CAIE FP2 2014 June — Question 5 13 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeSmall oscillations period
DifficultyChallenging +1.8 This is a challenging compound pendulum problem requiring: (1) calculation of moment of inertia for a composite body with non-uniform mass distribution (glass sheet plus four metal strips), using parallel axis theorem multiple times; (2) finding the center of mass of the composite system; (3) applying the small-angle approximation to derive SHM; (4) calculating the period. The multi-component setup, extensive algebraic manipulation, and integration of multiple mechanics concepts (rotational inertia, center of mass, SHM) make this substantially harder than typical A-level questions, though it follows a standard compound pendulum framework once set up.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x6.04d Integration: for centre of mass of laminas/solids

5 A uniform rectangular thin sheet of glass \(A B C D\), in which \(A B = 8 a\) and \(B C = 6 a\), has mass \(\frac { 3 } { 5 } M\). Each of the edges \(A B , B C , C D\) and \(D A\) has a thin strip of metal attached to it, as a border to the glass. The strips along \(A B\) and \(C D\) each have mass \(M\), and the strips along \(B C\) and \(D A\) each have mass \(\frac { 1 } { 3 } M\). Show that the moment of inertia of the whole object (glass and metal strips) about an axis through \(A\) perpendicular to the plane of the object is \(128 M a ^ { 2 }\). The object is free to rotate about this axis, which is fixed and smooth. The object hangs in equilibrium with \(C\) vertically below \(A\). It is displaced through a small angle and released from rest. Show that it will move in approximate simple harmonic motion and state the period of the motion.

Question 5:
MI of components about A:
AnswerMarks Guidance
Glass: \(\frac{3M}{5}\{\frac{1}{3}(5a)^2 + 25a^2\} = 20Ma^2\)M1 A1
\(AB\): \(M\{\frac{1}{3}(4a)^2 + (4a)^2\} = 64Ma^2/3\)B1
\(AD\): \(\frac{1}{3}M\{\frac{1}{3}(3a)^2 + (3a)^2\} = 4Ma^2\)B1 (M1 for \(BC\) or \(CD\))
\(BC\): \(\frac{1}{3}M\{\frac{1}{3}(3a)^2 + 73a^2\} = 76Ma^2/3\)M1 A1
\(CD\): \(M\{\frac{1}{3}(4a)^2 + 52a^2\} = 172Ma^2/3\)A1
Total MI:
AnswerMarks Guidance
\(I = 128Ma^2\) A.G.A1 Find total MI about \(A\)
State or imply total mass acts at mid-point of \(AC\)M1 OR can first find total MI about centre of mass
Equation of motion:
AnswerMarks Guidance
\(Id^2\theta/dt^2 = [-](49Mg/15)\ 5a\sin\theta\)M1 A1 Use equation of circular motion to find \(d^2\theta/dt^2\)
\(d^2\theta/dt^2 = -(49g/384a)\ \theta\)A1 Approximate \(\sin\theta\) by \(\theta\) and substitute for \(I\)
Period:
AnswerMarks Guidance
\(T = 2\pi\sqrt{(384a/49g)}\) Find period \(T = 2\pi/\omega\) with \(\omega = \sqrt{(49g/384a)}\)
or \((16\pi/7)\sqrt{(6a/g)}\)
or \(17.6\sqrt{(a/g)}\) (A.E.F.)B1
# Question 5:

**MI of components about A:**
| Glass: $\frac{3M}{5}\{\frac{1}{3}(5a)^2 + 25a^2\} = 20Ma^2$ | M1 A1 | |
| $AB$: $M\{\frac{1}{3}(4a)^2 + (4a)^2\} = 64Ma^2/3$ | B1 | |
| $AD$: $\frac{1}{3}M\{\frac{1}{3}(3a)^2 + (3a)^2\} = 4Ma^2$ | B1 | (M1 for $BC$ or $CD$) |
| $BC$: $\frac{1}{3}M\{\frac{1}{3}(3a)^2 + 73a^2\} = 76Ma^2/3$ | M1 A1 | |
| $CD$: $M\{\frac{1}{3}(4a)^2 + 52a^2\} = 172Ma^2/3$ | A1 | |

**Total MI:**
| $I = 128Ma^2$ A.G. | A1 | Find total MI about $A$ |
| State or imply total mass acts at mid-point of $AC$ | M1 | OR can first find total MI about centre of mass |

**Equation of motion:**
| $Id^2\theta/dt^2 = [-](49Mg/15)\ 5a\sin\theta$ | M1 A1 | Use equation of circular motion to find $d^2\theta/dt^2$ |
| $d^2\theta/dt^2 = -(49g/384a)\ \theta$ | A1 | Approximate $\sin\theta$ by $\theta$ and substitute for $I$ |

**Period:**
| $T = 2\pi\sqrt{(384a/49g)}$ | | Find period $T = 2\pi/\omega$ with $\omega = \sqrt{(49g/384a)}$ |
| or $(16\pi/7)\sqrt{(6a/g)}$ | | |
| or $17.6\sqrt{(a/g)}$ (A.E.F.) | B1 | |

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5 A uniform rectangular thin sheet of glass $A B C D$, in which $A B = 8 a$ and $B C = 6 a$, has mass $\frac { 3 } { 5 } M$. Each of the edges $A B , B C , C D$ and $D A$ has a thin strip of metal attached to it, as a border to the glass. The strips along $A B$ and $C D$ each have mass $M$, and the strips along $B C$ and $D A$ each have mass $\frac { 1 } { 3 } M$. Show that the moment of inertia of the whole object (glass and metal strips) about an axis through $A$ perpendicular to the plane of the object is $128 M a ^ { 2 }$.

The object is free to rotate about this axis, which is fixed and smooth. The object hangs in equilibrium with $C$ vertically below $A$. It is displaced through a small angle and released from rest. Show that it will move in approximate simple harmonic motion and state the period of the motion.

\hfill \mbox{\textit{CAIE FP2 2014 Q5 [13]}}