| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2008 |
| 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 compound pendulum problem requiring moment of inertia calculation using parallel axis theorem, then applying SHM equation for small oscillations. While it involves multiple steps and careful bookkeeping of distances, the techniques are routine for M5 students with no novel insight required. The 'show that' in part (a) provides a target to verify, making it slightly easier than open-ended problems. |
| Spec | 6.04e Rigid body equilibrium: coplanar forces6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(I = \frac{1}{3}m(9a)^2 + \frac{1}{2}(2m)a^2 + 2m(9a)^2\) | M1 A1 A1 | |
| \(= 27ma^2 + ma^2 + 162ma^2\) | ||
| \(= 190ma^2\) | A1* | *(4)* |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(M(L)\): \(mg\dfrac{9a}{2}\sin\theta + 2mg\cdot 9a\sin\theta = -190ma^2\ddot{\theta}\) | M1 A2 | |
| \(\ddot{\theta} = -\dfrac{9g}{76a}\sin\theta\) | M1 | |
| For small \(\theta\), \(\sin\theta \approx \theta \Rightarrow \ddot{\theta} = -\dfrac{9g}{76a}\theta\), so S.H.M. | A1 | |
| Period \(= 2\pi\sqrt{\dfrac{76a}{9g}} = \dfrac{4\pi}{3}\sqrt{\dfrac{19a}{g}}\) | DM1 A1 | Total: 11 |
## Question 5:
### Part (a):
| Working | Marks | Notes |
|---|---|---|
| $I = \frac{1}{3}m(9a)^2 + \frac{1}{2}(2m)a^2 + 2m(9a)^2$ | M1 A1 A1 | |
| $= 27ma^2 + ma^2 + 162ma^2$ | | |
| $= 190ma^2$ | A1* | ***(4)*** |
### Part (b):
| Working | Marks | Notes |
|---|---|---|
| $M(L)$: $mg\dfrac{9a}{2}\sin\theta + 2mg\cdot 9a\sin\theta = -190ma^2\ddot{\theta}$ | M1 A2 | |
| $\ddot{\theta} = -\dfrac{9g}{76a}\sin\theta$ | M1 | |
| For small $\theta$, $\sin\theta \approx \theta \Rightarrow \ddot{\theta} = -\dfrac{9g}{76a}\theta$, so S.H.M. | A1 | |
| Period $= 2\pi\sqrt{\dfrac{76a}{9g}} = \dfrac{4\pi}{3}\sqrt{\dfrac{19a}{g}}$ | DM1 A1 | **Total: 11** |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dadd5bac-b547-42dd-838e-60a786555472-3_303_1301_1089_390}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A pendulum $P$ is modelled as a uniform rod $A B$, of length $9 a$ and mass $m$, rigidly fixed to a uniform circular disc of radius $a$ and mass $2 m$. The end $B$ of the rod is attached to the centre of the disc, and the rod lies in the plane of the disc, as shown in Figure 1. The pendulum is free to rotate in a vertical plane about a fixed smooth horizontal axis $L$ which passes through the end $A$ and is perpendicular to the plane of the disc.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of $P$ about $L$ is $190 m a ^ { 2 }$.
The pendulum makes small oscillations about $L$.
\item By writing down an equation of motion for $P$, find the approximate period of these small oscillations.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2008 Q5 [11]}}