| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2007 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Energy methods for rotation |
| Difficulty | Challenging +1.2 This is a standard M5 rotation problem requiring energy conservation, moment of inertia calculation, and force analysis. While it involves multiple parts and careful bookkeeping of distances from the pivot, the techniques are routine for Further Maths mechanics: apply conservation of energy (part a is given), use τ=Iα for angular acceleration (part b), and resolve forces perpendicular to the rod (part c). The setup is more complex than basic mechanics but follows standard M5 procedures without requiring novel insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| MI of rod + particle \(= \frac{1}{12}\cdot 3m(2a)^2 + 3m\left(\frac{1}{2}a\right)^2 + m\left(\frac{3}{2}a\right)^2\) | M1, M1 | |
| \(= 4ma^2\) | A1 | |
| \(\frac{1}{2}\cdot 4ma^2\left(\omega^2 - \frac{g}{a}\right) = 3mg\frac{a}{2}(1-\cos\theta) + mg\frac{3a}{2}(1-\cos\theta)\) | M1 A ft A1 | |
| \(\omega^2 = \frac{g}{2a}(5-3\cos\theta)\) | DM1 A1 | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(4ma^2\ddot{\theta} = 3mg\frac{1}{2}a\sin\theta + mg\frac{3}{2}a\sin\theta\) | M1 A1 ft | |
| \(\Rightarrow \ddot{\theta} = \frac{3g\sin\theta}{4a}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(F + 4mg\cos\theta = 3m\cdot\frac{1}{2}a\omega^2 + m\cdot\frac{3}{2}a\omega^2\) | M1 A1 | |
| \(= 3ma\cdot\frac{g}{2a}(5-3\cos\theta)\) | DM1 | |
| \(\Rightarrow F = \frac{mg}{2}(15-17\cos\theta)\) | DM1 | |
| When \(\theta = \phi\), \(F=0 \Rightarrow \cos\phi = \frac{15}{17}\) | A1 | (5),(16) |
## Question 8:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| MI of rod + particle $= \frac{1}{12}\cdot 3m(2a)^2 + 3m\left(\frac{1}{2}a\right)^2 + m\left(\frac{3}{2}a\right)^2$ | M1, M1 | |
| $= 4ma^2$ | A1 | |
| $\frac{1}{2}\cdot 4ma^2\left(\omega^2 - \frac{g}{a}\right) = 3mg\frac{a}{2}(1-\cos\theta) + mg\frac{3a}{2}(1-\cos\theta)$ | M1 A ft A1 | |
| $\omega^2 = \frac{g}{2a}(5-3\cos\theta)$ | DM1 A1 | **(8)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $4ma^2\ddot{\theta} = 3mg\frac{1}{2}a\sin\theta + mg\frac{3}{2}a\sin\theta$ | M1 A1 ft | |
| $\Rightarrow \ddot{\theta} = \frac{3g\sin\theta}{4a}$ | A1 | **(3)** |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $F + 4mg\cos\theta = 3m\cdot\frac{1}{2}a\omega^2 + m\cdot\frac{3}{2}a\omega^2$ | M1 A1 | |
| $= 3ma\cdot\frac{g}{2a}(5-3\cos\theta)$ | DM1 | |
| $\Rightarrow F = \frac{mg}{2}(15-17\cos\theta)$ | DM1 | |
| When $\theta = \phi$, $F=0 \Rightarrow \cos\phi = \frac{15}{17}$ | A1 | **(5),(16)** |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5180a4e0-dafe-4595-a517-e3501f7aed40-5_533_584_292_703}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform rod $A B$ has mass $3 m$ and length $2 a$. It is free to rotate in a vertical plane about a smooth fixed horizontal axis through the point $X$ on the rod, where $A X = \frac { 1 } { 2 } a$. A particle of mass $m$ is attached to the rod at $B$. At time $t = 0$, the rod is vertical, with $B$ above $A$, and is given an initial angular speed $\sqrt { \frac { g } { a } }$. When the rod makes an angle $\theta$ with the upward vertical, the angular speed of the rod is $\omega$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item By using the principle of the conservation of energy, show that
$$\omega ^ { 2 } = \frac { g } { 2 a } ( 5 - 3 \cos \theta )$$
\item Find the angular acceleration of the rod when it makes an angle $\theta$ with the upward vertical.
When $\theta = \phi$, the resultant force of the axis on the rod is in a direction perpendicular to the rod.
\item Find $\cos \phi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2007 Q8 [16]}}