| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass problem requiring calculation of centre of mass coordinates for a composite wire shape, then using equilibrium conditions to find an angle. The geometry is straightforward (right angles, given lengths), and the method is routine: find COM using moments, then use tan θ = horizontal/vertical distance. Slightly easier than average due to the simple geometry and standard technique. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| portion | mass | \(x\) |
| \(AB\) | \(2a\rho\) | \(0\) |
| \(BC\) | \(3a\rho\) | \(\frac{3}{2}a\) |
| \(CD\) | \(a\rho\) | \(3a\) |
| total | \(6a\rho\) | \(\bar{x}\) |
| \(\rho =\) mass per unit area; \(x, y\) coords. taken horiz./vert. from \(B\) | M2 A2 | |
| \(\bar{x} = \frac{\frac{15}{2}a^2\rho}{6a\rho} = \frac{5a}{4}\) from \(AB\) | M1 A1 | |
| \(\bar{y} = \frac{\frac{5}{2}a^2\rho}{6a\rho} = \frac{5a}{12}\) from \(BC\) | M1 A1 | |
| (b) \(2a - \frac{5a}{12} = \frac{19a}{12}\) | A1 | |
| Diagram showing triangle with height \(\frac{19a}{12}\), base \(\frac{5a}{4}\) | ||
| \(\tan\theta = \frac{\frac{5a}{4}}{\frac{19a}{12}} = \frac{15}{19} \quad \theta = 38°\) (nearest degree) | M2 A1 | (12) |
**(a)** (i), (ii) Table with centre of mass calculations:
| portion | mass | $x$ | $y$ | $mx$ | $my$ |
|---------|------|-----|-----|------|------|
| $AB$ | $2a\rho$ | $0$ | $a$ | $0$ | $2a^2\rho$ |
| $BC$ | $3a\rho$ | $\frac{3}{2}a$ | $0$ | $\frac{9}{2}a^2\rho$ | $0$ |
| $CD$ | $a\rho$ | $3a$ | $\frac{1}{2}a$ | $3a^2\rho$ | $\frac{1}{2}a^2\rho$ |
| total | $6a\rho$ | $\bar{x}$ | $\bar{y}$ | $\frac{15}{2}a^2\rho$ | $\frac{5}{2}a^2\rho$ |
$\rho =$ mass per unit area; $x, y$ coords. taken horiz./vert. from $B$ | M2 A2 |
$\bar{x} = \frac{\frac{15}{2}a^2\rho}{6a\rho} = \frac{5a}{4}$ from $AB$ | M1 A1 |
$\bar{y} = \frac{\frac{5}{2}a^2\rho}{6a\rho} = \frac{5a}{12}$ from $BC$ | M1 A1 |
**(b)** $2a - \frac{5a}{12} = \frac{19a}{12}$ | A1 |
Diagram showing triangle with height $\frac{19a}{12}$, base $\frac{5a}{4}$ |
$\tan\theta = \frac{\frac{5a}{4}}{\frac{19a}{12}} = \frac{15}{19} \quad \theta = 38°$ (nearest degree) | M2 A1 | (12)4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ef2dd10c-5a3c-4868-af00-aaf7f2937d7e-3_222_350_242_788}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Figure 2 shows an earring consisting of a uniform wire $A B C D$ of length $6 a$ bent to form right angles at $B$ and $C$ such that $A B$ and $C D$ are of length $2 a$ and $a$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the distance of the centre of mass from
\begin{enumerate}[label=(\roman*)]
\item $\quad A B$,
\item $B C$.
The earring is to be worn such that it hangs in equilibrium suspended from the point $A$.
\end{enumerate}\item Find, to the nearest degree, the angle made by $A B$ with the downward vertical.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q4 [12]}}