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Question

Physics Question on System of Particles & Rotational Motion

A uniform bar of mass MM and Length LL is bent in the form of an equilateral triangle. Find the moment of inertia of the triangle about an axis passing through the centre of mass and perpendicular to the plane of the triangle.

A

ML2ML^2

B

ML2/2ML^2/2

C

ML2/27ML^2/27

D

ML2/54ML^2/54

Answer

ML2/54ML^2/54

Explanation

Solution

x=L6tan30x=\frac{L}{6} tan \,30^{\circ}
=138L=\frac{\sqrt{13}}{8}L
Moment of inertia about C.G of one rod =Ig+Mx2=I_{g}+Mx^{2}
=M3×(L3)212+M3×(318L)2=\frac{\frac{M}{3}\times\left(\frac{L}{3}\right)^{2}}{12}+\frac{M}{3}\times\left(\frac{\sqrt{3}}{18}L\right)^{2}

Moment of inertia of all 33 rods
=ML2162×3=\frac{ML^{2}}{162}\times3
=ML254=\frac{ML^{2}}{54}