M5 June 2013 (R) Q5
5.
A uniform circular lamina has radius \(2a\) and centre \(C\). The points \(P\), \(Q\), \(R\) and \(S\) on the lamina are the vertices of a square with centre \(C\) and \(CP = a\). Four circular discs, each of radius \(\dfrac{a}{2}\), with centres \(P\), \(Q\), \(R\) and \(S\), are removed from the lamina. The remaining lamina forms a template \(T\), as shown in Figure 1.
The radius of gyration of \(T\) about an axis through \(C\), perpendicular to \(T\), is \(k\).
(a) Show that \(k^2 = \dfrac{55a^2}{24}\) (7)
The template \(T\) is free to rotate in a vertical plane about a fixed smooth horizontal axis which is perpendicular to \(T\) and passes through a point on its outer rim.
(b) Write down an equation of rotational motion for \(T\) and deduce that the period of small oscillations of \(T\) about its stable equilibrium position is\[2\pi\sqrt{\left(\frac{151a}{48g}\right)}\] (8)
| Scheme | Marks |
|---|---|
| Mass ratio is 16:1. | B1 |
| \(I_{\text{Big Disc}} = \dfrac{1}{2}\,16m(2a)^2 = 32ma^2\) | M1 A1 |
| \(I_{\text{Small Disc}} = \dfrac{1}{2}m\left(\dfrac{1}{2}a\right)^2 + ma^2 = \dfrac{9}{8}ma^2\) | |
| \(I_T = 32ma^2 - 4 \times \dfrac{9}{8}ma^2 = \dfrac{55}{2}ma^2\) | M1 A1 |
| Mass of \(T = 16m - 4m = 12m\) | |
| \(k^2 = \dfrac{55}{2}ma^2 \div 12m = \dfrac{55a^2}{24}\) * | M1 A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(I_{Rim} = \dfrac{55ma^2}{24} + m(2a)^2 = \dfrac{151ma^2}{24}\) | M1 A1 |
| \(mg2a\sin\theta = -\dfrac{151ma^2}{24}\ddot{\theta}\) | M1 A1 A1 |
| For small \(\theta,\ \sin\theta \approx \theta\) | M1 |
| \(-\dfrac{48g}{151a}\theta = \ddot{\theta}\) | M1 |
| SHM; \(T = 2\pi\sqrt{\dfrac{151a}{48g}}\) * | A1 |
| (8) | |
| (15 marks) |