| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Find unknown parameter from period |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring: (1) application of parallel axis theorem to find moment of inertia of a composite body (disc + rod), (2) energy conservation with rotational kinetic energy, and (3) solving a resulting quadratic equation. The multi-step nature, combination of rotational dynamics concepts, and algebraic manipulation place it well above average difficulty, though the individual techniques are standard for Further Maths students. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| 3(i) | I = ⅓ M (3a)2 + M (3a)2 or (4/3) M (3a)2 [= 12 Ma2] | |
| AB | B1 | Find or state MI of rod AB about axis l |
| Answer | Marks | Guidance |
|---|---|---|
| disc | M1A1 | Find MI of disc about axis l |
| I = 13 M a2 + 2M x2 | A1 | Find MI of object about axis l |
| Answer | Marks | Guidance |
|---|---|---|
| 3(ii) | ½ I ω2 = Mg × 3a cos θ + 2Mg x cos θ | M1A1 |
| Answer | Marks |
|---|---|
| ω2 = 6(3a + 2x)g / 5(13 a2 + 2x2) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| x2 – 3ax + 2a2 = 0, x = a or 2a | M1A1 | Equate ω2 to 2g/5a and solve quadratic for x |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(i) ---
3(i) | I = ⅓ M (3a)2 + M (3a)2 or (4/3) M (3a)2 [= 12 Ma2]
AB | B1 | Find or state MI of rod AB about axis l
I = ½ 2 M a2 + 2M x2 [= Ma2 + 2M x2]
disc | M1A1 | Find MI of disc about axis l
I = 13 M a2 + 2M x2 | A1 | Find MI of object about axis l
4
--- 3(ii) ---
3(ii) | ½ I ω2 = Mg × 3a cos θ + 2Mg x cos θ | M1A1 | Find ω2 or angular speed ω at angle θ by energy,
with cos θ = 3/5
ω2 = 6(3a + 2x)g / 5(13 a2 + 2x2) | A1
6a(3a + 2x) = 2(13 a2 + 2x2)
x2 – 3ax + 2a2 = 0, x = a or 2a | M1A1 | Equate ω2 to 2g/5a and solve quadratic for x
5
Question | Answer | Marks | Guidance\includegraphics{figure_3}
A uniform disc, of radius $a$ and mass $2M$, is attached to a thin uniform rod $AB$ of length $6a$ and mass $M$. The rod lies along a diameter of the disc, so that the centre of the disc is a distance $x$ from $A$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find the moment of inertia of the object, consisting of disc and rod, about a fixed horizontal axis $l$ through $A$ and perpendicular to the plane of the disc. [4]
\end{enumerate}
The object is free to rotate about the axis $l$. The object is held with $AB$ horizontal and is released from rest. When $AB$ makes an angle $\theta$ with the vertical, where $\cos \theta = \frac{3}{5}$, the angular speed of the object is $\sqrt{\left(\frac{2g}{5a}\right)}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the possible values of $x$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2018 Q3 [9]}}