Question

A bead of mass \(m\) stays at point \(P(a, b)\) on a wire bent in the shape of a parabola \(y=4 c x^{2}\) and rotating with angular speed \(\omega\) (see figure).

The value of \(\omega\) is (neglect friction) :

• A. $\sqrt{\frac{2 gC }{ ab }}$
• B. $2 \sqrt{2 gC }$
• C. $\sqrt{\frac{2 g }{ C }}$
• D. $2 \sqrt{ gC }$

Answer 1

The Correct Option is B

For Particle to be in equilibrium, 

\(mg\,sin \theta = mx\,\omega^2cos\theta\)

⇒ \(tan \theta =\frac{\omega^2x}g\)

Also, \(y=4 c x^{2}\)

⇒ \(\frac{dy}{dx} = 8cx =\) Slope at point P = \(tan \theta\)

Equating both values of \(tan\theta\)  we get,

\(\frac{\omega^2x}g = 8cx\)

⇒ \(\omega^2 = 8cg\)

⇒ \(\omega = \sqrt{8cg} = 2\sqrt{2cg}\)

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