M5 June 2013 (R) Q4
4. Show, using integration, that the moment of inertia of a uniform solid right circular cone of mass \(M\), height \(h\) and base radius \(a\), about an axis through the vertex, parallel to the base, is
\[\frac{3M}{20}(a^2 + 4h^2)\][You may assume without proof that the moment of inertia of a uniform circular disc, of radius \(r\) and mass \(m\), about a diameter is \(\dfrac{1}{4}mr^2\).] (13)
| Scheme | Marks |
|---|---|
| \(\rho = \dfrac{3M}{\pi a^2 h}\) | B1 |
| \(\delta m = \pi y^2 \rho\,\delta x\) | M1 A1 |
| \(y = \dfrac{ax}{h}\) | M1 A1 |
| \(\delta I = \dfrac{1}{4}\delta m y^2 + \delta m x^2\) | M1 A1 |
| \(= \dfrac{3M}{4h^5}(a^2x^4 + 4h^2x^4)\,\delta x\) | M1 A1 |
| \(I = \dfrac{3M}{4h^5}\displaystyle\int_0^h (a^2x^4 + 4h^2x^4)\,\mathrm{d}x\) | M1 |
| \(= \dfrac{3M}{4h^5}\left[\dfrac{a^2x^5}{5} + \dfrac{4h^2x^5}{5}\right]_0^h\) | A2 |
| \(= \dfrac{3M}{20}(a^2 + 4h^2)\) * | A1 |
| (13) | |
| (13 marks) |