| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2017 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Small oscillations period |
| Difficulty | Challenging +1.8 This compound pendulum problem requires multiple advanced techniques: deriving moments of inertia through integration in two different orientations (perpendicular theorem application), then applying rotational dynamics for small oscillations. The integration setup is non-trivial, requiring careful coordinate geometry and the perpendicular axis theorem. Part (c) demands understanding of the rotational equation of motion and small angle approximations. This goes well beyond standard M5 questions and requires sophisticated problem-solving across multiple mechanics concepts. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\delta A = \frac{3x}{2}\delta x\) | M1 A1 | First M1 for attempt at area of strip; First A1 if correct |
| \(\delta m = \frac{3x}{2}\delta x \cdot \frac{m}{12a^2} = \frac{mx\,\delta x}{8a^2}\) | M1 | Second M1 for multiplying area by correct mass per unit area |
| \(\delta I = \frac{mx\,\delta x}{8a^2} \cdot x^2 = \frac{mx^3\,\delta x}{8a^2}\) | M1 | Third M1 for \(\delta m \cdot x^2\) or other appropriate \(\delta I\) |
| \(I_y = \int_0^{4a} \frac{mx^3\,dx}{8a^2} = 8ma^2\) | DM1 A1 (6) | Fourth DM1 (dep on 3rd M) for integrating between limits; Second A1 for printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\delta I = \frac{1}{3} \cdot \frac{mx\,\delta x}{8a^2}\left(\frac{3x}{4}\right)^2 = \frac{3mx^3\,\delta x}{128a^2}\) | M1 A1 | First M1 for using appropriate expression for \(\delta I\); First A1 for correct expression |
| \(I_y = \int_0^{4a} \frac{3mx^3\,dx}{128a^2} = 1.5ma^2\) | DM1 A1 (4) | Second DM1 (dep on 1st M) for integrating between limits; Second A1 for printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Perp axes: \(I = 1.5ma^2 + 8ma^2 = 9.5ma^2\) | M1 A1 | First M1 for use of perpendicular axes theorem; First A1 for correct MI |
| \(mg\frac{8a}{3}\sin\theta = -\frac{19ma^2}{2}\ddot{\theta}\) | M1 A1 | Second M1 for moments equation with usual rules; Second A1 for correct equation |
| For small \(\theta\): \(\ddot{\theta} = -\frac{16g}{57a}\theta\) | M1 | Third M1 for small angle approximation giving SHM form |
| \(T = 2\pi\sqrt{\frac{57a}{16g}}\) | A1 (6) | Third A1 for answer |
# Question 4:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\delta A = \frac{3x}{2}\delta x$ | M1 A1 | First M1 for attempt at area of strip; First A1 if correct |
| $\delta m = \frac{3x}{2}\delta x \cdot \frac{m}{12a^2} = \frac{mx\,\delta x}{8a^2}$ | M1 | Second M1 for multiplying area by correct mass per unit area |
| $\delta I = \frac{mx\,\delta x}{8a^2} \cdot x^2 = \frac{mx^3\,\delta x}{8a^2}$ | M1 | Third M1 for $\delta m \cdot x^2$ or other appropriate $\delta I$ |
| $I_y = \int_0^{4a} \frac{mx^3\,dx}{8a^2} = 8ma^2$ | DM1 A1 (6) | Fourth DM1 (dep on 3rd M) for integrating between limits; Second A1 for printed answer |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\delta I = \frac{1}{3} \cdot \frac{mx\,\delta x}{8a^2}\left(\frac{3x}{4}\right)^2 = \frac{3mx^3\,\delta x}{128a^2}$ | M1 A1 | First M1 for using appropriate expression for $\delta I$; First A1 for correct expression |
| $I_y = \int_0^{4a} \frac{3mx^3\,dx}{128a^2} = 1.5ma^2$ | DM1 A1 (4) | Second DM1 (dep on 1st M) for integrating between limits; Second A1 for printed answer |
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Perp axes: $I = 1.5ma^2 + 8ma^2 = 9.5ma^2$ | M1 A1 | First M1 for use of perpendicular axes theorem; First A1 for correct MI |
| $mg\frac{8a}{3}\sin\theta = -\frac{19ma^2}{2}\ddot{\theta}$ | M1 A1 | Second M1 for moments equation with usual rules; Second A1 for correct equation |
| For small $\theta$: $\ddot{\theta} = -\frac{16g}{57a}\theta$ | M1 | Third M1 for small angle approximation giving SHM form |
| $T = 2\pi\sqrt{\frac{57a}{16g}}$ | A1 (6) | Third A1 for answer |
---\begin{enumerate}
\item A uniform lamina $P Q R$ of mass $m$ is in the shape of an isosceles triangle, with $P Q = P R = 5 a$ and $Q R = 6 a$. The midpoint of $Q R$ is $T$.\\
(a) Show, using integration, that the moment of inertia of the lamina about an axis which passes through $P$ and is parallel to $Q R$, is $8 m a ^ { 2 }$.\\
(b) Show, using integration, that the moment of inertia of the lamina about an axis which passes through $P$ and $T$, is $1.5 m a ^ { 2 }$.\\[0pt]
[You may assume without proof that the moment of inertia of a uniform rod, of mass $m$ and length $2 l$, about an axis perpendicular to the rod through its midpoint is $\frac { 1 } { 3 } m l ^ { 2 }$ ]\\
(4)
\end{enumerate}
The lamina is now free to rotate in a vertical plane about a fixed smooth horizontal axis $A$ which passes through $P$ and is perpendicular to the plane of the lamina. The lamina makes small oscillations about its position of stable equilibrium.\\
(c) By writing down an equation of rotational motion for the lamina as it rotates about $A$, find the approximate period of these small oscillations.
\hfill \mbox{\textit{Edexcel M5 2017 Q4 [16]}}