| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2015 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: rigid body compound pendulum |
| Difficulty | Challenging +1.3 This is a standard M5/FM2 compound pendulum question requiring moment of inertia calculation, small oscillations formula, and energy conservation. While it involves multiple steps and careful bookkeeping of the composite system (rod + particle), the techniques are routine for Further Maths students: parallel axis theorem, standard SHM period formula, and impulse-energy methods. The 60° turning angle in part (b) is straightforward energy conservation with no novel insight required. |
| Spec | 6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moment of inertia of rod about \(A\): \(\frac{1}{3}(3m)(2a)^2 = 4ma^2\) | M1 A1 | Using \(\frac{1}{3}ML^2\) |
| Moment of inertia of particle about \(A\): \(2m(2a)^2 = 8ma^2\) | B1 | |
| Total \(I = 4ma^2 + 8ma^2 = 12ma^2\) | A1 | |
| Distance of centre of mass from \(A\): \(\bar{x} = \frac{3m \cdot a + 2m \cdot 2a}{5m} = \frac{7a}{5}\) | M1 A1 | Taking moments for combined system |
| \(T = 2\pi\sqrt{\frac{I}{Mgh}} = 2\pi\sqrt{\frac{12ma^2}{5mg \cdot \frac{7a}{5}}}\) | M1 | Applying \(T = 2\pi\sqrt{\frac{I}{Mgh}}\) |
| \(T = 2\pi\sqrt{\frac{12a}{7g}}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Angular velocity after impulse: \(J \cdot 2a = I\omega = 12ma^2\omega\) | M1 A1 | Angular impulse = change in angular momentum |
| \(\omega = \frac{J}{6ma}\) | A1 | |
| Using energy conservation, pendulum turns through \(60°\): | M1 | |
| Rise in height of centre of mass: \(\frac{7a}{5}(1 - \cos 60°) = \frac{7a}{5} \cdot \frac{1}{2} = \frac{7a}{10}\) | M1 A1 | Correct height change |
| \(\frac{1}{2}I\omega^2 = 5mg \cdot \frac{7a}{10}\) | M1 | Energy equation |
| \(\frac{1}{2}(12ma^2)\left(\frac{J}{6ma}\right)^2 = \frac{7mga}{2}\) | ||
| \(\frac{J^2}{6m} = \frac{7mga}{2}\) | ||
| \(J = a\sqrt{21m^2g} = m\sqrt{21ga}\) | A1 | cao |
# Question 6:
## Part (a): Period of small oscillations
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moment of inertia of rod about $A$: $\frac{1}{3}(3m)(2a)^2 = 4ma^2$ | M1 A1 | Using $\frac{1}{3}ML^2$ |
| Moment of inertia of particle about $A$: $2m(2a)^2 = 8ma^2$ | B1 | |
| Total $I = 4ma^2 + 8ma^2 = 12ma^2$ | A1 | |
| Distance of centre of mass from $A$: $\bar{x} = \frac{3m \cdot a + 2m \cdot 2a}{5m} = \frac{7a}{5}$ | M1 A1 | Taking moments for combined system |
| $T = 2\pi\sqrt{\frac{I}{Mgh}} = 2\pi\sqrt{\frac{12ma^2}{5mg \cdot \frac{7a}{5}}}$ | M1 | Applying $T = 2\pi\sqrt{\frac{I}{Mgh}}$ |
| $T = 2\pi\sqrt{\frac{12a}{7g}}$ | A1 | cao |
## Part (b): Finding J
| Answer/Working | Mark | Guidance |
|---|---|---|
| Angular velocity after impulse: $J \cdot 2a = I\omega = 12ma^2\omega$ | M1 A1 | Angular impulse = change in angular momentum |
| $\omega = \frac{J}{6ma}$ | A1 | |
| Using energy conservation, pendulum turns through $60°$: | M1 | |
| Rise in height of centre of mass: $\frac{7a}{5}(1 - \cos 60°) = \frac{7a}{5} \cdot \frac{1}{2} = \frac{7a}{10}$ | M1 A1 | Correct height change |
| $\frac{1}{2}I\omega^2 = 5mg \cdot \frac{7a}{10}$ | M1 | Energy equation |
| $\frac{1}{2}(12ma^2)\left(\frac{J}{6ma}\right)^2 = \frac{7mga}{2}$ | | |
| $\frac{J^2}{6m} = \frac{7mga}{2}$ | | |
| $J = a\sqrt{21m^2g} = m\sqrt{21ga}$ | A1 | cao |\begin{enumerate}
\item A pendulum is modelled as a uniform rod $A B$, of mass $3 m$ and length $2 a$, which has a particle of mass $2 m$ attached at $B$. The pendulum is free to rotate in a vertical plane about a fixed smooth horizontal axis $L$ which passes through $A$. The vertical plane is perpendicular to the axis $L$.\\
(a) Find the period of small oscillations of the pendulum about its position of stable equilibrium.
\end{enumerate}
The pendulum is hanging at rest in a vertical position, with $B$ below $A$, when it is given a horizontal impulse of magnitude $J$. The impulse acts at $B$ in a vertical plane which is perpendicular to the axis $L$.
Given that the pendulum turns through an angle of $60 ^ { \circ }$ before first coming to instantaneous rest,\\
(b) find $J$.\\
\hfill \mbox{\textit{Edexcel M5 2015 Q6 [16]}}