M5 June 2007 Q6
6.
A lamina \(S\) is formed from a uniform disc, centre \(O\) and radius \(2a\), by removing the disc of centre \(O\) and radius \(a\), as shown in Figure 2. The mass of \(S\) is \(M\).
(a) Show that the moment of inertia of \(S\) about an axis through \(O\) and perpendicular to its plane is \(\tfrac{5}{2}Ma^2\). (3)
The lamina is free to rotate about a fixed smooth horizontal axis \(L\). The axis \(L\) lies in the plane of \(S\) and is a tangent to its outer circumference, as shown in Figure 2.
(b) Show that the moment of inertia of \(S\) about \(L\) is \(\tfrac{21}{4}Ma^2\). (4)
\(S\) is displaced through a small angle from its position of stable equilibrium and, at time \(t = 0\), it is released from rest. Using the equation of motion of \(S\), with a suitable approximation,
(c) find the time when \(S\) first passes through its position of stable equilibrium. (6)
| Scheme | Marks |
|---|---|
| \(I_O = \dfrac{M}{3\pi a^2}\left(\tfrac{\pi}{2}(2a)^2(2a)^2 - \tfrac{\pi}{2}(a)^2(a)^2\right)\) | M1 A1 |
| \(= \dfrac{5Ma^2}{2}\) * | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(I_{\text{diameter}} = \tfrac{1}{2}\dfrac{5Ma^2}{2}\) (perp. axes) | M1 A1 |
| \(I_L = \dfrac{5Ma^2}{4} + M(2a)^2\) (parallel axes) | M1 |
| \(= \dfrac{21Ma^2}{4}\) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| M(\(L\)), \(-Mg2a\sin\theta = \dfrac{21Ma^2}{4}\ddot{\theta}\) | M1 A1 |
| \(\sin\theta \approx \theta \Rightarrow \ddot{\theta} = -\dfrac{8g}{21a}\theta\), so SHM | DM1 A1 |
| Time \(= \tfrac{1}{4}2\pi\sqrt{\dfrac{21a}{8g}}\) | DM1 |
| \(= \tfrac{\pi}{2}\sqrt{\dfrac{21a}{8g}}\) | A1 |
| (6) | |
| (13 marks) |