| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Small oscillations period |
| Difficulty | Challenging +1.2 This is a standard compound pendulum problem requiring moment of inertia calculation using parallel axis theorem, followed by routine application of SHM formulas for small oscillations. While it involves multiple steps and Further Maths content, the techniques are direct applications of standard results with no novel insight required—harder than typical A-level due to the FM topic but still a textbook exercise. |
| 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 |
| MI of sphere about diameter: \(I_C = (2/5) \cdot 3M(2a)^2 = [24Ma^2/5]\) | M1 | |
| MI of sphere about axis through \(O\): \(I_C + 3Ma^2 = [39Ma^2/5]\) | M1 | |
| MI of particle about axis through \(O\): \(M(3a)^2 = [45Ma^2/5]\) | B1 | |
| Sum MI of system about \(O\): \(I = 84Ma^2/5\) | A1 | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equation of motion: \(I\,d^2\theta/dt^2 = -3Mg\,a\sin\theta - Mg\cdot 3a\sin\theta\) | M1 A1 | A.E.F. |
| Put \(\sin\theta \approx \theta\) (SHM): \(I\,d^2\theta/dt^2 = -6Mga\,\theta\), \([d^2\theta/dt^2 = -(5g/14a)\theta]\) | M1 | |
| Period \(T\) from SHM formula: \(T = 2\pi/\sqrt{6Mga/(84Ma^2/5)} = 2\pi\sqrt{14a/5g}\) or \(10.5\sqrt{(a/g)}\) | M1 A1 | A.E.F. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use SHM formula: \(\theta = \alpha\cos\omega t\) | M1 | |
| Find time \(t\) to \(\theta = \frac{1}{2}\alpha\): \(t = (1/\omega)\cos^{-1}\frac{1}{2} = (1/\omega)(\pi/3) = (\pi/3)\sqrt{14a/5g}\) | M1 A1 | |
| Subtotal: 3 marks | Total: 12 marks |
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| MI of sphere about diameter: $I_C = (2/5) \cdot 3M(2a)^2 = [24Ma^2/5]$ | M1 | |
| MI of sphere about axis through $O$: $I_C + 3Ma^2 = [39Ma^2/5]$ | M1 | |
| MI of particle about axis through $O$: $M(3a)^2 = [45Ma^2/5]$ | B1 | |
| Sum MI of system about $O$: $I = 84Ma^2/5$ | A1 | A.G. |
**Subtotal: 4 marks**
## Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equation of motion: $I\,d^2\theta/dt^2 = -3Mg\,a\sin\theta - Mg\cdot 3a\sin\theta$ | M1 A1 | A.E.F. |
| Put $\sin\theta \approx \theta$ (SHM): $I\,d^2\theta/dt^2 = -6Mga\,\theta$, $[d^2\theta/dt^2 = -(5g/14a)\theta]$ | M1 | |
| Period $T$ from SHM formula: $T = 2\pi/\sqrt{6Mga/(84Ma^2/5)} = 2\pi\sqrt{14a/5g}$ or $10.5\sqrt{(a/g)}$ | M1 A1 | A.E.F. |
**Subtotal: 5 marks**
## Question 5(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use SHM formula: $\theta = \alpha\cos\omega t$ | M1 | |
| Find time $t$ to $\theta = \frac{1}{2}\alpha$: $t = (1/\omega)\cos^{-1}\frac{1}{2} = (1/\omega)(\pi/3) = (\pi/3)\sqrt{14a/5g}$ | M1 A1 | |
**Subtotal: 3 marks | Total: 12 marks**
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{96b6c92d-6d13-452f-84ec-37c45651b232-2_529_493_1667_826}
A uniform solid sphere with centre $C$, radius $2 a$ and mass $3 M$, is pivoted about a smooth horizontal axis and hangs at rest. The point $O$ on the axis is vertically above $C$ and $O C = a$. A particle $P$ of mass $M$ is attached to the sphere at its lowest point (see diagram). Show that the moment of inertia of the system about the axis through $O$ is $\frac { 84 } { 5 } M a ^ { 2 }$.
The system is released from rest with $O P$ making a small angle $\alpha$ with the downward vertical. Find\\
(i) the period of small oscillations,\\
(ii) the time from release until $O P$ makes an angle $\frac { 1 } { 2 } \alpha$ with the downward vertical for the first time.
\hfill \mbox{\textit{CAIE FP2 2011 Q5 [12]}}