Rank in order, from smallest to largest, the moments of inertia i1, i2, and i3 about axes through the centers of the rods. rank i1, i2, and i3 from smallest to largest. to rank items as equivalent, overlap them.

Ranking from the smallest to the largest moment of inertia ; I₂ < I₁ < I₃

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Given that the formula for moment of inertia

I = m₁r₂² + m₂r₂² + m₃r₃³ + -----

when we apply this formula to the moments of inertia

I₁ = m*(r/2)² + m*(r/2)² =

I₂ = 2m * (r/4)² + m*(r/4)² =

I₃ = (m/2)*r² + (m/2)*r² =  

Therefore from the given values gotten from the summation of the various moments of inertia when ranking  the moment of inertia from the smallest to the largest we will have  I₂ < I₁ < I₃

Hence we can conclude that ranking from the smallest to the largest moment of inertia is   I₂ < I₁ < I₃

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Given :

Pair 1 --

Mass -  m , m and

Radius -  R

Pair 2 --

Mass - 2m , m and

Radius -  

Pair 3 --

Mass -    and

Radius - 2R

Solution :

We know that moment of inertia is,

Now for pair 1,

---- (1)

Now for pair 2,

 ----- (2)

Now for pair 3,

 ------ (3)

So, from equation (1), (2) and (3) we get,

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