CAIE FP2 2009 June — Question 3 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2009
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeMI with removed region
DifficultyChallenging +1.8 This question requires applying parallel axis theorem to an annular lamina (non-trivial geometry), setting up energy conservation with rotational KE, and solving an inequality for the angular speed range. It combines multiple advanced mechanics concepts but follows a structured path with clear physical setup, making it challenging but accessible to well-prepared Further Maths students.
Spec6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces

3 \includegraphics[max width=\textwidth, alt={}, center]{15ed1dfc-8188-4e20-9c0b-ce31af35f0b6-2_513_711_890_717} A uniform lamina of mass \(m\) is bounded by concentric circles with centre \(O\) and radii \(a\) and \(2 a\). The lamina is free to rotate about a fixed smooth horizontal axis \(T\) which is tangential to the outer rim (see diagram). Show that the moment of inertia of the lamina about \(T\) is \(\frac { 21 } { 4 } m a ^ { 2 }\). When hanging at rest, with \(O\) vertically below \(T\), the lamina is given an angular speed \(\omega\) about \(T\). The lamina comes to instantaneous rest in the subsequent motion. Neglecting air resistance, find the set of possible values of \(\omega\).

Question 3:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(I_{2a} = m_{2a}a^2\) or \(I_a = \tfrac{1}{4}m_a a^2\)M1 Use perpendicular axes theorem for both discs (or lamina)
\(I = I_{2a} - I_a\ [= a^2(m_{2a} - \tfrac{1}{4}m_a)]\)M1 Combine to find MI of lamina about diameter (or \(T\))
\(I_T = I + 4ma^2\)M1 Use parallel axes theorem for lamina (or both discs)
\(m_{2a} = 4m/3\) and \(m_a = m/3\)B1 Find masses of both discs in terms of \(m\)
\(I_T = a^2(4 - \tfrac{1}{4})m/3 + 4ma^2\)
\(= 5ma^2/4 + 4ma^2 = 21ma^2/4\) A.G.A1 Combine to find MI of lamina about \(T\). Part mark: 5
\(\tfrac{1}{2}I_T\omega^2\) and \(4mga\)M1 A1 Relate initial KE to change in PE at highest point
\(\omega < \sqrt{32g/21a}\) Find set of values [or max. value] of \(\omega\) (A.E.F.)
*or* \(1.23\sqrt{(g/a)}\) *or* \(3.90/\sqrt{a}\)A1 Part mark: 3. Total: [8]
## Question 3:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $I_{2a} = m_{2a}a^2$ or $I_a = \tfrac{1}{4}m_a a^2$ | M1 | Use perpendicular axes theorem for both discs (or lamina) |
| $I = I_{2a} - I_a\ [= a^2(m_{2a} - \tfrac{1}{4}m_a)]$ | M1 | Combine to find MI of lamina about diameter (or $T$) |
| $I_T = I + 4ma^2$ | M1 | Use parallel axes theorem for lamina (or both discs) |
| $m_{2a} = 4m/3$ and $m_a = m/3$ | B1 | Find masses of both discs in terms of $m$ |
| $I_T = a^2(4 - \tfrac{1}{4})m/3 + 4ma^2$ | | |
| $= 5ma^2/4 + 4ma^2 = 21ma^2/4$ **A.G.** | A1 | Combine to find MI of lamina about $T$. Part mark: 5 |
| $\tfrac{1}{2}I_T\omega^2$ and $4mga$ | M1 A1 | Relate initial KE to change in PE at highest point |
| $\omega < \sqrt{32g/21a}$ | | Find set of values [or max. value] of $\omega$ (A.E.F.) |
| *or* $1.23\sqrt{(g/a)}$ *or* $3.90/\sqrt{a}$ | A1 | Part mark: 3. Total: **[8]** |

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\includegraphics[max width=\textwidth, alt={}, center]{15ed1dfc-8188-4e20-9c0b-ce31af35f0b6-2_513_711_890_717}

A uniform lamina of mass $m$ is bounded by concentric circles with centre $O$ and radii $a$ and $2 a$. The lamina is free to rotate about a fixed smooth horizontal axis $T$ which is tangential to the outer rim (see diagram). Show that the moment of inertia of the lamina about $T$ is $\frac { 21 } { 4 } m a ^ { 2 }$.

When hanging at rest, with $O$ vertically below $T$, the lamina is given an angular speed $\omega$ about $T$. The lamina comes to instantaneous rest in the subsequent motion. Neglecting air resistance, find the set of possible values of $\omega$.

\hfill \mbox{\textit{CAIE FP2 2009 Q3 [8]}}