A particle is describing circular motion in a | vedclass

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(A) Let $m$ be the mass of the particle,$r$ be the radius of the circular path,and $N$ be the normal reaction force from the cone surface.For the part

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increases as $h$ increases keeping $\alpha$ same

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(A) Let $m$ be the mass of the particle,$r$ be the radius of the circular path,and $N$ be the normal reaction force from the cone surface.For the particle in horizontal circular motion,the forces acting are gravity ($mg$ downwards) and normal reaction ($N$ perpendicular to the surface).The vertical component of the normal reaction balances the weight: $N \cos \alpha = mg$.The horizontal component of the normal reaction provides the centripetal force: $N \sin \alpha = m \omega^2 r$.Dividing the two equations: $	an \alpha = \frac{m \omega^2 r}{mg} = \frac{\omega^2 r}{g}$.Since $r = h 	an \alpha$,we have $	an \alpha = \frac{\omega^2 (h 	an \alpha)}{g}$.This simplifies to $\omega^2 = \frac{g}{h}$,or $\omega = \sqrt{\frac{g}{h}}$.The period of revolution $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{h}{g}}$.From this expression,$T \propto \sqrt{h}$.Therefore,as $h$ increases,the period of revolution $T$ increases.

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