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

PhysicsW.E.P.Vertical circular motionEasy2 minQB

PhysicsEasysingle choice

A small block of mass $m$ is released from rest from point $A$ inside a smooth hemisphere bowl of radius $R$ , which is fixed on ground such that $OA$ is horizontal. The ratio $(x)$ of magnitude of centripetal force and normal reaction on the block at any point $B$ varies with $θ$ as:

Options:

Answer:

Solution:

$\frac{m v^{2}}{R} = N - m g \sin θ$

$N = \frac{m v^{2}}{R} + m g \sin θ$

By energy conservation,

$m g R s i n θ = \frac{1}{2} m v^{2}$

$\frac{m v^{2}}{R} = 2 m g \sin θ$

$N = 3 m g \sin θ$

$R a t i o = \frac{m v^{2}}{R N} = \frac{2}{3}$ (constant)

$x = \frac{2}{3}$

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

⏱ 2mℹ️ Source: QB

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