question_answer

A uniform thin bar of mass 3 kg and length 0.9 m is bent to make an equilateral triangle. The moment of inertia of the triangle through its centre of mass and perpendicular to plane of triangle is :

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\[MI=\frac{M{{l}^{2}}}{6}+\frac{M{{l}^{2}}}{6}+\frac{M{{l}^{2}}}{6}\] \[=0.045\text{ }kg-{{m}^{2}}\]