A bead of mass m stays at point P(a, b) on a wire bent in the shape of a parabola y = 4Cx2 and rotating with angular speed ω (see figure). The value of ω is (neglect friction): (a) 2√(2gC) (b) 2√(gC) (c) √(2gC/ab) (d) √(2g/C)

Answer: (a). In the rotating frame, normal reaction balances mg and the centrifugal force mω² x, so the wire's slope satisfies tan θ = (ω² x)/(g). Slope of the parabola: dy/dx = 8Cx, so at P it is 8Ca. 8Ca = (ω² a)/(g) ⇒ ω² = 8gC So the answer is (a) ω = 2√(2gC).

A bead of mass m stays at point P(a, b) on a wire bent in the shape… | JEE Main 2020 (2 Sep, Shift 1)

Q14JEE Main 2020 · 2 SEP · SHIFT 1Difficult

Conical Pendulum & Funnel

A bead of mass m stays at point P(a, b) on a wire bent in the shape of a parabola y = 4Cx2 and rotating with angular speed $\omega$ (see figure). The value of $\omega$ is (neglect friction):

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Answer: (a)

In the rotating frame, normal reaction balances $mg$ and the centrifugal force $m\omega^2 x$ , so the wire's slope satisfies $\tan\theta = \frac{\omega^2 x}{g}$ .
Slope of the parabola: $\frac{dy}{dx} = 8Cx$ , so at P it is $8Ca$ .
$8Ca = \frac{\omega^2 a}{g} \Rightarrow \omega^2 = 8gC$
So the answer is (a) $\omega = 2\sqrt{2gC}$ .

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