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Question Number 18262 by Tinkutara last updated on 17/Jul/17
A point P is located above an inclined  plane. It is possible to reach the plane  by sliding under gravity down a straight  frictionless wire joining to some point  P ′ on the plane. How should P ′ be  chosen so as to minimize the time  taken?
$$\mathrm{A}\:\mathrm{point}\:{P}\:\mathrm{is}\:\mathrm{located}\:\mathrm{above}\:\mathrm{an}\:\mathrm{inclined} \\ $$$$\mathrm{plane}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{by}\:\mathrm{sliding}\:\mathrm{under}\:\mathrm{gravity}\:\mathrm{down}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{frictionless}\:\mathrm{wire}\:\mathrm{joining}\:\mathrm{to}\:\mathrm{some}\:\mathrm{point} \\ $$$${P}\:'\:\mathrm{on}\:\mathrm{the}\:\mathrm{plane}.\:\mathrm{How}\:\mathrm{should}\:{P}\:'\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{minimize}\:\mathrm{the}\:\mathrm{time} \\ $$$$\mathrm{taken}? \\ $$
Commented by Tinkutara last updated on 17/Jul/17
Commented by ajfour last updated on 18/Jul/17
Answered by ajfour last updated on 18/Jul/17
Perpendicular from P to incline  be D.  let time taken to reach P′  be t.  length of wire=Dsec ((π/2)−α−θ)                               =(D/(sin (θ+α)))   acceleration along wire=gsin θ  (D/(sin (θ+α)))=((gt^2 sin θ)/2)  t^2 =((2D)/(gsin θsin (θ+α))) >0     =((4D)/(cos α−cos (2θ+α)))  For t^2  to be minimum,   denominator is a maximum.  that is     cos (2θ−α)=−1                         2θ−α=π                            θ=(π/2)−(α/2) .  wire is then the angular bisector  of ⊥ from P to incline plane and  vertical line from P to incline plane.
$$\mathrm{Perpendicular}\:\mathrm{from}\:\mathrm{P}\:\mathrm{to}\:\mathrm{incline} \\ $$$$\mathrm{be}\:\mathrm{D}. \\ $$$$\mathrm{let}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{P}'\:\:\mathrm{be}\:\mathrm{t}. \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{wire}=\mathrm{Dsec}\:\left(\frac{\pi}{\mathrm{2}}−\alpha−\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{D}}{\mathrm{sin}\:\left(\theta+\alpha\right)} \\ $$$$\:\mathrm{acceleration}\:\mathrm{along}\:\mathrm{wire}=\mathrm{gsin}\:\theta \\ $$$$\frac{\mathrm{D}}{\mathrm{sin}\:\left(\theta+\alpha\right)}=\frac{\mathrm{gt}^{\mathrm{2}} \mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\mathrm{t}^{\mathrm{2}} =\frac{\mathrm{2D}}{\mathrm{gsin}\:\theta\mathrm{sin}\:\left(\theta+\alpha\right)}\:>\mathrm{0} \\ $$$$\:\:\:=\frac{\mathrm{4D}}{\mathrm{cos}\:\alpha−\mathrm{cos}\:\left(\mathrm{2}\theta+\alpha\right)} \\ $$$$\mathrm{For}\:\mathrm{t}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{be}\:\mathrm{minimum}, \\ $$$$\:\mathrm{denominator}\:\mathrm{is}\:\mathrm{a}\:\mathrm{maximum}. \\ $$$$\mathrm{that}\:\mathrm{is}\:\:\:\:\:\mathrm{cos}\:\left(\mathrm{2}\theta−\alpha\right)=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\theta−\alpha=\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\theta=\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}\:. \\ $$$$\mathrm{wire}\:\mathrm{is}\:\mathrm{then}\:\mathrm{the}\:\mathrm{angular}\:\mathrm{bisector} \\ $$$$\mathrm{of}\:\bot\:\mathrm{from}\:\mathrm{P}\:\mathrm{to}\:\mathrm{incline}\:\mathrm{plane}\:\mathrm{and} \\ $$$$\mathrm{vertical}\:\mathrm{line}\:\mathrm{from}\:\mathrm{P}\:\mathrm{to}\:\mathrm{incline}\:\mathrm{plane}. \\ $$
Commented by Tinkutara last updated on 19/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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