Spring on Loop

Initial spring potential energy:

$U_0 = \frac{1}{2} k (2R)^2 = 2k R^2$

Spring potential energy as a function of $\theta$ , where $\theta$ is the angle between the displacement vector (from origin) and the vertical.

$x = R \, sin\theta \\ y = R \, cos\theta \\ \Delta x = x = R \, sin\theta \\ \Delta y = y + R = R \, cos\theta + R \\ U = \frac{1}{2} k ((\Delta x)^2 + (\Delta y)^2) = \frac{1}{2} k (2R^2)(1 + cos\theta) = k R^2(1 + cos\theta)$

Change in spring potential energy:

$\Delta U = U_0 - U = 2k R^2 - k R^2(1 + cos\theta) = k R^2(1 - cos\theta)$

Equate $\Delta U$ to the kinetic energy:

$\Delta U = \frac{1}{2} mv^2 \implies mv^2 = 2k R^2(1 - cos\theta)$

Calculate centripetal force:

$F_c = \frac{mv^2}{R} = 2k R(1 - cos\theta)$

Spring force in the radial direction (by inspection):

$F_{sr} = \sqrt{2} kR \sqrt{1 + cos\theta} \,\, cos(\frac{\theta}{2})$

There is no reaction force when the centripetal force is equal to the spring force in the radial direction:

$2k R(1 - cos\theta) = \sqrt{2} kR \sqrt{1 + cos\theta} \,\, cos(\frac{\theta}{2}) \\ \sqrt{2}(1 - cos\theta) = \sqrt{1 + cos\theta} \,\, cos(\frac{\theta}{2}) = \sqrt{1 + cos\theta} \frac{\sqrt{1 + cos\theta}}{\sqrt{2}} \\ 2(1-cos\theta) = 1 + cos\theta \\ cos\theta = \frac{1}{3}$

Final stretch calculation:

$\sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(2R^2)(1 + cos\theta)} = \sqrt{(2R^2)(1 + \frac{1}{3})} = \boxed{R \sqrt{\frac{8}{3}}}$