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JEE Main 2016 · Physics · Kinematics and Laws of Motion · medium

A small bead slides without friction on a smooth circular wire hoop of radius 0.5 m kept in a vertical plane. The hoop rotates uniformly about its…

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A small bead slides without friction on a smooth circular wire hoop of radius 0.5 m0.5\ \mathrm{m} kept in a vertical plane. The hoop rotates uniformly about its vertical diameter. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the angular speed of the hoop for which the bead stays at rest relative to the hoop at a position where its radius makes an angle of 60∘60^\circ with the downward vertical is
  1. A. 20 rad s−1\sqrt{20}\ \mathrm{rad\,s^{-1}}
  2. B. 210 rad s−12\sqrt{10}\ \mathrm{rad\,s^{-1}}
  3. C. 10 rad s−110\ \mathrm{rad\,s^{-1}}
  4. D. 4.8 rad s−14.8\ \mathrm{rad\,s^{-1}}
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Answer: (B) 210 rad s−12\sqrt{10}\ \mathrm{rad\,s^{-1}}

The bead moves in a horizontal circle of radius r=Rsin⁡θr=R\sin\theta with θ=60∘\theta=60^\circ. The only forces are gravity and the normal reaction NN directed along the radius towards the centre of the hoop. Vertical: Ncos⁡θ=mgN\cos\theta=mg. Horizontal: Nsin⁡θ=mω2Rsin⁡θN\sin\theta=m\omega^2R\sin\theta, which gives N=mω2RN=m\omega^2R. Combining, cos⁡θ=gω2R\cos\theta=\dfrac{g}{\omega^2R}, so ω2=gRcos⁡60∘=100.5×0.5=40\omega^2=\dfrac{g}{R\cos60^\circ}=\dfrac{10}{0.5\times0.5}=40 and ω=210≈6.3 rad s−1\omega=2\sqrt{10}\approx6.3\ \mathrm{rad\,s^{-1}}. Setting ω2=g/R=20\omega^2=g/R=20 would hold only at θ=0\theta=0 or cos⁡θ=1\cos\theta=1, so the answer must be larger than 20\sqrt{20}.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake