Answer:
We
know, angular momentum \[L=I\omega \]and K.E. of rotation, \[K=\frac{1}{2}I{{\omega
}^{2}}=\frac{1}{2}\frac{{{I}^{2}}{{\omega }^{2}}}{I}=\frac{1}{2I}\left(
{{L}^{2}} \right)\]
When L is constant, \[K\propto
1/I\]
As \[{{I}_{A}}>{{I}_{B}}\] \[\therefore
\] \[{{K}_{A}}<{{K}_{B}}\] or \[{{K}_{B}}>{{K}_{A}}\]i.e.,
the body B has greater K.E. of rotation than the body A.
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