question_answer

Point masses 1, 2, 3 and 4 kg are lying at the point (0, 0, 0), (2, 0, 0), (0, 3, 0) and (-2, - 2, 0) respectively. The moment of inertia of this system about x-axis will be :

A)  \[43kg-{{m}^{2}}\]                                         

B)  \[34kg-{{m}^{2}}\]

C)  \[27\text{ }kg-{{m}^{2}}\]                                          

D)  \[72kg-{{m}^{2}}\]

Correct Answer: A

Solution :

                Moment of inertia of the whole system about the axis of rotation will be equal to the sum of the moments of inertia of all the particles. \[I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}\] \[\therefore \]  \[I={{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2}+{{m}_{3}}r_{3}^{2}+{{m}_{4}}r_{4}^{2}\]                 \[I=(1\times 0)+(2\times 0)+(3\times {{3}^{2}})+4{{(-2)}^{2}}\] \[I=0+0+27+16=43\text{ }kg-{{m}^{2}}\]