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MHT CET 2024 (11 May, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2024 (11 May, Shift 2)

Q15MHT CET 2024 · 11 MAY · SHIFT 2Medium

Centripetal Acceleration & Force

A particle of mass ' $m$ ' is performing uniform circular motion along a circular path of radius ' $r$ '. Its angular momentum about the axis passing through the centre and perpendicular to the plane is ' $L$ '. The kinetic energy of the particle is

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Answer: (a)

Angular momentum $L = mvr \Rightarrow v = \frac{L}{mr}$ .
$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{L^2}{m^2r^2}$
$KE = \frac{L^2}{2mr^2}$
So the answer is (a).

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