| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question requiring systematic application of parallel axis theorem and energy conservation. Part (i) involves calculating moment of inertia for multiple components using I = I_cm + md², which is routine for FM students. Part (ii) uses energy conservation (PE = KE) with center of mass calculations. While it requires careful bookkeeping across multiple objects and several calculation steps, the techniques are standard textbook methods with no novel insight required. The equilateral triangle geometry is given explicitly, reducing problem-solving demands. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks |
|---|---|
| 4(i) | I = ½ ma2 + 2 × {½ ma2 + m (4a)2 } |
| Answer | Marks | Guidance |
|---|---|---|
| [= (½ + 2 × {33/2)} ma2 = 67 ma2/2] | M1 A1 | Find MI of discs about axis l |
| Answer | Marks | Guidance |
|---|---|---|
| [= 13 ma2/9] | M1 A1 | Find MI of e.g. rod joining one of A,B or A,C about |
| Answer | Marks | Guidance |
|---|---|---|
| [= 37 ma2/9] | A1 | Find MI of rod joining B,C about axis l |
| I = (67/2 + 37/9 + 2 × 13/9) ma2 = 81 ma2/2 | A1 | Combine to find MI of object about axis l |
| Total: | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | h = 4a | B1 |
| ½ I ω2 = 4 mgh, ω2 = 64g /81a | M1 A1 FT | Find angular velocity ω when B below A by energy |
| Answer | Marks | Guidance |
|---|---|---|
| ω = (8/9) √(g/a) or 0⋅889 √(g/a) or 2⋅81/√a | A1 | (requires some simplification for this A1) |
| Total: | 4 |
Question 4:
--- 4(i) ---
4(i) | I = ½ ma2 + 2 × {½ ma2 + m (4a)2 }
discs
[= (½ + 2 × {33/2)} ma2 = 67 ma2/2] | M1 A1 | Find MI of discs about axis l
I or I = ⅓(⅓m) a2 + (⅓m) (2a)2 (AEF)
AB AC
[= 13 ma2/9] | M1 A1 | Find MI of e.g. rod joining one of A,B or A,C about
axis l
(M1 for finding MI of any of the 3 rods)
I = ⅓(⅓m) a2 + (⅓m) (2a√3)2 (AEF)
BC
[= 37 ma2/9] | A1 | Find MI of rod joining B,C about axis l
I = (67/2 + 37/9 + 2 × 13/9) ma2 = 81 ma2/2 | A1 | Combine to find MI of object about axis l
Total: | 6
--- 4(ii) ---
4(ii) | h = 4a | B1 | Find or state vertical change h of centre of mass
½ I ω2 = 4 mgh, ω2 = 64g /81a | M1 A1 FT | Find angular velocity ω when B below A by energy
(FT on I)
ω = (8/9) √(g/a) or 0⋅889 √(g/a) or 2⋅81/√a | A1 | (requires some simplification for this A1)
Total: | 4\includegraphics{figure_4}
Three identical uniform discs, $A$, $B$ and $C$, each have mass $m$ and radius $a$. They are joined together by uniform rods, each of which has mass $\frac{1}{3}m$ and length $2a$. The discs lie in the same plane and their centres form the vertices of an equilateral triangle of side $4a$. Each rod has one end rigidly attached to the circumference of a disc and the other end rigidly attached to the circumference of an adjacent disc, so that the rod lies along the line joining the centres of the two discs (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find the moment of inertia of this object about an axis $l$, which is perpendicular to the plane of the object and through the centre of disc $A$. [6]
\item The object is free to rotate about the horizontal axis $l$. It is released from rest in the position shown, with the centre of disc $B$ vertically above the centre of disc $A$.
Write down the change in the vertical position of the centre of mass of the object when the centre of disc $B$ is vertically below the centre of disc $A$. Hence find the angular velocity of the object when the centre of disc $B$ is vertically below the centre of disc $A$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2017 Q4 [10]}}