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Physics Question on Moment Of Inertia

Three identical thin rods each of length l and mass M are joined together to form a letter H. What is the moment of inertia of the system about one of the sides of H ?

A

Ml26\frac{Ml^{2}}{6}

B

Ml24\frac{Ml^{2}}{4}

C

23Ml2\frac{2}{3}Ml^{2}

D

43Ml2\frac{4}{3}Ml^{2}

Answer

43Ml2\frac{4}{3}Ml^{2}

Explanation

Solution

Moment of inertia of the system about rod A I=IA+IB+IC=0+(Ml212+Ml24)+Ml2I=I_{A}+I_{B}+I_{C}=0+\left(\frac{Ml^{2}}{12}+\frac{Ml^{2}}{4}\right)+Ml^{2} =43Ml2=\frac{4}{3}Ml^{2}