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Question

Physics Question on Moment Of Inertia

Three thin metal rods, each of mass MM and length LL, are welded to form an equilateral triangle. The moment of inertia of the composite structure about an axis passing through the centre of mass of the structure and perpendicular to its plane is

A

12ML2\frac{1}{2}ML^{2}

B

13ML2\frac{1}{3}ML^{2}

C

23ML2\frac{2}{3}ML^{2}

D

14ML2\frac{1}{4}ML^{2}

Answer

12ML2\frac{1}{2}ML^{2}

Explanation

Solution

Moment of inertia of rod BCB C about an axis perpendicular to plane of triangle ABCA B C and passing through the mid-point of rod BCB C (i.e., DD ) is l1=m212l_{1}=\frac{m^{2}}{12} From theorem of parallel axes, moment of inertia of this rod about the asked axis is l2=l1+Mr2l_{2}=l_{1}+M r^{2} =M212+M(23)=M26=\frac{M^{2}}{12}+M\left(\frac{}{2 \sqrt{3}}\right)=\frac{M^{2}}{6} \therefore Moment of inertia of all three rod is l=3l2l=3 l_{2} =3(M26)=M22=3\left(\frac{M^{2}}{6}\right)=\frac{M^{2}}{2}