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Q.
The angular momentum of a particle describing uniform circular motion is $L$. If its kinetic energy is halved and angular velocity doubled, its new angular momentum is
KEAMKEAM 2011System of Particles and Rotational Motion
Solution:
Angular momentum, $L = I \omega \ldots \ldots \ldots . .$ (i) Where, I is the moment of inertia Kinetic energy, $K =\frac{ 1 }{ 2 } I \omega ^{2}$
$\Longrightarrow K =\frac{1}{2}( I \times \omega ) \omega$
$\Longrightarrow K =\frac{1}{2} L \omega$
$($ from eqn(i)) $\left.L =\frac{ 2 K }{\omega} \ldots \ldots . ii \right)$
Now, given angular frequency is halved, so, let $\omega^{\prime}=\frac{\omega}{2}$ And the kinetic energy is doubled, so, $K ^{\prime}= 2 K$
If $L ^{\prime}$ is the new angular momentum, then using (ii) we can write, $L ^{\prime}=\frac{ 2 K ^{\prime}}{ \omega ^{\prime}}$
$\Longrightarrow L ^{\prime}=4 \frac{ 2 K }{ \omega }$
$\Longrightarrow L ^{\prime}= 4 L$
So, angular momentum becomes four times its original value.