CAIE FP2 2013 November — Question 1 6 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeComposite body MI calculation
DifficultyChallenging +1.2 This is a standard moment of inertia calculation requiring application of the parallel axis theorem to multiple components (three rods in different orientations plus a disc). While it involves several steps and careful bookkeeping of distances, the techniques are routine for Further Maths students who have studied this topic. The setup is clearly defined with no geometric ambiguity, making it a straightforward but moderately lengthy calculation.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass

1 \includegraphics[max width=\textwidth, alt={}, center]{b486decd-75b8-44bd-889f-2472f1163871-2_553_435_258_854} Three identical uniform rods, \(A B , B C\) and \(C D\), each of mass \(M\) and length \(2 a\), are rigidly joined to form three sides of a square. A uniform circular disc, of mass \(\frac { 2 } { 3 } M\) and radius \(a\), has the opposite ends of one of its diameters attached to \(A\) and \(D\) respectively. The disc and the rods all lie in the same plane (see diagram). Find the moment of inertia of the system about the axis \(A D\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(I_{AB} = \frac{1}{3}Ma^2 + Ma^2 = \frac{4Ma^2}{3}\)B1 Find MI of rod \(AB\) or \(CD\) about \(AD\)
\(I_{BC} = M(2a)^2 = 4Ma^2\)B1 Find MI of rod \(BC\) about \(AD\)
\(I_{Disc} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} Ma^2 = \frac{Ma^2}{6}\)M1 A1 Find MI of disc about \(AD\) by perpendicular axes
\(I = 2I_{AB} + I_{BC} + I_{Disc} = \frac{41}{6}Ma^2\)M1 A1 Find MI of system about \(A\)
Total: 6 marks
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{AB} = \frac{1}{3}Ma^2 + Ma^2 = \frac{4Ma^2}{3}$ | B1 | Find MI of rod $AB$ or $CD$ about $AD$ |
| $I_{BC} = M(2a)^2 = 4Ma^2$ | B1 | Find MI of rod $BC$ about $AD$ |
| $I_{Disc} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} Ma^2 = \frac{Ma^2}{6}$ | M1 A1 | Find MI of disc about $AD$ by perpendicular axes |
| $I = 2I_{AB} + I_{BC} + I_{Disc} = \frac{41}{6}Ma^2$ | M1 A1 | Find MI of system about $A$ |
| **Total: 6 marks** | | |

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\includegraphics[max width=\textwidth, alt={}, center]{b486decd-75b8-44bd-889f-2472f1163871-2_553_435_258_854}

Three identical uniform rods, $A B , B C$ and $C D$, each of mass $M$ and length $2 a$, are rigidly joined to form three sides of a square. A uniform circular disc, of mass $\frac { 2 } { 3 } M$ and radius $a$, has the opposite ends of one of its diameters attached to $A$ and $D$ respectively. The disc and the rods all lie in the same plane (see diagram). Find the moment of inertia of the system about the axis $A D$.

\hfill \mbox{\textit{CAIE FP2 2013 Q1 [6]}}