| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Composite body MI calculation |
| Difficulty | Challenging +1.8 This is a multi-part Further Maths mechanics question requiring: (1) proving a given moment of inertia result using standard formulas and superposition, (2) applying the parallel axis theorem, and (3) using energy conservation with rotational dynamics. While it involves several steps and Further Maths content (moments of inertia), each part follows standard procedures without requiring novel geometric insight or particularly complex algebra. The 'show that' structure and energy method are routine for this level. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| MI of one side of square about midpoint: \(\frac{1}{3}\left(\frac{M}{4}\right)\frac{a^2}{2} = \frac{Ma^2}{24}\) | M1 A1 | |
| MI of side of square (or of square) about \(O\): \(\frac{Ma^2}{24} + \left(\frac{M}{4}\right)\frac{a^2}{2} = \frac{Ma^2}{6}\) | M1 A1 | |
| MI of ring about \(O\): \(2Ma^2\) | B1 | |
| Sum to find MI of system about \(O\): \(I_O = 4 \times \frac{Ma^2}{6} + 2Ma^2 = 8\frac{Ma^2}{3}\) | A1 | A.G. |
| MI of system about axis through \(A\): \(I_A = \frac{1}{2}I_O + 3Ma^2 = 13\frac{Ma^2}{3}\) | M1 A1 | |
| State or imply that speed is max when \(AC\) vertical | M1 | |
| Use energy when \(AC\) vertical (or at general point): \(\frac{1}{2}(I_A + 4Ma^2)\omega^2 = [25\frac{Ma^2}{3}]\) or \(\frac{1}{2}I_A\omega^2 + \frac{1}{2}Mv^2\) | ||
| \(= 5Mga(1 + \cos 60°)\) or \(\frac{15Mga}{2}\) | M1 | A1 for each side of eqn, A.E.F. |
| A1 A1 | ||
| Substitute for \(I_A\) and find max speed \(v = 2a\omega\): \(\omega^2 = \frac{15Mga/2}{25Ma^2/6} = \frac{9g}{5a}\) | ||
| \(v = 6\sqrt{\frac{ga}{5}}\) or \(6\sqrt{2a}\) or \(8.49\sqrt{a}\) | M1 A1 | [6+2+6=14] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \(k\) by equating area under graph to 1: \(\frac{1}{2}k + k = 1\), \(k = \frac{2}{3}\) | M1 A1 | |
| Find \(f(x)\) for \(0 \leq x \leq 1\) and \(1 < x \leq 3\): \(kx\left[=\frac{2}{3}x\right]\), \(\frac{1}{2}k(3-x)\left[= 1 - \frac{1}{3}x\right]\) | M1 A1 | |
| Integrate to find \(F(x)\) for \(0 < x \leq 1\) and \(1 < x \leq 3\): \(\frac{1}{3}x^2\), \(x - \frac{1}{2} - \frac{x^2}{6}\) | M1 A1 | A.G. [6] |
| (i) Relate dist. fn. \(G(y)\) of \(Y\) to \(X\): \(G(y) = P(Y < y) = P(X^2 < y)\) | working may be omitted | |
| \(= P(X < y^{1/2}) = F(y^{1/2})\) | ||
| \(= \frac{1}{3}y\), \(y^{1/2} - \frac{1}{2} - \frac{y}{6}\) | M1 A1 | |
| Differentiate to find \(g(y)\): \(g(y) = \frac{1}{3}\) \((0 < y \leq 1)\) | ||
| \(\frac{1}{2}y^{-1/2} - \frac{1}{6}\) \((1 < y \leq 9)\) | ||
| \([0 \text{ otherwise}]\) | M1 A1 | [4] |
| (ii) Use \(G(m) = \frac{1}{2}\) to find eqn for median \(m\): \(m^{1/2} - \frac{1}{2} - \frac{m}{6} = \frac{1}{2}\) | M1 | |
| Solve quadratic for \(\sqrt{m}\): \((\sqrt{m})^2 - 6\sqrt{m} + 6 = 0\), \(\sqrt{m} = 3 \pm \sqrt{3}\) | M1 A1 | |
| Select value of \(m\) in interval: \(m = (3 - \sqrt{3})^2 = 12 - 6\sqrt{3}\) or \(1.61\) | B1 | [4+14] |
# Question 11:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| MI of one side of square about midpoint: $\frac{1}{3}\left(\frac{M}{4}\right)\frac{a^2}{2} = \frac{Ma^2}{24}$ | M1 A1 | |
| MI of side of square (or of square) about $O$: $\frac{Ma^2}{24} + \left(\frac{M}{4}\right)\frac{a^2}{2} = \frac{Ma^2}{6}$ | M1 A1 | |
| MI of ring about $O$: $2Ma^2$ | B1 | |
| Sum to find MI of system about $O$: $I_O = 4 \times \frac{Ma^2}{6} + 2Ma^2 = 8\frac{Ma^2}{3}$ | A1 | **A.G.** |
| MI of system about axis through $A$: $I_A = \frac{1}{2}I_O + 3Ma^2 = 13\frac{Ma^2}{3}$ | M1 A1 | |
| State or imply that speed is max when $AC$ vertical | M1 | |
| Use energy when $AC$ vertical (or at general point): $\frac{1}{2}(I_A + 4Ma^2)\omega^2 = [25\frac{Ma^2}{3}]$ or $\frac{1}{2}I_A\omega^2 + \frac{1}{2}Mv^2$ | | |
| $= 5Mga(1 + \cos 60°)$ or $\frac{15Mga}{2}$ | M1 | A1 for each side of eqn, A.E.F. |
| | A1 A1 | |
| Substitute for $I_A$ and find max speed $v = 2a\omega$: $\omega^2 = \frac{15Mga/2}{25Ma^2/6} = \frac{9g}{5a}$ | | |
| $v = 6\sqrt{\frac{ga}{5}}$ or $6\sqrt{2a}$ or $8.49\sqrt{a}$ | M1 A1 | **[6+2+6=14]** |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $k$ by equating area under graph to 1: $\frac{1}{2}k + k = 1$, $k = \frac{2}{3}$ | M1 A1 | |
| Find $f(x)$ for $0 \leq x \leq 1$ and $1 < x \leq 3$: $kx\left[=\frac{2}{3}x\right]$, $\frac{1}{2}k(3-x)\left[= 1 - \frac{1}{3}x\right]$ | M1 A1 | |
| Integrate to find $F(x)$ for $0 < x \leq 1$ and $1 < x \leq 3$: $\frac{1}{3}x^2$, $x - \frac{1}{2} - \frac{x^2}{6}$ | M1 A1 | **A.G.** **[6]** |
| **(i)** Relate dist. fn. $G(y)$ of $Y$ to $X$: $G(y) = P(Y < y) = P(X^2 < y)$ | | working may be omitted |
| $= P(X < y^{1/2}) = F(y^{1/2})$ | | |
| $= \frac{1}{3}y$, $y^{1/2} - \frac{1}{2} - \frac{y}{6}$ | M1 A1 | |
| Differentiate to find $g(y)$: $g(y) = \frac{1}{3}$ $(0 < y \leq 1)$ | | |
| $\frac{1}{2}y^{-1/2} - \frac{1}{6}$ $(1 < y \leq 9)$ | | |
| $[0 \text{ otherwise}]$ | M1 A1 | **[4]** |
| **(ii)** Use $G(m) = \frac{1}{2}$ to find eqn for median $m$: $m^{1/2} - \frac{1}{2} - \frac{m}{6} = \frac{1}{2}$ | M1 | |
| Solve quadratic for $\sqrt{m}$: $(\sqrt{m})^2 - 6\sqrt{m} + 6 = 0$, $\sqrt{m} = 3 \pm \sqrt{3}$ | M1 A1 | |
| Select value of $m$ in interval: $m = (3 - \sqrt{3})^2 = 12 - 6\sqrt{3}$ or $1.61$ | B1 | **[4+14]** |\begin{center}
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A rigid body is made from uniform wire of negligible thickness and is in the form of a square $A B C D$ of mass $M$ enclosed within a circular ring of radius $a$ and mass $2 M$. The centres of the square and the circle coincide at $O$ and the corners of the square are joined to the circle (see diagram). Show that the moment of inertia of the body about an axis through $O$, perpendicular to the plane of the body, is $\frac { 8 } { 3 } M a ^ { 2 }$.
Hence find the moment of inertia of the body about an axis $l$, through $A$, in the plane of the body and tangential to the circle.
A particle $P$ of mass $M$ is now attached to the body at $C$. The system is able to rotate freely about the fixed axis $l$, which is horizontal. The system is released from rest with $A C$ making an angle of $60 ^ { \circ }$ with the upward vertical. Find, in terms of $a$ and $g$, the greatest speed of $P$ in the subsequent motion.
\hfill \mbox{\textit{CAIE FP2 2011 Q11 EITHER}}