M5 June 2007 Q8
8.
A uniform rod \(AB\) has mass \(3m\) and length \(2a\). It is free to rotate in a vertical plane about a smooth fixed horizontal axis through the point \(X\) on the rod, where \(AX = \tfrac{1}{2}a\). A particle of mass \(m\) is attached to the rod at \(B\). At time \(t = 0\), the rod is vertical, with \(B\) above \(A\), and is given an initial angular speed \(\sqrt{\dfrac{g}{a}}\). When the rod makes an angle \(\theta\) with the upward vertical, the angular speed of the rod is \(\omega\), as shown in Figure 3.
(a) By using the principle of the conservation of energy, show that\[\omega^2 = \frac{g}{2a}(5 - 3\cos\theta).\] (8)
(b) Find the angular acceleration of the rod when it makes an angle \(\theta\) with the upward vertical. (3)
When \(\theta = \phi\), the resultant force of the axis on the rod is in a direction perpendicular to the rod.
(c) Find \(\cos\phi\). (5)
| Scheme | Marks |
|---|---|
| MI of rod + particle \(= \tfrac{1}{12}3m(2a)^2 + 3m\left(\tfrac{1}{2}a\right)^2, + m\left(\tfrac{3}{2}a\right)^2\) | M1, M1 |
| \(= 4ma^2\) | A1 |
| \(\tfrac{1}{2}4ma^2\left(\omega^2 - \dfrac{g}{a}\right) = 3mg\dfrac{a}{2}(1 - \cos\theta) + mg\dfrac{3a}{2}(1 - \cos\theta)\) | M1 Aft A1 |
| \(\omega^2 = \dfrac{g}{2a}(5 - 3\cos\theta)\) | DM1 A1 |
| (8) |
| Scheme | Marks |
|---|---|
| \(4ma^2\ddot{\theta} = 3mg\tfrac{1}{2}a\sin\theta + mg\tfrac{3}{2}a\sin\theta\) | M1 A1ft |
| \(\Rightarrow\ \ddot{\theta} = \dfrac{3g\sin\theta}{4a}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(F + 4mg\cos\theta = 3m\tfrac{1}{2}a\omega^2 + m\tfrac{3}{2}a\omega^2\) | M1 A1 |
| \(= 3ma\dfrac{g}{2a}(5 - 3\cos\theta)\) | DM1 |
| \(\Rightarrow\ F = \dfrac{mg}{2}(15 - 17\cos\theta)\) | DM1 |
| When \(\theta = \phi\), \(F = 0\) \(\Rightarrow\ \cos\phi = \tfrac{15}{17}\) | A1 |
| (5) | |
| (16 marks) |