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Rotational
Inertia

Understand how mass distribution determines an object's resistance to angular acceleration, and master the key formula \( I = \sum m_i r_i^2 \).

AP Physics 1

What Is Rotational Inertia?

In linear motion, mass tells you how hard it is to change an object's velocity. In rotational motion, rotational inertia (also called the moment of inertia) tells you how hard it is to change an object's angular velocity.

The symbol is \( I \), measured in \( \text{kg} \cdot \text{m}^2 \). A larger \( I \) means the object is harder to spin up or slow down.

The Rotational Analogy

Newton's second law for linear motion is \( F = ma \). The rotational version replaces every quantity with its rotational counterpart:

Linear
\( F = ma \)
Force, mass, acceleration
Rotational
\( \tau = I\alpha \)
Torque, inertia, angular accel.

Why Mass Distribution Matters

Mass near axis Small I Mass far from axis Large I → Same mass

Two objects with the same total mass can have very different rotational inertias. Moving mass farther from the axis increases \( I \) dramatically because distance is squared in the formula.

The Core Formula

For a system of point masses:

1
\( m_i \): the mass of the \( i \)-th particle (in kg)
2
\( r_i \): the perpendicular distance from the \( i \)-th particle to the axis of rotation (in m)
3
\( I \): the resulting rotational inertia (in \( \text{kg} \cdot \text{m}^2 \))

Because \( r \) is squared, doubling the distance from the axis quadruples the contribution to \( I \).

Visualizing \( I = \sum m_i r_i^2 \) for Point Masses

Axis 2 kg 0.3 m 3 kg 0.5 m 1 kg 0.8 m

Rotational Inertia of Common Shapes

For continuous objects, the sum becomes an integral. The AP Physics 1 exam provides these formulas, but you should recognize them:

Solid Cylinder
\( I = \tfrac{1}{2}MR^2 \)
Axis through center
Solid Sphere
\( I = \tfrac{2}{5}MR^2 \)
Axis through center
Thin Hoop
\( I = MR^2 \)
All mass at radius \( R \)

Comparing mass distributions

The disk and hoop picture is a separate geometry comparison with equal mass and radius. The pulley question below uses different values.

For equal mass 2 kg and radius 0.20 m about central axes, a solid disk has I = 0.040 kg·m² and a thin hoop has I = 0.080 kg·m².
For equal mass 2 kg and radius 0.20 m about central axes, a solid disk has I = 0.040 kg·m² and a thin hoop has I = 0.080 kg·m².

A massive-pulley example

This separate source example applies rotational inertia to a pulley and hanging block.

A solid cylindrical pulley of mass M = 4.0 kg and radius R = 0.20 m is mounted on a frictionless axle. A block of mass m = 3.0 kg hangs from a light string wrapped around the pulley. The system is released from rest. Find: (a) the angular acceleration of the pulley, (b) the linear acceleration of the block, and (c) the tension in the string.

Show answer and explanation

Step 1 — Identify the Rotational Inertia of the Pulley: The pulley is a solid cylinder rotating about its central axis, so I = ½MR² = ½(4.0 kg)(0.20 m)² = 0.080 kg·m². I = 0.080 kg·m² Step 2 — Apply Newton's Second Law to the Block: Taking downward as positive for the block, the forces on it are its weight mg (downward) and the string tension T (upward). Newton's second law gives: mg − T = ma, where a is the linear acceleration of the block. mg − T = ma …(1) Step 3 — Apply Newton's Second Law for Rotation to the Pulley: The only torque on the pulley comes from the tension T acting at radius R. Since the string unwinds, the torque is τ = TR. Applying τₙₑₜ = Iα: TR = Iα. TR = Iα …(2) Step 4 — Relate Linear and Angular Quantities: Because the string does not slip on the pulley, the tangential acceleration of the rim equals the linear acceleration of the block: a = Rα, so α = a/R. α = a / R Step 5 — Solve for the Linear Acceleration: From equation (2): T = Iα/R = I(a/R)/R = Ia/R². Substituting into equation (1): mg − Ia/R² = ma. Solving for a: a = mg / (m + I/R²). Plugging in: a = (3.0)(9.8) / (3.0 + 0.080/0.04) = 29.4 / (3.0 + 2.0) = 29.4 / 5.0 = 5.88 m/s². a ≈ 5.9 m/s² Step 6 — Find the Angular Acceleration: α = a/R = 5.88 / 0.20 = 29.4 rad/s². α ≈ 29 rad/s² Step 7 — Find the Tension in the String: T = Ia/R² = (0.080)(5.88)/(0.04) = 11.76 N. Alternatively, T = m(g − a) = 3.0(9.8 − 5.88) = 3.0(3.92) = 11.76 N. Both methods agree, confirming our result. T ≈ 11.8 N

A pulley with its rotational inertia exposed

An exposed pulley rotor supports a taut cord and a blue load suspended clear of the base.The falling load accelerates while cord tension supplies torque to the pulley. With no slip, the load acceleration and pulley angular acceleration satisfy a = Rα; tension is smaller than the load’s weight.

Acceleration with a massive pulley

Acceleration with a massive pulley. A 3 kg load drives a 4 kg solid pulley of radius 0.20 m. With I = 0.080 kg·m², the no-slip equations give a = 5.88 m/s², T = 11.76 N and α = 29.4 rad/s².A 3 kg load drives a 4 kg solid pulley of radius 0.20 m. With I = 0.080 kg·m², the no-slip equations give a = 5.88 m/s², T = 11.76 N and α = 29.4 rad/s².

Disk and hoop under equal torque

Disk and hoop under equal torque. A 4 kg disk and hoop, each of radius 0.25 m, experience the same 1 N·m net torque. Their inertias are 0.125 and 0.250 kg·m², so their angular accelerations are 8 and 4 rad/s².A 4 kg disk and hoop, each of radius 0.25 m, experience the same 1 N·m net torque. Their inertias are 0.125 and 0.250 kg·m², so their angular accelerations are 8 and 4 rad/s².

Mass vs Rotational Inertia

Mass \( m \) Rotational Inertia \( I \)
What it measures Resistance to linear acceleration Resistance to angular acceleration
Depends on Amount of matter only Amount of matter and its distribution
Units \( \text{kg} \) \( \text{kg} \cdot \text{m}^2 \)
Can change without adding mass? No Yes: move mass closer or farther from axis
Newton's 2nd Law form \( F = ma \) \( \tau = I\alpha \)

The Axis Changes Everything

Axis Center axis \( I = \tfrac{1}{12}ML^2 \) Axis End axis \( I = \tfrac{1}{3}ML^2 \) vs

The Parallel Axis Theorem

If you know the rotational inertia \( I_\text{cm} \) about the center of mass, you can find \( I \) about any parallel axis at distance \( d \):

1
\( I_\text{cm} \): rotational inertia about the center-of-mass axis
2
\( M \): total mass of the object
3
\( d \): distance between the center-of-mass axis and the new parallel axis

How to Solve Rotational Inertia Problems

1
Identify the axis of rotation. Every \( I \) value is defined relative to a specific axis.
2
List each mass and its distance \( r \) from the axis. For point masses, measure directly. For extended shapes, use the known formulas.
3
Apply the formula. For point masses: \( I = \sum m_i r_i^2 \). For shapes: use \( I = cMR^2 \) with the correct prefactor \( c \).
4
Combine contributions. If the system has multiple parts, find \( I \) for each part about the same axis, then add them.
5
Check units. The answer must be in \( \text{kg} \cdot \text{m}^2 \). If not, recheck your distances.

Example 1: Two Point Masses

Two masses are attached to a light rod and rotate about its center. Mass A = 4 kg at 0.5 m from the axis. Mass B = 2 kg at 1.0 m from the axis. Find \( I \).

Axis 4 kg 0.5 m 2 kg 1.0 m

Example 2: Hoop vs Disk Rolling Race

A thin hoop and a solid disk have the same mass \( M \) and radius \( R \). Both start from rest and roll down an incline. Which arrives first?

Hoop
\( I = MR^2 \)
Prefactor: 1
More inertia per mass
Solid Disk
\( I = \tfrac{1}{2}MR^2 \)
Prefactor: 0.5
Less inertia per mass

A tactile rolling race with visible mass distribution

A tabletop ramp with a hoop, disk and sphere, the latter two showing conceptual exposed interiors.For bodies rolling without slipping from the same initial center-of-mass height, mass distribution determines acceleration and the partition of kinetic energy. Equal elapsed time and equal height loss are separate comparisons.

Rolling race at equal elapsed time

Rolling race at equal elapsed time. Equal-mass, equal-radius bodies start together on a 30° ramp. At 0.6 s, the hoop travels 0.441 m, the disk 0.588 m and the solid sphere 0.630 m, with no slipping.Equal-mass, equal-radius bodies start together on a 30° ramp. At 0.6 s, the hoop travels 0.441 m, the disk 0.588 m and the solid sphere 0.630 m, with no slipping.

Example 3: Composite System

A solid disk of mass 6 kg and radius 0.4 m has a 2 kg point mass bolted at its rim. Find the total \( I \) about the disk's center axis.

1
Disk: \( I_\text{disk} = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(6)(0.4)^2 = 0.48\;\text{kg}\cdot\text{m}^2 \)
2
Point mass at rim: \( I_\text{point} = mr^2 = (2)(0.4)^2 = 0.32\;\text{kg}\cdot\text{m}^2 \)
3
Total: \( I = 0.48 + 0.32 = \) 0.80 \( \text{kg}\cdot\text{m}^2 \)

Common Mistakes to Avoid

✕ Forgetting the axis
Rotational inertia is not a fixed property: it changes whenever you choose a different axis of rotation.
✕ Forgetting to square \( r \)
The formula is \( mr^2 \), not \( mr \). Forgetting the square gives answers that are too small.
✕ Mixing up \( R \) and \( d \)
In the parallel axis theorem, \( d \) is the shift distance between axes, not the object's radius.

Summary of Key Formulas

Shape / Scenario Rotational Inertia \( I \) Axis
Point mass \( mr^2 \) Distance \( r \) from mass
Thin hoop (ring) \( MR^2 \) Through center, perpendicular
Solid disk / cylinder \( \tfrac{1}{2}MR^2 \) Through center, perpendicular
Solid sphere \( \tfrac{2}{5}MR^2 \) Through center
Thin rod (center) \( \tfrac{1}{12}ML^2 \) Through center, perpendicular
Thin rod (end) \( \tfrac{1}{3}ML^2 \) Through one end, perpendicular
Parallel axis theorem \( I_\text{cm} + Md^2 \) Any axis parallel to CM axis

Question 1 of 5

A 3 kg point mass is located 2 m from an axis of rotation. What is its rotational inertia?

Question 2 of 5

Two identical solid disks have the same mass and radius. Disk A spins about its center axis. Disk B spins about an axis at its rim. Which has greater rotational inertia?

Question 3 of 5

A figure skater pulls her arms inward during a spin. Her rotational inertia decreases. If no external torque acts, what happens to her angular velocity?

Question 4 of 5

A thin hoop and a solid sphere of equal mass and radius roll down the same incline from rest. Which reaches the bottom first?

Question 5 of 5

A uniform thin rod of mass \( M = 4\;\text{kg} \) and length \( L = 3\;\text{m} \) rotates about one end. What is its rotational inertia?