[Physics] Moment of inertia if a part is cut from original object

moment of inertiarotational-kinematics

Now if I want to find out the moment of inertia about a object say a square of lamina about axis passing through center of mass and perpendicular to it ,
then that would be
$ML^2 /12$
and now if I cut some part from it ,say another square of side say$ b $ then I think I should deduce moment of inertia of this square about the same axis and then subtract it

$ML^2 /12 -Mb^2 /12$ = moment of inertia of left over part

but proceeding ahead gives me incorrect results .

Then am I correct ? ( cause I used to do the same for calculation of center of mass when some part was cut from the original object) Can I apply the same principle here or its different?

Best Answer

Yes that's a perfectly good approach.

Suppose you have an object made up from two parts, $A$ and $B$, then the total moment of inertia is the sum of the moments of inertia of the two parts:

$$ I_{tot} = I_A + I_B $$

So if $A$ is the object with the square hole and $B$ is the square that fills the hole then it's quite correct so say:

$$ I_A = I_{tot} - I_B $$

However there are two things to consider. Firstly the moment of inertia of a square plate about a perpendicular axis through its centre is:

$$ I = \frac{ML^2}{6} $$

so that's a $6$ in the denominator not a $12$. Secondly if the square you cut out is not centred on the axis you need to calculate its moment of inertia using the parallel axis theorem.

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