Parallel Axis Theorem (Mass Moment of Inertia)

Parallel Axis Theorem

• Consider an element of mass dm having coordinates x,y,z with respect to origin as shown in Fig. 7.4.1. Then,

• Now consider two sets of parallel coordinate axes one passing through a point O and the other passing through the centre of gravity G of the body as shown in Fig. 7.4.2.

• Let be the coordinates of the centroid G with respect to O and x,y,z be the coordinates of dm with respect to O. If x', y', z' are coordinates of dm with respect to G.

• From equation (7.4.1),

Similarly,

Ix, Iy and Iz are moments of inertia about centroidal axes. is the perpendicular distance between the centroidal X' axis and the X axis. Similarly, is the perpendicular distance between Y' and Y axes, and, is the perpendicular distance between Z' and Z axes.

• Equations (7.4.4), (7.4.5) and (7.4.6) represent the parallel axis theorem which can be stated as follows:

• The mass moment of inertia of a body about any axis is equal to the sum of the mass moment of inertia about a parallel centroidal axis and the product of mass and square of the distance between the two parallel axes.

A general equation for the above theorem can be written as

where

I = Moment of inertia about a given axis,

IG = Moment of inertia about a parallel centroidal axis.

and

d = Distance between the two parallel axes

Solved Examples for Understanding

Example 7.4.1 

Determine the mass moment of inertia of a rod of length L and a small area of cross section A about an axis perpendicular to its length at its end.

Solution: 

Consider a small element of length dx at distance x from one end as shown in Fig. 7.4.3.

Volume of the element dV = A dx

Mass of the element dm = ρ A dx

Example 7.4.2 

Determine mass moment of inertia of a cylinder shown in Fig. 7.4.4 about X, Y and Z axes.

Solution: 

Consider the differential element to be a disc of radius R, thickness dx at distance x from G as shown in Fig. 7.4.4 (a).

For the differential element,

The moment of inertia of differential element about y' axis is

By parallel axis theorem,

Example 7.4.3 

Determine moment of inertia of a solid sphere of radius R and mass m about its centroidal axis.

Solution: 

Consider the differential element to be a disc of radius r and thickness dx at distance x from the origin as shown in Fig. 7.4.5.

Example 7.4.4 

A rectangular prism is shown in Fig. 7.4.6. The origin is at the geometric centre of the prism. The X, Y and Z-axes pass through the mid points of faces.

Derive the mass moment of inertia of the prism about the X-axis. 

Solution: 

Choose a slab perpendicular to X-axis as the differential element as shown in Fig. 7.4.6 (a).

The differential element is at distance x from the origin and has width dx.

The volume of differential element is,

dV = aL dx

Mass of differential element is

dm = ρ aL dx

As the X-axis is perpendicular to the plane of the disc,

Example 7.4.5 

Determine the moment of inertia of the cone shown in Fig. 7.4.7 about the X, Y and Z axes. Take mass of the cone as m.

Solution: 

Consider the differential element to be a disc at distance x from 0 and of width dx and radius r as shown in Fig. 7.4.7 (a). 

By similarity of triangles,

Moment of inertia of disc about X-axis is,

The moment of inertia of disc about y' axis is,

⸫ By parallel axes theorem, its moment of inertia about Y-axis is,

By symmetry,