M5 June 2011 Q7
7. 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 \(\dfrac{3}{10}Ma^2\).
[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 \(\dfrac{1}{2}mr^2\).] (10)
| Scheme | Marks |
|---|---|
| \(r_x = \dfrac{rx}{h}\) | M1A1 |
| \(\delta m = \pi r_x^{\,2}\delta x.\rho\) | M1 |
| \(= \pi\left(\dfrac{rx}{h}\right)^2\delta x.\dfrac{3M}{\pi r^2h}\) \(= \dfrac{3M}{h^3}x^2\delta x\) | A1 |
| \(\delta I = \dfrac{1}{2}\delta m\,r_x^{\,2}\) | M1A1 |
| \(= \dfrac{1}{2}\dfrac{3M}{h^3}x^2\delta x\left(\dfrac{rx}{h}\right)^2\) \(= \dfrac{3Mr^2}{2h^5}x^4\delta x\) | A1 (DM1) |
| \(I = \dfrac{3Mr^2}{2h^5}\displaystyle\int_0^h x^4\,dx\) | M1 |
| \(= \dfrac{3Mr^2}{2h^5}\left[\dfrac{x^5}{5}\right]_0^h\) | A1 |
| \(= \dfrac{3Mr^2}{10}\) | A1 |
| (10 marks) |