Theorems Of Perpendicular And Parallel Axes

These are two useful theorems relating to moment of inertia. We shall first discuss the theorem of perpendicular axes and its simple yet instructive application in working out the moments of inertia of some regular-shaped bodies.

Theorem of perpendicular axes 

Fig.1

Fig. 1: Theorem of perpendicular axes applicable to a planar body; x and y axes are two perpendicular axes in the plane and the z-axis is perpendicular to the plane.

This theorem is applicable to bodies which are planar. In practice this means the theorem applies to flat bodies whose thickness is very small compared to their other dimensions (e.g. length, breadth or radius). Fig. 1 illustrates the theorem. It states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

The figure shows a planar body. An axis perpendicular to the body through a point O is taken as the z-axis. Two mutually perpendicular axes lying in the plane of the body and concurrent with z-axis, i.e., passing through O, are taken as the x and y-axes. The theorem states that

$I_z=I_x+I_y$

Let us look at the usefulness of the theorem through an example.

Example 1

What is the moment of inertia of a disc about one of its diameters?

Answer We assume the moment of inertia of the disc about an axis perpendicular to it and through its centre to be known; it is $\frac{MR^2}{2}$, where M is the mass of the disc and R is its radius (Table 1).

Fig.2

Fig. 2 Moment of inertia of a disc about a diameter, given its moment of inertia about the perpendicular axis through its centre.

The disc can be considered to be a planar body. Hence the theorem of perpendicular axes is applicable to it. As shown in Fig. 2, we take three concurrent axes through the centre of the disc, O, as the x–, y– and z–axes; x– and y–axes lie in the plane of the disc and z–axis is perpendicular to it. By the theorem of perpendicular axes,

$I_z=I_x+I_y$

Now, x and y axes are along two diameters of the disc, and by symmetry the moment of inertia of the disc is the same about any diameter. Hence

$I_x=I_y$

and $I_z=2I_x$

but $I_z=\frac{MR^2}{2}$

So finally, $I_x=I_z/2=\frac{MR^2}{4}$

Thus the moment of inertia of a disc about any of its diameter is $\frac{MR^2}{4}$.

Find similarly the moment of inertia of a ring about any of its diameters. Will the theorem be applicable to a solid cylinder?

Theorem of parallel axes

Fig.3

Fig.3 The theorem of parallel axes The z and z′ axes are two parallel axes separated by a distance a; O is the centre of mass of the body, OO’ = a.

This theorem is applicable to a body of any shape. It allows to find the moment of inertia of a body about any axis, given the moment of inertia of the body about a parallel axis through the centre of mass of the body. We shall only state this theorem and not give its proof. We shall, however, apply it to a few simple situations which will be enough to convince us about the usefulness of the theorem. The theorem may be stated as follows:

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

As shown in the Fig. 3, z and z′ are two parallel axes, separated by a distance a. The z-axis passes through the centre of mass O of the rigid body. Then according to the theorem of parallel axes

$I_{z'}=I_z+Ma^2$

where $I_z$ and $I_{z'}$  are the moments of inertia of the body about the z and z′ axes respectively, M is the total mass of the body and a is the perpendicular distance between the two parallel axes.

Example 1

What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end?

Answer

For the rod of mass M and length l, $I=\frac{Ml^2}{12}$. Using the parallel axes theorem, $I'=I+Ma^2$ with a = l/2 we get,

$I'=M\frac{l^2}{12}+M\left(\frac{l}{2}\right)^2=\frac{ml^2}{3}$

We can check this independently since I is half the moment of inertia of a rod of mass 2M and length 2l about its midpoint,

$I'=2M.\frac{4l^2}{12} \times \frac{1}{2}=\frac{ml^2}{3}$

Example 2 

What is the moment of inertia of a ring about a tangent to the circle of the ring?

Answer 

The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring.

Fig.4

The distance between these two parallel axes is R, the radius of the ring. Using the parallel axes theorem,

$I_{tanget}=I_{dia}+MR^2$

$I_{tanget}=\frac{MR^2}{2}+MR^2=\frac{3}{2}MR^2$

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