The Physics Philes, lesson 49: Show Me the (Parallel-Axis Theorem) Proof!
For the past couple of weeks I’ve been discussing moment of inertia, what it is, and how to calculate it. This week I’ll try to explain the related parallel-axis theorem.
A body can have an infinite number of axes around which it can rotate. That means a body doesn’t have just one moment of inertia. But there is a relationship between the moment of inertia about an axis that runs through the center of mass and the moment of inertia about a parallel axis.
The parallel-axis theorem is stated as:
In English, the moment of inertia about the parallel axis is equal to the moment of inertia about the axis through the center of mass added to the product of the body’s mass and the square of the distance between the two axes. To prove this theorem, we need to draw a diagram.
OK, let me try explain this. We have two axes, one at the origin and one at point P. These are parallel to the z-axis. Let’s pretend that we take a very thin slice of the xy-plane, perpendicular to the z-axis. We’ll place the center of mass at the origin, so the coordinates are 0. Thus, the axis going through the center of mass has xy coordinates of (0,0). The coordinates of the axis through point P is (a,b). The distance between points O and P is indicated by d, where d^2 = a^2 + b^2. Our mass element is indicated by the green dot (m sub i).
OK, so that’s what the diagram is meant to convey. Now…THE PROOF!
The coordinates of the mass element is (xi,yi). The moment of inertia about the center of mass can be expressed as
And the moment of inertia about the axis at point P is
Because these equations don’t take the z-axis into account, we can extend these equations to encompass all particle on all slices. The moment of inertia through point P becomes the moment of inertia of the entire body for an axis through point P. The expanded equation for moment of inertia though point P is
A few pieces of that equation should look familiar. The first part is the equation for the moment of inertia for the axis at the center of mass. The second and third sums are proportional to the x- and y-coordinates at the center of mass. (Remember the definition of center of mass? Just add up all the products of each particle’s mass and position then divide by the sum of the mass of the particles.) Those two middle sums are equal to because we’ve taken the center of mass as being at the origin. The last term is the distance multiplied by the total mass. Thus, we’re left with the parallel-axis theorem. Ta-da!
A rigid body has a lower moment of inertia about an axis through the center of mass than its moment of inertia about any other parallel axis. That means it’s easier to start a body rotating about an axis through the center of mass.
Now let’s do a quick example problem. Part of a mechanical linkage has a mass of 3.6 kg. The moment of inertia about an axis 0.15 m from the center of mass is 0.132 kg · m^2. Let’s find the moment of inertia about a parallel axis through the center of mass.
Alrighty. We’re going to use the parallel-axis theorem. What parts of the equation do we know?
- moment of inertia about an axis that is not the center of mass = 0.132 kg · m^2
- mass = 3.6 kg
- distance = 0.15 m
Great! We can find the moment of inertia about the axis going through the center of mass. Just plug in the numbers. After rearranging the equation, we get:
The moment of inertia about an axis through the center of mass is less than that about the axis through another point. This is what we would expect.
Woot! We did it! Next week we’ll continue on with rotational motion with a discussion of torque. FUN TIMES AHEAD.
Featured image credit: Hash Milhan