A) \[\frac{I}{\omega }\]
B) \[I{{\omega }^{2}}\]
C) \[I\omega \]
D) none of these
Correct Answer: C
Solution :
Let a body be rotating about an axis with angular velocity \[\omega \]. Let a particle be at distance \[{{r}_{1}}\] from the axis of rotation, then its linear velocity is \[{{v}_{1}}={{r}_{1}}\omega \] If mass of particle be \[{{m}_{1}}\], then its linear momentum is \[{{m}_{1}}{{v}_{1}}\]. Moment of this momentum = momentum \[\times \] distance \[={{m}_{1}}{{v}_{1}}={{r}_{1}}={{m}_{1}}\,({{r}_{1}}\omega )\times {{r}_{1}}\] \[={{m}_{1}}r_{1}^{2}\omega \] Sum of linear moments of linear momenta of all the panicles, that is angular momentum is given by \[J={{m}_{1}}r_{1}^{2}\omega +{{m}_{2}}r_{2}^{2}\omega +...\] \[=\,({{m}_{1}}\,r_{1}^{2}+{{m}_{2}}r_{2}^{2}+....)\,\omega \] \[=\Sigma =m{{r}^{2}}\omega \] where, \[I=m{{r}^{2}}\] \[\therefore \] \[J=I\omega \]
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