question_answer

Three particles, each of mass m grams situated at the vertices of an equilateral triangle ABC of side I cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram\[-c{{m}^{2}}\]units will be

A)  \[(3/4)m{{l}^{2}}\]

B)  \[2m{{l}^{2}}\]

C)  \[(5/4)m{{l}^{2}}\]

D)  \[(3/2)m{{l}^{2}}\]

Correct Answer: C

Solution :

 Moment of inertia of the system about AX is given by Moment of inertia \[={{m}_{A}}r_{A}^{2}+{{m}_{B}}r_{B}^{2}+{{m}_{C}}r_{C}^{2}\] Moment of inertia \[=m{{(0)}^{2}}+m{{(l)}^{2}}+m{{(l\sin 30{}^\circ )}^{2}}\] \[=m{{l}^{2}}+\frac{m{{l}^{2}}}{4}=\frac{5}{4}m{{l}^{2}}\] Alternative: Moment of inertia of a system about a line OC perpendicular to AB, in the plane of ABC is \[{{I}_{CO}}=m\times 0+m\times {{\left( \frac{l}{2} \right)}^{2}}+m\times {{\left( \frac{l}{2} \right)}^{2}}\] \[\therefore \] \[{{I}_{CO}}=\frac{m{{l}^{2}}}{4}+\frac{m{{l}^{2}}}{4}=\frac{m{{l}^{2}}}{2}\] According to parallel-axis theorem \[{{I}_{AX}}={{I}_{CO}}+M{{x}^{2}}\] where\[x=\]distance of AX from CO, M = total mass of system \[{{I}_{AX}}=\frac{m{{l}^{2}}}{2}+3m\times {{\left( \frac{l}{2} \right)}^{2}}\] \[{{I}_{AX}}=\frac{m{{l}^{2}}}{2}+\frac{3m{{l}^{2}}}{4}=\frac{5}{4}m{{l}^{2}}\]