CAIE FP2 2012 November — Question 5 12 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeEnergy method angular speed
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics question requiring: (1) calculation of moment of inertia using parallel axis theorem for four rods in different orientations, (2) application of energy conservation in rotational motion with a non-trivial geometry, and (3) relating angular velocity to linear speed. The multi-step nature, geometric complexity of the compound pendulum, and need to track energy through rotation elevate this above standard SHM questions, though the techniques themselves are well-established for Further Maths students.
Spec6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass
5 Four identical uniform rods, each of mass \(m\) and length \(2 a\), are rigidly joined to form a square frame \(A B C D\). Show that the moment of inertia of the frame about an axis through \(A\) perpendicular to the plane of the frame is \(\frac { 40 } { 3 } m a ^ { 2 }\). The frame is suspended from \(A\) and is able to rotate freely under gravity in a vertical plane, about a horizontal axis through \(A\). When the frame is at rest with \(C\) vertically below \(A\), it is given an angular velocity \(\sqrt { } \left( \frac { 6 g } { 5 a } \right)\). Find the angular velocity of the frame when \(A C\) makes an angle \(\theta\) with the downward vertical through \(A\). When \(A C\) is horizontal, the speed of \(C\) is \(k \sqrt { } ( g a )\). Find the value of \(k\) correct to 3 significant figures.
Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
MI of rod \(AB\) (or \(AD\)) about \(A\): \(I_{AB} = \frac{1}{3}ma^2 + ma^2 = \frac{4}{3}ma^2\)B1
MI of rod \(BC\) (or \(CD\)) about \(A\): \(I_{BC} = \frac{1}{3}ma^2 + m \cdot 5a^2 = \frac{16}{3}ma^2\)M1
MI of frame about \(A\): \(I = 2(I_{AB} + I_{BC}) = \frac{40ma^2}{3}\)M1 A1 A.G.
Use energy to find angular velocity \(\omega\) at angle \(\theta\): \(\frac{1}{2}I\omega^2 = \frac{1}{2}I(6g/5a) - 4mg\cdot a\sqrt{2}(1-\cos\theta)\)M1 A2 lose A1 for one incorrect term
Substitute for \(I\) and simplify: \(\omega = \sqrt{\{(3g/5a)(2 - \sqrt{2}(1-\cos\theta))\}}\)M1 A1 (A.E.F.) Total part: [5]
Equate \(AC\cdot\omega\) to \(k\sqrt{ga}\) when \(\theta = 90°\): \(k\sqrt{ga} = 2\sqrt{2}a\sqrt{\{(3g/5a)(2-\sqrt{2})\}}\)M1 A1
\(k = 2\sqrt{\{6(2-\sqrt{2})/5\}} = 1.68\)A1 Total part: [3], Question total: [12] 
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| MI of rod $AB$ (or $AD$) about $A$: $I_{AB} = \frac{1}{3}ma^2 + ma^2 = \frac{4}{3}ma^2$ | B1 | |
| MI of rod $BC$ (or $CD$) about $A$: $I_{BC} = \frac{1}{3}ma^2 + m \cdot 5a^2 = \frac{16}{3}ma^2$ | M1 | |
| MI of frame about $A$: $I = 2(I_{AB} + I_{BC}) = \frac{40ma^2}{3}$ | M1 A1 | **A.G.** |
| Use energy to find angular velocity $\omega$ at angle $\theta$: $\frac{1}{2}I\omega^2 = \frac{1}{2}I(6g/5a) - 4mg\cdot a\sqrt{2}(1-\cos\theta)$ | M1 A2 | lose A1 for one incorrect term |
| Substitute for $I$ and simplify: $\omega = \sqrt{\{(3g/5a)(2 - \sqrt{2}(1-\cos\theta))\}}$ | M1 A1 | (A.E.F.) Total part: [5] |
| Equate $AC\cdot\omega$ to $k\sqrt{ga}$ when $\theta = 90°$: $k\sqrt{ga} = 2\sqrt{2}a\sqrt{\{(3g/5a)(2-\sqrt{2})\}}$ | M1 A1 | |
| $k = 2\sqrt{\{6(2-\sqrt{2})/5\}} = 1.68$ | A1 | Total part: [3], Question total: [12] |

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5 Four identical uniform rods, each of mass $m$ and length $2 a$, are rigidly joined to form a square frame $A B C D$. Show that the moment of inertia of the frame about an axis through $A$ perpendicular to the plane of the frame is $\frac { 40 } { 3 } m a ^ { 2 }$.

The frame is suspended from $A$ and is able to rotate freely under gravity in a vertical plane, about a horizontal axis through $A$. When the frame is at rest with $C$ vertically below $A$, it is given an angular velocity $\sqrt { } \left( \frac { 6 g } { 5 a } \right)$. Find the angular velocity of the frame when $A C$ makes an angle $\theta$ with the downward vertical through $A$.

When $A C$ is horizontal, the speed of $C$ is $k \sqrt { } ( g a )$. Find the value of $k$ correct to 3 significant figures.

\hfill \mbox{\textit{CAIE FP2 2012 Q5 [12]}}
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