# The moment of inertia of a thin rod about a normal axis through its centre is I . it is bent at centre such that ,the two parts are perpendicular to the axis .The moment of inertia of the system...

The moment of inertia of a thin rod about a normal axis through its centre is I . it is bent at centre such that ,the two parts are perpendicular to the axis .The moment of inertia of the system about the same axis is

A) 2I

B) I

C) I/2

D) 4I

*print*Print*list*Cite

Hello!

The moment of inertia first defined for a point mass (with respect to a chosen axis).

Then the definition is extended to finite systems of point masses as the **sum** of each point mass' moment of inertia.

Then it is extended to bodies with arbitrary distributed mass with the means of an integral.

Therefore the moment of inertia is additive: for any two bodies A and B (or two parts of a body)

`I(A+B)=I(A)+I(B),`

where A+B means the system of bodies A and B (or a single body). And of course for all `I`'s the axis must be the same.

Now to our case. Two parts of the body are the two halves. They have the same mass. Before the bending they had the same mutual arrangement with the axis, therefore their moments were the same (and each was the half of the moment of the entire rod).

After the bending, these conditions are preserved. Each half has the same mass and, if we imagine each of them separately, their mutual arrangement with the axis will be indistinguishable.

So the answer is **B**, **the moment of inertia will be the same**.