M5 June 2006 Q1
1.
(a) Prove, using integration, that the moment of inertia of a uniform rod, of mass \(m\) and length \(2a\), about an axis perpendicular to the rod through one end is \(\tfrac{4}{3}ma^2\). (3)
(b) Hence, or otherwise, find the moment of inertia of a uniform square lamina, of mass \(M\) and side \(2a\), about an axis through one corner and perpendicular to the plane of the lamina. (3)
| Scheme | Marks |
|---|---|
| \(I = \displaystyle\int_0^{2a}\frac{m}{2a}x^2\,\mathrm{d}x\) | M1 |
| \(= \dfrac{m}{2a}\left[\dfrac{x^3}{3}\right]_0^{2a}\) | A1 |
| \(= \dfrac{4}{3}ma^2\) * | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(I_x = I_y = \tfrac{4}{3}Ma^2\) (stretching rule) | M1 |
| \(I_z = I_x + I_y = \tfrac{8}{3}Ma^2\) (\(\perp^{\text{ar}}\) axes) | M1 A1 |
| (3) | |
| (6 marks) |
Notes
(Corrected from the printed mark scheme: the mass of the lamina is written as \(m\); the question gives it as \(M\).)