| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: rigid body compound pendulum |
| Difficulty | Hard +2.3 This compound pendulum problem requires calculating moment of inertia for a composite body using parallel axis theorem, finding the center of mass, deriving the SHM equation for small oscillations, and applying energy conservation for large amplitude motion. It demands strong mechanics understanding, multi-step calculations with algebraic manipulation, and goes beyond standard textbook exercises typical of Further Maths. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_{AB} = \frac{1}{3}\ 4m(3a)^2 + 4m(7a)^2 = 208ma^2\) | M1 A1 | State or find MI of rod \(AB\) about \(C\) |
| \(I_{disc} = \frac{1}{2}\ 8m(2a)^2 + 8m(2a)^2 = 48ma^2\) | M1 A1 | Find MI of disc about \(C\) |
| \(I = I_{AB} + I_{disc} = 256ma^2\) | A1 | Find MI of body about \(C\) |
| \(I\ \frac{d^2\theta}{dt^2} = [-](4 \times 7 + 8 \times 2)mga\sin\theta\) | M1 A1 | Use equation of circular motion to find \(\frac{d^2\theta}{dt^2}\) |
| \(\frac{d^2\theta}{dt^2} = -\frac{11g}{64a}\ \theta\) | A1 | Approximate \(\sin\theta\) by \(\theta\) and substitute for \(I\) |
| \(T = 16\pi\sqrt{\frac{a}{11g}}\) A.G. | B1 | Find period \(T = \frac{2\pi}{\omega}\) with \(\omega = \sqrt{\frac{11g}{64a}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}I\omega^2 = 4mg\ 7a(1 - \cos\theta) + 8mg\ 2a(1-\cos\theta)\) | M1 A1 | Use energy to find max. angular velocity \(\omega\) |
| \(\omega^2 = \frac{2(11mga)0.4}{256ma^2}\) | Substitute for \(I\) and \(\cos\theta\) and simplify | |
| \(= \frac{11g}{80a}\) | A1 | |
| \(v_A = 10a\omega = \sqrt{\frac{55ga}{4}}\) | M1 A1 | Find maximum speed of \(A\) (A.E.F.) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(x) = \frac{1}{3}(x-20)^3/24000 + \frac{1}{3}\ 20^3/24000\) | M1 | Find \(F(x)\) for \(0 < x < 60\) by integration |
| \(= (x-20)^3/72000 + \frac{1}{9}\) | A1 | |
| \(F(x) = 0\ (x \leq 0),\ 1\ (x \geq 60)\) | B1 | State \(F(x)\) for other values of \(x\) |
| \(G(t) = P(T < t) = P(60 - X < t) = P(X > 60 - t)\) | Find \(G(t)\) for \(0 < t < 60\) from \(X + T = 60\) | |
| \(= 1 - F(60-t)\) | ||
| \(= \frac{8}{9} - \frac{(40-t)^3}{72000}\) A.G. | M1 A1 | |
| \(\frac{8}{9} - \frac{(40-m)^3}{72000} = \frac{1}{2}\) | M1 | Formulate equation for median \(m\) of \(T\) |
| \((40-m)^3 = (\frac{8}{9} - \frac{1}{2})72000 = 28000\) | M1 | Find value of \(m\) |
| \(m = 40 - 28000^{1/3} = 9.63\) | A1 | |
| \(g(t) = (40-t)^2/24000\) | M1 A1 | Find \(g(t)\) for \(0 < t < 60\) |
| \(E(T) = \int t \cdot g(t)\, dt = \int(40^2t - 80t^2 + t^3)\, dt/24000\) | M1 | Find \(E(T)\) from \(\int t\, g(t)\, dt\) |
| \(= \left[\frac{1}{2}40^2t^2 - \frac{1}{3}80t^3 + \frac{1}{4}t^4\right]_0^{60}/24000\) | A1 | |
| \(= 120 - 240 + 135 = 15\) | A1 |
## Question 11(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{AB} = \frac{1}{3}\ 4m(3a)^2 + 4m(7a)^2 = 208ma^2$ | M1 A1 | State or find MI of rod $AB$ about $C$ |
| $I_{disc} = \frac{1}{2}\ 8m(2a)^2 + 8m(2a)^2 = 48ma^2$ | M1 A1 | Find MI of disc about $C$ |
| $I = I_{AB} + I_{disc} = 256ma^2$ | A1 | Find MI of body about $C$ |
| $I\ \frac{d^2\theta}{dt^2} = [-](4 \times 7 + 8 \times 2)mga\sin\theta$ | M1 A1 | Use equation of circular motion to find $\frac{d^2\theta}{dt^2}$ |
| $\frac{d^2\theta}{dt^2} = -\frac{11g}{64a}\ \theta$ | A1 | Approximate $\sin\theta$ by $\theta$ and substitute for $I$ |
| $T = 16\pi\sqrt{\frac{a}{11g}}$ **A.G.** | B1 | Find period $T = \frac{2\pi}{\omega}$ with $\omega = \sqrt{\frac{11g}{64a}}$ |
## Question 11(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}I\omega^2 = 4mg\ 7a(1 - \cos\theta) + 8mg\ 2a(1-\cos\theta)$ | M1 A1 | Use energy to find max. angular velocity $\omega$ |
| $\omega^2 = \frac{2(11mga)0.4}{256ma^2}$ | | Substitute for $I$ and $\cos\theta$ and simplify |
| $= \frac{11g}{80a}$ | A1 | |
| $v_A = 10a\omega = \sqrt{\frac{55ga}{4}}$ | M1 A1 | Find maximum speed of $A$ (A.E.F.) |
---
## Question 11(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = \frac{1}{3}(x-20)^3/24000 + \frac{1}{3}\ 20^3/24000$ | M1 | Find $F(x)$ for $0 < x < 60$ by integration |
| $= (x-20)^3/72000 + \frac{1}{9}$ | A1 | |
| $F(x) = 0\ (x \leq 0),\ 1\ (x \geq 60)$ | B1 | State $F(x)$ for other values of $x$ |
| $G(t) = P(T < t) = P(60 - X < t) = P(X > 60 - t)$ | | Find $G(t)$ for $0 < t < 60$ from $X + T = 60$ |
| $= 1 - F(60-t)$ | | |
| $= \frac{8}{9} - \frac{(40-t)^3}{72000}$ **A.G.** | M1 A1 | |
| $\frac{8}{9} - \frac{(40-m)^3}{72000} = \frac{1}{2}$ | M1 | Formulate equation for median $m$ of $T$ |
| $(40-m)^3 = (\frac{8}{9} - \frac{1}{2})72000 = 28000$ | M1 | Find value of $m$ |
| $m = 40 - 28000^{1/3} = 9.63$ | A1 | |
| $g(t) = (40-t)^2/24000$ | M1 A1 | Find $g(t)$ for $0 < t < 60$ |
| $E(T) = \int t \cdot g(t)\, dt = \int(40^2t - 80t^2 + t^3)\, dt/24000$ | M1 | Find $E(T)$ from $\int t\, g(t)\, dt$ |
| $= \left[\frac{1}{2}40^2t^2 - \frac{1}{3}80t^3 + \frac{1}{4}t^4\right]_0^{60}/24000$ | A1 | |
| $= 120 - 240 + 135 = 15$ | A1 | |\begin{center}
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\end{center}
A rigid body consists of a thin uniform rod $A B$, of mass $4 m$ and length $6 a$, joined at $B$ to a point on the circumference of a uniform circular disc, with centre $O$, mass $8 m$ and radius $2 a$. The point $C$ on the circumference of the disc is such that $B C$ is a diameter and $A B C$ is a straight line (see diagram). The body rotates about a smooth fixed horizontal axis through $C$, perpendicular to the plane of the disc. The angle between $C A$ and the downward vertical at time $t$ is denoted by $\theta$.\\
(i) Given that the body is performing small oscillations about the downward vertical, show that the period of these oscillations is approximately $16 \pi \sqrt { } \left( \frac { a } { 11 g } \right)$.\\
(ii) Given instead that the body is released from rest in the position given by $\cos \theta = 0.6$, find the maximum speed of $A$.
\hfill \mbox{\textit{CAIE FP2 2013 Q11 EITHER}}