A bead of mass m is located on a pa...
Question
A bead of mass m is located on a parabolic wire with its axis vertical and vertex directed towards downward as in figure and whose equation is x^2= ay. If the coeffi
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( mathrm{A} ) bead of mass ( mathrm{m} ) is located on a parabolic wire with its axis vertical and vertex directed towards downward as in figure and whose equation is ( x^{2}= ) ay. If the coefficient of friction is ( mu, ) the highest distance above the ( x ) -axis at which the particle will be in equilibrium is
(A) ( mu a )
(B) ( mu^{2} a )
(C) ( frac{1}{4} mu^{2} a )
(D) ( frac{1}{2} mu a )

JEE/Engineering Exams
Physics
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A bead of mass m is located on a parabolic wire with its axis vertical and vertex directed towards downward as in figure and whose equation is x^2= ay. If the coefficient of friction is μ, the highest distance above the x -axis at which the particle will be in equilibrium is
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tangent at any x distance would be ( => ) tanthetha ( =2 times / a )
Now
friction = umgcosthetha Balancing friction with ( mathrm{mg}^{*} ) sin(thetha) we get, ucosthetha = sinthetha
or
tan(thetha) ( =u ) ( 2 x / a=u )
( x=a u / 2 )
The highest point would be, ( y=(a u / 2)^{wedge} 2 / a=>a^{wedge} 2 u^{wedge} 2 / 4 a=>a u^{wedge} 2 / 4 )

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