A particle circulating in a circular path has angular momentum L if its rotational kinetic energy is halved and frequency is doubled then what will be angular momentum
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OR, Angular momentum = L = Iω Where, I is the moment of inertia and ω is the angular velocity. So frequency of revolution is, f = ω/2π => ω = 2πf Kinetic energy, K = ½ Iω^2 = ½ Lω = ½ (2πfL) = πfL …………..(1) Now, frequency is doubled, so, let f' = 2f = 2(ω/2π) = ω/π The kinetic energy is halved, so, K' = K/2 = ½ (πfL) If L' is the new angular momentum, then using (1) we can write, K'= πf/L' => ½ (πfL) = π(2f)L' => L' = L/4 So, angular momentum becomes one fourth of its original value