Question

A circular turn table of radius 0.5 m has a smooth groove in horizontal direction. A ball of mass 90 g is placed inside the groove along with a spring of spring constant 10⁴ N/m. The ball is at a distance 0.1 m from the centre when the turn table is at rest. On rotating the turn table with a constant angular velocity of 100 rad/ s, what is the distance through which the ball moves away from the centre?

Answer 1

$$\begin{gathered} Forceactingontheball=m{\omega }^{2}r \\ F=\frac{90}{1000}\times {100}^2\times 0.5=450N \\ Now, \\ F=kx(forthespring) \\ 450={10}^{4}x \\ x=0.045m \\ x=45mm=4.5cm. \end{gathered}$$
Therefore the ball moves a distance of 4.5 cm from the table.