# A particle performing uniform circular motion has  angular momentum L . its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is Option 1) Option 2) Option 3) Option 4)

As we learnt in

Rotational angular momentum -

$\vec{L}= I\vec{w}$

- wherein

$I$ = Moment of inertia about fixed axis of rotation

$w$ = angular velocity of rotation

We know that $L=I\omega$

$K.E_{not}=\frac{1}{2}I \omega^{2}$

$\frac{L}{K}_{not}=\frac{2I\omega}{I\omega^{2}}=>\frac{L}{K}=\frac{2}{\omega}=>L=\frac{2K}{\omega}$

$\therefore \frac{L_{1}}{L_{2}}=\frac{K_{1}}{K_{2}}\times \frac{\omega_{2}}{\omega_{1}}=2\times 2=4$

$\frac{L_{1}}{L_{2}}=4=>L_{2}=\frac{L_{1}}{4} = \frac{L}{4}$

Option 1)

This is the correct option.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

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