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Question

Physics Question on laws of motion

A particle is released on a vertical smooth semicircular track from point XX so that OXOX makes angle θ\theta from the vertical (see figure). The normal reaction of the track on the particle vanishes at point YY where OYOY makes angle ϕ\phi with the horizontal. Then :

A

sinϕ=cosθsin\,\phi = cos \,\theta

B

sinϕ=12cosθsin\,\phi =\frac{1}{2} cos \,\theta

C

sinϕ=23cosθsin\,\phi =\frac{2}{3} cos \,\theta

D

sinϕ=34cosθsin\,\phi =\frac{3}{4} cos \,\theta

Answer

sinϕ=23cosθsin\,\phi =\frac{2}{3} cos \,\theta

Explanation

Solution