Answer:
Angular momentum, \[L=I\omega \] K.E. of rotation, \[K=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\frac{{{I}^{2}}{{\omega }^{2}}}{I}=\frac{1}{2}\frac{{{L}^{2}}}{I}\] For constant L, \[K\propto \frac{1}{I}\] As \[{{I}_{A}}>{{I}_{B}}\] \[\therefore \] \[{{K}_{A}}<{{K}_{B}}\] or \[{{K}_{B}}>{{K}_{A.}}\] So body B has a greater rotational K.E.
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