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

Question

A particle is released on a vertical smooth semicircular track from point

X

so that,

O X

makes angle

θ

from the vertical (see figure). The normal reaction of the track on the particle vanishes at the point

Y

where

O Y

makes an angle

ϕ

with the horizontal. Then

Aicharya · Accepted Answer

m g R cos &#952; - R sin &#981; = 1 2 m V 2 &#160; &#160; . . . . . . . (1) On losing contact N =0&#8658; m g sin&#981; = m V 2 R &#160; &#160; . . . . . . (2) &#8658; m g R cos&#952; - sin&#981; = 1 2 m g R sin&#981; &#8658; 2 cos &#952; = 3 sin&#981;

A particle is released on a vertical smooth semicircular track from point X so that, O X makes angle θ…

Solution

A particle is released on a vertical smooth semicircular track from point X so that, O X makes angle &#952;…

Solution

A particle is released on a vertical smooth semicircular track from point X so that, O X makes angle θ…