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← JEE Main 2014 · 19 Apr · Shift 1
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JEE Main 2014 · 19 Apr · Shift 1, Physics question 5

A particle is released on a vertical smooth semicircular track from point X so that OX makes angle θ from the vertical (see figure). The normal reaction of the track on the particle vanishes at point Y where OY makes angle φ with the horizontal. Then :

A semicircular track (dome) with centre O on its flat base. OX makes angle θ with the vertical; OY, further down the right side, makes angle φ with the horizontal.
Choose one option
Answer and worked solution

Official answer: C

From X to Y the height drops by R(cos⁡θ−sin⁡ϕ)R(\cos\theta-\sin\phi), so v2=2gR(cos⁡θ−sin⁡ϕ)v^2=2gR(\cos\theta-\sin\phi).

At Y the radial component of gravity is mgsin⁡ϕmg\sin\phi; with N=0N=0, mgsin⁡ϕ=mv2Rmg\sin\phi=\dfrac{mv^2}R. So sin⁡ϕ=2cos⁡θ−2sin⁡ϕ\sin\phi=2\cos\theta-2\sin\phi:
sin⁡ϕ=23cos⁡θ.\sin\phi=\frac23\cos\theta.

Topics:Motion in a vertical circleConservation of mechanical energy
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