| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | MI with removed region |
| Difficulty | Challenging +1.8 This M5 question requires multiple sophisticated techniques: finding moment of inertia using parallel axis theorem for a composite body (disc with hole), locating the centre of mass of the lamina, then applying energy conservation with rotational kinetic energy. While the steps are systematic, it demands careful bookkeeping, understanding of advanced mechanics concepts beyond standard A-level, and multi-stage problem-solving typical of Further Maths Mechanics 5. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| Mass of disc removed \(= m\) | B1 | |
| \(\frac{1}{2}4m(4a)^2 + 4m(4a)^2\) | M1 A1 | |
| \(\frac{1}{2}m(2a)^2 + m(5a)^2\) | M1 A1 | |
| \(I = \frac{1}{2}4m(4a)^2 + 4m(4a)^2 - \left(\frac{1}{2}m(2a)^2 + m(5a)^2\right)\) | DM1 | |
| \(= 69ma^2\) * | A1 | Total: (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(4m \cdot 0 = 3m\bar{x} - ma\) | M1 | |
| \(\bar{x} = \frac{1}{3}a\) (from \(O\)) | A1 | |
| \(\frac{1}{2}69ma^2\Omega^2 = 3mg\left(4a - \frac{1}{3}a\right)\left(1-\cos\frac{2\pi}{3}\right)\) | M1 A2 | |
| \(\Omega = \sqrt{\frac{11g}{23a}}\) | A1 | Total: (6), Q Total: 13 |
## Question 2:
### Part (a):
| Working/Answer | Mark | Notes |
|---|---|---|
| Mass of disc removed $= m$ | B1 | |
| $\frac{1}{2}4m(4a)^2 + 4m(4a)^2$ | M1 A1 | |
| $\frac{1}{2}m(2a)^2 + m(5a)^2$ | M1 A1 | |
| $I = \frac{1}{2}4m(4a)^2 + 4m(4a)^2 - \left(\frac{1}{2}m(2a)^2 + m(5a)^2\right)$ | DM1 | |
| $= 69ma^2$ * | A1 | **Total: (7)** |
### Part (b):
| Working/Answer | Mark | Notes |
|---|---|---|
| $4m \cdot 0 = 3m\bar{x} - ma$ | M1 | |
| $\bar{x} = \frac{1}{3}a$ (from $O$) | A1 | |
| $\frac{1}{2}69ma^2\Omega^2 = 3mg\left(4a - \frac{1}{3}a\right)\left(1-\cos\frac{2\pi}{3}\right)$ | M1 A2 | |
| $\Omega = \sqrt{\frac{11g}{23a}}$ | A1 | **Total: (6), Q Total: 13** |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9fa0d04e-ff7b-46f8-a8ec-44393e383cdf-03_504_584_267_671}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform circular disc has mass $4 m$, centre $O$ and radius $4 a$. The line $P O Q$ is a diameter 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 $P Q$ where $Q R = 5 a$, 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 $69 m a ^ { 2 }$.
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,
\item find $\Omega$ in terms of $g$ and $a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2010 Q2 [13]}}