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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Vertical circular motionMedium2 minQB

PhysicsMediumsingle choice

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

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Solution:

$m g (R \cos θ - R \sin ϕ) = \frac{1}{2} m V^{2} . . . . . . . (1)$

On losing contact $N =0\Rightarrow m g sinϕ = \frac{m V^{2}}{R} . . . . . . (2)$

$\Rightarrow m g R (cosθ - sinϕ) = \frac{1}{2} m g R sinϕ$

$\Rightarrow 2 cos θ = 3 sinϕ$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Vertical circular motion

⏱ 2mℹ️ Source: QB

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