A) \[\sqrt{\frac{I{{\omega }^{2}}-m{{v}^{2}}}{I}}\]
B) \[\sqrt{\frac{(I+m{{R}^{2}}){{\omega }^{2}}-m{{v}^{2}}}{I}}\]
C) \[\frac{I\omega -mvR}{I}\]
D) \[\frac{(I\omega -m{{R}^{2}})\omega -mvR}{I}\]
Correct Answer: D
Solution :
[d] Let the angular velocity of disc after child jumps off. be \[\omega '\] \[\therefore \] From conservation of angular momentum \[(I+m{{R}^{2}})\omega =mvR+I\omega \]. \[\therefore \,\,\,\omega '=\frac{(I+m{{R}^{2}})\omega +mvR}{I}\]You need to login to perform this action.
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