Edexcel M5 2003 June — Question 6 18 marks

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2003
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeSmall oscillations period
DifficultyChallenging +1.8 This is a challenging M5 compound pendulum problem requiring: (a) deriving moment of inertia by integration (standard but non-trivial), (b) applying parallel axis theorem with multiple masses (algebraically demanding), and (c) setting up and solving SHM equation with small angle approximation. The multi-part structure, need for parallel axis theorem, and careful handling of the geometry make this substantially harder than average, though it follows a recognizable M5 template.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x6.04c Composite bodies: centre of mass
6. (a) Prove, using integration, that the moment of inertia of a uniform circular disc, of mass \(m\) and radius \(a\), about an axis through its centre \(O\) perpendicular to the plane of the disc is \(\frac { 1 } { 2 } m a ^ { 2 }\). The line \(A B\) is a diameter of the disc and \(P\) is the mid-point of \(O A\). The disc is free to rotate about a fixed smooth horizontal axis \(L\). The axis lies in the plane of the disc, passes through \(P\) and is perpendicular to \(O A\). A particle of mass \(m\) is attached to the disc at \(A\) and a particle of mass \(2 m\) is attached to the disc at \(B\).
(b) Show that the moment of inertia of the loaded disc about \(L\) is \(\frac { 21 } { 4 } m a ^ { 2 }\). At time \(t = 0 , P B\) makes a small angle with the downward vertical through \(P\) and the loaded disc is released from rest. By obtaining an equation of motion for the disc and using a suitable approximation,
(c) find the time when the loaded disc first comes to instantaneous rest. END
AnswerMarks
(a) \((\delta I) = (\rho)2\pi r\delta r \times r^2\)M1
Using \((\rho) = \frac{m}{\pi a^2}\)M1
Completion: \(I = \frac{2m}{a^2}\left[\frac{r^4}{4}\right]_0^a = \frac{1}{2}ma^2\) (*)M1 A1 (4)
Disc: Use of \(\perp\) axis theorem to find \(I_{L^*}\): \(I_{L^*} = \frac{1}{2}\left(\frac{1}{2}ma^2\right) = \frac{1}{4}ma^2\)M1
Use of parallel axis theorem: \(I_L = \frac{1}{4}ma^2 + m\left(\frac{a}{2}\right)^2 = \frac{1}{2}ma^2\)M1 A1
For loaded disc: \(I = \frac{1}{2}ma^2 + m\left(\frac{a}{2}\right)^2 + 2m\left(\frac{3a}{2}\right)^2 = \frac{21}{4}ma^2\) (*)M1 A1 cso (6)
(c) \(I\ddot{\theta} = \left\{mg\left(\frac{a}{2}\right)\sin\theta - mg\left(\frac{a}{2}\right)\sin\theta - 2mg\left(\frac{3a}{2}\right)\sin\theta\right\}\)M1 A1 A1
[A1 for signs, A1 "terms"]
\(\left[\frac{21}{4}ma^2\ddot{\theta} = -3mga\sin\theta\right]\)
For small angles \(\theta \approx \sin\theta\): \(\frac{21}{4}ma^2\ddot{\theta} = -3mga\theta\)M1
\(\ddot{\theta} = -\frac{4g}{7a}\theta\)A1 ft
\(\Rightarrow\) SHM with \(\omega^2 = \frac{4g}{7a}\)M1
Time = \(\frac{\pi}{\omega}\); \(-\pi\sqrt{\frac{7a}{4g}}\) or \(\frac{\pi}{2}\sqrt{\frac{7a}{g}}\)M1; A1 (8)
Total: (18 marks)
**(a)** $(\delta I) = (\rho)2\pi r\delta r \times r^2$ | M1 |

Using $(\rho) = \frac{m}{\pi a^2}$ | M1 |

Completion: $I = \frac{2m}{a^2}\left[\frac{r^4}{4}\right]_0^a = \frac{1}{2}ma^2$ (*) | M1 A1 (4) |

Disc: Use of $\perp$ axis theorem to find $I_{L^*}$: $I_{L^*} = \frac{1}{2}\left(\frac{1}{2}ma^2\right) = \frac{1}{4}ma^2$ | M1 |

Use of parallel axis theorem: $I_L = \frac{1}{4}ma^2 + m\left(\frac{a}{2}\right)^2 = \frac{1}{2}ma^2$ | M1 A1 |

For loaded disc: $I = \frac{1}{2}ma^2 + m\left(\frac{a}{2}\right)^2 + 2m\left(\frac{3a}{2}\right)^2 = \frac{21}{4}ma^2$ (*) | M1 A1 cso (6) |

**(c)** $I\ddot{\theta} = \left\{mg\left(\frac{a}{2}\right)\sin\theta - mg\left(\frac{a}{2}\right)\sin\theta - 2mg\left(\frac{3a}{2}\right)\sin\theta\right\}$ | M1 A1 A1 |

[A1 for signs, A1 "terms"] | |

$\left[\frac{21}{4}ma^2\ddot{\theta} = -3mga\sin\theta\right]$ | |

For small angles $\theta \approx \sin\theta$: $\frac{21}{4}ma^2\ddot{\theta} = -3mga\theta$ | M1 |

$\ddot{\theta} = -\frac{4g}{7a}\theta$ | A1 ft |

$\Rightarrow$ SHM with $\omega^2 = \frac{4g}{7a}$ | M1 |

Time = $\frac{\pi}{\omega}$; $-\pi\sqrt{\frac{7a}{4g}}$ or $\frac{\pi}{2}\sqrt{\frac{7a}{g}}$ | M1; A1 (8) |

**Total: (18 marks)**
6. (a) Prove, using integration, that the moment of inertia of a uniform circular disc, of mass $m$ and radius $a$, about an axis through its centre $O$ perpendicular to the plane of the disc is $\frac { 1 } { 2 } m a ^ { 2 }$.

The line $A B$ is a diameter of the disc and $P$ is the mid-point of $O A$. The disc is free to rotate about a fixed smooth horizontal axis $L$. The axis lies in the plane of the disc, passes through $P$ and is perpendicular to $O A$. A particle of mass $m$ is attached to the disc at $A$ and a particle of mass $2 m$ is attached to the disc at $B$.\\
(b) Show that the moment of inertia of the loaded disc about $L$ is $\frac { 21 } { 4 } m a ^ { 2 }$.

At time $t = 0 , P B$ makes a small angle with the downward vertical through $P$ and the loaded disc is released from rest. By obtaining an equation of motion for the disc and using a suitable approximation,\\
(c) find the time when the loaded disc first comes to instantaneous rest.

END

\hfill \mbox{\textit{Edexcel M5 2003 Q6 [18]}}
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