Rotational Inertia
Why is a baseball bat easier to swing when you choke up, gripping it closer to the barrel, than when you hold it by the very end of the handle? The bat has the same mass either way. What changed is where the mass sits relative to your hands, which are the axis. By the end of this page you will be able to explain that, and calculate it.
What rotational inertia is
In straight-line motion, mass tells you how hard it is to change an object's velocity. For spinning, the matching quantity is rotational inertia, $I$. It tells you how hard it is to change an object's angular velocity.
Two things matter: how much mass there is, and how far from the axis it sits. Notice that the same object can have different values of $I$ about different axes.
Rotational inertia plays the role of mass in the rotational form of Newton's second law, $\tau_{net} = I\alpha$. For the same torque, a larger $I$ gives a smaller angular acceleration $\alpha$.
Define rotational inertia in your own words. Then state which two properties of an object determine its rotational inertia about a given axis, and give the SI unit of…
Define rotational inertia in your own words. Then state which two properties of an object determine its rotational inertia about a given axis, and give the SI unit of rotational inertia.
Point masses: I = Σmr²
For a small object whose size is tiny compared with its distance from the axis (a point mass), the rotational inertia is
$$I = mr^2$$where $r$ is the distance from the axis. Mass counts once, but distance counts squared. Doubling $r$ makes $I$ four times larger. That is why far-away mass matters so much.
For several point masses, add the contributions:
A small ball of mass $0.50$ kg is attached to the end of a light rod of negligible mass, and the rod is $0.80$ m long. The system rotates about the other end of the rod.…
A small ball of mass $0.50$ kg is attached to the end of a light rod of negligible mass, and the rod is $0.80$ m long. The system rotates about the other end of the rod. Treating the ball as a point mass, calculate the rotational inertia of the system about that axis.
Two point masses, $2.0$ kg and $3.0$ kg, are fixed on a light rod at distances $0.40$ m and $0.20$ m from the axis of rotation, respectively. They lie on opposite sides…
Two point masses, $2.0$ kg and $3.0$ kg, are fixed on a light rod at distances $0.40$ m and $0.20$ m from the axis of rotation, respectively. They lie on opposite sides of the axis. Calculate the total rotational inertia of the two masses about the axis.
Four identical point masses of $0.25$ kg each sit at the corners of a square of side $0.60$ m, held by a frame of negligible mass. Find the rotational inertia of the…
Four identical point masses of $0.25$ kg each sit at the corners of a square of side $0.60$ m, held by a frame of negligible mass. Find the rotational inertia of the system about an axis perpendicular to the square and passing through its center. (Hint on geometry only: the distance from the center to a corner is half the diagonal.)
Mass distribution and extended objects
Real objects are not point masses, but the same idea holds: mass far from the axis raises $I$. Using calculus, physicists find $I$ for common shapes. In AP Physics 1 you are given these results, so you use them as formulas:
| Object (axis) | $I$ |
|---|---|
| Thin ring or hoop, about its central axis | $MR^2$ |
| Uniform solid disk, about its central axis | $\tfrac{1}{2}MR^2$ |
| Uniform thin rod, axis through center, perpendicular | $\tfrac{1}{12}ML^2$ |
| Uniform thin rod, axis through one end, perpendicular | $\tfrac{1}{3}ML^2$ |
The ring has the largest $I$ for a given $M$ and $R$, because all its mass is at the full distance $R$. The solid disk has some mass near the center, so its $I$ is only half as much.
Sofia has two disks of the same mass of $2.0$ kg and the same radius of $0.30$ m. Disk A is a uniform solid disk with $I = \tfrac{1}{2}MR^2$. Disk B has all its mass…
Sofia has two disks of the same mass of $2.0$ kg and the same radius of $0.30$ m. Disk A is a uniform solid disk with $I = \tfrac{1}{2}MR^2$. Disk B has all its mass concentrated in a thin ring at the rim, with $I = MR^2$. (a) Calculate $I$ for each disk about its central axis. (b) Explain which one is harder to start spinning and why, in terms of mass distribution.
Changing the axis: parallel-axis theorem
The value of $I$ depends on the axis. Suppose you know $I_{cm}$ about an axis through the center of mass. To find $I$ about a parallel axis a distance $d$ away, use the parallel-axis theorem.
Why does $I$ go up? Moving the axis away from the center of mass puts more of the mass farther from the axis, and the distance is squared. So $I_{cm}$ is always the smallest value among parallel axes.
A uniform thin rod of mass $M = 1.2$ kg and length $L = 1.0$ m has rotational inertia $I = \tfrac{1}{12}ML^2$ about an axis through its center, perpendicular to the rod.…
A uniform thin rod of mass $M = 1.2$ kg and length $L = 1.0$ m has rotational inertia $I = \tfrac{1}{12}ML^2$ about an axis through its center, perpendicular to the rod. Use the parallel-axis theorem to find the rotational inertia about a parallel axis through one end of the rod. Then state how many times larger it is than the center value.
A student builds a dumbbell from two $1.0$ kg point-like masses at the ends of a light rod of length $0.60$ m. She considers two axes, both perpendicular to the rod:…
A student builds a dumbbell from two $1.0$ kg point-like masses at the ends of a light rod of length $0.60$ m. She considers two axes, both perpendicular to the rod: axis 1 passes through the center of the rod, and axis 2 passes through one of the masses. (a) Calculate the rotational inertia about each axis. (b) Using the same applied torque of $0.50$ N$\cdot$m, find the angular acceleration for each axis. (c) Explain, using mass distribution, why the two results differ, and verify that your axis 2 value is consistent with the parallel-axis theorem.
Why it matters: pulling the arms in
Rotational inertia explains why a skater spins up easily with her arms pulled in. Her mass is the same, but her arms are close to the axis, so $I$ drops a lot. With a given torque, $\alpha = \tau/I$ becomes much larger.
A figure skater spins with her arms extended. Model her as a central cylinder of mass $50$ kg and radius $0.20$ m with $I = \tfrac{1}{2}MR^2$, plus two arms, each a…
A figure skater spins with her arms extended. Model her as a central cylinder of mass $50$ kg and radius $0.20$ m with $I = \tfrac{1}{2}MR^2$, plus two arms, each a point mass of $4.0$ kg located $0.80$ m from the axis. She then pulls her arms in so that each arm is $0.20$ m from the axis. (a) Calculate her total rotational inertia in each configuration. (b) Find the ratio of the two values. (c) Explain, without using angular momentum equations, why it is easier for her to produce a large angular acceleration with arms pulled in when a given torque acts.