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MHT CET 2019 (2 May, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2019 (2 May, Shift 1)

Q46MHT CET 2019 · 2 MAY · SHIFT 1Medium

Centripetal Acceleration & Force

The real force ' $F$ ' acting on a particle of mass $m$ ' performing circular motion acts along the radius of circle ' $r$ ' and is directed towards the centre of circle. The square root of magnitude of such force is ( $T =$ periodic time)

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Answer: (a)

Centripetal force $F = m\omega^2 r = \frac{4\pi^2 m r}{T^2}$ .
$\sqrt{F} = \frac{2\pi}{T}\sqrt{mr}$
So the answer is (a) $\frac{2\pi}{T}\sqrt{mr}$ .

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