JEE Advanced 2016 Paper 2 Physics: Paragraph 1

At equilibrium the right-hand side of the equation of motion is zero, so $\cos\theta_0=\dfrac{g}{\omega^2R}=\dfrac{10}{40\times0.5}=\dfrac12$ and $\theta_0=60^\circ$. The bead moves on a horizontal circle of radius $R\sin\theta_0$, so it needs a horizontal inward force $m\omega^2R\sin\theta_0$. The wire's normal force acts along the radius towards the centre, and its vertical component must support the weight: $N\cos\theta_0=mg$, giving $N=mg/\cos\theta_0=2mg$. The horizontal check agrees: $N\sin\theta_0=m\omega^2R\sin\theta_0$ gives $N=m\omega^2R=m\times40\times0.5=20m=2mg$. Option A would be the reaction if the hoop were not rotating and the bead sat at the lowest point. Rotation therefore doubles the normal reaction in this equilibrium, because the bead has to be pushed round a horizontal circle as well as held up against gravity.

Linearise $\ddot\theta=f(\theta)$ with $f=-\dfrac gR\sin\theta+\omega^2\sin\theta\cos\theta$ about $\theta_0$. Then $\ddot{\delta}=f'(\theta_0)\delta$ with $f'(\theta_0)=-\dfrac gR\cos\theta_0+\omega^2\cos2\theta_0$. Using $\cos\theta_0=g/(\omega^2R)$ the first term is $-\omega^2\cos^2\theta_0$, and $\cos2\theta_0=2\cos^2\theta_0-1$, so $f'(\theta_0)=-\omega^2\sin^2\theta_0$. The angular frequency of small oscillations is $\Omega=\omega\sin\theta_0=\sqrt{40}\times\dfrac{\sqrt3}{2}=\sqrt{30}\ \mathrm{rad\,s^{-1}}$, and $T=2\pi/\sqrt{30}=1.147\ \mathrm{s}\approx1.15\ \mathrm{s}$. The value $0.99\ \mathrm{s}$ is $2\pi/\omega$ and $1.41\ \mathrm{s}$ is the non-rotating value $2\pi\sqrt{R/g}$. The stability of this equilibrium follows from the negative sign of $f'(\theta_0)$, which is why the bead really does oscillate about it rather than run away; the value $1.99\ \mathrm{s}$ arises from dividing by $\omega\cos\theta_0$ instead.