This theorem is useful because if we know the moment of inertia of a shape, we can calculate its moment of inertia about any parallel axis by adding a correction factor.
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For a 2D object, the theorem states (area moment of inertia):
If we know the moment of inertia about an axis that passes through the centroid, then we can calculate moment of inertia about any other parallel axis. \[I=I_{centroid}+d^{2} A\]
\(I_{centroid}\) is the moment of inertia about the centroid \(A\) is the area of the 2D object \(d\) is the distance between the two axis |
For a body of mass the theorem states (mass moment of inertia):
If we know the moment of inertia about an axis that passes through the centre of mass, then we can calculate moment of inertia about any other parallel axis. \[I=I_{CofM}+d^{2} M\]
\(I_{CofM}\) is the moment of inertia about the centre of mass \(M\) is the mass of the body \(d\) is the distance between the two axis |
How Do You Use The Parallel Axis Theorem?
Consider the two cases below and be mindful of the location of the ring and the y-axis. We have formulas to calculate the inertia for a ring, yet the ring's distance away from the y-axis in case 2, will have an effect on the systems inertia about the y-axis.
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Case 1
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Case 2
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Think of the theorem as a trick to find the moment of inertia about any parallel axis we wish, just by using a correction factor. This is important because we are often given Moment of Inertia tables of common shapes and bodies.
So be smart and know how to use the tables. Below is a summary of the terminology and symbols for the ring:
- \(I_{xc}\) is the moment of inertia about the x-axis through the centroid (\(C\)). So the disk spinning in place, about the horizontal centroidal axis at (\(C\)).
- \(I_{yc}\) is the moment of inertia about the y-axis through the centroid (\(C\)). So the disk spinning in place, about the vertical centroidal axis at (\(C\))
- \(I_{x}\) is the moment of inertia about the x-axis. So the disk spins around the x-axis of the coordinate system shown in the table.
- \(I_{y}\) is the moment of inertia about the y-axis. So the disk spins around the y-axis of the coordinate system shown in the table.
Now you'll see that the parallel axis theorem is about leveraging the tables to find the moment of inertia for many different objects you may encounter.
Worked Example
A ring has an inner diameter of \(50 \: mm\), and an out diameter of \(100 \: mm\). What is the moment of inertia of a ring about the y-axis if its centroid is located at \(x=50 \: mm\)?
First, let's calculate the moment of inertia about the centroid axis using the table formula:
\[I_{y_{c}}=\frac{\pi\left(a^{4}-b^{4}\right)}{4}\]
Don't forget that we're given diameter but we need radius for our formula, so we need to divide the diameter by 2 to get radius:
\[I_{y_{c}}=\frac{\pi\left((\frac{100 \: mm}{2})^{4}-(\frac{50 \: mm}{2})^{4}\right)}{4}\]
\[I_{y_{c}}=4.6 \times 10^6 \: mm^4\]
Now, if the ring's centroid is located at \(y=50 \: mm\), we can apply the parallel axis theorem correction factor using the formula:
\[I=I_{centroid}+d^{2} A\]
\[I_{(50 \: mm)}=I_{y_{c}}+d^{2} A \]
\[I_{(50 \: mm)}= I_{y_{c}}+d^{2}\pi\left(a^{2}-b^{2}\right)\]
\[I_{(50 \: mm)}=4.6 \times 10^6 \: mm^4+(50 \: mm)^{2} \: \pi\left(\left(\frac{100 \: mm}{2}\right)^{2}-\left(\frac{50 \: mm}{2}\right)^{2}\right)\]
\[I_{(50 \: mm)}= 19.4 \times 10^6 \: mm^4 \]
The moment of inertia of a ring, about the y-axis if its centroid is located at \(y=50 \: mm\) is \(19.4 \times 10^6 \: mm^4\).
Wait, if the ring's centroid is at \(x=50 \: mm\), we are given that case in the table above:
First, let's calculate the moment of inertia about the centroid axis using the table formula:
\[I_{y_{c}}=\frac{\pi\left(a^{4}-b^{4}\right)}{4}\]
Don't forget that we're given diameter but we need radius for our formula, so we need to divide the diameter by 2 to get radius:
\[I_{y_{c}}=\frac{\pi\left((\frac{100 \: mm}{2})^{4}-(\frac{50 \: mm}{2})^{4}\right)}{4}\]
\[I_{y_{c}}=4.6 \times 10^6 \: mm^4\]
Now, if the ring's centroid is located at \(y=50 \: mm\), we can apply the parallel axis theorem correction factor using the formula:
\[I=I_{centroid}+d^{2} A\]
\[I_{(50 \: mm)}=I_{y_{c}}+d^{2} A \]
\[I_{(50 \: mm)}= I_{y_{c}}+d^{2}\pi\left(a^{2}-b^{2}\right)\]
\[I_{(50 \: mm)}=4.6 \times 10^6 \: mm^4+(50 \: mm)^{2} \: \pi\left(\left(\frac{100 \: mm}{2}\right)^{2}-\left(\frac{50 \: mm}{2}\right)^{2}\right)\]
\[I_{(50 \: mm)}= 19.4 \times 10^6 \: mm^4 \]
The moment of inertia of a ring, about the y-axis if its centroid is located at \(y=50 \: mm\) is \(19.4 \times 10^6 \: mm^4\).
Wait, if the ring's centroid is at \(x=50 \: mm\), we are given that case in the table above:
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\[I_{y}=\frac{5 \pi a^{4}}{4}-\pi a^{2} b^{2}-\frac{\pi b^{4}}{4}\]
\[I_{(50 \: mm)}=\frac{5 \pi (50 \: mm)^{4}}{4}-\pi (50 \: mm)^{2} (25 \: mm)^{2}-\frac{\pi (25 \: mm)^{4}}{4}\] \[I_{(50 \: mm)}= 19.4 \times 10^6 \: mm^4 \] |
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EXAM TIP: You'd be an absolutely crazy not to make a quick sketch. Sketches often save you a lot of time and reduce the risk of errors.
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Sometimes, you'll be forced to use the parallel axis theorem. If the ring's centroid is located at \(x=100 \: mm\), we must to apply the parallel axis theorem correction factor (since it's not in the table) using the formula: \[I=I_{centroid}+d^{2} A\] \[I_{(100 \: mm)}=I_{y_{centroid}}+d^{2} A \] |
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\[I_{(100 \: mm)}= I_{y_{c}}+d^{2}\pi\left(a^{2}-b^{2}\right)\]
\[I_{(100 \: mm)}=4.6 \times 10^6 \: mm^4+(100 \: mm)^{2} \: \pi\left(\left(\frac{100 \: mm}{2}\right)^{2}-\left(\frac{50 \: mm}{2}\right)^{2}\right)\]
\[I_{(100 \: mm)}= 63.5 \times 10^6 \: mm^4 \]
The moment of inertia of a ring, about the y-axis if its centroid is located at \(x=100 \: mm\) is \(63.5 \times 10^6 \: mm^4\). This makes sense as the moment of inertia has increased by moving the ring further away from the y-axis.
\[I_{(100 \: mm)}=4.6 \times 10^6 \: mm^4+(100 \: mm)^{2} \: \pi\left(\left(\frac{100 \: mm}{2}\right)^{2}-\left(\frac{50 \: mm}{2}\right)^{2}\right)\]
\[I_{(100 \: mm)}= 63.5 \times 10^6 \: mm^4 \]
The moment of inertia of a ring, about the y-axis if its centroid is located at \(x=100 \: mm\) is \(63.5 \times 10^6 \: mm^4\). This makes sense as the moment of inertia has increased by moving the ring further away from the y-axis.
EXAM TIP: Remember you can move your coordinate system anywhere you want to simplify calculations.
