In the previous section, we saw the theorem of perpendicular axes. In this section we will see theorem of parallel axes
1. Consider the body shown in fig.7.120(a) below
• A blue line passes through the body
• This blue line has two peculiarities:
(i) It passes through the center of mass (C) of the body
(ii) It is parallel to the z-axis of the reference frame
2. The body will have a unique value of I about the blue line
• Since the blue line is parallel to the z-axis, we will call that value Iz
3. If we know the value of Iz, we can find the I about any axis parallel to the blue line
• All we need to know are extra two items:
(i) The mass ‘M’ of the body
(ii) Distance ‘a’ between the blue line and the ‘new axis’
4. In fig.b, the new axis is shown in yellow color
• The yellow axis is:
♦ Parallel to the blue axis
♦ At a distance of ‘a’ from the blue axis
5. We want the moment of inertia about the yellow axis
• Let us denote it as Iz'
• Then according to the parallel axes theorem,
$\mathbf\small{I_{z'}=I_z+Ma^2}$
♦ We can take any axis passing through the C
• However, we can find the I only about those axes which are parallel to the 'axis passing through C'
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 it’s center of mass and the product of it’s mass and the square of the distance between the two parallel axes
• This can be elaborated in steps:
1. We want the I about the yellow line shown in the fig.7.121(a) below
• Let us denote this I as Iyellow
• So we want Iyellow
2. For that, we want 3 items:
(i) The I about the magenta line shown in fig.b
• Let us denote that I as Imagenta
• The magenta line should satisfy two conditions:
♦ It must pass through center of mass
♦ It must be parallel to our required yellow axis
(Note that, the magenta line which passes through the C need not be parallel to any of the coordinate axes)
(ii) The mass ‘M’ of the body
(iii) The distance ‘a’ between the yellow and magenta axes
3. Once we get the 3 items, we can calculate Iyellow as the sum of two items:
(i) Imagenta
(ii) Ma2
• That is., $\mathbf\small{I_{yellow}=I_{magenta}+Ma^2}$
1. Consider the body shown in fig.7.120(a) below
• A blue line passes through the body
 |
| Fig.7.120 |
(i) It passes through the center of mass (C) of the body
(ii) It is parallel to the z-axis of the reference frame
2. The body will have a unique value of I about the blue line
• Since the blue line is parallel to the z-axis, we will call that value Iz
3. If we know the value of Iz, we can find the I about any axis parallel to the blue line
• All we need to know are extra two items:
(i) The mass ‘M’ of the body
(ii) Distance ‘a’ between the blue line and the ‘new axis’
4. In fig.b, the new axis is shown in yellow color
• The yellow axis is:
♦ Parallel to the blue axis
♦ At a distance of ‘a’ from the blue axis
5. We want the moment of inertia about the yellow axis
• Let us denote it as Iz'
• Then according to the parallel axes theorem,
$\mathbf\small{I_{z'}=I_z+Ma^2}$
• Note that, for applying this theorem, the 'axis passing through C' need not be parallel to the z-axis
• However, we can find the I only about those axes which are parallel to the 'axis passing through C'
■ The theorem of parallel axes can be stated as follows:
• This can be elaborated in steps:
1. We want the I about the yellow line shown in the fig.7.121(a) below
 |
| Fig.7.121 |
• So we want Iyellow
2. For that, we want 3 items:
(i) The I about the magenta line shown in fig.b
• Let us denote that I as Imagenta
• The magenta line should satisfy two conditions:
♦ It must pass through center of mass
♦ It must be parallel to our required yellow axis
(Note that, the magenta line which passes through the C need not be parallel to any of the coordinate axes)
(ii) The mass ‘M’ of the body
(iii) The distance ‘a’ between the yellow and magenta axes
3. Once we get the 3 items, we can calculate Iyellow as the sum of two items:
(i) Imagenta
(ii) Ma2
• That is., $\mathbf\small{I_{yellow}=I_{magenta}+Ma^2}$
We will see some solved examples
Solved example 7.28
What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end?
Solution:
1. Fig.7.122 below shows the rod of mass M and length l
• The magenta line is perpendicular to the rod
• The magenta line passes through the C of the rod
2. We already know the moment of inertia of the rod about that magenta line
• Let us denote it as I
We have: $\mathbf\small{I=\frac{Ml^2}{12}}$
3. The yellow line is parallel to the magenta line
• It passes through one end of the rod
• We want the moment of inertia of the rod about this yellow line
• Let us denote it as I'
4. Applying theorem of parallel axes, we have: $\mathbf\small{I'=I+Ma^2}$
5. The distance 'a' between the magenta and yellow lines = $\mathbf\small{\frac{l}{2}}$
• Substituting the values, we get: $\mathbf\small{I'=\frac{Ml^2}{12}+M\left(\frac{l}{2}\right)^2=\frac{Ml^2}{3}}$
Solved example 7.29
What is the moment of inertia of a ring about a tangent to the circle of the ring?
Solution:
1. Fig.7.123 below shows the ring of mass M and radius R
• The magenta line lies in the plane of the ring
• The magenta line passes through the center of the ring
2. We already know the moment of inertia of the ring about that magenta line
• Let us denote it as I
We have: $\mathbf\small{I=\frac{MR^2}{2}}$
3. For any circle, infinite number of tangents can be drawn
• Two of them will be parallel to any given diameter
• In our present case, two tangents can be drawn parallel to the magenta line
• One of them is shown in yellow color in the fig.
• We want the moment of inertia of the ring about this yellow line
• Let us denote it as I'
4. Applying theorem of parallel axes, we have: $\mathbf\small{I'=I+Ma^2}$
5. The distance 'a' between the magenta and yellow lines = R
Substituting the values, we get: $\mathbf\small{I'=\frac{MR^2}{2}+MR^2=\frac{3MR^2}{2}}$
Solved example 7.30
Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be $\mathbf\small{\frac{2MR^2}{5}}$, where M is the mass of the sphere and R is the radius of the sphere
Solution:
1. Fig.7.124 below shows the ring of mass M and radius R
• The magenta line passes through the center of the sphere
2. We are given the moment of inertia of the sphere about that magenta line
• Let us denote it as I
We have: $\mathbf\small{I=\frac{2MR^2}{5}}$
3. For any sphere, infinite number of tangents can be drawn
• Two of them will be parallel to any given diameter
• In our present case, two tangents can be drawn parallel to the magenta line
• One of them is shown in yellow color in the fig.
• We want the moment of inertia of the sphere about this yellow line
• Let us denote it as I'
4. Applying theorem of parallel axes, we have: $\mathbf\small{I'=I+Ma^2}$
5. The distance 'a' between the magenta and yellow lines = R
Substituting the values, we get: $\mathbf\small{I'=\frac{2MR^2}{5}+MR^2=\frac{7MR^2}{5}}$
Solved example 7.31
Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be $\mathbf\small{\frac{MR^2}{4}}$, find its moment of inertia about an axis normal to the disc and passing through a point on its edge
Solution:
1. Fig.7.125(a) below shows the disc of mass M and radius R
• The red line lies in the plane of the disc
• The red line passes through the center of the disc
• So the red line is an extension of a diameter
2. We are given the moment of inertia of the disc about that red line
• Let us denote it as IRed
We have: $\mathbf\small{I_{Red}=\frac{MR^2}{4}}$
3. The green line lies in the plane of the disc
• The green line passes through the center of the disc
• So the green line is also an extension of a diameter
• Thus we get:
The I about green line (IGreen)= I about the red line (IRed) = $\mathbf\small{\frac{MR^2}{4}}$
4. Now, the green line is perpendicular to the red line
• Also, the magenta line passes through the point of intersection of the red and green lines
• So we can apply the theorem of perpendicular axes: IMagenta = IRed + IGreen
$\mathbf\small{\Rightarrow I_{Magenta}=\left(\frac{MR^2}{4}+\frac{MR^2}{4}\right)=\frac{MR^2}{2}}$
5. The magenta line passes through the point of intersection of red and green lines
• The red and green lines are extensions of diameters
• So the magenta line passes through the center of the disc
• Also it is perpendicular to the disc
• So we can write:
IMagenta is the moment of inertia about the axis which is perpendicular to the disc and also passing through the center
6. In fig.b, the yellow line is parallel to the disc
• Applying theorem of parallel axes, we get: IYellow = IMagenta + Ma2.
7. The distance 'a' between the magenta and yellow lines = R
• Substituting the values, we get: $\mathbf\small{I_{Yellow}=\left(\frac{MR^2}{2}+MR^2\right)=\frac{3MR^2}{2}}$
Solved example 7.28
What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end?
Solution:
1. Fig.7.122 below shows the rod of mass M and length l
 |
| Fig.7.122 |
• The magenta line passes through the C of the rod
2. We already know the moment of inertia of the rod about that magenta line
• Let us denote it as I
We have: $\mathbf\small{I=\frac{Ml^2}{12}}$
3. The yellow line is parallel to the magenta line
• It passes through one end of the rod
• We want the moment of inertia of the rod about this yellow line
• Let us denote it as I'
4. Applying theorem of parallel axes, we have: $\mathbf\small{I'=I+Ma^2}$
5. The distance 'a' between the magenta and yellow lines = $\mathbf\small{\frac{l}{2}}$
• Substituting the values, we get: $\mathbf\small{I'=\frac{Ml^2}{12}+M\left(\frac{l}{2}\right)^2=\frac{Ml^2}{3}}$
Solved example 7.29
What is the moment of inertia of a ring about a tangent to the circle of the ring?
Solution:
1. Fig.7.123 below shows the ring of mass M and radius R
 |
| Fig.7.123 |
• The magenta line passes through the center of the ring
2. We already know the moment of inertia of the ring about that magenta line
• Let us denote it as I
We have: $\mathbf\small{I=\frac{MR^2}{2}}$
3. For any circle, infinite number of tangents can be drawn
• Two of them will be parallel to any given diameter
• In our present case, two tangents can be drawn parallel to the magenta line
• One of them is shown in yellow color in the fig.
• We want the moment of inertia of the ring about this yellow line
• Let us denote it as I'
4. Applying theorem of parallel axes, we have: $\mathbf\small{I'=I+Ma^2}$
5. The distance 'a' between the magenta and yellow lines = R
Substituting the values, we get: $\mathbf\small{I'=\frac{MR^2}{2}+MR^2=\frac{3MR^2}{2}}$
Solved example 7.30
Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be $\mathbf\small{\frac{2MR^2}{5}}$, where M is the mass of the sphere and R is the radius of the sphere
Solution:
1. Fig.7.124 below shows the ring of mass M and radius R
 |
| Fig.7.124 |
2. We are given the moment of inertia of the sphere about that magenta line
• Let us denote it as I
We have: $\mathbf\small{I=\frac{2MR^2}{5}}$
3. For any sphere, infinite number of tangents can be drawn
• Two of them will be parallel to any given diameter
• In our present case, two tangents can be drawn parallel to the magenta line
• One of them is shown in yellow color in the fig.
• We want the moment of inertia of the sphere about this yellow line
• Let us denote it as I'
4. Applying theorem of parallel axes, we have: $\mathbf\small{I'=I+Ma^2}$
5. The distance 'a' between the magenta and yellow lines = R
Substituting the values, we get: $\mathbf\small{I'=\frac{2MR^2}{5}+MR^2=\frac{7MR^2}{5}}$
Solved example 7.31
Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be $\mathbf\small{\frac{MR^2}{4}}$, find its moment of inertia about an axis normal to the disc and passing through a point on its edge
Solution:
1. Fig.7.125(a) below shows the disc of mass M and radius R
 |
| Fig.7.125 |
• The red line passes through the center of the disc
• So the red line is an extension of a diameter
2. We are given the moment of inertia of the disc about that red line
• Let us denote it as IRed
We have: $\mathbf\small{I_{Red}=\frac{MR^2}{4}}$
3. The green line lies in the plane of the disc
• The green line passes through the center of the disc
• So the green line is also an extension of a diameter
• Thus we get:
The I about green line (IGreen)= I about the red line (IRed) = $\mathbf\small{\frac{MR^2}{4}}$
4. Now, the green line is perpendicular to the red line
• Also, the magenta line passes through the point of intersection of the red and green lines
• So we can apply the theorem of perpendicular axes: IMagenta = IRed + IGreen
$\mathbf\small{\Rightarrow I_{Magenta}=\left(\frac{MR^2}{4}+\frac{MR^2}{4}\right)=\frac{MR^2}{2}}$
5. The magenta line passes through the point of intersection of red and green lines
• The red and green lines are extensions of diameters
• So the magenta line passes through the center of the disc
• Also it is perpendicular to the disc
• So we can write:
IMagenta is the moment of inertia about the axis which is perpendicular to the disc and also passing through the center
6. In fig.b, the yellow line is parallel to the disc
• Applying theorem of parallel axes, we get: IYellow = IMagenta + Ma2.
7. The distance 'a' between the magenta and yellow lines = R
• Substituting the values, we get: $\mathbf\small{I_{Yellow}=\left(\frac{MR^2}{2}+MR^2\right)=\frac{3MR^2}{2}}$
In the next section, we will see kinematics of rotational motion about a fixed axis
