Consider the Fig. 8.4,

Let the mass moment of inertia about x- axis be I_{xx}

Mass moment of inertia about y- axis be I_{yy}

Mass moment of inertia about z- axis be I_{zz}

So, it can be written as-

I_{xx} = _{x}^{2} dm = ^{2} + z^{2}) dm

I_{yy} = _{y}^{2} dm = ^{2} + z^{2}) dm

I_{zz} = _{z}^{2} dm = ^{2} + y^{2}) dm

Suppose z is negligible as compare to x and y, then z = 0

Hence,

I_{xx} = ^{2} dm

= ^{2} x pt dA

= pt ^{2} dA

- I
_{x}

Similarly,

I_{yy} = pt. I_{y}

Now,

I_{xx} + I_{yy} = pt (I_{x} + I_{y})

= pt. I_{z}

Hence,

I_{xx} + I_{yy} = I_{zz}

Considering the above Equations, 8.8 is for thin flat plate and whenever we calculate differential mass dm for element of thickness dz

Thus, it can be modified as-

dI_{xx} + dI_{yy} = dI_{zz}

**Example 8.1:**

Suppose a slender rod of length L, mass m which is perpendicular to the axis where x-axis is passes through it as shown in Fig. P-8.1.

Find its mass moment of inertia

**Solution:**

Take, dm = m/ L dx

Thus,

The mass moment of inertia about y-axis can be calculated as,

I_{yy} = ^{2} dm

= m/ L x 1/ 3 x L^{3}

= mL^{2}/ 3

**Links of Previous Main Topic:-**

- Introduction concept of equilibrium of rigid body
- Friction introduction
- Introduction about distributed forces
- Area moments of inertia in rectangular and polar coordinates
- Mass moment of inertia introduction
- Radius of gyration in mass moments of inertia second moment of mass
- Transfer of axis
- Mass moment of inertia about x axis y axis z axis

**Links of Next Mechanical Engineering Topics:-**