question_answer

The moment of inertia of two rotating bodies A and B are \[{{I}_{A}}\] and \[{{I}_{B}}\] . \[({{I}_{A}}>{{I}_{B}})\] and their angular momenta are equal. Which one has greater K.E.?

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.