| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Equation of motion angular acceleration |
| Difficulty | Standard +0.8 This is a standard mechanics problem involving rotational dynamics with coupled linear and angular motion. It requires setting up equations for both the particle (F=ma) and disc (τ=Iα), relating linear and angular quantities (a=rα), and using energy methods or kinematics for part (ii). While systematic, it demands understanding of multiple concepts (moments of inertia, torque, tension as a constraint force) and careful algebraic manipulation, placing it moderately above average difficulty. |
| Spec | 3.04a Calculate moments: about a point6.02k Power: rate of doing work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T \times 0.4 = 0.2\,\frac{d^2\theta}{dt^2}\) | M1 | Eqn of motion for disc |
| \(1.5g - T = 1.5 \times 0.4\,\frac{d^2\theta}{dt^2}\) | M1 | Eqn of motion for particle |
| \(1.5g = (0.6 + 0.5)\,\frac{d^2\theta}{dt^2}\) | M1 | Eliminate \(T\) |
| \(\frac{d^2\theta}{dt^2} = \frac{15g}{11}\) or \(13.6\ \text{rad s}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER Integrate: \(\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2 = \frac{15g}{11}\theta + c\) | M1 | |
| Apply initial conditions, \(\theta = \frac{\pi}{6}\): \(\left(\frac{d\theta}{dt}\right)^2 = \frac{5\pi g}{11}\) or \(14.3\) | A1 | |
| OR Use energy: \(\frac{1}{2}(0.2)\left(\frac{d\theta}{dt}\right)^2 + \frac{1}{2}(1.5)(0.4\,\frac{d\theta}{dt})^2 = 1.5g \times 0.4 \times \frac{\pi}{6}\) | (M1) | |
| \(\left(\frac{d\theta}{dt}\right)^2 = \frac{5\pi g}{11}\) or \(14.3\) | (A1) | |
| \(v = 0.4\,\frac{d\theta}{dt} = 51\ \text{m s}^{-1}\) | B1 | Find speed of particle |
| Subtotal: 3 | Total: [7] |
## Question 2(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T \times 0.4 = 0.2\,\frac{d^2\theta}{dt^2}$ | M1 | Eqn of motion for disc |
| $1.5g - T = 1.5 \times 0.4\,\frac{d^2\theta}{dt^2}$ | M1 | Eqn of motion for particle |
| $1.5g = (0.6 + 0.5)\,\frac{d^2\theta}{dt^2}$ | M1 | Eliminate $T$ |
| $\frac{d^2\theta}{dt^2} = \frac{15g}{11}$ or $13.6\ \text{rad s}^{-2}$ | A1 | |
**S.R.:** M1 only for $1.5g \times 0.4 = 0.2\,\frac{d^2\theta}{dt^2}$; $[\frac{d^2\theta}{dt^2} = 30,\ (\frac{d\theta}{dt})^2 = 10\pi,\ v = 2.24]$
**Subtotal: 4**
---
## Question 2(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **EITHER** Integrate: $\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2 = \frac{15g}{11}\theta + c$ | M1 | |
| Apply initial conditions, $\theta = \frac{\pi}{6}$: $\left(\frac{d\theta}{dt}\right)^2 = \frac{5\pi g}{11}$ or $14.3$ | A1 | |
| **OR** Use energy: $\frac{1}{2}(0.2)\left(\frac{d\theta}{dt}\right)^2 + \frac{1}{2}(1.5)(0.4\,\frac{d\theta}{dt})^2 = 1.5g \times 0.4 \times \frac{\pi}{6}$ | (M1) | |
| $\left(\frac{d\theta}{dt}\right)^2 = \frac{5\pi g}{11}$ or $14.3$ | (A1) | |
| $v = 0.4\,\frac{d\theta}{dt} = 51\ \text{m s}^{-1}$ | B1 | Find speed of particle |
**Subtotal: 3 | Total: [7]**
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\includegraphics[max width=\textwidth, alt={}, center]{34024618-0ff9-44a1-ac57-d4d7e8a3655e-2_431_421_881_861}
A uniform disc of radius 0.4 m is free to rotate without friction in a vertical plane about a horizontal axis through its centre. The moment of inertia of the disc about the axis is $0.2 \mathrm {~kg} \mathrm {~m} ^ { 2 }$. One end of a light inextensible string is attached to a point on the rim of the disc and the string is wound round the rim. The other end of the string is attached to a particle of mass 1.5 kg which hangs freely (see diagram). The system is released from rest. Find\\
(i) the angular acceleration of the disc,\\
(ii) the speed of the particle when the disc has turned through an angle of $\frac { 1 } { 6 } \pi$.
\hfill \mbox{\textit{CAIE FP2 2012 Q2 [7]}}