| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Energy method angular speed |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring: (1) calculation of moment of inertia using parallel axis theorem for composite bodies (rod + lamina), (2) application of rotational dynamics with energy methods, and (3) solving for angular acceleration and velocity. While the techniques are standard for FM2, the multi-body system and careful geometric setup make it substantially harder than typical A-level questions. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_{rod} = (4/3)m(3a)^2 = 12ma^2\) | B1 | Find MI of rod about \(A\) |
| \(I_{lamina} = 2ma^2 + 64ma^2 = 66ma^2\) | M1 A1 | Find MI of circular lamina about \(A\) |
| \(I = I_{rod} + I_{lamina} = 78ma^2\) A.G. | A1 | Find MI of rod plus lamina about \(A\) |
| \(I' = I + (2m)(6a)^2 = 150ma^2\) | B1 | Find new MI including particle about \(A\) |
| \(I'd^2\theta/dt^2 = [-]\,mg(3 + 8 + 12)a\sin\theta\) | M1 A1 | Use equation of circular motion (using \(I\) can earn B0 M1 A1 A0) |
| \(d^2\theta/dt^2 = -(23g/150a)\sin\theta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}I'\omega^2 = \frac{1}{2}I'\omega_0^2 + (3+8+12)mga\) | M1 A1 | Use energy when \(AB\) vertical to find \(\omega\) (or by integration) |
| \(\omega^2 = 9g/25a + 46mga/150ma^2\) | ||
| \(\omega = \sqrt{2g/3a}\) or \(0.816\sqrt{g/a}\) or \(2.58/\sqrt{a}\) (AEF) | A1 | using \(I\) can earn M1 A1 A0 |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{rod} = (4/3)m(3a)^2 = 12ma^2$ | B1 | Find MI of rod about $A$ |
| $I_{lamina} = 2ma^2 + 64ma^2 = 66ma^2$ | M1 A1 | Find MI of circular lamina about $A$ |
| $I = I_{rod} + I_{lamina} = 78ma^2$ **A.G.** | A1 | Find MI of rod plus lamina about $A$ |
| $I' = I + (2m)(6a)^2 = 150ma^2$ | B1 | Find new MI including particle about $A$ |
| $I'd^2\theta/dt^2 = [-]\,mg(3 + 8 + 12)a\sin\theta$ | M1 A1 | Use equation of circular motion (using $I$ can earn B0 M1 A1 A0) |
| $d^2\theta/dt^2 = -(23g/150a)\sin\theta$ | A1 | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}I'\omega^2 = \frac{1}{2}I'\omega_0^2 + (3+8+12)mga$ | M1 A1 | Use energy when $AB$ vertical to find $\omega$ (or by integration) |
| $\omega^2 = 9g/25a + 46mga/150ma^2$ | | |
| $\omega = \sqrt{2g/3a}$ or $0.816\sqrt{g/a}$ or $2.58/\sqrt{a}$ (AEF) | A1 | using $I$ can earn M1 A1 A0 |
**Total: 11 marks**
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{71a3b842-9d31-4c25-b894-ca6d1f47d84b-3_319_794_255_678}
A uniform rod $A B$, of mass $m$ and length $6 a$, is rigidly attached at $B$ to a point on the circumference of a uniform circular lamina of mass $m$, radius $2 a$ and centre $O$. The lamina and the rod are in the same vertical plane, and $A B O$ is a straight line (see diagram). Show that the moment of inertia of the system about an axis $l$ through $A$ perpendicular to the plane of the lamina is $78 m a ^ { 2 }$.
A particle of mass $2 m$ is now attached at $B$ and the system is free to rotate in a vertical plane about the fixed axis $l$ which is horizontal. Initially $A B$ is horizontal, with $O$ moving downwards and the system having angular velocity $\frac { 3 } { 5 } \sqrt { } \left( \frac { g } { a } \right)$. At time $t , A B$ makes an angle $\theta$ with the downward vertical through $A$.\\
(i) Find, in terms of $a , g$ and $\theta$, an expression for $\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } }$.\\
(ii) Find the angular velocity of the system when $B$ is vertically below $A$.
\hfill \mbox{\textit{CAIE FP2 2012 Q5 [11]}}