M3 January 2012 Q4
4. A light elastic string \(AB\) has natural length 0.8 m and modulus of elasticity 19.6 N. The end \(A\) is attached to a fixed point. A particle of mass 0.5 kg is attached to the end \(B\). The particle is moving with constant angular speed \(\omega\) rad s\(^{-1}\) in a horizontal circle whose centre is vertically below \(A\). The string is inclined at 60\(^\circ\) to the vertical.
(a) Show that the extension of the string is 0.4 m. (5)
(b) Find the value of \(\omega\). (5)
| Scheme | Marks |
|---|---|
| \(\uparrow\ \ T\cos 60^\circ = 0.5g,\ \ T = g \qquad (1)\) | M1, A1 |
| Extension in the string \(= x,\ T = \dfrac{\lambda x}{a} = \dfrac{19.6x}{0.8}\) | B1 |
| Using (1), \(\qquad g = 24.5x,\quad x = 0.4\) m * | M1, A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\rightarrow\ T\sin 60^\circ = 0.5 \times r \times \omega^2 \qquad (2)\) | M1 A1 |
| Using (2) \(\qquad g\sin 60^\circ = 0.5 \times (0.8 + 0.4)\sin 60^\circ\omega^2\) | M1 A1 |
| \(\omega^2 = \dfrac{2g}{1.2},\quad \omega = \sqrt{\dfrac{5g}{3}} \quad (4.04 \text{ or } 4.0)\) | A1 |
| (5) | |
| (10 marks) |