M5 June 2007 Q4
4.
A region \(R\) is bounded by the curve \(y^2 = 4ax\) \((y \gt 0)\), the \(x\)-axis and the line \(x = a\) \((a \gt 0)\), as shown in Figure 1. A uniform solid \(S\) of mass \(M\) is formed by rotating \(R\) about the \(x\)-axis through \(360^\circ\). Using integration, prove that the moment of inertia of \(S\) about the \(x\)-axis is \(\tfrac{4}{3}Ma^2\).
(You may assume without proof that the moment of inertia of a uniform disc, of mass \(m\) and radius \(r\), about an axis through its centre perpendicular to its plane is \(\tfrac{1}{2}mr^2\).)
| Scheme | Marks |
|---|---|
| \(V = \pi\displaystyle\int_0^a 4ax\,\mathrm{d}x\) | M1 |
| \(= 2\pi a^3\) | A1 |
| \(\delta m = \dfrac{M}{2\pi a^3}.\pi 4ax\,\delta x \qquad \left(= \dfrac{2M}{a^2}x\,\delta x\right)\) | M1 |
| \(\delta I = \tfrac{1}{2}\dfrac{2M}{a^2}x\,\delta x.y^2 = \dfrac{4M}{a}x^2\,\delta x\) | M1 A1 |
| \(I = \dfrac{4M}{a}\displaystyle\int_0^a x^2\,\mathrm{d}x = \tfrac{4}{3}Ma^2\) | DM1 A1 |
| (7) |