Prove MI by integration

A question is this type if and only if it asks the student to derive a moment of inertia formula using integration (e.g. for discs, spheres, cones, rods, laminae, solids of revolution), where the primary task is the integration proof itself.

31 questions · Challenging +1.4

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OCR M4 2006 June Q3
8 marks Standard +0.8
The region bounded by the \(x\)-axis, the lines \(x = 1\) and \(x = 2\) and the curve \(y = \frac{1}{x^2}\) for \(1 \leq x \leq 2\), is occupied by a uniform lamina of mass 24 kg. The unit of length is the metre. Find the moment of inertia of this lamina about the \(x\)-axis. [8]
Edexcel M5 Q7
10 marks Challenging +1.2
Prove, using integration, that the moment of inertia of a uniform solid right circular cone, of mass \(M\) and base radius \(a\), about its axis is \(\frac{3}{10}Ma^2\). [You may assume, without proof, that the moment of inertia of a uniform circular disc, of mass \(m\) and radius \(r\), about an axis through its centre and perpendicular to its plane is \(\frac{1}{2}mr^2\).] [10]
Edexcel M5 Q7
16 marks Challenging +1.8
  1. A uniform lamina of mass \(m\) is in the shape of a triangle \(ABC\). The perpendicular distance of \(C\) from the line \(AB\) is \(h\). Prove, using integration, that the moment of inertia of the lamina about \(AB\) is \(\frac{1}{6}mh^2\). [7]
  1. Deduce the radius of gyration of a uniform square lamina of side \(2a\), about a diagonal. [3]
The points \(X\) and \(Y\) are the mid-points of the sides \(RQ\) and \(RS\) respectively of a square \(PQRS\) of side \(2a\). A uniform lamina of mass \(M\) is in the shape of \(PQXYS\).
  1. Show that the moment of inertia of this lamina about \(XY\) is \(\frac{79}{84}Ma^2\). [6]
Edexcel M5 2006 June Q1
6 marks Standard +0.3
  1. Prove, using integration, that the moment of inertia of a uniform rod, of mass \(m\) and length \(2a\), about an axis perpendicular to the rod through one end is \(\frac{4}{3}ma^2\). [3]
  2. Hence, or otherwise, find the moment of inertia of a uniform square lamina, of mass \(M\) and side \(2a\), about an axis through one corner and perpendicular to the plane of the lamina. [3]
Edexcel M5 2012 June Q7
16 marks Challenging +1.2
  1. A uniform lamina of mass \(m\) is in the shape of a triangle \(ABC\). The perpendicular distance of \(C\) from the line \(AB\) is \(h\). Prove, using integration, that the moment of inertia of the lamina about \(AB\) is \(\frac{1}{6}mh^2\). [7]
  2. Deduce the radius of gyration of a uniform square lamina of side \(2a\), about a diagonal. [3]
The points \(X\) and \(Y\) are the mid-points of the sides \(RQ\) and \(RS\) respectively of a square \(PQRS\) of side \(2a\). A uniform lamina of mass \(M\) is in the shape of \(PQXYS\).
  1. Show that the moment of inertia of this lamina about \(XY\) is \(\frac{79}{84}Ma^2\). [6]
CAIE FP2 2013 November Q3
Standard +0.8
3 hours
Additional Materials:
Answer Booklet/Paper
Graph Paper
List of Formulae (MF10) \section*{READ THESE INSTRUCTIONS FIRST} If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet.
Write your Centre number, candidate number and name on all the work you hand in.
Write in dark blue or black pen.
You may use a soft pencil for any diagrams or graphs.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
Where a numerical value is necessary, take the acceleration due to gravity to be \(10 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
The use of a calculator is expected, where appropriate.
Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit.
You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together.
[0pt] The number of marks is given in brackets [ ] at the end of each question or part question.