OCR MEI M2 2010 January — Question 4 18 marks

Exam BoardOCR MEI
ModuleM2 (Mechanics 2)
Year2010
SessionJanuary
Marks18
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeL-shaped or composite rectangular lamina
DifficultyStandard +0.3 This is a standard multi-part centre of mass question requiring systematic application of the formula for composite shapes. While it involves multiple components (lamina, wire, 3D bin), each part follows routine procedures: decompose into rectangles/line segments, find individual centroids, apply the weighted average formula. The 'show that' parts provide checkpoints. No novel geometric insight or complex integration required—just careful bookkeeping across several steps.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
4 In this question, coordinates refer to the axes shown in the figures and the units are centimetres.
Fig. 4.1 shows a lamina KLMNOP shaded. The lamina is made from uniform material and has the dimensions shown. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f2aaae62-a5f3-47da-afa5-1dd4b37ea2d6-5_512_442_468_532} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f2aaae62-a5f3-47da-afa5-1dd4b37ea2d6-5_510_460_470_1153} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
\end{figure}
  1. Show that the \(x\)-coordinate of the centre of mass of this lamina is 26 and calculate the \(y\)-coordinate. A uniform thin heavy wire KLMNOPQ is bent into the shape of part of the perimeter of the lamina KLMNOP with an extension of the side OP to Q, as shown in Fig. 4.2.
  2. Show that the \(x\)-coordinate of the centre of mass of this wire is 23.2 and calculate the \(y\)-coordinate. The wire is freely suspended from Q and hangs in equilibrium.
  3. Draw a diagram indicating the position of the centre of mass of the hanging wire and calculate the angle of QO with the vertical. A wall-mounted bin with an open top is shown in Fig. 4.3. The centre part has cross-section KLMNOPQ; the two ends are in the shape of the lamina KLMNOP. The ends are made from the same uniform, thin material and each has a mass of 1.5 kg . The centre part is made from different uniform, thin material and has a total mass of 7 kg . \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{f2aaae62-a5f3-47da-afa5-1dd4b37ea2d6-5_499_540_2017_804} \captionsetup{labelformat=empty} \caption{Fig. 4.3}
    \end{figure}
  4. Calculate the \(x\) - and \(y\)-coordinates of the centre of mass of the bin.
4(i)
AnswerMarks Guidance
\(4000\left(\frac{\bar{x}}{5}\right) = 4800\left(\frac{30}{40}\right) - 800\left(\frac{50}{20}\right)\)M1 Any complete method for c.m.
so \(\bar{x} = 26\)
AnswerMarks Guidance
\(\bar{y} = 44\)A1, E1, A1 Either one RHS term correct or one component of both RHS terms correct; [SC 2 for correct \(\bar{y}\) seen if M 0]; cao
4(ii)
AnswerMarks Guidance
\(250\left(\frac{\bar{x}}{\bar{y}}\right)\)M1 Any complete method for c.m.
\(= 110\left(\frac{0}{55}\right) + 40\left(\frac{20}{0}\right) + 40\left(\frac{40}{20}\right) + 20\left(\frac{50}{40}\right) + 40\left(\frac{60}{60}\right)\)B1, B1 Any 2 edges correct mass and c.m. or any 4 edges correct with mass and x or y c.m. coordinate correct; At most one consistent error
\(\bar{x} = 23.2\)
AnswerMarks Guidance
\(\bar{y} = 40.2\)E1, A1 (blank)
4(iii)
AnswerMarks Guidance
![Diagram showing point Q with vertical dashed line through it, point G at distance 23.2 horizontally and 40.2 vertically from origin O, with angle marked as 40.2, and horizontal distance marked as 23.2. Label N on horizontal axis.]B1 Indicating c.m. vertically below Q
(blank)B1 Clearly identifying correct angle (may be implied) and lengths
Angle is \(\arctan\left(\frac{23.2}{110-40.2}\right)\)M1 Award for \(\arctan\left(\frac{b}{a}\right)\) where \(b = 23.2\) and \(a = 69.8\) or 40.2 or where \(b = 69.8\) or 40.2 and \(a = 23.2\). Allow use of their value for y only.
\(= 18.3856...\) so 18.4° (3 s.f.)A1 cao
4(iv)
AnswerMarks Guidance
\(10\left(\frac{\bar{x}}{\bar{y}}\right) = 2 \times 1.5 \times \left(\frac{26}{44}\right) + 7\left(\frac{23.2}{40.2}\right)\)M1, B1, A1 Combining the parts using masses; Using both ends; All correct
\(\bar{x} = 24.04\) so 24.0 (3 s.f.)
AnswerMarks Guidance
\(\bar{y} = 41.34\) so 41.3 (3 s.f.)A1, F1 cao; FT their y values only 
## 4(i)
$4000\left(\frac{\bar{x}}{5}\right) = 4800\left(\frac{30}{40}\right) - 800\left(\frac{50}{20}\right)$ | M1 | Any complete method for c.m.
so $\bar{x} = 26$
$\bar{y} = 44$ | A1, E1, A1 | Either one RHS term correct or one component of both RHS terms correct; [SC 2 for correct $\bar{y}$ seen if M 0]; cao

## 4(ii)
$250\left(\frac{\bar{x}}{\bar{y}}\right)$ | M1 | Any complete method for c.m.
$= 110\left(\frac{0}{55}\right) + 40\left(\frac{20}{0}\right) + 40\left(\frac{40}{20}\right) + 20\left(\frac{50}{40}\right) + 40\left(\frac{60}{60}\right)$ | B1, B1 | Any 2 edges correct mass and c.m. or any 4 edges correct with mass and x or y c.m. coordinate correct; At most one consistent error
$\bar{x} = 23.2$
$\bar{y} = 40.2$ | E1, A1 | (blank)

## 4(iii)
![Diagram showing point Q with vertical dashed line through it, point G at distance 23.2 horizontally and 40.2 vertically from origin O, with angle marked as 40.2, and horizontal distance marked as 23.2. Label N on horizontal axis.] | B1 | Indicating c.m. vertically below Q
(blank) | B1 | Clearly identifying correct angle (may be implied) and lengths
Angle is $\arctan\left(\frac{23.2}{110-40.2}\right)$ | M1 | Award for $\arctan\left(\frac{b}{a}\right)$ where $b = 23.2$ and $a = 69.8$ or 40.2 or where $b = 69.8$ or 40.2 and $a = 23.2$. Allow use of their value for y only.
$= 18.3856...$ so 18.4° (3 s.f.) | A1 | cao

## 4(iv)
$10\left(\frac{\bar{x}}{\bar{y}}\right) = 2 \times 1.5 \times \left(\frac{26}{44}\right) + 7\left(\frac{23.2}{40.2}\right)$ | M1, B1, A1 | Combining the parts using masses; Using both ends; All correct
$\bar{x} = 24.04$ so 24.0 (3 s.f.)
$\bar{y} = 41.34$ so 41.3 (3 s.f.) | A1, F1 | cao; FT their y values only
4 In this question, coordinates refer to the axes shown in the figures and the units are centimetres.\\
Fig. 4.1 shows a lamina KLMNOP shaded. The lamina is made from uniform material and has the dimensions shown.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f2aaae62-a5f3-47da-afa5-1dd4b37ea2d6-5_512_442_468_532}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f2aaae62-a5f3-47da-afa5-1dd4b37ea2d6-5_510_460_470_1153}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}

(i) Show that the $x$-coordinate of the centre of mass of this lamina is 26 and calculate the $y$-coordinate.

A uniform thin heavy wire KLMNOPQ is bent into the shape of part of the perimeter of the lamina KLMNOP with an extension of the side OP to Q, as shown in Fig. 4.2.\\
(ii) Show that the $x$-coordinate of the centre of mass of this wire is 23.2 and calculate the $y$-coordinate.

The wire is freely suspended from Q and hangs in equilibrium.\\
(iii) Draw a diagram indicating the position of the centre of mass of the hanging wire and calculate the angle of QO with the vertical.

A wall-mounted bin with an open top is shown in Fig. 4.3. The centre part has cross-section KLMNOPQ; the two ends are in the shape of the lamina KLMNOP.

The ends are made from the same uniform, thin material and each has a mass of 1.5 kg . The centre part is made from different uniform, thin material and has a total mass of 7 kg .

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f2aaae62-a5f3-47da-afa5-1dd4b37ea2d6-5_499_540_2017_804}
\captionsetup{labelformat=empty}
\caption{Fig. 4.3}
\end{center}
\end{figure}

(iv) Calculate the $x$ - and $y$-coordinates of the centre of mass of the bin.

\hfill \mbox{\textit{OCR MEI M2 2010 Q4 [18]}}
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