M3 June 2010 Q5
5.
A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is fixed at the point \(O\). The particle is initially held with \(OP\) horizontal and the string taut. It is then projected vertically upwards with speed \(u\), where \(u^2 = 5ag\). When \(OP\) has turned through an angle \(\theta\) the speed of \(P\) is \(v\) and the tension in the string is \(T\), as shown in Figure 5.
(a) Find, in terms of \(a\), \(g\) and \(\theta\), an expression for \(v^2\). (3)
(b) Find, in terms of \(m\), \(g\) and \(\theta\), an expression for \(T\). (4)
(c) Prove that \(P\) moves in a complete circle. (3)
(d) Find the maximum speed of \(P\). (2)
| Scheme | Marks |
|---|---|
| Energy: \(\quad mga\sin\theta = \dfrac{1}{2}m \times 5ag - \dfrac{1}{2}mv^2\) | M1 A1 |
| \(v^2 = 5ag - 2ag\sin\theta\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Eqn of motion along radius: \(T + mg\sin\theta = \dfrac{mv^2}{a}\) | M1 A1 |
| \(T = \dfrac{m}{a}(5ag - 2ag\sin\theta) - mg\sin\theta\) | M1 |
| \(T = mg(5 - 3\sin\theta)\) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| At \(C\), \(\ \theta = 90^\circ\) \(T = mg(5 - 3) = 2mg\) | M1 A1 |
| \(T > 0 \quad \therefore P\) reaches \(C\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Max speed at lowest point \((\theta = 270^\circ;\qquad v^2 = 5ag - 2ag\sin 270)\) | M1 |
| \(v^2 = 5ag + 2ag\) \(v = \sqrt{(7ag)}\) | A1 |
| (2) | |
| (12 marks) |