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 :

Question

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) sin φ = cos θ (B) sin φ =(1/2) cos θ (C) sin φ =(2/3) cos θ (D) sin φ =(3/4) cos θ

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0dd8fcfd80554bbe0629283670e10fc0-.png" /> ( mv 2/ r )= mg sin φ dots(i) mg r cos θ=(1/2) mv 2+rgsin φ ( v 2/ rg )=2 cos θ-2 sin φ dots(ii) sin φ=2 cos θ-2 sin φ 3 sin φ=2 cos θ sin φ=(2/3) cos φ

Q. 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 $O Y$ makes angle $ϕ$ with the horizontal. Then :

$s in ϕ = cos θ$

$s in ϕ = \frac{1}{2} cos θ$

$s in ϕ = \frac{2}{3} cos θ$

$s in ϕ = \frac{3}{4} cos θ$

$\frac{m v ^{2}}{r} = m g sin ϕ \dots (i)$
$m g r cos θ = \frac{1}{2} m v^{2} + r g s in ϕ$
$\frac{v ^{2}}{r g} = 2 cos θ - 2 sin ϕ \dots (ii)$
$sin ϕ = 2 cos θ - 2 sin ϕ$
$3 sin ϕ = 2 cos θ$
$sin ϕ = \frac{2}{3} cos ϕ$

Q. 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 :

sinϕ=cosθ

sinϕ=21​cosθ

sinϕ=32​cosθ

sinϕ=43​cosθ