| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Small oscillations period |
| Difficulty | Challenging +1.3 This is a standard compound pendulum problem from Further Maths requiring application of parallel axis theorem and SHM approximation for small angles. While it involves multiple steps (calculating moment of inertia, deriving equation of motion, applying small angle approximation), these are well-practiced techniques in FM2. The 'show that' format provides a target to work towards, reducing problem-solving demand compared to open-ended questions. |
| Spec | 6.04a Centre of mass: gravitational effect6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| \(A_{ABCD} = 24a^2\), \(A_{EFGH} = 8a^2\), \(A_{Frame} = 16a^2\); \(M_{ABCD} = 3m/2\) and \(m_{EFGH} = m/2\) | B1 | Use areas to find masses \(M_{ABCD}\) and \(m_{EFGH}\) |
| \(I_{ABCD} = \frac{1}{3}M_{ABCD}((3a)^2 + (2a)^2) = \frac{13}{3}M_{ABCD}a^2 = \frac{13}{2}ma^2\) | B1 | Find MI of \(ABCD\) about axis at centre \(Q\) |
| \(I_{EFGH} = \frac{1}{3}m_{EFGH}((2a)^2 + a^2) = \frac{5}{6}m_{EFGH}a^2 = \frac{5}{6}ma^2\) | B1 | Find MI of \(EFGH\) about axis at centre \(Q\) |
| \(I_{ABCD} - I_{EFGH} + m \times (6a)^2 = \frac{121}{2} - \frac{113}{6})ma^2\) or \((\frac{17}{3} + 36)ma^2 = \frac{125}{3}ma^2\) | M1 A1 | Find MI of frame about axis at \(O\); result also follows from combining four rectangular parts |
| \(I_{Object} = \frac{11m}{12} \times (4a)^2 = \frac{44}{3}ma^2\) | B1 | Find MI of small object about axis at \(O\) |
| \(I = (\frac{13}{2} - \frac{5}{6} + 36 + \frac{44}{3})ma^2 = \frac{169}{3}ma^2\) AG | A1 | Combine to verify MI of system about axis at \(O\) |
| Answer | Marks | Guidance |
|---|---|---|
| \([-]I\frac{d^2\theta}{dt^2}\) or \([-]I\alpha = mg \times 6a\sin\theta + \frac{11}{12}mg \times 4a\sin\theta\) or \((23/12)mg \times (116/23)a\sin\theta = \frac{29}{3}mga\sin\theta\) | M1 A1 | Use eqn of circular motion to relate \(\frac{d^2\theta}{dt^2}\) to \(\sin\theta\), where \(\theta\) is angle of \(QO\) with vertical |
| \(\frac{d^2\theta}{dt^2}\) or \(\alpha = -\frac{29g}{169a}\theta\) or \(-(0.172\frac{g}{a})\theta\) | M1 A1 | Approximate \(\sin\theta\) by \(\theta\) to show SHM (M0 if wrong sign or \(\cos\theta \approx \theta\) used) |
| \(T = 2\pi/\sqrt{29g/169a} = 26\pi\sqrt{a/29g}\) or \(15.2\sqrt{a/g}\) or \(4.80\sqrt{a}\) (AEF) | A1 | Find period \(T\) from \(T = 2\pi/\omega\) (A1 requires some simplification) |
## Question 5(i):
$A_{ABCD} = 24a^2$, $A_{EFGH} = 8a^2$, $A_{Frame} = 16a^2$; $M_{ABCD} = 3m/2$ and $m_{EFGH} = m/2$ | B1 | Use areas to find masses $M_{ABCD}$ and $m_{EFGH}$
$I_{ABCD} = \frac{1}{3}M_{ABCD}((3a)^2 + (2a)^2) = \frac{13}{3}M_{ABCD}a^2 = \frac{13}{2}ma^2$ | B1 | Find MI of $ABCD$ about axis at centre $Q$
$I_{EFGH} = \frac{1}{3}m_{EFGH}((2a)^2 + a^2) = \frac{5}{6}m_{EFGH}a^2 = \frac{5}{6}ma^2$ | B1 | Find MI of $EFGH$ about axis at centre $Q$
$I_{ABCD} - I_{EFGH} + m \times (6a)^2 = \frac{121}{2} - \frac{113}{6})ma^2$ or $(\frac{17}{3} + 36)ma^2 = \frac{125}{3}ma^2$ | M1 A1 | Find MI of frame about axis at $O$; result also follows from combining four rectangular parts
$I_{Object} = \frac{11m}{12} \times (4a)^2 = \frac{44}{3}ma^2$ | B1 | Find MI of small object about axis at $O$
$I = (\frac{13}{2} - \frac{5}{6} + 36 + \frac{44}{3})ma^2 = \frac{169}{3}ma^2$ AG | A1 | Combine to verify MI of system about axis at $O$
**Total: 7**
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## Question 5(ii):
$[-]I\frac{d^2\theta}{dt^2}$ or $[-]I\alpha = mg \times 6a\sin\theta + \frac{11}{12}mg \times 4a\sin\theta$ or $(23/12)mg \times (116/23)a\sin\theta = \frac{29}{3}mga\sin\theta$ | M1 A1 | Use eqn of circular motion to relate $\frac{d^2\theta}{dt^2}$ to $\sin\theta$, where $\theta$ is angle of $QO$ with vertical
$\frac{d^2\theta}{dt^2}$ or $\alpha = -\frac{29g}{169a}\theta$ or $-(0.172\frac{g}{a})\theta$ | M1 A1 | Approximate $\sin\theta$ by $\theta$ to show SHM (**M0** if wrong sign or $\cos\theta \approx \theta$ used)
$T = 2\pi/\sqrt{29g/169a} = 26\pi\sqrt{a/29g}$ or $15.2\sqrt{a/g}$ or $4.80\sqrt{a}$ (AEF) | A1 | Find period $T$ from $T = 2\pi/\omega$ (**A1** requires some simplification)
**Total: 5**
---(i) Show that the moment of inertia of the system, consisting of frame and small object, about an axis through $O$ perpendicular to the plane of the frame, is $\frac { 169 } { 3 } m a ^ { 2 }$.\\
(ii) Show that small oscillations of the system about this axis are approximately simple harmonic and state their period.\\
\hfill \mbox{\textit{CAIE FP2 2017 Q5 [12]}}