CAIE FP2 2008 November — Question 1 5 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2008
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeComposite body MI calculation
DifficultyChallenging +1.2 This is a multi-step Further Maths mechanics problem requiring the parallel axis theorem applied to three separate wire segments, with careful geometric calculation of distances. While systematic, it demands more setup and coordination than a typical A-level question, placing it moderately above average difficulty.
Spec6.04d Integration: for centre of mass of laminas/solids

1 A uniform wire, of length \(24 a\) and mass \(m\), is bent into the form of a triangle \(A B C\) with angle \(A B C = 90 ^ { \circ }\), \(A B = 6 a\) and \(B C = 8 a\) (see diagram). Find the moment of inertia of the wire about an axis through \(A\) perpendicular to the plane of the wire.

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(I_{AB} = (6m/24)(\frac{1}{3}3^2 + 3^2)a^2 = 3ma^2\)M1 M1 for either \(I_{AB}\) or \(I_{AC}\)
\(I_{AC} = (10m/24)(\frac{1}{3}5^2 + 5^2)a^2\)
\(= (125/9)ma^2\)A1
\(I_{BC} = (8m/24)(\frac{1}{3}4^2 + 6^2 + 4^2)a^2\) M1 for \(I_{BC}\)
\(= (172/9)ma^2\)M1 A1
\(I = (324/9)ma^2 = 36ma^2\)A1 Total: 5
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_{AB} = (6m/24)(\frac{1}{3}3^2 + 3^2)a^2 = 3ma^2$ | M1 | M1 for either $I_{AB}$ or $I_{AC}$ |
| $I_{AC} = (10m/24)(\frac{1}{3}5^2 + 5^2)a^2$ | | |
| $= (125/9)ma^2$ | A1 | |
| $I_{BC} = (8m/24)(\frac{1}{3}4^2 + 6^2 + 4^2)a^2$ | | M1 for $I_{BC}$ |
| $= (172/9)ma^2$ | M1 A1 | |
| $I = (324/9)ma^2 = 36ma^2$ | A1 | **Total: 5** |

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1

A uniform wire, of length $24 a$ and mass $m$, is bent into the form of a triangle $A B C$ with angle $A B C = 90 ^ { \circ }$, $A B = 6 a$ and $B C = 8 a$ (see diagram). Find the moment of inertia of the wire about an axis through $A$ perpendicular to the plane of the wire.

\hfill \mbox{\textit{CAIE FP2 2008 Q1 [5]}}