Mass vs Rotational Inertia: The Difference

Mass measures how hard it is to change an object's straight-line motion, and an object has exactly one value of it. Rotational inertia measures how hard it is to change its rotation, and it depends on the axis as well as the object, so one object has a different value for every axis.

AP Physics: Unit 5 (topics 2.6 Gravitational Force, 5.4 Rotational Inertia, 5.6 Newton's Second Law in Rotational Form). Mass and rotational inertia are defined in different units of AP Physics 1. Mass sits in Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods, under Topic 2.6 at learning objective 2.6.D, describe inertial and gravitational mass, carrying EK 2.6.D.1 through 2.6.D.3: inertial mass or inertia is a property that determines how much an object's motion resists changes when interacting with another object, gravitational mass is related to the force of attraction between two systems with mass, and the two have been experimentally verified to be equivalent. Rotational inertia sits in Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent over about 15 to 20 class periods. Topic 5.4 has two learning objectives: 5.4.A, describe the rotational inertia of a rigid system relative to a given axis of rotation, carrying EK 5.4.A.1 through 5.4.A.3, and 5.4.B, describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass, carrying EK 5.4.B.1 and the parallel axis theorem at EK 5.4.B.2. Both I = sum of m_i r_i squared and I' = I_cm + M d squared are printed on the AP Physics 1 equation sheet. Two boundary statements apply to Topic 5.4: calculations are limited to systems of five or fewer objects arranged in a two-dimensional configuration, and students do not need to know the rotational inertia of extended rigid systems as these will be provided within the exam, though they should have a qualitative understanding of the factors that affect it, for example that rotational inertia is greater when mass is farther from the axis, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius. Suggested skills are 1.A, 2.A, 2.D and 3.C for Topic 2.6, and 1.B, 2.B, 2.C, 3.A and 3.B for Topic 5.4.

One number per object, against one number per axis

The analogy between the two is real and the AP Physics 1 CED builds on it. The place it breaks is the only thing you need to remember: mass belongs to the object, and rotational inertia belongs to an object plus a chosen axis.

EK 2.6.D.1 defines the linear quantity: objects have inertial mass, or inertia, a property that determines how much an object's motion resists changes when interacting with another object. Nothing in that sentence refers to anything outside the object. Put a $1.8$ kg wheel on a balance and it reads $1.8$ kg whichever way round you set it down.

EK 5.4.A.1 defines the rotational quantity: rotational inertia measures a rigid system's resistance to changes in rotation and is related to the mass of the system and the distribution of that mass relative to the axis of rotation. The last five words are the whole page. The axis is not context; it is an input.

The learning objective says it as well, which is unusual and worth noticing. LO 5.4.A: describe the rotational inertia of a rigid system relative to a given axis of rotation. The phrase "relative to a given axis" is in the objective itself, not buried in an essential knowledge statement, because a rotational inertia quoted without an axis is not an answer.

So the question "what is the rotational inertia of this rod?" has no answer. "What is the rotational inertia of this rod about an axis through its center, perpendicular to the rod?" has one. Worked example one takes a single two-sphere system and produces four different rotational inertias from it without changing the object or its mass by a gram.

A vocabulary note that matters for searching and for reading older textbooks: the CED calls this quantity rotational inertia every time. The phrase "moment of inertia" does not appear in the AP Physics 1 CED at all, nor in the AP Physics C: Mechanics CED, while "rotational inertia" appears dozens of times in each and is the label the equation sheet's symbol key uses for $I$ . They are the same quantity under two names, and the current courses use the newer one.

Mass vs rotational inertia, side by side

Question you are asking	Mass	Rotational inertia
Symbol	$m$ , or $M$ for a whole system	$I$
SI unit	$\text{kg}$	$\text{kg}\cdot\text{m}^2$
How many values one object has	Exactly one	One for every axis
Depends on where the axis is	No	Yes
Depends on which direction the axis points	No	Yes, EK 5.4.B.1 says "in a given plane"
What it resists	A change in translational motion	A change in rotation
Its second law	$\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ (EK 2.5.A.2)	$\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}}$ (EK 5.6.A.2)
Its kinetic energy	$K = \frac{1}{2}mv^2$	$K = \frac{1}{2}I\omega^2$
Its momentum	$\vec{p} = m\vec{v}$	$L = I\omega$
Also appears in a force law	Yes, gravitational mass, EK 2.6.D.2	No, there is no gravitational analogue
For a point object at distance $r$	$m$ , whatever $r$ is	$mr^2$
For several objects	Add the masses	Add the rotational inertias about the same axis, EK 5.4.A.3
Minimum value	Not applicable	Smallest for an axis through the center of mass, EK 5.4.B.1
CED essential knowledge	2.6.D.1, 2.6.D.2, 2.6.D.3	5.4.A.1 through 5.4.A.3, 5.4.B.1, 5.4.B.2

Three rows deserve unpacking.

The unit row is the distinction encoded in symbols. $\text{kg}\cdot\text{m}^2$ carries a length squared, and that length is the distance from the axis. A quantity whose unit contains a distance cannot be a property of the object alone, because a distance has to be measured from somewhere.

The "which direction the axis points" row is the one people miss even after learning the first. EK 5.4.B.1 is worded as a rigid system's rotational inertia in a given plane being at a minimum when the rotational axis passes through the system's center of mass. Moving the axis sideways changes $I$ ; tilting it changes $I$ too. Worked example one includes an axis along the rod itself, where the same two spheres give a rotational inertia of zero in the point-object model.

The "also appears in a force law" row is a genuine asymmetry, not a gap in the analogy. Mass does two jobs in this course. EK 2.6.D.1 gives it the inertial job and EK 2.6.D.2 gives gravitational mass the job of setting the force of attraction between two systems with mass, and EK 2.6.D.3 says inertial mass and gravitational mass have been experimentally verified to be equivalent. Rotational inertia has only the inertial job. There is no rotational analogue of $\lvert \vec{F}_g \rvert = \frac{Gm_1m_2}{r^2}$ on any AP sheet.

The case that separates them: move the axis, keep the object

Two $0.40$ kg spheres are fixed to the ends of a light rod $1.20$ m long. The mass of the system is $0.80$ kg in every row below and never changes. Only the axis moves.

Axis	Distance of each sphere from the axis	Rotational inertia
Through the center, perpendicular to the rod	$0.60$ m and $0.60$ m	$0.288\ \text{kg}\cdot\text{m}^2$
$0.30$ m off center, perpendicular to the rod	$0.30$ m and $0.90$ m	$0.360\ \text{kg}\cdot\text{m}^2$
Through one sphere, perpendicular to the rod	$0$ and $1.20$ m	$0.576\ \text{kg}\cdot\text{m}^2$
Along the rod itself	$0$ and $0$	$0$ in the point-object model

One object, one mass, four rotational inertias spanning a factor of infinity. A balance cannot distinguish any of these four situations. A torque can distinguish all of them.

The second and third rows are computed two ways in worked example one, once from $I = \sum m_i r_i^2$ and once from the parallel axis theorem, and they agree. The last row is the point-object model's answer, and it is worth reading with EK 5.1.A.1.i in mind: a rigid system is one that holds its shape but in which different points on the system move in different directions during rotation, and a rigid system cannot be modeled as an object. Real spheres have a finite radius and so a small nonzero rotational inertia about that axis; the Topic 5.4 boundary statement says students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam.

The reverse experiment separates them just as cleanly. Take a hoop and a solid disk of the same mass and the same radius. The masses are identical, so under the same net force they accelerate identically. Under the same net torque about their centers they do not: the hoop's angular acceleration is half the disk's. The CED uses this pair itself in the Topic 5.4 boundary statement, saying students should have a qualitative understanding of the factors that affect rotational inertia, for example how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius. Worked example three runs the numbers.

The parallel axis theorem is how much the axis matters

The axis dependence is not a vague caution. AP Physics 1 gives you an equation for it, and both the CED and the equation sheet carry it.

EK 5.4.B.2: the parallel axis theorem uses the following equation to relate the rotational inertia of a rigid system about any axis that is parallel to an axis through its center of mass:

I' = I_{\text{cm}} + Md^2

Here $I_{\text{cm}}$ is the rotational inertia about an axis through the center of mass, $M$ is the total mass of the system, and $d$ is the distance between the two parallel axes. That equation is printed on the [AP Physics 1 equation sheet](/formulas/ap-physics-1), in the rotational block, and it is also on the AP Physics C: Mechanics sheet. You do not have to derive it or recall it.

Read what it tells you about the comparison.

$Md^2$ is never negative, so $I' \geq I_{\text{cm}}$ always. That is exactly EK 5.4.B.1: a rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass. The theorem is why the minimum exists, and the minimum is at the center of mass rather than anywhere else.
The mass appears in the correction term. So mass does matter to rotational inertia, and heavily, but only in combination with a distance. Double the mass and you double $I$ at a fixed geometry; move the axis and you change $I$ with the mass untouched.
The dependence on $d$ is quadratic. Shift the axis by twice the distance and the correction grows by four.

There is one condition it is easy to lose. The reference axis has to pass through the center of mass. The theorem relates any axis to a parallel axis through the center of mass, not any axis to any other axis. If you have $I$ about some off-center axis and you want $I$ about a different off-center axis, you go back to the center of mass first and then out again.

The theorem is also the tool that makes the exam's promise usable. The Topic 5.4 boundary statement says the rotational inertia of extended rigid systems will be provided within the exam. What gets provided is typically the center-of-mass value, and the parallel axis theorem is how you move it to the axis the question actually uses. Worked example two does exactly that for a rod pivoted at one end, and shows that skipping the step makes the angular acceleration four times too large.

How much of the analogy actually holds

The CED is explicit that the rotational quantities are analogues, so it is worth being exact about how far that goes rather than treating the analogy as a slogan.

EK 5.1.A.4: angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships. Note what that statement covers: the kinematic quantities. It does not say mass and rotational inertia are analogous, and it does not need to, because the dynamics equations show the correspondence directly.

Translational statement, with its CED source	Rotational counterpart, with its CED source
$\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ , EK 2.5.A.2	$\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}}$ , EK 5.6.A.2
Velocity constant when $\sum \vec{F} = 0$ , EK 2.4.A.3	Angular velocity constant when the net torque is zero, EK 5.5.A.1.iii
$K = \frac{1}{2}mv^2$ , EK 3.1.A.1	$K = \frac{1}{2}I\omega^2$ , sheet, rotational block
$\vec{p} = m\vec{v}$ , EK 4.1.A.1	$L = I\omega$ , sheet, rotational block

Every line swaps $m$ for $I$ and a linear quantity for its angular counterpart. In that sense the analogy is exact, and EK 5.6.A.2 spells out the consequence in words: the angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system, which is precisely the role mass plays in the translational law.

Where it stops:

One value against many. Already covered. This is the important one.
No gravitational role. Mass appears in $\lvert \vec{F}_g \rvert = \frac{Gm_1m_2}{r^2}$ ; rotational inertia appears in no force law.
Additivity works differently. Masses simply add. Rotational inertias add only when they are computed about the same axis, which EK 5.4.A.3 states as $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$ . Adding two rotational inertias quoted about different axes is meaningless.
The system cannot always be collapsed to a point. EK 2.1.A.2 lets you treat a system as a single object when the properties or interactions of its constituent objects are not important, and EK 2.1.B.3 says a system can be modeled as a singular object located at the system's center of mass. That move is available for translational problems and it destroys the information a rotational problem needs. EK 5.1.A.1.i says it outright: a rigid system cannot be modeled as an object.

That last point is the deepest form of the distinction. Mass survives the collapse of a system to a point at its center of mass. Rotational inertia does not exist after that collapse, because the distances it depends on have been thrown away.

When they behave alike, and why that lulls you

Three situations let you treat rotational inertia as though it were a mass, and all three are common enough to build a habit on.

A single fixed axis, throughout the problem. If the axle never moves, $I$ is one number for the whole question, and it behaves exactly like a mass in every equation you write. Most first rotational problems are like this, which is why the axis dependence can go unnoticed for a whole unit.

An exam that hands you the number. The Topic 5.4 boundary statement says the rotational inertia of extended rigid systems will be provided, so a value arrives already computed and already attached to an axis you did not choose. Take the number, and the choice that produced it is invisible.

Proportional reasoning about mass alone. Double the mass of a system without changing its shape or its axis and $I$ doubles, exactly as a mass would. Every $r_i$ in $\sum m_i r_i^2$ is unchanged, so the sum scales with the masses. That works, and it teaches the wrong lesson if it is the only manipulation you ever do.

The distinction reappears in four exam-shaped places.

A question that changes the pivot. A rod swung from its end and from its center is the archetype, and the answers differ by a factor of four.
A question that compares two shapes of the same mass. Hoop against disk, and the ratio is $2$ .
A question about a person or a skater changing shape. Pulling the arms in reduces every $r_i$ , so $I$ falls while the mass is untouched. The CED's Topic 5.6 sample activity asks students to research the rotational inertia of a human body with arms outstretched and with arms pulled in, then analyze footage of a skater to see whether angular momentum is conserved.
A question that asks which of two objects reaches the bottom of a ramp first. Masses are irrelevant to the answer and rotational inertias decide it, which is covered under Topic 6.5, Rolling.

The reflex worth building: never write an $I$ without writing the axis next to it. In words, in the margin, before the number. It costs one phrase and it is the entire difference between the two quantities.

Where the confusion costs a mark

Each of these is a specific scoring error.

Quoting a rotational inertia without stating the axis. LO 5.4.A asks for the rotational inertia relative to a given axis of rotation, so an unlabeled $I$ is an incomplete answer, and on a multi-part question it makes the later parts unmarkable.
Using a provided center-of-mass value at an off-center pivot. This is the expensive one, because it produces a plausible number. For a uniform rod pivoted at one end it makes the angular acceleration four times too large, as worked example two shows.
Applying the parallel axis theorem from the wrong reference axis. $I' = I_{\text{cm}} + Md^2$ requires that $I_{\text{cm}}$ be about an axis through the center of mass. Starting from an off-center value and adding another $Md^2$ compounds the error.
Using a radius where the theorem wants the axis separation. $d$ is the distance between the two parallel axes, not a radius of the object. For a rod pivoted at its end, $d = L/2$ , not $L$ .
Forgetting to square the distance. $I = \sum m_i r_i^2$ is quadratic, so halving a distance quarters the contribution. This is also why the units are $\text{kg}\cdot\text{m}^2$ .
Adding rotational inertias computed about different axes. EK 5.4.A.3 adds rotational inertias of each object about that axis, one shared axis.
Assuming the heavier object is harder to spin. Two systems of the same mass can differ by a factor of two, and a lighter compact object can be harder to spin than a heavier spread-out one. Compare $\sum m_i r_i^2$ , not the masses.
Assuming the mass changed when the rotational inertia did. A skater pulling their arms in changes $I$ and not $m$ . Saying they became lighter is wrong and it usually comes with a wrong conservation statement attached.
Trying to compute the rotational inertia of a continuous body. The Topic 5.4 boundary statement limits AP Physics 1 to systems of five or fewer objects arranged in a two-dimensional configuration, and says the rotational inertia of extended rigid systems will be provided. If a question expects you to integrate, you have misread it.
Treating the system as a point at its center of mass and then asking about rotation. EK 5.1.A.1.i says a rigid system cannot be modeled as an object, and the collapse throws away the distances $I$ is built from.

What the CED asks, and how the exam frames it

The two quantities are defined in two different units of AP Physics 1.

Mass is defined in Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. It sits under Topic 2.6, Gravitational Force, at learning objective 2.6.D: describe inertial and gravitational mass. That objective carries exactly three pieces of essential knowledge, 2.6.D.1 through 2.6.D.3: the definition of inertial mass, the definition of gravitational mass, and the experimental finding that the two are equivalent. Suggested skills for Topic 2.6 are 1.A, 2.A, 2.D and 3.C.

Rotational inertia is defined in Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent of the multiple-choice section over about 15 to 20 class periods. Topic 5.4, Rotational Inertia, has two learning objectives. 5.4.A asks you to describe the rotational inertia of a rigid system relative to a given axis of rotation, carrying EK 5.4.A.1 through 5.4.A.3. 5.4.B asks you to describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass, carrying EK 5.4.B.1 and 5.4.B.2, the parallel axis theorem. Suggested skills for Topic 5.4 are 1.B, 2.B, 2.C, 3.A and 3.B.

Topic 5.4 carries two boundary statements, and both shape what you will be asked. The first: AP Physics 1 only expects students to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration. The second, in full because the second half is the part people drop: students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam, and students should have a qualitative understanding of the factors that affect rotational inertia, for example how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius.

On the equation sheet, the two quantities sit in the symbol key one after the other. The translational column lists $m$ as mass; the rotational column lists $m$ as mass, $M$ as mass and $I$ as rotational inertia. The equations that matter here are $I = \sum m_i r_i^2$ , $I' = I_{\text{cm}} + Md^2$ , $\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$ , $K = \frac{1}{2}I\omega^2$ and $L = I\omega$ . The single-object form $I = mr^2$ from EK 5.4.A.2 is a CED relevant equation and is not printed on the sheet; the summation form is what appears, and it reduces to $mr^2$ for one object anyway.

A CED sample instructional activity for Topic 5.3 is the cheapest demonstration of this distinction there is: give students a hard-boiled egg and a raw egg with no identifying marks, have them spin each one and touch it briefly on the top to stop it, and they should conclude that the raw egg is noticeably more difficult to start or stop. Same mass, same shell, different mass distribution.

For the routines, rotational kinematics has the angular equations and the torque-to-angular-acceleration method, and how to calculate torque has the $\tau = rF\sin\theta$ side. The CED framing is on Topic 5.4, Topic 5.6 and Topic 2.6, and the torque and rotational motion practice set has problems to work.

One object, one mass, four rotational inertias

Two spheres of mass $0.40$ kg each are fixed to the ends of a rigid rod of negligible mass and length $1.20$ m. Treat the spheres as point objects. Find the total mass and the rotational inertia about each of four axes: through the center perpendicular to the rod, $0.30$ m from the center perpendicular to the rod, through one sphere perpendicular to the rod, and along the rod itself. Then check two of the answers with the parallel axis theorem.

Total mass first, because it is the constant in this example: $M = 0.40 + 0.40 = 0.80\ \text{kg}$ . This value will not change in any line below, and a balance would return it for every one of the four axes.
Axis through the center, perpendicular to the rod. Each sphere is $0.60$ m from the axis. Using $I = \sum m_i r_i^2$ from EK 5.4.A.3: $I_{\text{cm}} = (0.40)(0.60)^2 + (0.40)(0.60)^2 = 0.144 + 0.144 = 0.288\ \text{kg}\cdot\text{m}^2$ . The axis passes through the center of mass, since the two masses are equal and symmetrically placed.
Axis $0.30$ m from the center, perpendicular to the rod. Now the spheres are at $0.60 - 0.30 = 0.30$ m and $0.60 + 0.30 = 0.90$ m. $I = (0.40)(0.30)^2 + (0.40)(0.90)^2 = 0.036 + 0.324 = 0.360\ \text{kg}\cdot\text{m}^2$ .
Axis through one sphere, perpendicular to the rod. That sphere sits on the axis, so its $r$ is zero and it contributes nothing. The other is the full $1.20$ m away. $I = (0.40)(0)^2 + (0.40)(1.20)^2 = 0 + 0.576 = 0.576\ \text{kg}\cdot\text{m}^2$ .
Axis along the rod itself. Both point objects lie on the axis, so both have $r = 0$ and $I = 0$ . In the point-object model that is exact. Real spheres of finite radius have a small nonzero value here, and the Topic 5.4 boundary statement says such extended values will be provided on the exam rather than calculated.
Check with the parallel axis theorem, EK 5.4.B.2. For the axis $0.30$ m off center: $I' = I_{\text{cm}} + Md^2 = 0.288 + (0.80)(0.30)^2 = 0.288 + 0.072 = 0.360\ \text{kg}\cdot\text{m}^2$ , matching the direct sum exactly.
For the axis through one sphere, the axis separation is $d = 0.60$ m: $I' = 0.288 + (0.80)(0.60)^2 = 0.288 + 0.288 = 0.576\ \text{kg}\cdot\text{m}^2$ , matching again. Note that at this distance the correction term happens to equal $I_{\text{cm}}$ , which is a coincidence of these numbers, not a rule.
Line the four up. Mass: $0.80$ kg, $0.80$ kg, $0.80$ kg, $0.80$ kg. Rotational inertia: $0.288$ , $0.360$ , $0.576$ and $0$ $\text{kg}\cdot\text{m}^2$ . The mass column has one entry repeated four times. The rotational inertia column has four different entries, and the smallest of the three perpendicular axes is the one through the center of mass, as EK 5.4.B.1 requires.

Mass is $0.80$ kg for all four axes. Rotational inertia is $0.288\ \text{kg}\cdot\text{m}^2$ about the center, $0.360\ \text{kg}\cdot\text{m}^2$ about an axis $0.30$ m off center, $0.576\ \text{kg}\cdot\text{m}^2$ about one sphere, and $0$ about the rod itself in the point-object model. The parallel axis theorem reproduces the second and third values from the first. Mass named the object; rotational inertia named the object and the axis.

A rod pivoted at its end, and the cost of skipping the theorem

A uniform rod of mass $2.4$ kg and length $1.50$ m is pivoted at one end and held horizontally, then released. An exam would provide the rod's rotational inertia about a perpendicular axis through its center as $I_{\text{cm}} = \frac{1}{12}ML^2$ . Find $I_{\text{cm}}$ , the rotational inertia about the end pivot, the net torque at the moment of release, and the initial angular acceleration. Then find what the angular acceleration would come out as if you used $I_{\text{cm}}$ by mistake. Take the direction of rotation as positive and use $g = 9.8\ \text{m/s}^2$ .

Rotational inertia about the center: $I_{\text{cm}} = \frac{1}{12}(2.4)(1.50)^2 = \frac{1}{12}(2.4)(2.25) = \frac{5.4}{12} = 0.45\ \text{kg}\cdot\text{m}^2$ .
The pivot is not at the center of mass, so this is exactly the situation LO 5.4.B names. The two axes are parallel and separated by half the length: $d = \frac{L}{2} = 0.75\ \text{m}$ .
Apply EK 5.4.B.2: $I' = I_{\text{cm}} + Md^2 = 0.45 + (2.4)(0.75)^2 = 0.45 + (2.4)(0.5625) = 0.45 + 1.35 = 1.80\ \text{kg}\cdot\text{m}^2$ . The rod is four times harder to angularly accelerate about its end than about its center, and its mass did not change.
Cross-check symbolically, since the standard end-pivot result should fall out: $I' = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2 = \frac{1}{3}(2.4)(2.25) = 1.80\ \text{kg}\cdot\text{m}^2$ . Matches.
Net torque at release. The gravitational force acts at the center of mass by EK 2.6.A.1.iii, which is $0.75$ m from the pivot, and the rod is horizontal so the force is perpendicular to that position vector, making $\sin\theta = 1$ in $\tau = rF\sin\theta$ : $\tau = (0.75)(2.4)(9.8) = 17.64\ \text{N}\cdot\text{m}$ .
Angular acceleration from EK 5.6.A.2: $\alpha = \frac{\tau_{\text{net}}}{I'} = \frac{17.64}{1.80} = 9.8\ \text{rad/s}^2$ .
Now the error. Using $I_{\text{cm}} = 0.45\ \text{kg}\cdot\text{m}^2$ instead gives $\alpha = \frac{17.64}{0.45} = 39.2\ \text{rad/s}^2$ , four times too large, exactly the ratio $\frac{I'}{I_{\text{cm}}} = 4$ . Nothing in that number looks wrong on its own, which is what makes this the expensive version of the mistake.
A physical check that catches it. The tangential acceleration of the far tip is $a_T = \alpha L$ from EK 5.2.A.2. With the correct $\alpha$ : $a_T = (9.8)(1.50) = 14.7\ \text{m/s}^2$ , which is $1.5g$ , so the tip does accelerate downward faster than a freely falling object, a real and well-known result. With the wrong $\alpha$ : $a_T = (39.2)(1.50) = 58.8\ \text{m/s}^2$ , which is $6g$ , and no torque from gravity alone can produce that.

$I_{\text{cm}} = 0.45\ \text{kg}\cdot\text{m}^2$ , $I' = 1.80\ \text{kg}\cdot\text{m}^2$ about the end pivot, net torque $17.64\ \text{N}\cdot\text{m}$ , and $\alpha = 9.8\ \text{rad/s}^2$ , giving the tip a tangential acceleration of $14.7\ \text{m/s}^2$ , that is $1.5g$ . Using the center-of-mass value at the end pivot returns $39.2\ \text{rad/s}^2$ , four times too large. The mass was $2.4$ kg throughout.

Hoop against disk: identical masses, different rotational inertias

A hoop and a solid disk each have mass $1.8$ kg and radius $0.25$ m. An exam would provide $I_{\text{hoop}} = MR^2$ and $I_{\text{disk}} = \frac{1}{2}MR^2$ about an axis through the center, perpendicular to the plane. Find each rotational inertia. Then find the translational acceleration each would get from a net force of $4.5$ N, the angular acceleration each would get from a net torque of $0.90\ \text{N}\cdot\text{m}$ , and the rotational kinetic energy of each at $\omega = 12\ \text{rad/s}$ .

Rotational inertias. Hoop: $I = MR^2 = (1.8)(0.25)^2 = (1.8)(0.0625) = 0.1125\ \text{kg}\cdot\text{m}^2$ . Disk: $I = \frac{1}{2}MR^2 = \frac{1}{2}(0.1125) = 0.05625\ \text{kg}\cdot\text{m}^2$ . The ratio is exactly $2$ .
Why the hoop is larger, in the CED's own words from the Topic 5.4 boundary statement: rotational inertia is greater when mass is farther from the axis of rotation. All of the hoop's mass sits at $R$ ; the disk's mass is spread from the center out to $R$ , so its average squared distance is smaller.
Translational test. Newton's second law from EK 2.5.A.2 uses the mass, not the rotational inertia: $a = \frac{\sum F}{m} = \frac{4.5}{1.8} = 2.5\ \text{m/s}^2$ for the hoop and $2.5\ \text{m/s}^2$ for the disk. Identical, because their masses are identical. No experiment involving a net force and a straight line can tell them apart.
Rotational test. EK 5.6.A.2 uses the rotational inertia: hoop, $\alpha = \frac{0.90}{0.1125} = 8.0\ \text{rad/s}^2$ ; disk, $\alpha = \frac{0.90}{0.05625} = 16.0\ \text{rad/s}^2$ . Different by a factor of two under the same torque, which is the mirror image of the previous step.
Energy test at the same angular speed. Hoop: $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.1125)(12)^2 = \frac{1}{2}(0.1125)(144) = 8.1\ \text{J}$ . Disk: $K = \frac{1}{2}(0.05625)(144) = 4.05\ \text{J}$ . Spinning both to $12\ \text{rad/s}$ costs twice as much energy for the hoop.
Collect the comparison. Mass: $1.8$ kg and $1.8$ kg. Translational acceleration under the same force: $2.5\ \text{m/s}^2$ and $2.5\ \text{m/s}^2$ . Radius: $0.25$ m and $0.25$ m. Rotational inertia: $0.1125$ and $0.05625\ \text{kg}\cdot\text{m}^2$ . Everything a balance or a spring scale could measure is identical, and everything involving an axis differs by a factor of two.

Hoop $0.1125\ \text{kg}\cdot\text{m}^2$ , disk $0.05625\ \text{kg}\cdot\text{m}^2$ , a ratio of $2$ . Under $4.5$ N both accelerate at $2.5\ \text{m/s}^2$ ; under $0.90\ \text{N}\cdot\text{m}$ the hoop reaches $8.0\ \text{rad/s}^2$ and the disk $16.0\ \text{rad/s}^2$ . At $12\ \text{rad/s}$ the hoop holds $8.1$ J and the disk $4.05$ J. Same mass, same radius, and every rotational answer differs by a factor of two.

Frequently asked questions

What is the difference between mass and rotational inertia?

Mass measures how strongly an object resists a change in its straight-line motion, and an object has exactly one value of it. Rotational inertia measures how strongly a rigid system resists a change in its rotation, and it depends on the axis as well as on the object, so one object has a different value for every axis you could spin it about. The AP Physics 1 CED builds the axis into the definition at essential knowledge 5.4.A.1, which says rotational inertia is related to the mass of the system and the distribution of that mass relative to the axis of rotation, and into the learning objective 5.4.A, which asks for the rotational inertia relative to a given axis of rotation.

Does rotational inertia depend on the axis of rotation?

Yes, and that is the single most important thing about it. Moving the axis changes the distance of every part of the system from it, and rotational inertia is the sum of each mass multiplied by the square of its distance from the axis, so the value changes. Tilting the axis changes it too. A pair of 0.40 kg spheres on the ends of a 1.20 m rod has a rotational inertia of 0.288 kg m squared about an axis through the center, 0.576 kg m squared about an axis through one sphere, and zero about the rod itself in the point-object model, while its mass stays at 0.80 kg throughout. Quoting a rotational inertia without naming the axis is an incomplete answer.

Is the parallel axis theorem in AP Physics 1?

Yes. It is required course content at essential knowledge 5.4.B.2, under learning objective 5.4.B, which asks students to describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass. The equation, I prime equals I about the center of mass plus M d squared, is also printed on the AP Physics 1 equation sheet in the rotational block, as well as on the AP Physics C: Mechanics sheet, so it does not have to be recalled. In the equation, M is the total mass of the system and d is the distance between the two parallel axes. The reference axis must pass through the center of mass.

Why is rotational inertia smallest about the center of mass?

Because the parallel axis theorem adds a term that can never be negative. It says the rotational inertia about any axis equals the value about a parallel axis through the center of mass plus the total mass times the square of the distance between the two axes, and a mass times a squared distance is always zero or positive. So the center-of-mass axis is the floor, reached only when that distance is zero. The AP Physics 1 CED states the result separately at essential knowledge 5.4.B.1: a rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass.

Is rotational inertia the same as moment of inertia?

They are two names for the same quantity, the sum of each mass in the system multiplied by the square of its distance from the axis. The current AP courses use rotational inertia: the phrase moment of inertia does not appear in the AP Physics 1 Course and Exam Description or in the AP Physics C: Mechanics description, while rotational inertia appears dozens of times in each, and the equation sheet's symbol key defines the symbol I as rotational inertia. Older textbooks and most engineering references say moment of inertia. Use rotational inertia in a free-response answer, because that is the language the course uses.

Can two objects with the same mass have different rotational inertias?

Yes, and the CED uses exactly this comparison. A hoop and a solid disk of the same mass and the same radius have rotational inertias in the ratio two to one about an axis through the center, because all of the hoop's mass sits at the full radius while the disk's mass is spread from the center outward. The Topic 5.4 boundary statement says students should have a qualitative understanding of the factors that affect rotational inertia, for example how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius. Under the same net force both would accelerate identically; under the same net torque they would not.

Do you have to calculate rotational inertia on the AP Physics 1 exam?

Only for small collections of objects. The Topic 5.4 boundary statement says AP Physics 1 only expects students to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration, using the sum of each mass times the square of its distance from the axis. For extended rigid bodies such as rods, disks and spheres, the same boundary statement says students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. What you do need is the parallel axis theorem, because the provided value is usually about the center of mass while the question often pivots the body somewhere else.