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Question

(C) For a particle moving in a horizontal circle on the inner surface of a smooth cone,the forces acting on the particle are its weight $mg$ (downward

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) For a particle moving in a horizontal circle on the inner surface of a smooth cone,the forces acting on the particle are its weight $mg$ (downward) and the normal reaction $N$ (perpendicular to the surface of the cone).The vertical component of the normal reaction balances the weight: $N \cos 	heta = mg$.The horizontal component of the normal reaction provides the necessary centripetal force: $N \sin 	heta = \frac{mv^2}{r}$,where $r$ is the radius of the circular path.Dividing the two equations: $	an 	heta = \frac{v^2}{rg}$.From the geometry of the cone,the radius $r$ at height $h$ is given by $r = h 	an 	heta$.Substituting $r$ into the equation: $	an 	heta = \frac{v^2}{(h 	an 	heta) g}$.This simplifies to $v^2 = gh 	an^2 	heta$,or $v = \sqrt{gh} 	an 	heta$.Given $h = 10 \ cm = 0.1 \ m$,$g = 10 \ m/s^2$,and assuming $	heta = 45^\circ$ (so $	an 45^\circ = 1$):$v = \sqrt{10 	imes 0.1} 	imes 1 = \sqrt{1} = 1 \ m/s$.

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