| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: rigid body compound pendulum |
| Difficulty | Challenging +1.8 This is a challenging M5 compound pendulum problem requiring moment of inertia for rotation about a non-central axis (parallel axis theorem), energy conservation, angular dynamics, and force resolution at instantaneous rest. It demands multiple sophisticated techniques and careful geometric reasoning, placing it well above average difficulty but within reach of strong Further Maths students. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| MI of sphere about \(L = \frac{2}{5}ma^2 + ma^2 = \frac{7}{5}ma^2\) | B1 | |
| Energy: \(\frac{1}{2}\times\frac{7}{5}ma^2\times\frac{18g}{7a} - \frac{1}{2}\times\frac{7}{5}ma^2\times\omega^2 = mga(1-\cos\theta)\) | M1 A2, 1, 0 | |
| \(\Rightarrow \frac{7}{10}a\omega^2 = \frac{8}{10}g + g\cos\theta\) | ||
| \(\omega^2 = \frac{g}{7a}(8+10\cos\theta) = \frac{2g}{7a}(4+5\cos\theta)\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{7}{5}ma^2\ddot{\theta} = -mga\sin\theta \Rightarrow \ddot{\theta} = -\frac{5g}{7a}\sin\theta\) | M1 A1 | (2) |
| [or \(2\omega\dot{\omega} = -\frac{10g}{7a}\sin\theta\times\omega \Rightarrow \dot{\omega} = -\frac{5g}{7a}\sin\theta\)] | alternative method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dot{\theta} = 0\) when \(\cos\theta = -\frac{4}{5}\) | B1 | |
| \(Y - mg\cos\theta = ma\dot{\theta}^2\) | M1 | radial equation |
| \(\dot{\theta} = 0\), \(\cos\theta = -\frac{4}{5} \Rightarrow Y = -\frac{4mg}{5}\) | M1 A1 | |
| \(X - mg\sin\theta = ma\ddot{\theta}\) | M1 | tangential equation |
| \(\dot{\theta} = 0\) and \(\cos\theta = -\frac{4}{5} \Rightarrow \sin\theta = \frac{3}{5}\) | M1 | |
| \(\Rightarrow \ddot{\theta} = -\frac{3g}{7a}\) | ||
| \(\Rightarrow X = \frac{3mg}{5} - \frac{3mg}{7} = \frac{6mg}{35}\) | A1 | |
| Magnitude of force \(= \sqrt{X^2 + Y^2}\) | M1 | |
| \(= mg\left[\left(\frac{6}{35}\right)^2 + \left(\frac{4}{5}\right)^2\right]^{\frac{1}{2}}\) | ||
| \(\approx 0.818\,mg\) | A1 | (10) (17) |
# Question 7:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| MI of sphere about $L = \frac{2}{5}ma^2 + ma^2 = \frac{7}{5}ma^2$ | B1 | |
| Energy: $\frac{1}{2}\times\frac{7}{5}ma^2\times\frac{18g}{7a} - \frac{1}{2}\times\frac{7}{5}ma^2\times\omega^2 = mga(1-\cos\theta)$ | M1 A2, 1, 0 | |
| $\Rightarrow \frac{7}{10}a\omega^2 = \frac{8}{10}g + g\cos\theta$ | | |
| $\omega^2 = \frac{g}{7a}(8+10\cos\theta) = \frac{2g}{7a}(4+5\cos\theta)$ | A1 | (5) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{7}{5}ma^2\ddot{\theta} = -mga\sin\theta \Rightarrow \ddot{\theta} = -\frac{5g}{7a}\sin\theta$ | M1 A1 | (2) |
| [or $2\omega\dot{\omega} = -\frac{10g}{7a}\sin\theta\times\omega \Rightarrow \dot{\omega} = -\frac{5g}{7a}\sin\theta$] | | alternative method |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dot{\theta} = 0$ when $\cos\theta = -\frac{4}{5}$ | B1 | |
| $Y - mg\cos\theta = ma\dot{\theta}^2$ | M1 | radial equation |
| $\dot{\theta} = 0$, $\cos\theta = -\frac{4}{5} \Rightarrow Y = -\frac{4mg}{5}$ | M1 A1 | |
| $X - mg\sin\theta = ma\ddot{\theta}$ | M1 | tangential equation |
| $\dot{\theta} = 0$ and $\cos\theta = -\frac{4}{5} \Rightarrow \sin\theta = \frac{3}{5}$ | M1 | |
| $\Rightarrow \ddot{\theta} = -\frac{3g}{7a}$ | | |
| $\Rightarrow X = \frac{3mg}{5} - \frac{3mg}{7} = \frac{6mg}{35}$ | A1 | |
| Magnitude of force $= \sqrt{X^2 + Y^2}$ | M1 | |
| $= mg\left[\left(\frac{6}{35}\right)^2 + \left(\frac{4}{5}\right)^2\right]^{\frac{1}{2}}$ | | |
| $\approx 0.818\,mg$ | A1 | (10) **(17)** |7. A uniform sphere, of mass $m$ and radius $a$, is free to rotate about a smooth fixed horizontal axis $L$ which forms a tangent to the sphere. The sphere is hanging in equilibrium below the axis when it receives an impulse, causing it to rotate about $L$ with an initial angular velocity of $\sqrt { \frac { 18 g } { 7 a } }$.
Show that, when the sphere has turned through an angle $\theta$,
\begin{enumerate}[label=(\alph*)]
\item the angular speed $\omega$ of the sphere is given by $\omega ^ { 2 } = \frac { 2 g } { 7 a } ( 4 + 5 \cos \theta )$,
\item the angular acceleration of the sphere has magnitude $\frac { 5 g } { 7 a } \sin \theta$.
\item Hence find the magnitude of the force exerted by the axis on the sphere when the sphere comes to instantaneous rest for the first time.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q7 [17]}}