M5 June 2012 Q7
7.
(a) A uniform lamina of mass \(m\) is in the shape of a triangle \(ABC\). The perpendicular distance of \(C\) from the line \(AB\) is \(h\). Prove, using integration, that the moment of inertia of the lamina about \(AB\) is \(\dfrac{1}{6}mh^2\). (7)
(b) Deduce the radius of gyration of a uniform square lamina of side \(2a\), about a diagonal. (3)
The points \(X\) and \(Y\) are the mid-points of the sides \(RQ\) and \(RS\) respectively of a square \(PQRS\) of side \(2a\). A uniform lamina of mass \(M\) is in the shape of \(PQXYS\).
(c) Show that the moment of inertia of this lamina about \(XY\) is \(\dfrac{79}{84}Ma^2\). (6)
| Scheme | Marks |
|---|---|
| \(\rho = \dfrac{2m}{bh}\) | B1 |
| \(\delta m = \rho\dfrac{b(h - x)}{h}\delta x\) \(= \dfrac{2m}{h^2}(h - x)\delta x\) | M1 |
| \(\delta I = \dfrac{2m}{h^2}(h - x)x^2\delta x\) | A1 |
| \(I = \displaystyle\int_0^h \dfrac{2m}{h^2}(h - x)x^2\,\mathrm{d}x = \dfrac{2m}{h^2}\left[\dfrac{hx^3}{3} - \dfrac{x^4}{4}\right]_0^h\) | M1 A1 |
| \(= \dfrac{1}{6}mh^2\) PRINTED ANSWER | DM1 A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(I = 2 \times \dfrac{1}{6}m(a\sqrt{2})^2 = \dfrac{2}{3}ma^2\) | B1 |
| \(k = \sqrt{\dfrac{I}{M}} = \sqrt{\dfrac{\frac{2}{3}ma^2}{2m}} = \dfrac{a}{\sqrt{3}}\) | M1 A1 |
| (3) |
| Scheme | Marks |
|---|---|
| MI of square about \(QS = \dfrac{1}{3}\dfrac{8M}{7}a^2 = \dfrac{8M}{21}a^2\) | M1 A1 |
| MI of square about \(XY = \dfrac{8M}{21}a^2 + \dfrac{8M}{7}\left(\dfrac{a\sqrt{2}}{2}\right)^2\) \(= \dfrac{20Ma^2}{21}\) | M1 A1 |
| Hence, \(I_{PQXYS} = \dfrac{20Ma^2}{21} - \dfrac{1}{6}\dfrac{M}{7}\left(\dfrac{a}{\sqrt{2}}\right)^2 = \dfrac{79Ma^2}{84}\) PRINTED | M1 A1 |
| (6) | |
| (16 marks) |