| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Angular speed and period |
| Difficulty | Challenging +1.8 This M5 question requires multiple advanced techniques: parallel axis theorem applied to a composite body (disc with hole), finding center of mass of composite lamina, and energy conservation with rotational motion. Part (a) involves systematic application of moment of inertia formulas with careful geometry, while part (b) requires setting up energy equations for rotational dynamics. The multi-step nature, composite body analysis, and integration of several M5 concepts makes this significantly harder than average, though the methods are standard for this module. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04b Find centre of mass: using symmetry |
\includegraphics{figure_2}
**Figure 1**
A uniform circular disc has mass $4m$, centre $O$ and radius $4a$. The line $POQ$ is a diameter of the disc. A circular hole of radius $2a$ is made in the disc with the centre of the hole at the point $R$ on $PQ$ where $QR = 5a$, as shown in Figure 1.
The resulting lamina is free to rotate about a fixed smooth horizontal axis $L$ which passes through $Q$ and is perpendicular to the plane of the lamina.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of the lamina about $L$ is $69ma^2$.
[7]
\end{enumerate}
The lamina is hanging at rest with $P$ vertically below $Q$ when it is given an angular velocity $\Omega$. Given that the lamina turns through an angle $\frac{2\pi}{3}$ before it first comes to instantaneous rest,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find $\Omega$ in terms of $g$ and $a$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q2 [13]}}