4 One end of a light elastic string of natural length 0.2 m and modulus of elasticity 100 N is attached to a fixed point \(A\). The other end is attached to a particle of mass 5 kg . The particle moves with angular speed \(\omega\) radians per second in a horizontal circle with the centre vertically below \(A\). The string makes an angle \(\theta\) with the vertical.
- By considering the horizontal component of the tension in the string, show that the tension in the string is \(( 1 + 5 x ) \omega ^ { 2 } \mathrm {~N}\), where \(x\) is the extension, in metres, of the string.
- (a) By considering vertical forces and also Hooke's law, deduce that \(\cos \theta = \frac { 1 } { 10 x }\).
(b) Show that \(\omega > \frac { 10 \sqrt { 3 } } { 3 }\). - When the value of \(\omega\) is \(5 \sqrt { 2 }\), find the radius of the circular motion.