Question:

A uniform bar of mass $ M $ and Length $ L $ 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.

Updated On: Jun 14, 2022
  • $ ML^2 $
  • $ ML^2/2 $
  • $ ML^2/27 $
  • $ ML^2/54 $
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The Correct Option is D

Solution and Explanation

$x=\frac{L}{6} tan \,30^{\circ} $
$=\frac{\sqrt{13}}{8}L$
Moment of inertia about C.G of one rod $=I_{g}+Mx^{2}$
$=\frac{\frac{M}{3}\times\left(\frac{L}{3}\right)^{2}}{12}+\frac{M}{3}\times\left(\frac{\sqrt{3}}{18}L\right)^{2}$


Moment of inertia of all $3$ rods
$=\frac{ML^{2}}{162}\times3$
$=\frac{ML^{2}}{54}$
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Concepts Used:

System of Particles and Rotational Motion

  1. The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
  2. The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
  3. The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.