CAIE FP2 2012 November — Question 2 7 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeEquation of motion angular acceleration
DifficultyStandard +0.8 This is a standard Further Maths mechanics problem involving moments of inertia, requiring application of rotational dynamics equations (τ = Iα) and energy methods. While it involves multiple steps and careful handling of the string constraint (v = rω), the approach is methodical and follows well-established techniques taught in FM2. The calculations are straightforward once the correct equations are set up, making it moderately challenging but not requiring novel insight.
Spec6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work

2 \includegraphics[max width=\textwidth, alt={}, center]{d3e9a568-a9ea-483e-8e65-90fdc4a69781-2_431_421_881_861} A uniform disc of radius 0.4 m is free to rotate without friction in a vertical plane about a horizontal axis through its centre. The moment of inertia of the disc about the axis is \(0.2 \mathrm {~kg} \mathrm {~m} ^ { 2 }\). One end of a light inextensible string is attached to a point on the rim of the disc and the string is wound round the rim. The other end of the string is attached to a particle of mass 1.5 kg which hangs freely (see diagram). The system is released from rest. Find
  1. the angular acceleration of the disc,
  2. the speed of the particle when the disc has turned through an angle of \(\frac { 1 } { 6 } \pi\).

Question 2(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(T \times 0.4 = 0.2\frac{d^2\theta}{dt^2}\)M1 Eqn of motion for disc
\(1.5g - T = 1.5 \times 0.4\frac{d^2\theta}{dt^2}\)M1 Eqn of motion for particle
\(1.5g = (0.6 + 0.5)\frac{d^2\theta}{dt^2}\)M1 Eliminate \(T\)
\(\frac{d^2\theta}{dt^2} = 15g/11\) or \(13.6\) rad s\(^{-2}\)A1
S.R.: M1 only for \(1.5g \times 0.4 = 0.2\frac{d^2\theta}{dt^2}\); \([\frac{d^2\theta}{dt^2} = 30, (\frac{d\theta}{dt})^2 = 10\pi, v = 2.24]\)
Total: 4 marks
Question 2(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
EITHER
\(\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2 = \frac{15g}{11}\theta + c\)M1 Integrate to find \(\left(\frac{d\theta}{dt}\right)^2\)
\(\left(\frac{d\theta}{dt}\right)^2 = 5\pi g/11\) or \(14.3\)A1 Apply initial conds. and \(\theta = \pi/6\)
OR
\(\frac{1}{2}(0.2)\left(\frac{d\theta}{dt}\right)^2 + \frac{1}{2}(1.5)(0.4\frac{d\theta}{dt})^2 = 1.5g \times 0.4 \times \pi/6\)(M1) Use energy to find \(\left(\frac{d\theta}{dt}\right)^2\)
\(\left(\frac{d\theta}{dt}\right)^2 = 5\pi g/11\) or \(14.3\)(A1) Simplify
\(v = 0.4\frac{d\theta}{dt} = 51\) m s\(^{-1}\)B1 Find speed of particle
Total: 3 marks [7]
## Question 2(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T \times 0.4 = 0.2\frac{d^2\theta}{dt^2}$ | M1 | Eqn of motion for disc |
| $1.5g - T = 1.5 \times 0.4\frac{d^2\theta}{dt^2}$ | M1 | Eqn of motion for particle |
| $1.5g = (0.6 + 0.5)\frac{d^2\theta}{dt^2}$ | M1 | Eliminate $T$ |
| $\frac{d^2\theta}{dt^2} = 15g/11$ or $13.6$ rad s$^{-2}$ | A1 | |

**S.R.:** M1 only for $1.5g \times 0.4 = 0.2\frac{d^2\theta}{dt^2}$; $[\frac{d^2\theta}{dt^2} = 30, (\frac{d\theta}{dt})^2 = 10\pi, v = 2.24]$

**Total: 4 marks**

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## Question 2(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| **EITHER** | | |
| $\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2 = \frac{15g}{11}\theta + c$ | M1 | Integrate to find $\left(\frac{d\theta}{dt}\right)^2$ |
| $\left(\frac{d\theta}{dt}\right)^2 = 5\pi g/11$ or $14.3$ | A1 | Apply initial conds. and $\theta = \pi/6$ |
| **OR** | | |
| $\frac{1}{2}(0.2)\left(\frac{d\theta}{dt}\right)^2 + \frac{1}{2}(1.5)(0.4\frac{d\theta}{dt})^2 = 1.5g \times 0.4 \times \pi/6$ | (M1) | Use energy to find $\left(\frac{d\theta}{dt}\right)^2$ |
| $\left(\frac{d\theta}{dt}\right)^2 = 5\pi g/11$ or $14.3$ | (A1) | Simplify |
| $v = 0.4\frac{d\theta}{dt} = 51$ m s$^{-1}$ | B1 | Find speed of particle |

**Total: 3 marks [7]**

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2\\
\includegraphics[max width=\textwidth, alt={}, center]{d3e9a568-a9ea-483e-8e65-90fdc4a69781-2_431_421_881_861}

A uniform disc of radius 0.4 m is free to rotate without friction in a vertical plane about a horizontal axis through its centre. The moment of inertia of the disc about the axis is $0.2 \mathrm {~kg} \mathrm {~m} ^ { 2 }$. One end of a light inextensible string is attached to a point on the rim of the disc and the string is wound round the rim. The other end of the string is attached to a particle of mass 1.5 kg which hangs freely (see diagram). The system is released from rest. Find\\
(i) the angular acceleration of the disc,\\
(ii) the speed of the particle when the disc has turned through an angle of $\frac { 1 } { 6 } \pi$.

\hfill \mbox{\textit{CAIE FP2 2012 Q2 [7]}}