M3 June 2011 Q4
4.
A light inextensible string has its ends attached to two fixed points \(A\) and \(B\). The point \(A\) is vertically above \(B\) and \(AB = 7a\). A particle \(P\) of mass \(m\) is fixed to the string and moves in a horizontal circle of radius \(3a\) with angular speed \(\omega\). The centre of the circle is \(C\) where \(C\) lies on \(AB\) and \(AC = 4a\), as shown in Figure 4. Both parts of the string are taut.
(a) Show that the tension in \(AP\) is \(\dfrac{5}{7}m\left(3a\omega^2 + g\right)\). (8)
(b) Find the tension in \(BP\). (2)
(c) Deduce that \(\omega \geqslant \dfrac{1}{2}\sqrt{\left(\dfrac{g}{a}\right)}\). (2)
| Scheme | Marks |
|---|---|
| \(\cos\theta = \dfrac{4}{5} \quad\) or \(\quad \sin\theta = \dfrac{3}{5}\) | B1 |
| R(vert) \(\quad T_B\cos 45 + mg = T_A\cos\theta\) \(\dfrac{1}{\sqrt{2}}T_B + mg = \dfrac{4}{5}T_A\) | M1 A1 |
| R(horiz) \(\quad T_A\sin\theta + T_B\cos 45 = m \times 3a\omega^2\) \(\dfrac{3}{5}T_A + \dfrac{1}{\sqrt{2}}T_B = 3ma\omega^2\) | M1 A1=A1 |
| \(\dfrac{3}{5}T_A - mg = 3ma\omega^2 - \dfrac{4}{5}T_a\) | M1 |
| \(\dfrac{7}{5}T_A = 3ma\omega^2 + mg\) \(T_A = \dfrac{5}{7}m\left(3a\omega^2 + g\right)\) * | A1 |
| (8) |
| Scheme | Marks |
|---|---|
| \(T_b = \sqrt{2}\left(\dfrac{4}{5}T_a - mg\right)\) \(= \sqrt{2}\left(\dfrac{4}{7}m\left(3a\omega^2 + g\right) - mg\right)\) | M1 |
| \(= \dfrac{3\sqrt{2}}{7}m\left(4a\omega^2 - g\right)\) oe | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(T_b \geqslant 0 \Rightarrow 4a\omega^2 \geqslant g\) | M1 |
| \(\omega^2 \geqslant \dfrac{g}{4a}\) \(\omega \geqslant \dfrac{1}{2}\sqrt{\dfrac{g}{a}}\) * | A1 |
| (Allow strict inequalities in (c).) | |
| (2) | |
| (12 marks) |