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MHT CET 2024 (9 May, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from MHT CET 2024 (9 May, Shift 2)

Q23MHT CET 2024 · 9 MAY · SHIFT 2Medium

Angular Momentum & its Conservation

A particle of mass ' m ' is rotating in a circular path of radius ' $r$ '. Its angular momentum is ' $L$ ' The centripetal force acting on it is ' $F$ '. The relation between ' $F$ ', ' $L$ ', ' $r$ ' and ' $m$ ' is

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Answer: (c)

$L = mvr \Rightarrow v = \dfrac{L}{mr}$
$F = \dfrac{mv^2}{r} = \dfrac{L^2}{mr^3}$
So $\dfrac{L^2}{m} = Fr^3$ : answer (c).

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