Question

4 thin rods each of length l and mass m are joined to form square.what is the mi about an axis along it's one diagonal?

Answer 1

heya......


1.We have to use the theorem of parallel axes: I = ICM +Ma2 where ICM is the the moment of inertia about an axis through the centre of mass of the rod and I is the moment of inertia about a parallel axis at distance a.

2.The moment of inertia of a thin rod of mass M and length L about an axis through the mid point of the rod and perpendicular to its length is ML2/12. The moment of inertia of the rod about a parallel axis at a distance L/2 (which is the distance of the centre of the frame from each ro.is ML2/12
+ M (L/2)2 = ML2/3
You must remember that moment of inertia is a scalar quantity. Since there are four rods in the square frame, the moment of inertia of the entire frame is four times the moment of inertia of one rod. Therefore Moment of inertia of one part is (4/3) ML2. 

But diagonals are also mutually perpendicular.Hence 

ID + ID = (4/3) ML^2

Hence ID = (2/3) ML^2

tysm......#gozmit

Answer 2

Hey mate!

Here's your answer!!

We have to use the theorem of parallel axes: I = ICM +Ma2 where ICM is the the moment of inertia about an axis through the centre of mass of the rod and I is the moment of inertia about a parallel axis at distance a.

The moment of inertia of a thin rod of mass M and length L about an axis through the mid point of the rod and perpendicular to its length is ML2/12. The moment ofinertia of the rod about a parallel axis at a distance L/2 (which is the distance of the centre of the frame from each ro.is ML2/12 + M (L/2)2 = ML2/3

You must remember that moment of inertia is a scalar quantity. Since there are four rods in the square frame, the moment of inertia of the entire frame is four times the moment of inertia of one rod. Therefore Moment of inertia of one part is (4/3) ML2.

But diagonals are also mutually perpendicular.
Hence
ID + ID = (4/3) ML²
Hence ID = (2/3) ML²

$hope \: it \: helps \: you.$
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