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Question: A particle describes a horizontal circle at the mouth of a funnel type vessel as shown in figure. Th...

A particle describes a horizontal circle at the mouth of a funnel type vessel as shown in figure. The surface of the funnel is frictionless. The velocity vv of the particle in terms of rr and θ\theta will be

A

v=rg/tan⁡ 6muθv = \sqrt{rg/\tan\mspace{6mu}\theta}

B

v=rg 6mutan⁡θv = \sqrt{rg\mspace{6mu}\tan\theta}

C

v=rg 6mucot⁡θv = \sqrt{rg\mspace{6mu}\cot\theta}

D

v=rg/cot⁡θv = \sqrt{rg}/\cot\theta

Answer

v=rg 6mucot⁡θv = \sqrt{rg\mspace{6mu}\cot\theta}

Explanation

Solution

For uniform circular motion of a particle mv2r=Rcos⁡θ\frac{mv^{2}}{r} = R\cos\theta ….(i)

and mg=Rsin⁡θmg = R\sin\theta ….(ii)

Dividing (i) by (ii)

v2rg=cot⁡θ⇒v=rgcot⁡θ\frac{v^{2}}{rg} = \cot\theta \Rightarrow v = \sqrt{rg\cot\theta}