The axis of reference is axis about which moment of area is taken. Most of times it is eir standard x or y axis or centeroidal axis.

The moment of inertia of a triangular section about an axis passing through C.G. and parallel to base is bh³/36.

Polar moment of inertia:

Moment of inertia about an axis perpendicular to plane of an area is known as polar moment of inertia.

Let Oz be axis perpendicular to axis Ox and Oy passing through origin O.

Moment of inertia of area of plane lamina about Oz axis is given by:

${\mathrm{I}}_{\mathrm{z}\mathrm{z}}=∫{\mathrm{r}}^2\mathrm{d}\mathrm{A}=∫\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)\mathrm{d}\mathrm{A}=∫{\mathrm{x}}^2\mathrm{d}\mathrm{A}+∫{\mathrm{y}}^2\mathrm{d}\mathrm{A}={\mathrm{I}}_{\mathrm{x}\mathrm{x}}+{\mathrm{I}}_{\mathrm{y}\mathrm{y}}$

The moment of inertia of an area about an axis perpendicular to its plane (polar moment of inertia) at any point O is equal to sum of moments of inertia about any two mutually perpendicular axes passing through same point O and lying in plane of area.

The moment of inertia of a circle, I = ?D⁴/64 = 491.07 cm⁴
The area of circle = 78.57 cm,
Radius of gyration = (I/A)^1/2 = 2.5 cm.

The centre of gravity of a uniform rod lies at its middle point. The whole weight of rod acts through its middle point.

Concept:

The CG of a semicircular plate of  r radius, from its base, is

$\stackrel{¯}{y}=\frac{4r}{3\mathrm{π}}$

Calculation:

Given:

r = 33 cm

$\stackrel{¯}{y}=\frac{4r}{3\mathrm{π}}=\frac{4×33}{3×\frac{22}{7}}$

y? = 14 cm

∴ C.G. of a semicircular plate of 66 cm diameter, from its base, is 14 cm.

Additional Information

C.G. of various plain lamina are shown below in table. Here x?  & y?  represent distance of C.G. from x and y-axis respectively.

Circle
Semicircle
Triangle
Cone
Rectangle
Quarter Circle
Solid hemisphere

Explanation:

Planar moment of inertia:

Planar moment of inertia describes how an area is distributed relative to a reference axis. This is important because it specifies area’s resistance to bending.  Planar moment of inertia is denoted with letter I.

Polar Moment of Inertia:

Polar moment of inertia is analogous to planar moment of inertia but is applicable to a cylindrical object and describes its resistance to torsion (twisting due to an applied torque). Polar moment of inertia is sometimes denoted with letter J or I_P.

The relation can be derived from Perpendicular axis orem.

Perpendicular axis orem: 

This orem states that moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in plane of body.

It can be expressed as Izz = Ixx + Iyy

Here Ixx, Iyy and Izz are moment of inertia of an object along X, Y and Z axis respectively out of which

Ixx and Iyy  are planar moments of inertia and Izz is Polar moment of inertia for given figure.

For solid shaft cross-section:

Planar moment of inertia

$I=\frac{\mathrm{Ï€}{D}^4}{64}{m}^4$

Polar Moment of Inertia

$I_P=\frac{\mathrm{Ï€}{D}^4}{64}+\frac{\mathrm{Ï€}{D}^4}{64}$

$I_P=\frac{\mathrm{Ï€}{D}^4}{32}{m}^4$

Hence, $\frac{I_p}{I}=2$

Additional Information

The equation for polar moment of inertia is essentially same as that for planar moment of inertia, but in case of polar moment, distance is measured to an axis parallel to area’s cross-section, instead of I, but its units are same as those for planar moment of inertia i.e. m⁴.

Area of triangle = 20, area of rectangle = 50
The center of gravity is given by, y = (a₁y₁ + a₂y₂) / (a₁ + a₂) = (20×10/3 + 50×5) / (20+50) = 4.52cm.
This will on for y-axis.
For x-axis, The center of gravity is given by, x = (a₁x₁ + a₂x₂) / (a₁ + a₂) = (20×6.33 + 50×2.5) / (20+50) = 3.59cm.
So center of gravity will be at (2.33, 4.33).

The center of gravity of a rod will be on its center. Here it will be at 10cm.

The formula of moment of inertia is, MOI = ar² where
M = mass, a = area, l = length, r = distance.