• Suppose, moment of inertia (MOI) of a body about a given line AB is to be calculated.
• CZ is the line parallel to AB passing through center of mass C of the body.
• \(I\) and \( I_{CM}\)  are the moment of inertia of the body about AB and CZ, respectively.

• The parallel - axis theorem states :

           \( I = I_{CM} + Md^2\)

where, d is the perpendicular distance between AB and CZ and M is the mass of the body.

Note :

• \(I_{CM}\) is the moment of inertia about an axis passing through the center of mass.
• The two axes taken, must be parallel.

(using parallel axis theorem)

• The moment of inertia of a uniform rod about a perpendicular axis passing through its center of mass is

            \(I_0=\dfrac {ML^2}{12}\) ....(1)

• To calculate moment of inertia of the rod about the axis AB (perpendicular to the rod passing through an end), parallel-axis theorem is used.

• By parallel- axis theorem, \(I=I_0+Md^2\)

The distance between the two parallel axes considered is,

\( d =\) \(\dfrac {L}{2}\)  ... (2)

 \(I=\dfrac {ML^2}{12}\)+ \(M\)\(\left(\dfrac {L}{2}\right)^2\)       ... [from (1) & (2)]

\(I\) = \(\dfrac {ML^2}{3}\) 

• Consider an element as a rod of thickness \(dx\) of a square sheet of mass M and length L. Mass of the element is given by :

 \(dm=\dfrac {Mass \;of \;body} {Area \;of \;body}{(Area\; of \;element)}\) 

 \(dm=\dfrac{M}{L^2}(Ldx)\)

 \(dm=\dfrac{M}{L}(dx)\)  ...(1)

• Moment of inertia of the element about an axis passing through its center of mass (z'-axis) is

 \(dI_0=\dfrac{dmL^2}{12}\)

• If the moment of inertia of all the elements about the axis of rotation of the body is known, then on integrating it, the moment of inertia of the whole body about the axis is obtained.
• For the calculation of moment of inertia of element about z-axis, parallel- axis theorem can be used.

• Thus, \(I_z=dI_0+Md^2\)  

       ⇒  \(dI=\dfrac{dmL^2}{12}+dmx^2\) 

• Moment of inertia of square sheet,

 \(I=\int dI\)

=\(\int \left(\dfrac {L^2}{12}+ x^2\right)dm\)

= \(\dfrac{M}{L}\) \(\int\limits_{–L/2}^{L/2}\)\(\left(\dfrac {L^2}{12}+ x^2\right)dx\) ...[from(1)]

\(I\) = \(\dfrac{M}{L}\) \(\left[\dfrac {L^2x}{12}+ \dfrac{x^3}{3}\right]_{-L/2}^{L/2}\)

 \(I=\dfrac{ML^2}{6}\)

• Consider a disk of mass M and radius R.
• As the disk is of uniform thickness, then

     mass (M) = Density \(\left(\rho\right)\) × volume (V)

 \(M=\rho\times(A\times\,t)\)

 \(M=\rho\)  \(\times\,\pi{R}^2\,t\) ... (1)

• A small disk of mass 'm' is removed from it.

Mass , m = Density × Volume

 \(m=\rho\times(A_1\times\,t)\)

 \(m=\rho\) \(\times\,\pi{r}^2×t\) 

\(\rho\) and t for smaller disk are same as that of large disk

       \(\therefore\)  \(m=\dfrac{M}{\pi{R^2×t}} ×\pi r^2×t\)...[from(1)]

 \(m=\dfrac{Mr^2}{R^2}\) or   \(\dfrac {m}{M}\)= \(\left(\dfrac{r}{R}\right)^2\)

• If the moment of inertia of the small disk and the remaining disk is \( I'\) and \( I\), respectively

then,

Moment of inertia of the whole disk about the axis CB is 

\(I_0=I+I' \)

\(\because\,I_0=\dfrac{MR^2}{2}\), \(I'=\dfrac{mr^2}{2}+md^2\)

\(\therefore\, I=I_0-I'\)

\(\Rightarrow I=\dfrac{MR^2}{2}\)– \(\dfrac{mr^2}{2}-md^2\)

Substituting the value of m

\(I=\dfrac{MR^2}{2}\) –  \(\dfrac{Mr^2}{R^2}\)\(\left(\dfrac {r^2}{2}+d^2\right)\)

For ring

• Moment of inertia of a ring of mass M and radius R about its axis perpendicular to the plane, passing through its center is

?            \(I_0=MR^2\) ...(1)

• Then, moment of inertia \((I)\) about the axis AB passing through its circumference is given by,

using parallel-axis theorem,

\(I=I_0+Md^2\)

• For ring, d = R  ... (2)

Thus,  \( I = MR^2 + M(R)^2 \)    [ from (1) and (2)]

         \( I = 2MR^2\)

For disk 

• Moment of inertia of a disk of mass M and radius R about its axis perpendicular to the plane, passing through its center is 

\(I_0=\dfrac {MR^2}{2}\) ...(3)

• Then, moment of inertia (I) about the axis AB passing through the circumference is given by,

using parallel-axis theorem,

\(I=I_0+Md^2\)

• For disk, d = R ... (4)

Thus,  \(I=\dfrac{MR^2}{2}+M(R)^2\)... [from equation (3) and (4)]

          \(I=\dfrac {3MR^2}{2}\)

•  Consider a disk of mass M₁ and radius R₁ and a ring of mass M₂ and radius R₂.

• The moment of inertia of a disk about the perpendicular axis (X₁Y₁) passing through center of mass is

\(I_{o_1}=\dfrac{M_1R_1^2}{2}\)

• By using parallel axis theorem, moment of inertia (\(I_1\)) of a disk about axis AB is

\(I=I_{o_1}+Md^2\)

 \(I_1=\dfrac{M_1R_1^2}{2}\) + \(M_1R_1^2\)

\(I_1=\dfrac {3}{2}M_1R_1^2\) ...(1)

• The moment of inertia  of a ring about the perpendicular axis (X₂Y₂) passing through the center of mass is

  \(I_{o_2}=M_2R_2^2\)

• By using parallel axis theorem, the moment of inertia (\(I_2\)) of a ring about axis AB is

\(I_2=I_{o_2}+Md^2\)

  \(I_2=M_2R_2^2\) + \(M_2R_2^2\)

\(I_2=2M_2R_2^2\)  ... (2)

• Thus, the moment of inertia of the combination of ring and disk about the given axis AB is the sum of moment of inertia of individual bodies about AB.

Mathematically,

\(I=I_1+I_2\)

 \(I=\dfrac {3}{2}M_1R_1^2\) + \(2M_2R_2^2\)