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# A particle performs uniform circular motion with an angular momentum L. If the angular frequency of the particle is doubled and kinetic energy is halved, its angular momentum becomes:A) $4L$B) $2L$C) $\dfrac{L}{2}$D) $\dfrac{L}{4}$

Last updated date: 13th Jul 2024
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Hint: Angular momentum is defined as the measure of the rotational momentum of the rotating body which is equal to the product of the angular velocity of the system and the moment of the inertia to the axis. Angular momentum is a vector quantity.

Complete step by step solution:
Given data:
Initial angular momentum = L
Initial angular frequency = $\omega$
Initial Kinetic energy = k
Final angular frequency, ${\omega ‘} = 2\omega$
Final Kinetic energy, ${k’} = \dfrac{k}{2}$
Final angular momentum =?
We know that angular momentum is given by the formula, $L = mvr$
And also we know that, $v = r\omega$
Thus L can be written as, $L = m\omega {r^2}$
The kinetic energy is given by the formula, $k = \dfrac{1}{2}m{v^2}$
We can also write angular momentum in terms of Inertia and angular frequency as, $L = I\omega$
Thus kinetic energy becomes, $k = \dfrac{1}{2}I{\omega ^2}$
$\therefore$Final Inertia, ${I’} = \dfrac{1}{8}I$
Thus substituting the value of ${I’}, {\omega '}$, we get,
$\Rightarrow {L'} = \dfrac{1}{8}I \times 2\omega = \dfrac{1}{4}I\omega$ $\Rightarrow \left( {\because {L'} = {I'}{\omega '}} \right)$
$\Rightarrow {L'} = \dfrac{L}{4}$ $\left( {\because L = I\omega } \right)$
Thus the final angular momentum becomes, ${L'} = \dfrac{L}{4}$.

Hence the correct option is D.