Vertical Circle

Classical Mechanics Level 4

A wire is shaped in form of circular arc of a fixed radius as shown in the figure ( 0 < θ < π 2 0<\theta <\frac { \pi }{ 2 } ). A small mass is given horizontal velocity v v from point O such that it travels on this smooth wire and reaches point A and then travels in the air to reach point B . Find the value of θ \theta such that v v is minimum.

If θ \theta can be represented as π n \frac{\pi}{n} , select n n .

This problem is originally part of set Mechanics problems by Abhishek Sharma .

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8 6 4 3

Abhishek Sharma by Abhishek Sharma , 22, India

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1 solution

Rohit Ner
Dec 14, 2015


Applying work energy theorem at bottom most point and point of projection
1 2 m ( v ′ 2 − v 0 2 ) = − m g R ( 1 + cos ⁡ θ ) v ′ 2 = v 0 2 − 2 g R ( 1 + cos ⁡ θ ) 2 R sin ⁡ θ = v ′ 2 sin ⁡ 2 θ g v ′ 2 = g R sec ⁡ θ v 0 2 − 2 g R ( 1 + cos ⁡ θ ) = g R sec ⁡ θ 2 v 0 d v 0 d θ = − 2 g R sin ⁡ θ + g R sec ⁡ θ tan ⁡ θ − 2 g R sin ⁡ θ + g R sec ⁡ θ tan ⁡ θ = 0 [ ∵ d v 0 d θ = 0 ] cos ⁡ θ = 1 2 \begin{aligned} \frac{1}{2}m\left({v'}^2-{{v}_{0}}^2\right)&=-mgR\left(1+\cos\theta\right)\\{v'}^2&={{v}_{0}}^2-2gR\left(1+\cos\theta\right)\\2R\sin\theta&=\frac{{v'}^2\sin2\theta}{g}\\{v'}^2&=gR\sec\theta\\ {{v}_{0}}^2-2gR\left(1+\cos\theta\right)&=gR\sec\theta\\2{v}_{0}\frac{d{v}_{0}}{d\theta}&=-2gR\sin\theta+gR\sec\theta\tan\theta\\-2gR\sin\theta+gR\sec\theta\tan\theta&=0 \left[\because \frac{d{v}_{0}}{d\theta}=0\right]\\\cos\theta&=\frac{1}{\sqrt{2}} \end{aligned} θ = π 4 \Huge\color{#3D99F6}{\boxed{\theta=\frac{\pi}{4}}}

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