| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2007 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | MI with removed region |
| Difficulty | Challenging +1.2 This is a standard compound pendulum problem requiring moment of inertia calculations (using parallel axis theorem), setting up the equation of motion for small oscillations, and finding the period. While it involves multiple steps and Further Maths content (M5), the techniques are routine applications of standard formulas with no novel insight required. The 'show that' structure guides students through the solution. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(I_O = \frac{M}{3\pi a^2}\left(\frac{\pi}{2}(2a)^2(2a)^2 - \frac{\pi}{2}(a)^2(a)^2\right)\) | M1 A1 | |
| \(= \frac{5Ma^2}{2}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(I_{\text{diameter}} = \frac{1}{2}\cdot\frac{5Ma^2}{2}\) (perpendicular axes) | M1 A1 | |
| \(I_L = \frac{5Ma^2}{4} + M(2a)^2\) (parallel axes) | M1 | |
| \(= \frac{21Ma^2}{4}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(M(L)\): \(-Mg\cdot 2a\sin\theta = \frac{21Ma^2}{4}\ddot{\theta}\) | M1 A1 | |
| \(\sin\theta \approx \theta \Rightarrow \ddot{\theta} = -\frac{8g}{21a}\theta\), so SHM | DM1 A1 | |
| Time \(= \frac{1}{4}\cdot 2\pi\sqrt{\frac{21a}{8g}}\) | DM1 | |
| \(= \frac{\pi}{2}\sqrt{\frac{21a}{8g}}\) | A1 | (6),(13) |
## Question 6:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $I_O = \frac{M}{3\pi a^2}\left(\frac{\pi}{2}(2a)^2(2a)^2 - \frac{\pi}{2}(a)^2(a)^2\right)$ | M1 A1 | |
| $= \frac{5Ma^2}{2}$ | A1 | **(3)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $I_{\text{diameter}} = \frac{1}{2}\cdot\frac{5Ma^2}{2}$ (perpendicular axes) | M1 A1 | |
| $I_L = \frac{5Ma^2}{4} + M(2a)^2$ (parallel axes) | M1 | |
| $= \frac{21Ma^2}{4}$ | A1 | **(4)** |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $M(L)$: $-Mg\cdot 2a\sin\theta = \frac{21Ma^2}{4}\ddot{\theta}$ | M1 A1 | |
| $\sin\theta \approx \theta \Rightarrow \ddot{\theta} = -\frac{8g}{21a}\theta$, so SHM | DM1 A1 | |
| Time $= \frac{1}{4}\cdot 2\pi\sqrt{\frac{21a}{8g}}$ | DM1 | |
| $= \frac{\pi}{2}\sqrt{\frac{21a}{8g}}$ | A1 | **(6),(13)** |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5180a4e0-dafe-4595-a517-e3501f7aed40-4_419_773_196_664}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A lamina $S$ is formed from a uniform disc, centre $O$ and radius $2 a$, by removing the disc of centre $O$ and radius $a$, as shown in Figure 2. The mass of $S$ is $M$.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of $S$ about an axis through $O$ and perpendicular to its plane is $\frac { 5 } { 2 } M a ^ { 2 }$.\\
(3)
The lamina is free to rotate about a fixed smooth horizontal axis $L$. The axis $L$ lies in the plane of $S$ and is a tangent to its outer circumference, as shown in Figure 2.
\item Show that the moment of inertia of $S$ about $L$ is $\frac { 21 } { 4 } M a ^ { 2 }$.\\
(4)\\
$S$ is displaced through a small angle from its position of stable equilibrium and, at time $t = 0$, it is released from rest. Using the equation of motion of $S$, with a suitable approximation,
\item find the time when $S$ first passes through its position of stable equilibrium.\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2007 Q6 [13]}}