(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,
\(F = kx = \left(\frac{mg}R\right)\left(\frac R4\right)= \frac{mg}4\)
The free body diagram of the ring will be as shown.
Here, F = kx = \(\frac {mg}4\)
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
Normal reaction N : Net force along y-axis on the ring just after the release will be zero.