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Rotational Motion question

2009 · Q162
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Rotational Motion question

2009 · Q162

NEETPhysicsRotational MotionMCQ+4 / −1
Four identical thin rods each of mass M and length lll, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is
  1. A
    23Ml2{2 \over 3}M{l^2}32​Ml2
  2. B
    133Ml2{{13} \over 3}M{l^2}313​Ml2
  3. C
    13Ml2{1 \over 3}M{l^2}31​Ml2
  4. D
    43Ml2{4 \over 3}M{l^2}34​Ml2
View written solutionFree

Correct answer: D

As per theorem of parallel axes :

I = Icm + Ma2

where Icm = moment of inertia about an axis through centre of mass of rod
I = moment of inertia about parallel axis at distance ‘a’

Now moment of inertia of thin rod of mass M and length L about an axis through the midpoint of the rod and perpendicular to its length is ML2/12.
The moment of inertia of a rod about parallel axis at distance L/2 will be :

ML2/12 + M(L/2)2 = ML2/3

As there are 4 rods in square frame, so moment of inertia of entire frame is 4 times the moment of inertia of 1 rod, so it will be (4/3) ML2

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