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Physics Question on System of Particles & Rotational Motion

A bead of mass mm stays at point P(a,b)P(a, b) on a wire bent in the shape of a parabola y=4cx2y=4 c x^{2} and rotating with angular speed ω\omega (see figure).
A bead of mass m stays at point Pa,b on a wire bent in the shape of a parabola y=4Cx2
The value of ω\omega is (neglect friction) :

A

2gCab\sqrt{\frac{2 gC }{ ab }}

B

22gC2 \sqrt{2 gC }

C

2gC\sqrt{\frac{2 g }{ C }}

D

2gC2 \sqrt{ gC }

Answer

22gC2 \sqrt{2 gC }

Explanation

Solution

A bead of mass m stays at point Pa,b on a wire bent in the shape of a parabola y=4Cx2

For Particle to be in equilibrium,

mg sinθ=mx ω2cosθmg\,sin \theta = mx\,\omega^2cos\theta

⇒ tanθ=ω2xgtan \theta =\frac{\omega^2x}g

Also, y=4cx2y=4 c x^{2}

⇒ dydx=8cx=\frac{dy}{dx} = 8cx = Slope at point P = tanθtan \theta

Equating both values of tanθtan\theta we get,

ω2xg=8cx\frac{\omega^2x}g = 8cx

⇒ ω2=8cg\omega^2 = 8cg

⇒ ω=8cg=22cg\omega = \sqrt{8cg} = 2\sqrt{2cg}

Therefore, The correct option is (B): 22cg2\sqrt{2cg}