| 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 +1.8 This compound pendulum problem requires multiple advanced techniques: parallel axis theorem for a sphere, combining moments of inertia, deriving the SHM equation from torque considerations, and applying small-angle approximations. The calculation is multi-step with potential for algebraic errors, and part (ii) requires understanding SHM phase relationships. While systematic, it demands solid grasp of rotational dynamics beyond standard A-level, typical of Further Mechanics questions. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| 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 | |
| Total MI: \(I = 84Ma^2/5\) | A1 | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Equation of motion: \(I\,d^2\theta/dt^2 = -3Mga\sin\theta - Mg\cdot 3a\sin\theta\) | 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 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 |
|---|---|---|
| Working/Answer | Marks | Notes |
| Use SHM formula: \(\theta = \alpha\cos\omega t\) | M1 | |
| Find \(t\) when \(\theta = \frac{1}{2}\alpha\): \(t = (1/\omega)\cos^{-1}\frac{1}{2} = (1/\omega)(\pi/3) = (\pi/3)\sqrt{(14a/5g)}\) | M1 A1 | |
| Subtotal: 3 marks | Total Q5: 12 marks |
## Question 5:
| Working/Answer | Marks | Notes |
|---|---|---|
| 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 | |
| Total MI: $I = 84Ma^2/5$ | A1 | A.G. |
**Subtotal: 4 marks**
## Question 5(i):
| Working/Answer | Marks | Notes |
|---|---|---|
| Equation of motion: $I\,d^2\theta/dt^2 = -3Mga\sin\theta - Mg\cdot 3a\sin\theta$ | 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 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):
| Working/Answer | Marks | Notes |
|---|---|---|
| Use SHM formula: $\theta = \alpha\cos\omega t$ | M1 | |
| Find $t$ when $\theta = \frac{1}{2}\alpha$: $t = (1/\omega)\cos^{-1}\frac{1}{2} = (1/\omega)(\pi/3) = (\pi/3)\sqrt{(14a/5g)}$ | M1 A1 | |
**Subtotal: 3 marks | Total Q5: 12 marks**
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{0d4a352c-4eda-45b4-9284-60c6fc680f02-2_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]}}