xaktly | Physics | Rotation

Moment of inertia

Rotational motion

The sections on rotational motion are best tackled in this order:

What the heck is a   moment,   anyway?

Our goal in this section is to learn about a specific aspect of rotational motion, the moment of inertia. But before we do that, it might help to learn a bit about the concept of a "moment" in mathematics. In the physics sense, it doesn't mean "a bit of time."

In mathematics, the word moment is a measure – or one of a set of measures or properties – that describe the shape of some distribution. A distribution can be a probability distribution or (often) a mass distribution. The moment of inertia describes how mass is distributed in a rotating object.

Moments in probability

Here is the familiar Gaussian (bell-shaped curve) distribution. It could represent any group of experimental measurements of some value, such as a length, a mass or a temperature.

The first moment of the distribution is the sum of all measurements divided by the number of measurements, or the mean (or average).

$$\bar{x} = \frac{1}{n} \sum_{i = 1}^n x_i$$

The second moment is called the variance (sigma2, σ2),

$$\sigma^2 = \frac{1}{n} \sum_{i = 1}^n (x_i - \bar{x})^2$$

In the figure, the square root of the variance, called the standard deviation (σ), is shown. It is a measure of the width of the probability distribution. A small standard deviation means that most of the data are clustered closely around the mean, or that the data is relatively precise. For a Gaussian distribution like the one shown, 68% of the measurements will lie between ±σ of the mean. We sometimes worry about what's between ±2σ and ±3σ as well.

The third moment of a probability distribution is called skewness. It is rarely used in common statistical analysis, but it is also a measure of the shape of a probability distribution, particularly when it's not as symmetric as the one shown above.

Center of mass, centroid

The center of mass of an object is the balance point, a single point from which, if suspended, the object would balance. The center of mass is the first moment of the mass distribution of a stationary object.

For point masses on a line, the center of mass is the sum of the linear position of each mass multiplied by its mass, and then that sum is divided by the total mass. Here's the formula for n masses.

$$c = \frac{1}{M} \sum_{i = 1}^n m_i r_i$$

Here's an example, three masses in a line, and for simplicity we'll assume that they're point masses (all of the mass of each concentrated at the center of the sphere):

The center of mass formula gives us:

$$ \require{cancel} c = \frac{1}{13 \, \cancel{Kg}} [4(0) + 1(5) + 8(22)] \, \cancel{Kg}\cdot m$$

Working the arithmetic gives the location of the center of mass:

$$c = \frac{182 \, \cancel{Kg}\cdot m}{13 \, \cancel{Kg}} = 13.9 \; m$$

Here it is in a picture. The center of mass, at 19.9 meters, is the balance point, the first moment of this mass distribution. You might think of it like this: Where do you pick up a shovel? – near the heavy end, where it will balance, right?

The mass moment can extend in more than one dimension. In two dimensions we call the center of mass the centroid. For a general 2-D figure, assuming mass to be evenly distributed, we use calculus (which we won't go into in this section) to find the centroid.

A nifty example of the usefulness of the centroid is the population centroid of the 48 contiguous United States. This figure shows the location of the population centroid, in eastern Tennessee. It's calculated by assuming a flat US of uniform thickness, and that each person has the same mass. Then the centroid would be the balance point.

Companies like Federal Express™ locate their central shipping hubs near the centroid to economize on costs; on average, the hub is closest to the most people (Note: there are generally more people in the east and south than north and west).

Moment of inertia

Now we can think of the moment of inertia as the first moment of the mass of a rotating or revolving object. In problems involving rotational motion, we substitute the moment of inertia for mass. In rotational motion, it's not only the mass that matters, but where it is located relative to the axis of rotation.

For example, if two spheres of mass 1Kg and with the same radius, but one hollow and one solid, are set rolling down a ramp, the solid sphere will roll faster. More on that later. First a simple example of calculationg a moment of inertia.

Moment of inertia of point masses in a line

The moment of inertia of n point masses on a line is

$$I = \sum_{i = 1}^n m_i r_i^2$$

where the ri are the distances between each mass and the center of mass, and the mi are the masses. Let's use the formula to calculate the moment of inertia for the three-mass problem we did above. Here is the picture with the distances from the center of mass indicated:

The formula gives us the moment of inertia:

$$ \begin{align} I &= 4(13.6)^2 + 1(8.9)^2 + 8*8.1)^2 \\[5pt] &= 1377 \; Kg \cdot m^2 \end{align}$$

Now even this moment of inertia isn't quite accurate; recall that we collapsed all of the mass of these spheres into single points on the line. In reality, some of that mass is distributed off of the line. We'll learn more about moments of inertia of more complicated objects, and of what use they are, below.

Reduced mass

The moment of inertia of two point masses turning about their common center of mass (see above) is

$$I = m_1 r_1^2 + m_2r_2^2$$

This can be reduced to a simpler form, that of a single mass rotating by using the reduced mass, μ:

$$I = \mu R^2$$

The reduced mass of two point masses is defined as:

$$\mu = \left( \frac{1}{m_1} + \frac{1}{m_2} \right)^{-1} = \frac{m_1 m_2}{m_1 + m_2}$$

The second equality is obtained by finding a common denominator for the two fractions inside the parenthesis, then taking the reciprocal.

Often, no rationale is given for how the reduced mass is obtained, so here is some. You can skip this, but it can also be enlightening. Your choice:

First let's define our system with these coordinates and masses. The point c is the center of mass of the system.

We first want to locate that center of mass. That is defined below, and we use coordinates 0 and R for the locations of our masses; those are the ri in the sum:

$$ \begin{align} c &= \frac{1}{m_1 + m_2} \sum m_i r_i \\[5pt] &= \frac{1}{m_1 + m_2} (m_1 r_1 + m_2 r_2) \\[5pt] &= \frac{1}{m_1 + m_2} (m_1 (0) + m_2 (R)) \\[5pt] &= \frac{m_2R}{m_1 + m_2} \end{align}$$

Note that R is the distance between the masses. Now that we have the center of mass, we go back to the moment of inertia for two point masses:

$$I = \sum m_i r_i^2 = m_1 r_1^2 + m_2 r_2^2$$

This time, the r1 and r2 are the distances of masses 1 & 2 from the center of mass. Plugging in the center of mass coordinate, they are:

$$r_1 = m \left( \frac{m_2R}{m_1 + m_2} \right)^2$$


$$r_2 = m_2 \left( R - \frac{m_2R}{m_1 + m_2} \right)^2$$

Now we can plug those into the moment-of-inertia equation:

$$I = m_1\left( \frac{m_2R}{m_1 + m_2} \right)^2 + m_2 \left( R - \frac{m_2R}{m_1 + m_2} \right)^2$$

In the second term on the right, we can factor out R:

$$I = m_1\left( \frac{m_2R}{m_1 + m_2} \right)^2 + m_2 \left( R \left( 1 - \frac{m_2}{m_1 + m_2}\right) \right)^2$$

and remove R2 from the parentheses:

Now the term inside the dashed box above can be modified, by finding a common denominator with 1, to give:

$$\left( 1 - \frac{m_2 R}{m_1 + m_2} \right)^2 = \left( \frac{m_1 + \cancel{m_2} - \cancel{m_2}}{m_1 + m_2} \right)^2$$

The m2's subtract away, and we replace that second term with what remains, multiplied by the m2R2 from above:

$$I = \frac{m_1 m_2^2 R^2}{(m_1 + m_2)^2} + \frac{m_1^2 m_2 R^2}{(m_1 + m_2)^2}$$

Now these terms have a common denominator. If we factor (m1 + m2) from the numerator, and divide that into the numerator, we get:

$$I = \frac{m_1 m_2 R^2 \cancel{(m+1 + m_2)}}{(m_1 + m_2)^{\cancel{2}}}$$

Then finally, the moment of inertia is our reduced mass multiplied by the distance between the original masses:

$$I = \frac{m_1 m_2}{m_1 + m_2} R^2 = \mu R^2$$

Reduced mass

The reduced mass of two point masses separated by a distance R is

$$\mu = \left( \frac{1}{m_1} + \frac{1}{m_2} \right)^{-1} = \frac{m_1 m_2}{m_1 + m_2}$$

The moment of inertia for such a system is then

$$I = \mu R^2$$

The reduced mass allows us to reduce a two-body rotation problem to a single particle revolving about a fixed point.

Moments of inertia – more complicated objects

The moment of inertia of two masses connected by a mass-less rod is the sum of the masses multiplied by the square of the distance between each mass and the center of mass.

The moment of inertia of the two-mass system shown is

$$I = m_1 r_1^2 + m_2 r_2^2$$

and in general, the moment of inertia of n masses in a line, where the ri are the distances from the center of mass, is the sum:

$$I = \sum_{i = 1}^n \, m_i r_i^2$$

The moments of inertia of more complicated objects are more difficult to compute. They rely on integral calculus, and I won't present their derivations here. I'll save that for another section. Here, however, are some of the results.

The moment of inertia of a cylindrical rod thin enough that its width is a small fraction of its length, and rotating about its center of mass, is

If we rotate that same rod about one of its ends, the moment of inertia is:

In these figures, we assume that the solids are of uniform density throughout. Changes in density would change the moment of inertia.

The moment of inertia of a solid sphere of radius R is

If we make that a hollow sphere with a thin shell the moment is

Finally, the moment of inertia of a cylinder about its central (z in the figure) axis is different from those about the other two axes passing through the center of mass of the cylinder. They are:


The Greek alphabet


Creative Commons License   optimized for firefox
xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.