5.4 - Rotational Inertia

Introduction: The Rotational Version of Mass

Rotational inertia (also called moment of inertia) is the rotational stand-in for mass.

Symbol:

I

(capital I).

Units: kg·m².

What it means: rotational inertia measures how hard it is to: - Get an object spinning. - Stop an object from spinning. - Change how it’s spinning.

The analogy to linear motion: - Mass (

m

) resists changes in linear motion. - Rotational inertia (

I

) resists changes in rotational motion.

Rotational Inertia Depends On Two Things

I \text{ depends on: } \begin{cases} \text{Mass of the object} \\ \text{How that mass is spread out relative to the axis} \end{cases}

The key insight: unlike mass, which is just a fixed property, rotational inertia depends on where the axis of rotation sits and how the mass is arranged around that axis.

Point Mass

For a single point mass at distance

r

from an axis:

I = mr^2

where: -

I

= rotational inertia (kg·m²). -

m

= mass (kg). -

r

= perpendicular distance from the axis to the mass (m).

Important: that

r

is the perpendicular distance from the axis, not just any old distance.

Multiple Point Masses (System of Particles)

For a system of several point masses, the total rotational inertia is just the sum of the individual ones:

I_{total} = \sum_{i} m_i r_i^2 = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + ...

About signs: the distance

r

always ends up positive in

I = mr^2

because it gets squared. You might use signed coordinates when locating things, but the squaring makes every contribution positive.

Example: two planets orbiting a star. - Planet 1 at distance

r_1

from the star contributes

m_1 r_1^2

. - Planet 2 at distance

r_2

contributes

m_2 r_2^2

. - Total:

I = m_1 r_1^2 + m_2 r_2^2

This is the approach from AP Physics 1 for discrete systems of point masses.

Rigid Objects (Continuous Mass Distribution)

For rigid objects where the mass is spread out continuously, you need calculus:

I = \int r^2 \, dm

where: -

dm

= an infinitesimally small chunk of mass. -

r

= the distance of that chunk from the axis.

This part is new in AP Physics C, and it means chopping the object into infinitely small pieces.

Linear Mass Density and dm

To actually evaluate

I = \int r^2 \, dm

, you need to rewrite

dm

in terms of a position variable.

Linear Mass Density

Linear mass density (

\lambda

) is mass per unit length:

\lambda = \frac{m}{L}

Units: kg/m.

For non-uniform objects,

\lambda

changes with position:

\lambda(x) = \frac{dm}{dx}

Solving for dm:

dm = \lambda(x) \, dx

And that’s what goes into the integral:

I = \int r^2 \, dm = \int r^2 \lambda(x) \, dx

Note: the variable might be

x

y

z

, or whatever position coordinate the geometry calls for.

Deriving I for a Uniform Rod

For a uniform rod of mass

M

and length

L

, the mass is spread evenly, so

\lambda = \frac{M}{L}

is a constant and

dm = \frac{M}{L}\,dx

. Put

x = 0

at the axis and measure

x

along the rod. Each piece’s distance from the axis is

|x|

, and squaring it takes care of the sign.

Axis through the center: the rod runs from

x = -\frac{L}{2}

x = \frac{L}{2}

I_{cm} = \int_{-L/2}^{L/2} x^2\,\frac{M}{L}\,dx = \frac{M}{L}\left[\frac{x^3}{3}\right]_{-L/2}^{L/2} = \frac{M}{L}\left(\frac{L^3}{24} + \frac{L^3}{24}\right) = \frac{1}{12}ML^2

Axis a quarter of the way from one end: now the rod runs from

x = -\frac{L}{4}

x = \frac{3L}{4}

I = \int_{-L/4}^{3L/4} x^2\,\frac{M}{L}\,dx = \frac{M}{3L}\left[\left(\frac{3L}{4}\right)^3 - \left(-\frac{L}{4}\right)^3\right] = \frac{M}{3L}\cdot\frac{28L^3}{64} = \frac{7}{48}ML^2

Check it with the parallel axis theorem (further down this page): the new axis is

d = \frac{L}{4}

from the center, so

I = \frac{1}{12}ML^2 + M\left(\frac{L}{4}\right)^2 = \frac{4}{48}ML^2 + \frac{3}{48}ML^2 = \frac{7}{48}ML^2

. ✓ Moving the axis away from the center makes

I

bigger: about

0.083ML^2

at the center,

0.146ML^2

a quarter of the way in, and

0.333ML^2

at the end.

With numbers: a uniform 0.150 kg meter stick (

L = 1.00

m) spinning about its center has

I = \frac{1}{12}(0.150\text{ kg})(1.00\text{ m})^2 = 0.0125

kg·m².

Deriving I for a Hoop, Disk, and Annular Ring

Round objects spinning about their central axis can be built out of thin rings that all share the same center. Every bit of a thin ring of radius

r

is the same distance

r

from the axis, so a ring of mass

dm

adds

r^2\,dm

to the rotational inertia.

Thin hoop or thin cylindrical shell: all of the mass

M

sits at the same radius

R

, so

R^2

comes out of the integral:

I = \int r^2\,dm = R^2\int dm = MR^2

Uniform solid disk: the mass is spread evenly over the face, so use the surface mass density (mass per area),

\sigma = \frac{M}{\pi R^2}

. Slice the disk into thin rings of radius

r

and width

dr

. Unrolled, each ring is a thin strip

2\pi r

long and

dr

wide, so its area is

dA = 2\pi r\,dr

and its mass is

dm = \sigma(2\pi r\,dr)

I = \int_0^R r^2\,\sigma(2\pi r\,dr) = 2\pi\sigma\int_0^R r^3\,dr = 2\pi\sigma\,\frac{R^4}{4} = \frac{\pi R^4}{2}\cdot\frac{M}{\pi R^2} = \frac{1}{2}MR^2

A disk has half the rotational inertia of a hoop with the same mass and radius because its mass is spread from the center out to the rim instead of all sitting at the rim.

Annular ring (a disk with a hole, like a washer): same setup, but the rings only run from the inner radius

R_1

to the outer radius

R_2

, and

\sigma = \frac{M}{\pi(R_2^2 - R_1^2)}

I = 2\pi\sigma\int_{R_1}^{R_2} r^3\,dr = \frac{\pi\sigma}{2}\left(R_2^4 - R_1^4\right) = \frac{\pi\sigma}{2}\left(R_2^2 - R_1^2\right)\left(R_2^2 + R_1^2\right) = \frac{1}{2}M\left(R_1^2 + R_2^2\right)

Check the limits: with no hole (

R_1 = 0

) this becomes the disk,

\frac{1}{2}MR_2^2

, and when the ring gets very thin (

R_1 \to R_2

) it becomes the hoop,

MR^2

With numbers: a 0.0200 kg washer with inner radius 0.0100 m and outer radius 0.0250 m has

I = \frac{1}{2}(0.0200\text{ kg})\left[(0.0100\text{ m})^2 + (0.0250\text{ m})^2\right] = 7.25 \times 10^{-6}

kg·m².

Common Rotational Inertia Formulas

The AP exam gives you the rotational inertia formulas when you need them, unless the whole point is to derive one.

Here are the usual suspects:

Object	Axis	Rotational Inertia
Point mass	Distance $r$ from axis	$I = mr^2$
Thin rod	Through center, perpendicular	$I = \frac{1}{12}mL^2$
Thin rod	Through end, perpendicular	$I = \frac{1}{3}mL^2$
Thin hoop/ring	Through center, perpendicular to plane	$I = mr^2$
Solid disk/cylinder	Through center, along axis	$I = \frac{1}{2}mr^2$
Hollow cylinder/hoop	Through center, along axis	$I = mr^2$
Solid sphere	Through center	$I = \frac{2}{5}mr^2$
Hollow sphere	Through center	$I = \frac{2}{3}mr^2$

The pattern: they all look like

I = \beta m r^2

, where

\beta

is some number (like

\frac{1}{2}

\frac{2}{5}

) that depends on how the mass is spread out.

Mass Distribution and Rotational Inertia

The core principle: for two objects with the same mass and size, whichever one has its mass farther from the axis has the bigger rotational inertia.

Think of it as “hugging the axis”: - Mass bunched up near the axis means smaller

I

, so it’s easier to spin. - Mass pushed out far from the axis means bigger

I

, so it’s harder to spin.

Comparison with the same mass $m$ and radius $r$ :

Object	Rotational Inertia	Mass Distribution
Solid disk	$I = \frac{1}{2}mr^2$	Mass spread through the interior (smallest $I$ )
Hollow cylinder	$I = mr^2$	All the mass out at the edge (biggest $I$ )

Racing down a ramp: when objects roll without slipping (Unit 6), the one whose rotational inertia is the smaller fraction of

mr^2

gets to the bottom first, no matter its mass or radius. - The solid sphere (

I = \frac{2}{5}mr^2

) wins. - The hollow sphere (

I = \frac{2}{3}mr^2

) is slower.

Hollow objects always lose to solid ones of the same shape because their mass sits far from the axis.

Minimum Rotational Inertia

For any rigid object, the smallest possible rotational inertia happens when the axis runs through the center of mass.

Any other axis gives you a bigger value.

Example: a rod. - Through the center:

I = \frac{1}{12}mL^2

(the minimum). - Through the end:

I = \frac{1}{3}mL^2 = 4 \times \frac{1}{12}mL^2

(bigger).

Parallel Axis Theorem

The parallel axis theorem lets you find the rotational inertia about any axis that’s parallel to one through the center of mass.

The Formula

I = I_{cm} + md^2

where: -

I

= rotational inertia about the new axis. -

I_{cm}

= rotational inertia about a parallel axis through the center of mass. -

m

= total mass. -

d

= perpendicular distance between the two parallel axes.

When to use it: whenever you know

I_{cm}

and need

I

about some other axis.

Example: a rod of length

L

. -

I_{cm} = \frac{1}{12}mL^2

(through the center). - For an axis through one end,

d = \frac{L}{2}

I_{end} = I_{cm} + md^2 = \frac{1}{12}mL^2 + m\left(\frac{L}{2}\right)^2

I_{end} = \frac{1}{12}mL^2 + \frac{1}{4}mL^2 = \frac{1}{12}mL^2 + \frac{3}{12}mL^2 = \frac{4}{12}mL^2 = \frac{1}{3}mL^2

✓

Which matches the table.

Example Problem: Rod with a Ball on the End

Problem: A uniform rod of mass 1.50 kg and length 0.800 m has a small 0.500 kg ball (treat it as a point mass) fixed to one end. Find the system’s rotational inertia about three axes, all perpendicular to the rod: (a) through the bare end O, (b) through the system’s center of mass, and (c) through the rod’s midpoint.

Solution

(a) Add the pieces. The rod about its end is

\frac{1}{3}ML^2

, and the ball is a point mass 0.800 m from O:

I_O = \tfrac{1}{3}(1.50)(0.800)^2 + (0.500)(0.800)^2 = 0.320 + 0.320 = 0.640\text{ kg}\cdot\text{m}^2

(b) Find the center of mass, then use the parallel axis theorem. Measured from O, the rod’s mass is centered at its midpoint (0.400 m) and the ball is at 0.800 m:

x_{cm} = \frac{(1.50)(0.400) + (0.500)(0.800)}{1.50 + 0.500} = \frac{1.00}{2.00} = 0.500\text{ m}

The O axis is

d = 0.500

m from the center of mass, and the total mass is 2.00 kg, so

I_O = I_{cm} + Md^2

gives

I_{cm} = I_O - Md^2 = 0.640 - (2.00)(0.500)^2 = 0.140\text{ kg}\cdot\text{m}^2

(c) Midpoint axis. The rod’s midpoint is 0.100 m from the center of mass:

I_{mid} = I_{cm} + Md^2 = 0.140 + (2.00)(0.100)^2 = 0.160\text{ kg}\cdot\text{m}^2

Check by adding the pieces directly: the rod about its own center is

\frac{1}{12}(1.50)(0.800)^2 = 0.0800

, and the ball, 0.400 m away, adds

(0.500)(0.400)^2 = 0.0800

, for a total of 0.160 kg·m². ✓

The read: - The axis through the center of mass gives the smallest value, 0.140 kg·m². Every other parallel axis gives more. - The parallel axis theorem only works when one of the two axes passes through the center of mass. Jumping straight from the midpoint to the end with

d = 0.400

m gives

0.160 + (2.00)(0.400)^2 = 0.480

kg·m², which is wrong (the right answer is 0.640 kg·m²).

Example Problem: Deriving Rotational Inertia

Problem: a tapered rod of length

L

and total mass

M

has a non-uniform linear mass density:

\lambda(x) = \gamma x^2

where

\gamma = \frac{3M}{L^3}

and

x

is the distance from point P at the left end of the rod. Using integral calculus, show that the rotational inertia

I

about an axis through P and perpendicular to the page is:

I = \frac{3}{5}ML^2

Solution

Step 1: Write the rotational inertia integral

For a continuous object:

I = \int r^2 \, dm

Here, the distance from the axis (point P) to any mass element is just

x

, so

r = x

I = \int_0^L x^2 \, dm

Step 2: Get dm in terms of x

Linear mass density is

\lambda = \frac{dm}{dx}

Solving for

dm

dm = \lambda \, dx

Step 3: Plug in the given $\lambda(x)$

dm = \gamma x^2 \, dx

Step 4: Substitute into the integral

I = \int_0^L x^2 (\gamma x^2) \, dx

I = \int_0^L \gamma x^4 \, dx

Step 5: Pull the constant $\gamma$ out front

I = \gamma \int_0^L x^4 \, dx

Step 6: Integrate with the power rule

\int x^4 \, dx = \frac{x^5}{5}

I = \gamma \left[ \frac{x^5}{5} \right]_0^L

Step 7: Evaluate the definite integral

I = \gamma \left( \frac{L^5}{5} - \frac{0^5}{5} \right)

I = \gamma \cdot \frac{L^5}{5}

Step 8: Substitute $\gamma = \frac{3M}{L^3}$

I = \frac{3M}{L^3} \cdot \frac{L^5}{5}

I = \frac{3M L^5}{5L^3}

I = \frac{3M L^2}{5}

I = \frac{3}{5}ML^2

Answer: we’ve shown

I = \frac{3}{5}ML^2

✓

What it tells you: this rod carries more of its mass toward the right end, which is farther from P, so its rotational inertia is on the large side. That

\frac{3}{5}

coefficient is what the lopsided distribution buys you.

Important Considerations for Integration Problems

Limits of Integration

The limits come from where the axis sits relative to the object.

The usual cases:

Axis at the left end: $\int_0^L ... dx$

Axis at the center: $\int_{-L/2}^{L/2} ... dx$

Axis a quarter of the way from the left: $\int_{-L/4}^{3L/4} ... dx$

Always work out the limits from the geometry and the axis placement.

The General Pattern

Rotational inertia formulas basically always boil down to:

I = (\text{some number}) \times M \times (\text{length})^2

That number depends on: - The shape of the object. - How the mass is distributed. - Where the axis is.

Common Mistakes

Forgetting that I depends on the axis. Rotational inertia is NOT a fixed property like mass. Move the axis and it changes.

Using diameter instead of radius. The formulas use radius $r$ , not diameter. If you’re handed a diameter, cut it in half first.

Not squaring the distance. It’s $I = mr^2$ , not $I = mr$ . Square it every time.

Wrong limits of integration. Pin down where the axis is and set your limits from there. The axis location sets the origin of your coordinate system.

Forgetting to substitute for $\gamma$ or other constants. In derivation problems your final answer has to be in terms of the given variables ( $M$ , $L$ , and so on), not leftovers like $\gamma$ .

Botching the power rule. $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ . For $x^4$ that’s $\frac{x^5}{5}$ , not $\frac{x^5}{4}$ .

Adding rotational inertias wrong. With the parallel axis theorem it’s $I = I_{cm} + md^2$ , not $I = I_{cm} + I_{new}$ .

Thinking hollow objects are faster. In ramp races, the object whose $I$ is the smaller fraction of $mr^2$ wins. Hollow objects have a bigger fraction (their mass sits out at the rim), so they’re slower.

Confusing center of mass with geometric center. For non-uniform objects these can be different spots. The minimum $I$ is about an axis through the center of mass, which isn’t always the geometric center.

Not realizing calculus is needed. Continuous bodies (rods, disks, spheres) call for calculus. Discrete point masses just need a summation.

Wrong expression for dm. It’s $dm = \lambda dx$ , where $\lambda$ is the linear mass density. If $\lambda$ varies, sub in the function $\lambda(x)$ .

Forgetting what $r$ means in the integral. In $\int r^2 dm$ , the $r$ is the distance from the axis of rotation to the mass element $dm$ , which depends on your coordinate system and axis choice.

Not simplifying before substituting. Pull the constants out of the integral first. It makes the integration cleaner and cuts down on errors.

Failing dimensional analysis. Check yourself: rotational inertia is in kg·m², so your answer should be a mass times a length squared.

Misusing the table formulas. Those formulas are for specific axes, usually through the center of mass. If the problem uses a different axis, either apply the parallel axis theorem or derive it from scratch.

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