A) \[0.030\text{ }kg-{{m}^{2}}\]
B) \[0.015\text{ }kg-{{m}^{2}}\]
C) \[0.090\text{ }kg-{{m}^{2}}\]
D) \[0.045\text{ }kg-{{m}^{2}}\]
Correct Answer: D
Solution :
From parallel axis theorem, \[I={{I}_{CM}}+M{{a}^{2}}\] where\[{{I}_{CM}}\]is moment of inertia about COM and a is distance of mass from axis of rotation. Moment of inertia of thin bar is\[\frac{M{{l}^{2}}}{12}\]. \[\therefore \] \[I=\frac{M{{l}^{2}}}{12}+M{{a}^{2}}\]You need to login to perform this action.
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