| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particle attached to lamina - find mass/position |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving a composite lamina (disc with circular hole removed) and then a particle attachment. Part (a) is routine application of the formula for composite bodies using moments about a point. Part (b) requires finding the new centre of mass with the particle attached and using the equilibrium condition that the centre of mass lies vertically below the suspension point. While it involves multiple steps, all techniques are standard M2 material with no novel insight required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\pi(4a)^2}{4} \quad \frac{\pi(2a)^2}{1} \quad \frac{(\pi(4a)^2 - \pi(2a)^2)}{3}\) | B1 | Correct mass ratios |
| \(\frac{4a}{2a}\) ... \(\frac{}{x}\) | B1 | Distance of c of m from \(P\) (or from a point on \(OP\)) |
| \((4 \times 4a) - (1 \times 2a) = 3\frac{}{x}\) | M1 | Moments about axis through \(P\), or about a parallel axis then convert the answer to distance from \(P\). Condone a sign slip. Answer given – check working carefully. Must reach positive answer legitimately |
| \(\frac{14a}{3} = x\) * | A1 | |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(OG = 4a\tan\alpha = \frac{10a}{3} \left(\Rightarrow PG = \frac{2a}{3}\right)\) | M1 | Vertical through \(S\) cuts \(OP\) at \(G\). Use trig to find the position of \(G\) on \(OP\) |
| A1 | \(OG = \frac{10a}{3}, OG = \frac{22a}{3}\) or \(PG = \frac{2a}{3}\) seen or implied | |
| \(M(P), (m + km)g.\frac{2a}{3}\cos\alpha = mg.\frac{14a}{3}\cos\alpha\) | M1 | Take moments about a point on \(OP\) – terms should be dimensionally consistent. Masses must be associated with the appropriate distances, which might be incorrectly evaluated or not yet found – e.g. accept with \(OG\). Must have the right terms but condone trig confusion. Also condone absence of trig |
| \(M(G): km \times \frac{2}{3}a = m \times \left(\frac{10}{3}a + \frac{2}{3}a\right) = 4ma\) | M1 | |
| \(M(O): m(1+k) \times \frac{10}{3}a + m \times \frac{2}{3}a = km \times 4a\) | ||
| \(M(C): \frac{12}{3}a \times (1 + k)m = \frac{14}{3}a \times km\) | A1 | cso (C is the position of the original centre of mass) |
| \(M(O): \frac{22}{3}a \times m(1 + k) = \frac{10}{3}a \times m + 8a \times km\) | ||
| \(k = 6\) | A1 | cso. See next page for more alternatives… |
| (5) | ||
| 9 |
**(a)**
| $\frac{\pi(4a)^2}{4} \quad \frac{\pi(2a)^2}{1} \quad \frac{(\pi(4a)^2 - \pi(2a)^2)}{3}$ | B1 | Correct mass ratios |
| $\frac{4a}{2a}$ ... $\frac{}{x}$ | B1 | Distance of c of m from $P$ (or from a point on $OP$) |
| $(4 \times 4a) - (1 \times 2a) = 3\frac{}{x}$ | M1 | Moments about axis through $P$, or about a parallel axis then convert the answer to distance from $P$. Condone a sign slip. Answer given – check working carefully. Must reach positive answer legitimately |
| $\frac{14a}{3} = x$ * | A1 | |
| | (4) | |
**(b)**
| $OG = 4a\tan\alpha = \frac{10a}{3} \left(\Rightarrow PG = \frac{2a}{3}\right)$ | M1 | Vertical through $S$ cuts $OP$ at $G$. Use trig to find the position of $G$ on $OP$ |
| | A1 | $OG = \frac{10a}{3}, OG = \frac{22a}{3}$ or $PG = \frac{2a}{3}$ seen or implied |
| $M(P), (m + km)g.\frac{2a}{3}\cos\alpha = mg.\frac{14a}{3}\cos\alpha$ | M1 | Take moments about a point on $OP$ – terms should be dimensionally consistent. Masses must be associated with the appropriate distances, which might be incorrectly evaluated or not yet found – e.g. accept with $OG$. Must have the right terms but condone trig confusion. Also condone absence of trig |
| $M(G): km \times \frac{2}{3}a = m \times \left(\frac{10}{3}a + \frac{2}{3}a\right) = 4ma$ | M1 | |
| $M(O): m(1+k) \times \frac{10}{3}a + m \times \frac{2}{3}a = km \times 4a$ | | |
| $M(C): \frac{12}{3}a \times (1 + k)m = \frac{14}{3}a \times km$ | A1 | cso (C is the position of the original centre of mass) |
| $M(O): \frac{22}{3}a \times m(1 + k) = \frac{10}{3}a \times m + 8a \times km$ | | |
| $k = 6$ | A1 | cso. See next page for more alternatives… |
| | (5) |
| | 9 | |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{12cd7355-f632-4a84-825f-a269851c6ec4-06_796_789_276_566}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform circular disc has centre $O$ and radius 4a. The lines $P Q$ and $S T$ are perpendicular diameters of the disc. A circular hole of radius $2 a$ is made in the disc, with the centre of the hole at the point $R$ on $O P$ where $O R = 2 a$, to form the lamina $L$, shown shaded in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of $L$ from $P$ is $\frac { 14 a } { 3 }$.
The mass of $L$ is $m$ and a particle of mass $k m$ is now fixed to $L$ at the point $P$. The system is now suspended from the point $S$ and hangs freely in equilibrium. The diameter $S T$ makes an angle $\alpha$ with the downward vertical through $S$, where $\tan \alpha = \frac { 5 } { 6 }$.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q4 [9]}}