A bead of mass m is located on a parabolic wire with its axis vertical and vertex direction towards downward as in figure and whose equation is $x^{2} = a y$ . 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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$μ^{2} a$

$μ^{2} a / 4$

$2 μ a$

$μ a$

answer is C.

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Detailed Solution

For the sliding not to occur when
$\tan θ \leq μ \tan θ = \frac{d y}{d x} = \frac{2 x}{a} = \frac{2 \sqrt{y}}{\sqrt{a}} = 2 \sqrt{\frac{y}{a}} ∴ 2 \sqrt{\frac{y}{a}} \leq μ o r y \leq \frac{a μ^{2}}{4}$

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