| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Small oscillations period |
| Difficulty | Challenging +1.2 This is a standard M5 compound pendulum problem requiring moment of inertia calculation using perpendicular axis theorem and parallel axis theorem, followed by small angle approximation for period. While it involves multiple steps and Further Maths content, it follows a well-established template with no novel insight required—harder than typical A-level due to the topic but routine for M5 students. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_{AB}\) about \(L\): rod \(AB\) has length \(2a\), mass \(m\), axis through one end: \(I_{AB} = \frac{1}{3}m(2a)^2 = \frac{4ma^2}{3}\) | M1 | Using \(\frac{1}{3}ml^2\) for rod about end |
| \(I_{BC}\) about \(L\): rod \(BC\) has length \(2a\), mass \(m\), axis through \(B\) at distance \(2a\) from \(L\). \(I_{BC} = m(2a)^2 + \frac{1}{3}m(2a)^2 = 4ma^2 + \frac{4ma^2}{3} = \frac{16ma^2}{3}\) | M1 | Using parallel axis theorem correctly |
| \(I_{total} = \frac{4ma^2}{3} + \frac{16ma^2}{3} = \frac{20ma^2}{3}\) | A1 | Completion (given answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Centre of mass of \(AB\): distance \(a\) below \(L\) along \(AB\) | B1 | |
| Centre of mass of \(BC\): distance \(2a\) below \(L\) (at \(B\)) horizontally distance \(a\) from \(B\) | M1 | Finding position of combined centre of mass |
| Combined centre of mass: horizontal distance \(\bar{x} = \frac{m \cdot 0 + m \cdot a}{2m} = \frac{a}{2}\), vertical distance below \(L\): \(\bar{y} = \frac{m \cdot a + m \cdot 2a}{2m} = \frac{3a}{2}\) | A1 | Correct coordinates of centre of mass |
| Distance of centre of mass from \(L\): \(d = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{3a}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{9a^2}{4}} = \frac{a\sqrt{10}}{2}\) | A1 | |
| Equation of motion: \(\frac{20ma^2}{3}\ddot{\theta} = -2mg \cdot \frac{a\sqrt{10}}{2}\sin\theta\) | M1 A1 | Correct equation of rotational motion with restoring torque |
| For small oscillations \(\sin\theta \approx \theta\): \(\ddot{\theta} = -\frac{3g\sqrt{10}}{20a}\theta\) | M1 | Small angle approximation |
| \(\omega^2 = \frac{3g\sqrt{10}}{20a}\), Period \(T = 2\pi\sqrt{\frac{20a}{3g\sqrt{10}}}\) | A1 | Accept equivalent simplified forms |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{AB}$ about $L$: rod $AB$ has length $2a$, mass $m$, axis through one end: $I_{AB} = \frac{1}{3}m(2a)^2 = \frac{4ma^2}{3}$ | M1 | Using $\frac{1}{3}ml^2$ for rod about end |
| $I_{BC}$ about $L$: rod $BC$ has length $2a$, mass $m$, axis through $B$ at distance $2a$ from $L$. $I_{BC} = m(2a)^2 + \frac{1}{3}m(2a)^2 = 4ma^2 + \frac{4ma^2}{3} = \frac{16ma^2}{3}$ | M1 | Using parallel axis theorem correctly |
| $I_{total} = \frac{4ma^2}{3} + \frac{16ma^2}{3} = \frac{20ma^2}{3}$ | A1 | Completion (given answer) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre of mass of $AB$: distance $a$ below $L$ along $AB$ | B1 | |
| Centre of mass of $BC$: distance $2a$ below $L$ (at $B$) horizontally distance $a$ from $B$ | M1 | Finding position of combined centre of mass |
| Combined centre of mass: horizontal distance $\bar{x} = \frac{m \cdot 0 + m \cdot a}{2m} = \frac{a}{2}$, vertical distance below $L$: $\bar{y} = \frac{m \cdot a + m \cdot 2a}{2m} = \frac{3a}{2}$ | A1 | Correct coordinates of centre of mass |
| Distance of centre of mass from $L$: $d = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{3a}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{9a^2}{4}} = \frac{a\sqrt{10}}{2}$ | A1 | |
| Equation of motion: $\frac{20ma^2}{3}\ddot{\theta} = -2mg \cdot \frac{a\sqrt{10}}{2}\sin\theta$ | M1 A1 | Correct equation of rotational motion with restoring torque |
| For small oscillations $\sin\theta \approx \theta$: $\ddot{\theta} = -\frac{3g\sqrt{10}}{20a}\theta$ | M1 | Small angle approximation |
| $\omega^2 = \frac{3g\sqrt{10}}{20a}$, Period $T = 2\pi\sqrt{\frac{20a}{3g\sqrt{10}}}$ | A1 | Accept equivalent simplified forms |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f932d7cb-1299-41d1-8248-cfbf639795ed-08_613_649_221_644}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform piece of wire $A B C$, of mass $2 m$ and length $4 a$, is bent into two straight equal portions, $A B$ and $B C$, which are at right angles to each other, as shown in Figure 1. The wire rotates freely in a vertical plane about a fixed smooth horizontal axis $L$ which passes through $A$ and is perpendicular to the plane of the wire.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of the wire about $L$ is $\frac { 20 m a ^ { 2 } } { 3 }$
\item By writing down an equation of rotational motion for the wire as it rotates about $L$, find the period of small oscillations of the wire about its position of stable equilibrium.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2016 Q5 [11]}}