CAIE FP2 2013 June — Question 5 13 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeMI with removed region
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring multiple sophisticated techniques: calculating moment of inertia using parallel axis theorem for a composite lamina, finding the center of mass of an irregular shape, and applying energy conservation for rotational motion with a non-trivial inequality condition. While the steps are systematic, the problem demands careful geometric reasoning, algebraic manipulation, and understanding of when complete revolutions occur (minimum speed at top position). This is significantly harder than standard A-level mechanics but follows established FM2 patterns.
Spec6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall

5 \includegraphics[max width=\textwidth, alt={}, center]{7fcedc6d-8dc1-4159-8a72-be0f6a3f659b-3_355_693_260_726} \(A B C D\) is a uniform rectangular lamina of mass \(m\) in which \(A B = 4 a\) and \(B C = 2 a\). The lines \(A C\) and \(B D\) intersect at \(O\). The mid-points of \(O A , O B , O C , O D\) are \(E , F , G , H\) respectively. The rectangle \(E F G H\), in which \(E F = 2 a\) and \(F G = a\), is removed from \(A B C D\) (see diagram). The resulting lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis through \(A\) and perpendicular to the plane of \(A B C D\). Show that the moment of inertia of this lamina about the axis is \(\frac { 85 } { 16 } m a ^ { 2 }\). The lamina hangs in equilibrium under gravity with \(C\) vertically below \(A\). The point \(C\) is now given a speed \(u\). Given that the lamina performs complete revolutions, show that $$u ^ { 2 } > \frac { 192 \sqrt { } 5 } { 17 } a g .$$

Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(I_{ABCD} = \frac{1}{3}m\{(2a)^2 + a^2\} = 5ma^2/3\)B1 Find MI of \(ABCD\) about \(O\)
\(I_{EFGH} = \frac{1}{3}(\frac{1}{4}m)\{a^2 + (\frac{1}{2}a)^2\} = 5ma^2/48\)M1 A1 Find MI of \(EFGH\) about \(O\)
\(I_O = I_{ABCD} - I_{EFGH} = 25ma^2/16\)M1 A1 *EITHER*: Find MI of final lamina about \(O\)
\(I_A = I_O + \frac{3}{4}m \times 5a^2 = 85ma^2/16\) A.GM1 A1, A1 Find MI of final lamina about \(A\)
\(I'_{ABCD} = I_{ABCD} + 5ma^2 = 20ma^2/3\)(M1 A1) *OR*: Find MI of \(ABCD\) about \(A\)
\(I'_{EFGH} = I_{EFGH} + \frac{1}{4}m \times 5a^2 = 65ma^2/48\)(M1 A1) Find MI of \(EFGH\) about \(A\)
\(I_A = I'_{ABCD} - I'_{EFGH} = 85ma^2/16\) A.G(B1) Find MI of final lamina about \(A\)
State or use \(u = (\sqrt{20})\,a\omega\)B1
\(\frac{1}{2}I_A\,\omega_{min}^2 = \frac{3}{4}mg \times (\sqrt{20})\,a\)M1 A1 Use energy when \(C\) above \(A\) to find \(\omega_{min}\)
\(u_{min}^2 = 20(32/85)(\frac{3}{4}\sqrt{20})\,ag = (192\sqrt{5}/17)\,ag\) A.G.M1 A1 Hence find \(u_{min}^2\)
Total: 13 marks
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{ABCD} = \frac{1}{3}m\{(2a)^2 + a^2\} = 5ma^2/3$ | B1 | Find MI of $ABCD$ about $O$ |
| $I_{EFGH} = \frac{1}{3}(\frac{1}{4}m)\{a^2 + (\frac{1}{2}a)^2\} = 5ma^2/48$ | M1 A1 | Find MI of $EFGH$ about $O$ |
| $I_O = I_{ABCD} - I_{EFGH} = 25ma^2/16$ | M1 A1 | *EITHER*: Find MI of final lamina about $O$ |
| $I_A = I_O + \frac{3}{4}m \times 5a^2 = 85ma^2/16$ **A.G** | M1 A1, A1 | Find MI of final lamina about $A$ |
| $I'_{ABCD} = I_{ABCD} + 5ma^2 = 20ma^2/3$ | (M1 A1) | *OR*: Find MI of $ABCD$ about $A$ |
| $I'_{EFGH} = I_{EFGH} + \frac{1}{4}m \times 5a^2 = 65ma^2/48$ | (M1 A1) | Find MI of $EFGH$ about $A$ |
| $I_A = I'_{ABCD} - I'_{EFGH} = 85ma^2/16$ **A.G** | (B1) | Find MI of final lamina about $A$ |
| State or use $u = (\sqrt{20})\,a\omega$ | B1 | |
| $\frac{1}{2}I_A\,\omega_{min}^2 = \frac{3}{4}mg \times (\sqrt{20})\,a$ | M1 A1 | Use energy when $C$ above $A$ to find $\omega_{min}$ |
| $u_{min}^2 = 20(32/85)(\frac{3}{4}\sqrt{20})\,ag = (192\sqrt{5}/17)\,ag$ **A.G.** | M1 A1 | Hence find $u_{min}^2$ |

**Total: 13 marks**

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\includegraphics[max width=\textwidth, alt={}, center]{7fcedc6d-8dc1-4159-8a72-be0f6a3f659b-3_355_693_260_726}\\
$A B C D$ is a uniform rectangular lamina of mass $m$ in which $A B = 4 a$ and $B C = 2 a$. The lines $A C$ and $B D$ intersect at $O$. The mid-points of $O A , O B , O C , O D$ are $E , F , G , H$ respectively. The rectangle $E F G H$, in which $E F = 2 a$ and $F G = a$, is removed from $A B C D$ (see diagram). The resulting lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis through $A$ and perpendicular to the plane of $A B C D$. Show that the moment of inertia of this lamina about the axis is $\frac { 85 } { 16 } m a ^ { 2 }$.

The lamina hangs in equilibrium under gravity with $C$ vertically below $A$. The point $C$ is now given a speed $u$. Given that the lamina performs complete revolutions, show that

$$u ^ { 2 } > \frac { 192 \sqrt { } 5 } { 17 } a g .$$

\hfill \mbox{\textit{CAIE FP2 2013 Q5 [13]}}