Edexcel M2 2007 January — Question 3 10 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2007
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeLamina with removed circle/semicircle
DifficultyStandard +0.8 This is a multi-step centre of mass problem requiring: (a) composite body calculation with negative mass for removed disc, coordinate geometry to find position; (b) moments equilibrium with the template rotating about a point. Requires careful setup of coordinate systems, understanding of composite bodies, and torque balance. More demanding than standard M2 questions due to the geometry and two-part nature, but follows established techniques without requiring novel insight.
Spec6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_1} Figure 1 shows a template \(T\) made by removing a circular disc, of centre \(X\) and radius 8 cm, from a uniform circular lamina, of centre \(O\) and radius 24 cm. The point \(X\) lies on the diameter \(AOB\) of the lamina and \(AX = 16\) cm. The centre of mass of \(T\) is at the point \(G\).
  1. Find \(AG\). [6]
The template \(T\) is free to rotate about a smooth fixed horizontal axis, perpendicular to the plane of \(T\), which passes through the mid-point of \(OB\). A small stud of mass \(\frac{1}{4}m\) is fixed at \(B\), and \(T\) and the stud are in equilibrium with \(AB\) horizontal. Modelling the stud as a particle,
  1. find the mass of \(T\) in terms of \(m\). [4]

(a) Mass Ratios: Large \(24^2\), Small \(8^2\), Template \(512\) (c.1810, c.200, c.1610)
Anything in ratio 9:1:8
\(M(A)\) \(9 \times 24 = 16 \times 1 + 8x\)
AnswerMarks Guidance
\(\bar{x} = 25\) (cm) exactB1, B1ft, M1*, A1, DM1*, A1 6 marks
(b) \(M(\text{axis})\) \(11M = 12 \times \frac{1}{4}m\) fit their \(\bar{x}\)
\((36 - \bar{x})M = 12 \times \frac{1}{4}m\)
AnswerMarks Guidance
\(M = \frac{3}{11}m\) (o.e.e.)M1 †, A1ft, DM1 †, A1 4 marks; 10 marks total
(a) **Mass Ratios:** Large $24^2$, Small $8^2$, Template $512$ (c.1810, c.200, c.1610)
Anything in ratio 9:1:8
$M(A)$ $9 \times 24 = 16 \times 1 + 8x$
$\bar{x} = 25$ (cm) exact | B1, B1ft, M1*, A1, DM1*, A1 | 6 marks

(b) $M(\text{axis})$ $11M = 12 \times \frac{1}{4}m$ fit their $\bar{x}$
$(36 - \bar{x})M = 12 \times \frac{1}{4}m$
$M = \frac{3}{11}m$ (o.e.e.) | M1 †, A1ft, DM1 †, A1 | 4 marks; 10 marks total

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\includegraphics{figure_1}

Figure 1 shows a template $T$ made by removing a circular disc, of centre $X$ and radius 8 cm, from a uniform circular lamina, of centre $O$ and radius 24 cm. The point $X$ lies on the diameter $AOB$ of the lamina and $AX = 16$ cm. The centre of mass of $T$ is at the point $G$.

\begin{enumerate}[label=(\alph*)]
\item Find $AG$. [6]
\end{enumerate}

The template $T$ is free to rotate about a smooth fixed horizontal axis, perpendicular to the plane of $T$, which passes through the mid-point of $OB$. A small stud of mass $\frac{1}{4}m$ is fixed at $B$, and $T$ and the stud are in equilibrium with $AB$ horizontal. Modelling the stud as a particle,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the mass of $T$ in terms of $m$. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2007 Q3 [10]}}