When you take a hard boiled egg and a raw egg of the same size out of the refrigerator, they weigh almost the same. When you spin them both on the counter and then stop each one with a fingertip, one of them stops and stays stopped, and the other one starts turning again under your finger. Do you think the two eggs have the same mass? If so, what else is different about them?

What is different is where the mass is and how freely it can move, and that is the quantity we are after. In linear motion, mass was the whole of an object’s resistance to being sped up. In rotation it is only half the answer.

1. Mass alone does not decide

Two wheels can have the same mass and the same radius and still be very different to get spinning. A bicycle rim with all its mass out at the edge is harder to start than a solid disk of the same mass, and when you stop pushing, the rim is also harder to stop.

The name for this resistance is rotational inertia, written \(I\). It plays exactly the role for rotation that mass played for straight line motion, and it depends on two things rather than one: how much mass there is, and how far that mass sits from the axis.

The second of those is the one that is easy to forget and hard to ignore once you have seen it. Moving mass outward raises the rotational inertia quickly, and the simulation puts two systems of equal mass next to each other so you can watch the difference without doing any arithmetic.

2. One small mass, and the square that matters

The simplest possible case is a single small mass on the end of a light rod, turning about the other end.

\(I = mr^{2}\)kilogram meters squared

The mass enters once and the distance enters twice. That square is the whole reason rotational inertia behaves so differently from mass. When you double \(m\), the rotational inertia doubles. When you double \(r\), it does not double, it becomes four times larger.

So a small mass far out can matter more than a large mass close in, and the simulation plots \(I\) against \(r\) so you can see that the relationship is a curve rather than a straight line.

3. A collection of masses is a sum

Real systems are made of many pieces at many distances. When you have more than one mass, you work out the rotational inertia of each one about the axis and add them.

\(I_{\text{tot}} = \sum I_i = \sum m_i r_i^{2}\)add the contributions, not the distances

Each piece carries its own distance, and no shortcut lets you add the masses first and use some average distance afterward. The square sits inside the sum, and no shortcut survives that.

In AP Physics 1 and 2 you will be asked to do this for five masses or fewer, arranged in a plane. For anything more complicated the rotational inertia is given to you, and what you are expected to have is a feel for which arrangement gives the larger value.

4. The axis through the center of mass is the cheapest one

A rigid system does not have one rotational inertia. It has a different one for every axis you might turn it about, and a problem is not fully stated until the axis is named.

Among all the parallel axes you could choose in a given direction, one of them gives the smallest rotational inertia, and it is the axis through the center of mass. Any other parallel axis gives a larger value.

This is not a special fact to memorize separately. It falls out of the sum: moving the axis away from the center of mass moves most of the mass further from it, and distance counts twice over. The simulation plots the rotational inertia against the position of the axis so you can watch the curve bottom out, and the bottom lands exactly on the center of mass every time.

5. Moving the axis away from the center of mass

When you know the rotational inertia about the center of mass and you want it about some other parallel axis, you do not have to start the whole calculation again. There is one equation that moves it for you.

\(I’ = I_{\text{cm}} + Md^{2}\)\(d\) is the distance between the two axes

Here \(M\) is the total mass of the system and \(d\) is the perpendicular distance between the new axis and the parallel axis through the center of mass. The new value is always larger, which agrees with what the last section showed, and the extra term is exactly what you would get if the whole mass sat at the center of mass.

Two conditions have to hold before you may use it, and both are easy to miss. The two axes must be parallel to each other, and one of them must pass through the center of mass. The equation says nothing about two arbitrary axes.

The simulation works the new value out twice, once from this equation and once by summing over the body directly, so you can watch two different calculations land on the same number.