| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2010 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Small oscillations period |
| Difficulty | Challenging +1.8 This M5 compound pendulum question requires: (a) a non-trivial integration to find moment of inertia about a parallel axis through the apex of a triangle, (b) applying perpendicular axis theorem and parallel axis theorem to find MOI about D, then using rotational dynamics with torque = Iα, and (c) small angle approximation for SHM period. The integration setup requires careful coordinate geometry, and the multi-step application of MOI theorems with rotational SHM is beyond standard A-level core content, placing it well above average difficulty. |
| Spec | 6.04d 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 | Mark | Notes |
| \(\delta A = \frac{8x}{3}\delta x\) | M1 A1 | |
| \(\delta m = \frac{8x}{3}\delta x \cdot \frac{m}{12a^2}\) or \(\delta m = \frac{8x}{3}\delta x \cdot \rho\) | DM1 | |
| \(\delta I = \frac{8x}{3}\delta x \cdot \frac{m}{12a^2} \cdot x^2 \left(= \frac{2m}{9a^2}x^3\delta x\right)\) | A1 | |
| \(I = \int_0^{3a} \frac{2m}{9a^2}x^3\,dx\) | M1 | |
| \(= \frac{2m}{9a^2}\left[\frac{x^4}{4}\right]_0^{3a}\) | ||
| \(= \frac{9ma^2}{2}\) * | A1 | Total: (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(I_A = \frac{9ma^2}{2} + \frac{8ma^2}{3} = \frac{43ma^2}{6}\) (perp axes rule) | M1 A1 | |
| \(I_A = I_G + m(2a)^2\) (parallel axes rule) | DM1 A1 | |
| \(I_D = I_G + ma^2\) (parallel axes rule) | A1 | |
| \(I_D = \frac{43ma^2}{6} - 3ma^2 = \frac{25ma^2}{6}\) | A1 | |
| \(mga\sin\theta = -\frac{25ma^2}{6}\ddot{\theta}\) | M1 | |
| \(\ddot{\theta} = -\frac{6g}{25a}\sin\theta\) | A1 | Total: (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| For small \(\theta\), \(\ddot{\theta} = -\frac{6g}{25a}\theta\) SHM | M1 | |
| \(T = 2\pi\sqrt{\frac{25a}{6g}} = 5\pi\sqrt{\frac{2a}{3g}}\) | A1 | Total: (2), Q Total: 16 |
## Question 3:
### Part (a):
| Working/Answer | Mark | Notes |
|---|---|---|
| $\delta A = \frac{8x}{3}\delta x$ | M1 A1 | |
| $\delta m = \frac{8x}{3}\delta x \cdot \frac{m}{12a^2}$ or $\delta m = \frac{8x}{3}\delta x \cdot \rho$ | DM1 | |
| $\delta I = \frac{8x}{3}\delta x \cdot \frac{m}{12a^2} \cdot x^2 \left(= \frac{2m}{9a^2}x^3\delta x\right)$ | A1 | |
| $I = \int_0^{3a} \frac{2m}{9a^2}x^3\,dx$ | M1 | |
| $= \frac{2m}{9a^2}\left[\frac{x^4}{4}\right]_0^{3a}$ | | |
| $= \frac{9ma^2}{2}$ * | A1 | **Total: (6)** |
### Part (b):
| Working/Answer | Mark | Notes |
|---|---|---|
| $I_A = \frac{9ma^2}{2} + \frac{8ma^2}{3} = \frac{43ma^2}{6}$ (perp axes rule) | M1 A1 | |
| $I_A = I_G + m(2a)^2$ (parallel axes rule) | DM1 A1 | |
| $I_D = I_G + ma^2$ (parallel axes rule) | A1 | |
| $I_D = \frac{43ma^2}{6} - 3ma^2 = \frac{25ma^2}{6}$ | A1 | |
| $mga\sin\theta = -\frac{25ma^2}{6}\ddot{\theta}$ | M1 | |
| $\ddot{\theta} = -\frac{6g}{25a}\sin\theta$ | A1 | **Total: (8)** |
### Part (c):
| Working/Answer | Mark | Notes |
|---|---|---|
| For small $\theta$, $\ddot{\theta} = -\frac{6g}{25a}\theta$ SHM | M1 | |
| $T = 2\pi\sqrt{\frac{25a}{6g}} = 5\pi\sqrt{\frac{2a}{3g}}$ | A1 | **Total: (2), Q Total: 16** |
---\begin{enumerate}
\item A uniform lamina $A B C$ of mass $m$ is in the shape of an isosceles triangle with $A B = A C = 5 a$ and $B C = 8 a$.\\
(a) Show, using integration, that the moment of inertia of the lamina about an axis through $A$, parallel to $B C$, is $\frac { 9 } { 2 } m a ^ { 2 }$.
\end{enumerate}
The foot of the perpendicular from $A$ to $B C$ is $D$. The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis which passes through $D$ and is perpendicular to the plane of the lamina. The lamina is released from rest when $D A$ makes an angle $\alpha$ with the downward vertical. It is given that the moment of inertia of the lamina about an axis through $A$, perpendicular to $B C$ and in the plane of the lamina, is $\frac { 8 } { 3 } m a ^ { 2 }$.\\
(b) Find the angular acceleration of the lamina when $D A$ makes an angle $\theta$ with the downward vertical.
Given that $\alpha$ is small,\\
(c) find an approximate value for the period of oscillation of the lamina about the vertical.
\hfill \mbox{\textit{Edexcel M5 2010 Q3 [16]}}