A bead of mass m stays at point P(a, b) on a wire bent in the shape of a parabola $$y = 4Cx^2$$ and rotating with angular speed $$\omega$$ (see figure). The value of $$\omega$$ is (neglect friction):

Equation of the parabolic wire profile: $$y = 4Cx^2$$

Slope of the tangent at any point on the parabola: $$\tan\theta = \frac{dy}{dx} = 8Cx$$

At point $$P(a,b)$$, the horizontal distance from the axis of rotation is $$x = a$$: $$\tan\theta = 8Ca$$

Balancing forces in the rotating frame of the wire: $$N\sin\theta = m\omega^2 a$$

$$N\cos\theta = mg$$

Dividing the horizontal equilibrium equation by the vertical one:

$$\tan\theta = \frac{m\omega^2 a}{mg} = \frac{\omega^2 a}{g}$$

$$8Ca = \frac{\omega^2 a}{g} \implies \omega^2 = 8gC$$

$$\omega = \sqrt{8gC} = 2\sqrt{2gC}$$