M5 June 2008 Q7
7. A uniform square lamina \(ABCD\), of mass \(2m\) and side \(3a\sqrt{2}\), is free to rotate in a vertical plane about a fixed smooth horizontal axis \(L\) which passes through \(A\) and is perpendicular to the plane of the lamina. The moment of inertia of the lamina about \(L\) is \(24ma^2\).
The lamina is at rest with \(C\) vertically above \(A\). At time \(t = 0\) the lamina is slightly displaced. At time \(t\) the lamina has rotated through an angle \(\theta\).
(a) Show that \[2a\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 = g(1 - \cos\theta).\] (4)
(b) Show that, at time \(t\), the magnitude of the component of the force acting on the lamina at \(A\), in a direction perpendicular to \(AC\), is \(\tfrac{1}{2}mg\sin\theta\). (7)
When the lamina reaches the position with \(C\) vertically below \(A\), it receives an impulse which acts at \(C\), in the plane of the lamina and in a direction which is perpendicular to the line \(AC\). As a result of this impulse the lamina is brought immediately to rest.
(c) Find the magnitude of the impulse. (5)
| Scheme | Marks |
|---|---|
| \(\tfrac{1}{2}24ma^2\dot{\theta}^2 = 2mg.3a(1 - \cos\theta)\) | M1 A1 A1 |
| \(\Rightarrow\ 2a\dot{\theta}^2 = g(1 - \cos\theta)\) * | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| (\(\searrow\)) \(2mg\sin\theta - X = 2m3a\ddot{\theta}\) | M1 A2 |
| M(\(L\)), \(2mg.3a\sin\theta = 24ma^2\ddot{\theta}\) | M1 A1 |
| \(\Rightarrow\ X = 2mg\sin\theta - 6ma\left(\dfrac{g\sin\theta}{4a}\right)\) | DM1 |
| \(= \dfrac{mg\sin\theta}{2}\) * | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(\theta = \pi\): \(2a\dot{\theta}^2 = g(1 - \cos\pi)\) | M1 |
| \(\dot{\theta} = \sqrt{\dfrac{g}{a}}\) | A1 |
| \(6a.I = 24ma^2\sqrt{\dfrac{g}{a}}\) | M1 A1 |
| \(\Rightarrow\ I = 4m\sqrt{ag}\) | A1 |
| (5) | |
| (16 marks) |