(a) CP = CO = Radius of circle (R)
∴ ∠CPO = ∠POC = 60°
∴ ∠OCP is also 60°
Therefore, DOCP is an equilateral triangle.
Natural length of spring is 3R/4.
∴ Extension in the spring x = R – \(\frac{3R}{4} = \frac R4\)
⇒ Spring force, F = kx = \(\left(\frac {m g}4\right)\) and N = Normal reaction.
(b) Tangential acceleration aT the ring will move forwards the x-axis just after the release. So, net force along x-axis
Therefore, tangential acceleration of the ring,
Normal reaction N Net force along y-axis on the ring just after the release will be zero.