| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: rigid body compound pendulum |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring parallel axis theorem for both disc and rod, center of mass calculation, and SHM period derivation with small angle approximation. The multi-component system and algebraic manipulation to find k elevate it significantly above standard A-level, though the techniques are systematic once identified. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | I = ½ × ½ 3M (2a)2 [= 3 Ma2] | |
| disc | M1 | Find MI of disc parallel to axis l at its centre using perpendicular |
| Answer | Marks | Guidance |
|---|---|---|
| disc disc | M1A1 | Find MI of disc about axis l |
| Answer | Marks | Guidance |
|---|---|---|
| rod | B1 | Find or state MI of rod AB about axis l |
| I = (78 + 3k) Ma2 = 3 Ma2 (26 + k) AG | A1 | Find MI of object about axis l |
| Answer | Marks | Guidance |
|---|---|---|
| disc | (M1) | SC: Invalidly applying theorems in wrong order to disc (max 2) |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | [–] I d2θ/dt2 = 3Mg × 5a sin θ + (kMg × (3a/2) sin θ | |
| [ = 3 (5 + ½ k) Mga sin θ ] | M1A1 | Use eqn of circular motion to find d2θ/dt2 where θ is angle of rod |
| Answer | Marks | Guidance |
|---|---|---|
| d2θ/dt2 = – {g (10 + k) / 2a(26 + k)} θ (AEF) | M1A1 | Approximate sin θ by θ to show SHM |
| Answer | Marks | Guidance |
|---|---|---|
| 2π / √{ (10 + k) / 2(26 + k)} = 4π | M1 | Find k by equating period T to 4π√(a/g) [so ω = ½√(g/a) ] |
| 2 (26 + k) = 4 (10 + k), k = 6 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | I = ½ × ½ 3M (2a)2 [= 3 Ma2]
disc | M1 | Find MI of disc parallel to axis l at its centre using perpendicular
axis theorem (M0 if theorem not used)
I ′ = I + 3M (5a)2 = 78 Ma2
disc disc | M1A1 | Find MI of disc about axis l
I = ⅓ kM (3a/2)2 + kM (3a/2)2 = 3k Ma2
rod | B1 | Find or state MI of rod AB about axis l
I = (78 + 3k) Ma2 = 3 Ma2 (26 + k) AG | A1 | Find MI of object about axis l
(A0 if inadequate explanation)
SC: I ′ = ½ {½ 3M (2a)2 + 3M (5a)2 } = (81/2) Ma2
disc | (M1) | SC: Invalidly applying theorems in wrong order to disc (max 2)
5
--- 5(ii) ---
5(ii) | [–] I d2θ/dt2 = 3Mg × 5a sin θ + (kMg × (3a/2) sin θ
[ = 3 (5 + ½ k) Mga sin θ ] | M1A1 | Use eqn of circular motion to find d2θ/dt2 where θ is angle of rod
with vertical
d2θ/dt2 = – {g (10 + k) / 2a(26 + k)} θ (AEF) | M1A1 | Approximate sin θ by θ to show SHM
(M0 if cos θ ≈ θ used or sign wrong; A0 if LHS unclear)
2π / √{ (10 + k) / 2(26 + k)} = 4π | M1 | Find k by equating period T to 4π√(a/g) [so ω = ½√(g/a) ]
2 (26 + k) = 4 (10 + k), k = 6 | A1
6
Question | Answer | Marks | GuidanceAn object is formed from a uniform circular disc, of radius $2a$ and mass $3M$, and a uniform rod $AB$, of length $4a$ and mass $kM$, where $k$ is a constant. The centre of the disc is $O$. The end $B$ of the rod is rigidly joined to a point on the circumference of the disc so that $OBA$ is a straight line. The fixed horizontal axis $l$ is in the plane of the object, passes through $A$ and is perpendicular to $AB$.
\begin{enumerate}[label=(\roman*)]
\item Show that the moment of inertia of the object about the axis $l$ is $3Ma^2(26 + k)$.
[5]
\item The object is free to rotate about $l$.
Show that small oscillations of the object about $l$ are approximately simple harmonic. Given that the period of these oscillations is $4\pi\sqrt{\frac{a}{g}}$, find the value of $k$.
[6]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2018 Q5 [11]}}