10.4 Moment of Inertia from Mass Distribution

The race down the ramp

A solid ball and a hollow ball sit at the top of a ramp. Same mass. Same radius. Same starting position. You release them at the same instant.

Which one reaches the bottom first?

If you have only met $F = ma$, this question seems impossible. Both balls have the same mass. Gravity pulls on them equally. The same force acts on the same mass --- shouldn't they accelerate at the same rate?

But they don't. The solid ball wins, every single time.

The difference has nothing to do with how much mass each ball has. It has everything to do with where that mass sits. The hollow ball has all its mass pushed out to the rim. The solid ball has mass spread throughout its interior, closer to the center. That difference in mass distribution makes the hollow ball harder to spin up --- and since both balls must spin as they roll, the one that's harder to spin falls behind.

The property that captures this idea --- a single number that tells you how the mass is distributed relative to a rotation axis --- is called moment of inertia. It is the central concept of this section.

Prediction

Before you read further, commit to an answer.

A solid disk and a hoop have the same mass $M$ and the same radius $R$. Both can rotate about an axis through their centers.

Which has a larger moment of inertia? And what does your answer mean for how easy each one is to spin up from rest?

Think about where the mass is in each object. The hoop has all its mass at distance $R$ from the center. The solid disk has mass spread from the center all the way out to $R$.

Hold your prediction in mind. We will return to it.

Why "where" matters as much as "how much"

Here is the guiding question for this section: Why does where the mass is matter as much as how much mass there is?

You already have intuition for this, even if you have not formalized it. Think about spinning in an office chair. Tuck your arms in close to your body, and you spin easily --- a small push gets you rotating quickly. Now extend your arms straight out, holding something heavy in each hand. The same push barely gets you moving. You didn't gain or lose any mass. You just moved it farther from the axis.

The farther mass is from the rotation axis, the harder it is to start (or stop) that mass rotating. And this is not a small effect. Double the distance from the axis and the contribution to rotational inertia quadruples, because the dependence is on $r^2$, not $r$.

Exploration: redistribute the mass

[Interactive: Mass Distribution Explorer. A top-down view of a rotating disk with adjustable mass placement. A slider labeled "Mass Distribution" moves continuously between two extremes: all mass concentrated at the center (left end of slider) and all mass concentrated at the rim (right end of slider).

As the student moves the slider, two quantities update in real time:

Moment of inertia $I$, displayed as a number and a bar chart.

Angular acceleration $\alpha$, computed for a fixed applied torque ($\tau = 1\;\text{N}!\cdot!\text{m}$), displayed as both a number and a spinning animation whose speed reflects $\alpha$.

A visual overlay shows the mass distribution on the disk: a shaded region that shifts between a dense central blob and a thin outer ring as the slider moves.

Guided prompts appear alongside the interactive:

"Move all the mass to the rim. What value does $I$ take? How fast does the object spin up?"

"Now move all the mass to the center. What happens to $I$? What happens to $\alpha$?"

"Find a slider position in between. How does $I$ scale as you move mass outward?"

"Notice: the total mass never changes. Only its position changes. Yet the angular acceleration can differ by a large factor."]

Spend some time with this. The key observation: the same total mass can produce very different moments of inertia depending on how it is distributed relative to the axis. Mass near the axis contributes little to $I$. Mass far from the axis contributes a lot.

Concept reveal: the moment of inertia

Here is the mathematical definition that captures everything you just explored.

For a collection of point masses $m_1, m_2, \ldots, m_n$ at distances $r_1, r_2, \ldots, r_n$ from the rotation axis:

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

Each mass contributes $m_i r_i^2$ to the total. The farther a mass is from the axis, the more it contributes --- and the contribution grows as the square of the distance.

For a continuous object, the sum becomes an integral:

$$I = \int r^2\, dm$$

where $r$ is the distance from the axis to the mass element $dm$, and the integral runs over the entire object.

Moment of inertia is rotational inertia. Just as mass measures an object's resistance to translational acceleration (how hard it is to push), moment of inertia measures an object's resistance to angular acceleration (how hard it is to spin up or slow down). A large $I$ means the object resists changes in its rotational motion. A small $I$ means it spins up easily.

But there is a critical difference between mass and moment of inertia. Mass is a property of the object alone --- it does not depend on how you push it or where you stand. Moment of inertia depends on the axis of rotation. The same object can have different moments of inertia about different axes. Spin a baton about its center, and $I$ is relatively small. Spin it about one end, and $I$ is much larger --- because the far half of the baton is now at a greater distance from the axis.

Return to your prediction. The hoop has all its mass at distance $R$, so $I_{\text{hoop}} = MR^2$. The solid disk has mass distributed from $r = 0$ to $r = R$, with most of it closer to the center than $R$, so $I_{\text{disk}} = \tfrac{1}{2}MR^2$. The hoop has twice the moment of inertia of the disk, even though they have the same mass and radius. The hoop is harder to spin up.

Connection to translational dynamics

This is a good moment to step back and see the analogy clearly.

Translational motion	Rotational motion
Force $F$	Torque $\tau$
Mass $m$	Moment of inertia $I$
Acceleration $a$	Angular acceleration $\alpha$
$F = ma$	$\tau = I\alpha$

In $F = ma$, mass tells you how hard it is to accelerate an object in a straight line. In $\tau = I\alpha$, moment of inertia tells you how hard it is to angularly accelerate an object around an axis. The structure of the two equations is identical.

But the analogy has a boundary. Mass is a single number that belongs to the object. Moment of inertia depends on the axis you choose. Move the axis, and $I$ changes --- even though the object itself has not changed at all. This axis-dependence is the new ingredient that makes rotational dynamics richer than translational dynamics.

Variation: same mass, different distributions

To sharpen your intuition, compare three objects that all have the same mass $M$ and the same outer radius $R$, rotating about an axis through their center.

Object	Mass distribution	Moment of inertia
Solid cylinder	Mass spread uniformly throughout	$I = \tfrac{1}{2}MR^2$
Hollow cylinder (thick-walled)	Mass concentrated between an inner and outer radius	$I$ between $\tfrac{1}{2}MR^2$ and $MR^2$
Thin ring (hoop)	All mass at distance $R$ from the center	$I = MR^2$

What changed? What stayed the same?

The total mass did not change. The outer radius did not change. The only thing that varied was how the mass is distributed between the center and the rim. And that single change doubled the moment of inertia from the solid cylinder to the thin ring.

The thin ring has the largest $I$ because every bit of its mass sits at the maximum distance from the axis. The solid cylinder has the smallest $I$ because much of its mass is close to the center, where $r$ is small and $r^2$ is even smaller. The hollow cylinder falls in between --- its mass is pushed outward, but not all the way to the rim.

[Interactive: Shape Comparison. Three rotating objects --- solid cylinder, hollow cylinder, thin ring --- shown side by side. All have the same mass and outer radius. A fixed torque is applied to each. Students watch the resulting angular accelerations. The solid cylinder spins up fastest; the thin ring spins up slowest. A bar chart shows the moment of inertia for each, and the relationship $\alpha = \tau / I$ is displayed, making the inverse relationship between $I$ and $\alpha$ visible.]

This comparison makes a concrete prediction you can test: if all three objects roll down the same ramp, the solid cylinder reaches the bottom first, the hollow cylinder second, and the thin ring last. The race down the ramp is really a race between moments of inertia.

A worked example: discrete masses

Let's compute a moment of inertia from scratch. Three point masses are attached to a light rod that rotates about one end:

$m_1 = 2\;\text{kg}$ at $r_1 = 0.1\;\text{m}$ from the axis
$m_2 = 3\;\text{kg}$ at $r_2 = 0.3\;\text{m}$ from the axis
$m_3 = 1\;\text{kg}$ at $r_3 = 0.5\;\text{m}$ from the axis

Applying the definition:

$$I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2$$

$$I = (2)(0.1)^2 + (3)(0.3)^2 + (1)(0.5)^2$$

$$I = 0.02 + 0.27 + 0.25 = 0.54\;\text{kg}!\cdot!\text{m}^2$$

Notice something: $m_3$ is the lightest mass, but it contributes almost as much to $I$ as the heaviest mass $m_2$, because it is much farther from the axis. The $r^2$ factor amplifies the contribution of distant mass.

Practice

Layer 1: Concrete

Four point masses are arranged on a massless cross-shaped frame that rotates about its center:

$m_1 = 1\;\text{kg}$ at $r_1 = 0.2\;\text{m}$ (north)

$m_2 = 1\;\text{kg}$ at $r_2 = 0.2\;\text{m}$ (south)

$m_3 = 2\;\text{kg}$ at $r_3 = 0.4\;\text{m}$ (east)

$m_4 = 2\;\text{kg}$ at $r_4 = 0.4\;\text{m}$ (west)

Compute the moment of inertia about the center of the cross.

Check your answer

Apply the definition directly: $$I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + m_4 r_4^2$$ $$I = (1)(0.2)^2 + (1)(0.2)^2 + (2)(0.4)^2 + (2)(0.4)^2$$ $$I = 0.04 + 0.04 + 0.32 + 0.32 = 0.72\;\text{kg}\!\cdot\!\text{m}^2$$ Notice that the two 2 kg masses at 0.4 m dominate the moment of inertia, contributing $0.64\;\text{kg}\!\cdot\!\text{m}^2$ out of $0.72\;\text{kg}\!\cdot\!\text{m}^2$ total --- nearly 89%. This is the $r^2$ factor at work: doubling the distance from 0.2 m to 0.4 m quadruples each mass's contribution, and the outer masses are also twice as heavy.

Layer 2: Pattern

All of the following objects have the same mass $M$ and the same radius $R$. Rank them by moment of inertia about an axis through their center, from smallest to largest:

(A) A solid sphere: $I = \tfrac{2}{5}MR^2$

(B) A solid cylinder (disk): $I = \tfrac{1}{2}MR^2$

(C) A thin spherical shell: $I = \tfrac{2}{3}MR^2$

(D) A thin ring (hoop): $I = MR^2$

Rank A through D from smallest to largest moment of inertia. Then explain the pattern in one sentence.

Check your answer

Ranking from smallest to largest $I$: $$A < B < C < D$$ $$\tfrac{2}{5}MR^2 < \tfrac{1}{2}MR^2 < \tfrac{2}{3}MR^2 < MR^2$$ The pattern: **the more mass is concentrated at the outer edge (far from the axis), the larger the moment of inertia.** The solid sphere has mass distributed throughout its interior, much of it close to the center, so its $I$ is smallest. The thin ring has all its mass at the maximum distance $R$, so its $I$ is largest. The solid cylinder and thin shell fall in between, with the shell ranking higher because its mass is pushed farther outward on average than the cylinder's.

Layer 3: Structure

Why does the moment of inertia depend on the axis of rotation?

A uniform rod has $I = \tfrac{1}{12}ML^2$ about its center, but $I = \tfrac{1}{3}ML^2$ about one end --- four times larger.

Explain, in terms of the definition $I = \int r^2\,dm$, why shifting the axis from the center to the end changes $I$ so dramatically. What is $r$ measuring in each case, and why does that change matter?

Check your answer

In the formula $I = \int r^2\,dm$, the variable $r$ is the perpendicular distance from each mass element to the chosen axis. When you change the axis, every mass element gets a new value of $r$ --- even though the object itself has not moved. When the axis is at the **center** of the rod, mass extends from $r = 0$ (at the center) to $r = L/2$ (at each end). The farthest any mass element sits from the axis is $L/2$. When the axis is at **one end**, mass extends from $r = 0$ (at the axis end) to $r = L$ (at the far end). Half the rod is now at distances between $L/2$ and $L$ from the axis --- distances that did not exist in the center-axis case. Because $r$ enters as $r^2$, this distant mass contributes disproportionately. The far end of the rod, at distance $L$, contributes four times as much per unit mass as a point at $L/2$. This is why $I$ depends on the axis: the axis determines what $r$ means for every piece of mass, and since $r$ enters squared, even modest shifts in the axis can produce large changes in $I$.

Layer 4: Debug

A student needs to find the moment of inertia of a uniform disk rotating about a point on its rim (not its center). The student looks up the moment of inertia of a disk about its center --- $I_{\text{center}} = \tfrac{1}{2}MR^2$ --- and uses that value directly.

The correct answer is $I_{\text{rim}} = \tfrac{3}{2}MR^2$.

What concept did the student miss, and how does it fix the calculation?

Check your answer

The student missed the **parallel axis theorem**. This theorem states that if you know the moment of inertia about an axis through the center of mass, you can find it about any parallel axis a distance $d$ away using: $$I = I_{\text{cm}} + Md^2$$ For the disk rotating about a point on its rim, the new axis is a distance $d = R$ from the center of mass. So: $$I_{\text{rim}} = I_{\text{cm}} + MR^2 = \tfrac{1}{2}MR^2 + MR^2 = \tfrac{3}{2}MR^2$$ The student's error was treating the center-of-mass moment of inertia as if it applied to a different axis. The moment of inertia about the center is always the *minimum* for any parallel axis --- moving the axis away from the center always increases $I$, because you are effectively placing more mass at greater distances from the new axis. The parallel axis theorem is the tool that corrects this. Once you know $I_{\text{cm}}$, you can find $I$ about any parallel axis without re-doing the integral from scratch.

Reflection

Think about what you have explored in this section.

How does moment of inertia capture "where the mass is" in a single number?

A moment of inertia is a sum (or integral) of $r^2\,dm$ --- every piece of mass, weighted by the square of its distance from the axis. That single number encodes the entire spatial distribution of mass, at least as it matters for rotation. Two objects with different shapes, different sizes, and different internal structures can have the same $I$ if their mass happens to be distributed in a way that produces the same weighted sum.

Consider: is any information lost when you collapse a full mass distribution down to one number? What aspects of the distribution does $I$ not capture?

Looking Ahead

You now have the three ingredients of rotational dynamics: torque (Section 10.1), rotational kinematics (Chapter 9), and moment of inertia (this section). Torque is the cause. Angular acceleration is the effect. Moment of inertia is what connects them --- the measure of how stubbornly an object resists being spun.

In the next section, we bring these ingredients together into a single equation: $\sum \tau = I\alpha$. This is Newton's second law for rotation. It will let you predict exactly how an object's rotation changes when forces are applied --- not just qualitatively ("it's harder to spin"), but quantitatively ("the angular acceleration is 4.2 rad/s$^2$"). Everything you learned about mass distribution in this section feeds directly into that equation.