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MHT CET 2024 (2 May, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2024 (2 May, Shift 1)

Q10MHT CET 2024 · 2 MAY · SHIFT 1Easy

Centripetal Acceleration & Force

A particle is performing uniform circular motion along the circumference of the circle of diameter 1 m with frequency 4 Hz. The acceleration of the particle in $\text{m/s}^2$ is

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Answer: (d)

$r = 0.5$ m, $\omega = 2\pi n = 8\pi$ rad/s.
$a = \omega^2 r = 64\pi^2 \times 0.5 = 32\pi^2$ m/s $^2$ .
So the answer is (d) $32\pi^2$ .

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