These are two useful theorems relating to the 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

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 the perpendicular axis and lying in the plane of the body.

This theorem is applicable only to the plane bodies. Let \(X\) and \(Y\)-axes be chosen in the plane of the body and \(Z\)-axis perpendicular to this plane, three axes being mutually perpendicular. Then the theorem states that
\(I_z=I_x+I_y \dots(7.36)\)

Proof:

Consider an arbitrary particle \(P\) of the body (figure 7.29). Let \(P Q\) and \(P R\) be the perpendiculars from \(P\) on the \(X\) and the \(Y\)-axes respectively. Also \(P O\) is the perpendicular from \(P\) to the \(Z\)-axis. Thus, the moment of inertia of the body about the Z-axis is
\(
\begin{aligned}
I_z & =\sum_i m_i(P O)^2=\sum_i m_i\left(P Q^2+O Q^2\right) \\
& =\sum_i m_i\left(P Q^2+P R^2\right) \\
& =\sum_i m_i(P Q)^2+\sum_i m_i(P R)^2 \\
& =I_x+I_y
\end{aligned}
\)

Example 7.20: Find the moment of inertia of a uniform ring of mass \(M\) and radius \(R\) about a diameter.

Solution:

Let \(A B\) and \(C D\) be two mutually perpendicular diameters of the ring. Take them as \(X\) and \(Y\)-axes and the line perpendicular to the plane of the ring through the centre as the \(Z\)-axis. The moment of inertia of the ring about the \(Z\)-axis is \(I=M R^2\). As the ring is uniform, all of its diameters are equivalent and so \(I_x=I_y\). From the perpendicular axes theorem,
\(
I_z=I_x+I_y \text {. Hence } I_x=\frac{I_z}{2}=\frac{M R^2}{2} \text { }
\)

Example 7.21: What is the moment of inertia of a disc about one of its diameters?

We assume the moment of inertia of the disc about an axis perpendicular to it and through its centre to be known; it is \(M R^2 / 2\), where \(M\) is the mass of the disc and \(R\) is its radius.

The disc can be considered to be a planar body. Hence the theorem of perpendicular axes is applicable to it. As shown in Fig. 7.30, 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
and
\(
\begin{aligned}
& I_x=I_y \\
& I_z=2 I_x \\
& I_z=M R^2 / 2
\end{aligned}
\)
But
\(I_z=M R^2 / 2\)
So finally, \(I_x=I_z / 2=M R^2 / 4\)
Thus the moment of inertia of a disc about any of its diameter is \(M R^2 / 4\).

Theorem of parallel axes

It states that 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. 

Suppose we have to obtain the moment of inertia of a body about a given line \(A B\) (figure 7.31). Let \(C\) be the centre of mass of the body and let \(C Z\) be the line parallel to \(A B\) through \(C\). Let \(I\) and \(I_0\) be the moments of inertia of the body about \(A B\) and \(C Z\) respectively. The parallel axes theorem states that
\(
I=I_0+M d^2
\)
where \(d\) is the perpendicular distance between the parallel lines \(A B\) and \(C Z\) and \(M\) is the mass of the body.

Take \(C\) to be the origin and \(C Z\) the \(Z\)-axis. Let \(C A\) be the perpendicular from \(C\) to \(A B\). Take \(C A\) to be the \(X\)-axis. As \(C A=d\), the coordinates of \(A\) are \((d, 0,0)\).
Let \(P\) be an arbitrary particle of the body with the coordinates \(\left(x_i, y_i, z_i\right)\). Let \(P Q\) and \(P R\) be the perpendiculars from \(P\) to \(C Z\) and \(A B\) respectively. Note that \(P\) may not be in the plane containing \(C Z\) and \(A B\). We have \(C Q=z_i\). Also \(A R=C Q=z_i\). Thus, the point \(Q\) has coordinates \(\left(0,0, z_i\right)\) and the point \(R\) has coordinates \(\left(d, 0, z_i\right)\).
\(
\begin{aligned}
I & =\sum_i m_i(P R)^2 \\
& =\sum_i m_i\left[\left(x_i-d\right)^2+\left(y_i-0\right)^2+\left(z_i-z_i\right)^2\right] \\
& =\sum_i m_i\left(x_i^2+y_i^2+d^2-2 x_i d\right) \\
& =\sum_i m_i\left(x_i^2+y_i^2\right)+\sum_i m_i d^2-2 d \sum_i m_i x_i \ldots(i)
\end{aligned}
\)
We have
\(
\sum_i m_i x_i=M X_{C M}=0 .
\)
The moment of inertia about \(C Z\) is,
\(
\begin{aligned}
I_0 & =\sum_i m_i(P Q)^2 \\
& =\sum_i m_i\left[\left(x_i-0\right)^2+\left(y_i-0\right)^2+\left(z_i-z_i\right)^2\right] \\
& =\sum_i m_i\left(x_i^2+y_i^2\right)
\end{aligned}
\)
From (i),
\(
I=I_0+\sum m_i d^2=I_0+M d^2
\)

Example 7.22: What is the moment of inertia of a rod of mass \(M[latex], length [latex]l\) about an axis perpendicular to it through one end?

Solution:

For the rod of mass \(M\) and length \(l\), \(I=M l^2 / 12\). Using the parallel axes theorem, \(I^{\prime}=I+M d^2\) with \(d=l / 2\) we get,
\(
I^{\prime}=M \frac{l^2}{12}+M\left(\frac{l}{2}\right)^2=\frac{M l^2}{3}
\)
We can check this independently since \(I\) is half the moment of inertia of a rod of mass \(2 M\) and length \(2 l\) about its midpoint,
\(
I^{\prime}=2 M \cdot \frac{4 l^2}{12} \times \frac{1}{2}=\frac{M l^2}{3}
\)

Example 7.23: What is the moment of inertia of a ring about a tangent to the circle of the ring?

Solution:

The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. The distance between these two parallel axes is \(R\), the radius of the ring. Using the parallel axes theorem,

\(
I_{\text {tangent }}=I_{d i a}+M R^2=\frac{M R^2}{2}+M R^2=\frac{3}{2} M R^2
\)

Example 7.24: Find the moment of inertia of a solid cylinder of mass \(M\) and radius \(R\) about a line parallel to the axis of the cylinder and on the surface of the cylinder.

Solution:

The moment of inertia of the cylinder about its axis \(=\frac{M R^2}{2}\)
Using parallel axes theorem
\(
I=I_0+M R^2=\frac{M R^2}{2}+M R^2=\frac{3}{2} M R^2 .
\)
Similarly, the moment of inertia of a solid sphere about a tangent is
\(
\frac{2}{5} M R^2+M R^2=\frac{7}{5} M R^2 
\)

Radius of Gyration

The radius of gyration \(k\) of a body about a given line is defined by the equation
\(
I=M k^2
\)
where \(I\) is its moment of inertia about the given line and \(M\) is its total mass. It is the radius of a ring with the given line as the axis such that if the total mass of the body is distributed on the ring, it will have the same moment of inertia \(I\). For example, the radius of gyration of a uniform disc of radius \(r\) about its axis is \(r / \sqrt{ } 2\).