In the previous section, we saw the basics about radius of gyration. In this section we will see theorem of perpendicular axes
1. The theorem of perpendicular axes is applicable to planar bodies
• In planar bodies, the thickness will be very small when compared to other dimensions
2. For example, consider a rectangular wooden plank having the following dimensions:
Length 50 cm, width 35 cm and thickness 0.75 cm
• Here the thickness is very small when compared to the length and width
• The plank will indeed appear to be planar to us
• If the thickness is a higher value like 20 cm, we would not call it a 'plank'. We would call it a 'block of wood'
2. Another example: Consider a circular disc of radius 25 cm and thickness 1.5 cm
• Here, the thickness is very small when compared to the radius
• The disc will indeed appear to be planar to us
• If the thickness is a higher value like 35 cm, we would not call it a 'disc'. We would call it a 'cylinder'
3. Consider the planar body shown in brown color in fig.7.117(a) below:
• A red line lies on it's surface
4. If we consider that red line as an axis, the body will possess a unique value of I about that axis
• Let us set up the frame of reference in such a way that, the x-axis is parallel to the red line
• Then we can denote the I taken about the red line as Ix
5. A green line is drawn perpendicular to the red line
• This green line also lies on the surface of the body
• Since the green line is perpendicular to the red line, it will be parallel to the y-axis
• We can denote the I about the green line as Iy
6. In fig.b, a blue line is drawn
• This blue line passes through the point of intersection of the red and green lines
• Also it is perpendicular to both red and green lines
• So obviously, the blue line is parallel to the z-axis
• We can denote the I about the blue line as Iz
■ Then, according to the theorem of perpendicular axes: Iz = Ix + Iy
7. The theorem states that:
The moment of inertia of a planar body (lamina) about an axis perpendicular to it's plane is equal to the sum of it's moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body
8. The above statement can be elaborated as follows:
(i) We want the I of a planar body about an 'axis L' which is perpendicular to the plane of the body
In the fig.7.117(b), this axis is shown in blue color
(ii) This required I can be obtained as the sum of two items (a) and (b) given below:
(a) The I of the body about an 'axis 1' which satisfies two conditions:
♦ 'axis 1' must lie on the plane of the body
♦ 'axis 1' must pass through the 'point of intersection of the axis L and the body'
• In fig.b, this axis is shown in red color
(b) The I of the body about an 'axis 2' which satisfies two conditions:
♦ 'axis 2' must lie on the plane of the body
♦ 'axis 1' must pass through the 'point of intersection of the axes L and 1'
• In fig.b, this axis is shown in green color
(iii) We denoted the I mentioned in ii(a) as Ix
• We denoted the I mentioned in ii(b) as Iy
• So their sum is Ix + Iy
■ So the required I about the 'axis perpendicular to the plane of the body' is Ix + Iy
Solved example 7.26
What is the moment of inertia of a circular disc about one of it's diameters?
Solution:
1. In the fig.7.118(a) below, a red line lies on the surface of a circular disc
• Also, this line passes through the center of the disc
• So obviously, the red line is the extension of a diameter of the disc
■ If we obtain the I of the disc about this red line, we will get the required answer
2. Assume that, the red line is parallel to the x-axis
• Then we can denote the I about the red line as Ix
3. A green line is present in the fig.
• Like the red line,
♦ The green line also lies in the plane of the disc
♦ The green line also passes through the center of the disc
• So obviously, like the red line, the green line is also the extension of a diameter of the disc
■ If we obtain the I of the disc about this green line, we will get the required answer
4. Also, the green line is perpendicular to the red line
• Since the green line is perpendicular to the red line, it will be parallel to the y-axis
• Then we can denote the I about the green line as Iy
5. Now draw a blue line through the point of intersection of the red and green lines
• This blue line must be perpendicular to the disc
• So the blue line is parallel to the z-axis
• Then we can denote the I about the blue line as Iz
7. Here all the conditions required for the application of the theorem of perpendicular axes are satisfied
• So we can write: Iz = Ix + Iy
8. Any diameter will split the disc into two equal parts
• In other words, all the diameters split the disc symmetrically
• So I about all diameters will be the same
• Both the red and green lines are extensions of diameters.
• Thus we get: Ix = Iy
■ Thus the result in (7) becomes: Iz = 2Ix
9. We know the I of a circular disc about an axis which is perpendicular and passing through the center. It is $\mathbf\small{\frac{MR^2}{2}}$
• So in our present case, $\mathbf\small{I_z=\frac{MR^2}{2}}$
10. Substituting these values in (8), we get: $\mathbf\small{2I_x=\frac{MR^2}{2}}$
$\mathbf\small{\Rightarrow I_x=\frac{MR^2}{4}}$
Solved example 7.27
What is the moment of inertia of a ring about one of it's diameters?
Solution:
1. Let the red and green lines in fig.7.118(a) above, pass through the center of a ring
• Then we can write:
♦ The I of the ring about the red line is Ix
♦ The I of the ring about the green line is Iy
• Since those lines are extensions of diameters, both the moments of inertia will be equal
• So we get: Ix = Iy
2. The blue line is perpendicular to both the red and green lines
• Also it passes through the point of intersection of the red and green lines
(Note that this point of intersection, is actually, the center of the ring)
• This blue line is parallel to the z-axis and so, we can denote the I about it as Iz
3. We know the I of a circular ring about an axis which is perpendicular and passing through the center. It is $\mathbf\small{MR^2}$
1. The theorem of perpendicular axes is applicable to planar bodies
• In planar bodies, the thickness will be very small when compared to other dimensions
2. For example, consider a rectangular wooden plank having the following dimensions:
Length 50 cm, width 35 cm and thickness 0.75 cm
• Here the thickness is very small when compared to the length and width
• The plank will indeed appear to be planar to us
• If the thickness is a higher value like 20 cm, we would not call it a 'plank'. We would call it a 'block of wood'
2. Another example: Consider a circular disc of radius 25 cm and thickness 1.5 cm
• Here, the thickness is very small when compared to the radius
• The disc will indeed appear to be planar to us
• If the thickness is a higher value like 35 cm, we would not call it a 'disc'. We would call it a 'cylinder'
3. Consider the planar body shown in brown color in fig.7.117(a) below:
 |
| Fig.7.117 |
4. If we consider that red line as an axis, the body will possess a unique value of I about that axis
• Let us set up the frame of reference in such a way that, the x-axis is parallel to the red line
• Then we can denote the I taken about the red line as Ix
5. A green line is drawn perpendicular to the red line
• This green line also lies on the surface of the body
• Since the green line is perpendicular to the red line, it will be parallel to the y-axis
• We can denote the I about the green line as Iy
6. In fig.b, a blue line is drawn
• This blue line passes through the point of intersection of the red and green lines
• Also it is perpendicular to both red and green lines
• So obviously, the blue line is parallel to the z-axis
• We can denote the I about the blue line as Iz
■ Then, according to the theorem of perpendicular axes: Iz = Ix + Iy
7. The theorem states that:
The moment of inertia of a planar body (lamina) about an axis perpendicular to it's plane is equal to the sum of it's moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body
8. The above statement can be elaborated as follows:
(i) We want the I of a planar body about an 'axis L' which is perpendicular to the plane of the body
In the fig.7.117(b), this axis is shown in blue color
(ii) This required I can be obtained as the sum of two items (a) and (b) given below:
(a) The I of the body about an 'axis 1' which satisfies two conditions:
♦ 'axis 1' must lie on the plane of the body
♦ 'axis 1' must pass through the 'point of intersection of the axis L and the body'
• In fig.b, this axis is shown in red color
(b) The I of the body about an 'axis 2' which satisfies two conditions:
♦ 'axis 2' must lie on the plane of the body
♦ 'axis 1' must pass through the 'point of intersection of the axes L and 1'
• In fig.b, this axis is shown in green color
(iii) We denoted the I mentioned in ii(a) as Ix
• We denoted the I mentioned in ii(b) as Iy
• So their sum is Ix + Iy
■ So the required I about the 'axis perpendicular to the plane of the body' is Ix + Iy
We will see a solved example based on the theorem
What is the moment of inertia of a circular disc about one of it's diameters?
Solution:
1. In the fig.7.118(a) below, a red line lies on the surface of a circular disc
• Also, this line passes through the center of the disc
• So obviously, the red line is the extension of a diameter of the disc
 |
| Fig.7.118 |
2. Assume that, the red line is parallel to the x-axis
• Then we can denote the I about the red line as Ix
3. A green line is present in the fig.
• Like the red line,
♦ The green line also lies in the plane of the disc
♦ The green line also passes through the center of the disc
• So obviously, like the red line, the green line is also the extension of a diameter of the disc
■ If we obtain the I of the disc about this green line, we will get the required answer
4. Also, the green line is perpendicular to the red line
• Since the green line is perpendicular to the red line, it will be parallel to the y-axis
• Then we can denote the I about the green line as Iy
5. Now draw a blue line through the point of intersection of the red and green lines
• This blue line must be perpendicular to the disc
• So the blue line is parallel to the z-axis
• Then we can denote the I about the blue line as Iz
7. Here all the conditions required for the application of the theorem of perpendicular axes are satisfied
• So we can write: Iz = Ix + Iy
8. Any diameter will split the disc into two equal parts
• In other words, all the diameters split the disc symmetrically
• So I about all diameters will be the same
• Both the red and green lines are extensions of diameters.
• Thus we get: Ix = Iy
■ Thus the result in (7) becomes: Iz = 2Ix
9. We know the I of a circular disc about an axis which is perpendicular and passing through the center. It is $\mathbf\small{\frac{MR^2}{2}}$
• So in our present case, $\mathbf\small{I_z=\frac{MR^2}{2}}$
10. Substituting these values in (8), we get: $\mathbf\small{2I_x=\frac{MR^2}{2}}$
$\mathbf\small{\Rightarrow I_x=\frac{MR^2}{4}}$
Solved example 7.27
What is the moment of inertia of a ring about one of it's diameters?
Solution:
1. Let the red and green lines in fig.7.118(a) above, pass through the center of a ring
• Then we can write:
♦ The I of the ring about the red line is Ix
♦ The I of the ring about the green line is Iy
• Since those lines are extensions of diameters, both the moments of inertia will be equal
• So we get: Ix = Iy
2. The blue line is perpendicular to both the red and green lines
• Also it passes through the point of intersection of the red and green lines
(Note that this point of intersection, is actually, the center of the ring)
• This blue line is parallel to the z-axis and so, we can denote the I about it as Iz
3. We know the I of a circular ring about an axis which is perpendicular and passing through the center. It is $\mathbf\small{MR^2}$
4. Applying theorem of perpendicular axes, we get: Iz = Ix + Iy
• Substituting the values, we get: $\mathbf\small{2I_x=MR^2}$
$\mathbf\small{\Rightarrow I_x=\frac{MR^2}{2}}$
Consider the cylinder shown in fig.7.119 below:
1. The red and green lines are perpendicular to each other
• The red and green lines pass through the center of the cylinder
• The red and green lines lie on a plane, which is parallel to the top and bottom circular faces
• So they are extensions of diameters
2. We can write:
• The moments of inertia about the red and green lines are Ix and Iy respectively
3. Now, the blue line is the 'axis of the cylinder'
• Obviously, it will pass through the point of intersection of the red and green lines
4. In this case, the theorem of perpendicular axes is not applicable
• In other words, the sum (Ix + Iy) is not equal to Iz
• This is because, a cylinder is not a planar body
 |
| Fig.7.119 |
• The red and green lines pass through the center of the cylinder
• The red and green lines lie on a plane, which is parallel to the top and bottom circular faces
• So they are extensions of diameters
2. We can write:
• The moments of inertia about the red and green lines are Ix and Iy respectively
3. Now, the blue line is the 'axis of the cylinder'
• Obviously, it will pass through the point of intersection of the red and green lines
4. In this case, the theorem of perpendicular axes is not applicable
• In other words, the sum (Ix + Iy) is not equal to Iz
• This is because, a cylinder is not a planar body
In the next section, we will see theorem of parallel axes
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