Answer:
Rotational K.E., \[K=\frac{1}{2}I{{\omega }^{2}}\] \[\therefore \] \[I=\frac{2K}{{{\omega }^{2}}}\] Angular momentum, \[L=I\omega =\frac{2K}{{{\omega }^{2}}}.\omega =\frac{2K}{\omega }\] When angular frequency is doubled and kinetic energy is halved, the angular momentum becomes, \[L'=\frac{2(K/2)}{2\omega }=\frac{1}{4}.\frac{2K}{\omega }=\frac{\mathbf{L}}{\mathbf{4}}\]
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