M5 June 2008 Q6
6. A uniform solid right circular cylinder has mass \(M\), height \(h\) and radius \(a\). Find, using integration, its moment of inertia about a diameter of one of its circular ends.
[You may assume without proof that the moment of inertia of a uniform circular disc, of mass \(m\) and radius \(a\), about a diameter is \(\tfrac{1}{4}ma^2\).]
| Scheme | Marks |
|---|---|
| \(\delta m = \pi a^2\delta x.\dfrac{M}{\pi a^2h} = \dfrac{M\delta x}{h}\) | M1 A1 |
| \(\delta I = \dfrac{1}{4}\delta m.a^2 + \delta m.x^2\) | M1 A1 |
| \(= \dfrac{M}{4h}(a^2 + 4x^2)\delta x\) | M1 A1 |
| \(I = \displaystyle\int_0^h\frac{M}{4h}(a^2 + 4x^2)\,\mathrm{d}x\) | M1 A1 |
| \(= \dfrac{M}{4h}\left[a^2x + \dfrac{4}{3}x^3\right]_0^h\) | M1 |
| \(= \dfrac{M}{4}\left(a^2 + \dfrac{4}{3}h^2\right)\) | |
| \(= \dfrac{M}{12}(3a^2 + 4h^2)\) | A1 |
| (10 marks) |