IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

Now the axis passes through one of the corners => it will be common to the corner of 2 rods.

So, moment of inertia of these 2 rods = 2/3ml²

Consider the third rod :

The moment of inertia of the third rod about its center is ml²/12.

the distance of the center of the rod from the corner =

3l/2

Using parallel axis theorem, MI of this rod about the corner

= ml²/12 + 3ml²/4 = 10ml²/12.

So net moment of inertia = 2/3ml² + 10ml²/12
= 18ml²/12.

To find radius of gyration:

Net mass of the system = 3m. Let radius of gyration be K.

Now we know that Moment of inertia = MK²

so 18ml²/12 = 3mK²

which gives K = l/

2

Hope that helped!

Cheers!

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