A uniform circular disc of radius ‘R’ is rolling without slipping on a rough horizontal surface with a constant acceleration ‘a’. Then the radius of curvature of trajectory of point ‘A’ of the disc relative to the ground at the given instant as shown in the figure is

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$2 \sqrt{2} R$

$\sqrt{2 R}$

answer is B.

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Detailed Solution

$∴$ $a_{A (n)} = \frac{ω^{2} R}{\sqrt{2}} = \frac{V^{2}}{\sqrt{2} R}$ Velocity of point ‘A’ $V_{A} = \sqrt{V^{2} + ω^{2} R^{2}} = v \sqrt{2}$ Normal acceleration of point A,
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$a_{A} (n) = ω^{2} R \cos 45 ° + α R \cos 45 ° - a \cos 45 °$

radius of curvature of trajectory of point ‘A’ relative to the ground is

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