geometry relating G to A |
| Position of A: \((0, 16)\); G at \((25.37, 2.07)\) | ||
| Horizontal distance \(= 25.37\), vertical distance \(= 16 - 2.07 = 13.93\) | A1 | |
| \(\tan\alpha = \frac{25.37}{13.93}\) | M1 | |
| \(\alpha \approx 61.2°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking moments about A with OB vertical | M1 | |
| \(F \times 72\cos\alpha' ... \) using horizontal distances from A | ||
| \(12 \times \bar{x}_{horiz} = F \times 72\) (where distances are horizontal from A with OB vertical) | A1 | |
| \(F = \frac{12 \times 25.37}{72} \approx 4.23\) N | A1 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Semi-circle: $r=8$, $\theta = \frac{\pi}{2}$ | M1 | Correct identification of parameters |
| $CG = \frac{8\sin(\pi/2)}{\pi/2} = \frac{8}{\pi/2} = \frac{16}{\pi}$ | A1 | |
| Centre C is at $(0,0)$; G is $\frac{16}{\pi}$ to the left, so $x = -\frac{16}{\pi}$; $y$-coordinate of centre $= 8$ | A1 | Confirm both coordinates |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Length of semi-arc $= \pi \times 8 = 8\pi$ | B1 | |
| Length of OB $= 72$ | B1 | |
| $\bar{x}$: taking moments about $y$-axis | M1 | Correct method for composite |
| $8\pi \times \left(-\frac{16}{\pi}\right) + 72 \times 36 = (8\pi + 72)\bar{x}$ | A1 | Correct moment equation for $x$ |
| $\bar{x} = \frac{-128 + 2592}{8\pi + 72} = \frac{2464}{8\pi + 72} \approx 25.37$ | A1 | |
| $\bar{y}$: $8\pi \times 8 + 72 \times 0 = (8\pi + 72)\bar{y}$ | M1 | |
| $\bar{y} = \frac{64\pi}{8\pi + 72} \approx 2.07$ | A1 | Both values confirmed to 2 d.p. |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagram: G shown vertically below A, with horizontal and vertical distances marked | B1 | |
| Horizontal distance of G from A $= 25.37 - 0 = 25.37$ (but A is at $x=0$; OB along $x$-axis, A at top of arc) | M1 | Correct geometry relating G to A |
| Position of A: $(0, 16)$; G at $(25.37, 2.07)$ | | |
| Horizontal distance $= 25.37$, vertical distance $= 16 - 2.07 = 13.93$ | A1 | |
| $\tan\alpha = \frac{25.37}{13.93}$ | M1 | |
| $\alpha \approx 61.2°$ | A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking moments about A with OB vertical | M1 | |
| $F \times 72\cos\alpha' ... $ using horizontal distances from A | | |
| $12 \times \bar{x}_{horiz} = F \times 72$ (where distances are horizontal from A with OB vertical) | A1 | |
| $F = \frac{12 \times 25.37}{72} \approx 4.23$ N | A1 | |
---
2 The position of the centre of mass, $G$, of a uniform wire bent into the shape of an arc of a circle of radius $r$ and centre C is shown in Fig. 2.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d4aeab2-332a-442f-b1e7-0bbf8a945f0f-3_325_1132_365_669}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
(i) Use this information to show that the centre of mass, G , of the uniform wire bent into the shape of a semi-circular arc of radius 8 shown in Fig. 2.2 has coordinates $\left( - \frac { 16 } { \pi } , 8 \right)$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d4aeab2-332a-442f-b1e7-0bbf8a945f0f-3_586_871_1016_806}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
A walking-stick is modelled as a uniform rigid wire. The walking-stick and coordinate axes are shown in Fig. 2.3. The section from O to A is a semi-circular arc and the section OB lies along the $x$-axis. The lengths are in centimetres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d4aeab2-332a-442f-b1e7-0bbf8a945f0f-3_394_958_1937_552}
\captionsetup{labelformat=empty}
\caption{Fig. 2.3}
\end{center}
\end{figure}
(ii) Show that the coordinates of the centre of mass of the walking-stick are ( $25.37,2.07$ ), correct to two decimal places.
The walking-stick is now hung from a shelf as shown in Fig. 2.4. The only contact between the walking-stick and the shelf is at A .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d4aeab2-332a-442f-b1e7-0bbf8a945f0f-4_339_374_388_842}
\captionsetup{labelformat=empty}
\caption{Fig. 2.4}
\end{center}
\end{figure}
(iii) When the walking-stick is in equilibrium, OB is at an angle $\alpha$ to the vertical.
Draw a diagram showing the position of the centre of mass of the walking-stick in relation to A .\\
Calculate $\alpha$.\\
(iv) The walking-stick is now held in equilibrium, with OB vertical and A still resting on the shelf, by means of a vertical force, $F \mathrm {~N}$, at B . The weight of the walking-stick is 12 N . Calculate $F$.
\hfill \mbox{\textit{OCR MEI M2 2007 Q2 [17]}}