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Q. A particle performing uniform circular motion has angular momentum $L$. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is

AIEEEAIEEE 2003System of Particles and Rotational Motion

Solution:

Angular momentum of a particle performing uniform circular motion
$L = Iω$
Kinetic energy, $K=\frac{1}{2}\omega^{2}$
Therefore, $L=\frac{2K}{\omega^{2}}\omega=\frac{2K}{\omega}$
$\frac{L_{1}}{L_{2}}=\frac{K_{1}\omega_{2}}{K_{2}\omega_{1}}$
$\frac{L_{1}}{L_{2}}=2\times2=4
L_{1}=\frac{L}{4}$