# Moment of inertia if a part is cut from original object

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?

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.

• okay I was wrong with the denominator , I got it . Thanks Commented Jan 10, 2017 at 15:59
• This is incorrect. Care must be taken to define $M$ here in order to do this. Commented Jan 12, 2023 at 21:00

Moments of inertia do not add up or subtract like areas.

Take the example of a hollow cylinder, with inner diameter $$d_I$$ and outer diameter $$d_O$$.

The MMOI of a solid rod with diameter $$d$$ is $$I_{\rm solid} = \tfrac{m}{2} \left( \tfrac{d}{2} \right)^2 = m \tfrac{d^2}{8} \tag{1}$$

So you might be tempted to write the MMOI of a hollow cylinder as

$$I_{\rm hollow} = m \tfrac{d_O^2}{8} - m \tfrac{d_I^2}{8}$$

But this is wrong

$$I_{\rm hollow} = m \left( \tfrac{d_O^2}{8} + \tfrac{d_I^2}{8} \right) \tag{2}$$

But how?

This is because you subtract the volumes of the two solids and because they have the same density the final mass $$m$$ that appears in the formula is not the same as the $$m$$ in the formula for the solid rod.

The mass of a hollow rod is

$$m = \int_{d_I/2}^{d_O/2} \rho \ell (2 \pi r)\,{\rm d}r = \rho \pi \ell \tfrac{d_O^2-d_I^2}{4} \tag{3}$$

which is used to find the density

$$\rho = \frac{4 m}{\pi \ell ( d_O^2-d_I^2) } \tag{4}$$

Now the MMOI can be calculated from the geometry and the density

$$I_{\rm hollow} = \int_{d_I/2}^{d_O/2} \rho \ell r^2 (2 \pi r)\,{\rm d}r = \rho 2 \pi \ell \left( \tfrac{d_O^2}{64} - \tfrac{d_I^2}{64} \right) \tag{5}$$

Now combine (4) and (5) and simplify to get (2)

This calculation hinges on the following algebra

$$\tfrac{d_o^4-d_I^4}{d_O^2-d_I^2} = \tfrac{ (d_O^2-d_I^2)(d_O^2+d_I^2)}{d_O^2-d_I^2} = (d_O^2+d_I^2)$$

So no, you cannot intuitively add/subtract MMOI of solids like you can with areas and volumes.

If you take care to use the appropriate mass, then you can do this

$$I_{\rm hollow} = m_O \tfrac{d_O^2}{8} - m_I \tfrac{d_I^2}{8} \tag{6}$$

this is identical to (2) given that $$m = m_O - m_I$$.

Also note that if the MMOI you are adding or subtracting is at an offset parallel axis, then you must account for the perpendicular offset $$h$$. For example:

$$I_{\rm sum} = I_1 + \left( I_2 + m_2 h_2^2 \right)$$

To add or subtract moments of inertia of a global element, the different parts of this element must be calculated according to THE SAME AXIS; a void area could, for instance, be subtracted!

If those initial inertias are not considered with respect to the same axis, you can use the Huygens-Steiner theorem to change them to the same axis.