Question

Three rods each of mass m and length I are joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the plane of triangle.

Answer 1

Solution :

Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the mid-point of rod BC (i.e., D) is

 

 

I₁ = ml^2/12

This is moment of inertia of one rod 

 

From theorem of parallel axes, moment of inertia of this rod about the asked axis is

 

   I₁= m/²/12

 

I₂ = I₁ + mr² =ml^2/12+m(l/2√3)=ml^2/6

Moment of inertia of all three rods=3×I₂ 

I=3×ml^2/6

 

 

I= m/²/2 this is moment of inertia of all three rods at centre of equilateral traingle 

 

 

 

 

The moment of inertia at centre of equilateral traingle by all three rods =ML^2/2