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\[\frac{M{{l}^{2}}}{12}\]. \[\therefore \] \[I=\frac{M{{l}^{2}}}{12}+M{{a}^{2}}\] From \[\Delta OAO,\] \[\tan {{30}^{o}}=\frac{a}{l/2}\] \[\Rightarrow \] \[a=\frac{l}{2}\tan {{30}^{o}}\] \[\therefore \] \[I=\frac{M{{l}^{2}}}{6}\] \[\therefore \]Total\[MI=\frac{M{{l}^{2}}}{6}+\frac{M{{l}^{2}}}{6}+\frac{M{{l}^{2}}}{6}\] \[=0.045\text{ }kg-{{m}^{2}}\]You need to login to perform this action.
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