M5 June 2014 (R) Q4
4. A uniform solid sphere has mass \(M\) and radius \(a\). Prove, using integration, that the moment of inertia of the sphere about a diameter is \(\dfrac{2Ma^2}{5}\)
[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 \(\dfrac{1}{2}mr^2\).] (8)
| Scheme | Marks |
|---|---|
| \(\delta V = \pi y^2\delta x\) | B1 |
| \(\delta m = \pi y^2\delta x \cdot \dfrac{3M}{4\pi a^3} = \dfrac{3M}{4a^3}y^2\delta x\) | M1 |
| \(\delta I = \tfrac{1}{2}\delta m\,y^2 = \dfrac{3M}{8a^3}y^4\delta x\) | M1 A1 |
| \(y = \sqrt{a^2 - x^2}\) | B1 |
| \(\delta I = \dfrac{3M}{8a^3}\left(\sqrt{a^2 - x^2}\right)^4\delta x = \dfrac{3M}{8a^3}(a^2 - x^2)^2\delta x\) | M1 |
| \(I = \dfrac{3M}{8a^3}\displaystyle\int_{-a}^{a}(a^4 - 2a^2x^2 + x^4)\,\mathrm{d}x\) | M1 |
| \(= \dfrac{2Ma^2}{5}\) ** | A1 |
| (8 marks) |