The angular versions of position, velocity, and acceleration
The Big Idea
Every linear motion concept has a rotational counterpart. Angle (θ) replaces position, angular velocity (ω) replaces velocity, angular acceleration (α) replaces acceleration. The kinematic equations have the exact same form — you just swap the variables.
Key Equations
Angular Velocity
ω = Δθ/Δt. Units: rad/s.
Angular Acceleration
α = Δω/Δt. Units: rad/s².
Kinematics 1
ω = ω₀ + αt
Kinematics 2
θ = θ₀ + ω₀t + ½αt²
Example
A spinning wheel
A wheel starts at rest and undergoes constant angular acceleration of 2 rad/s² for 5 seconds. What is its final angular velocity?
ω = ω₀ + αt = 0 + (2)(5) = 10 rad/s
How many radians did it rotate through? θ = ½αt² = ½(2)(25) = 25 rad (about 4 revolutions, since 2π ≈ 6.28).
Topic 5.2
Connecting Linear and Rotational Motion
The radius r is the bridge between linear and angular
The Big Idea
A point on a rotating object has BOTH angular quantities (ω, α) AND linear quantities (v, a). They're connected through the radius: v = rω, a_t = rα, s = rθ. A point farther from the axis moves faster tangentially for the same rotation. The tip of a propeller blade moves much faster than a point near the hub — even though both have the same ω.
The Bridge Equations
Arc Length
s = rθ (θ in radians)
Tangential Velocity
v = rω
Tangential Acceleration
a_t = rα
Centripetal Acceleration
a_c = ω²r = v²/r
Example
A car's tires
A car drives at 20 m/s. Its tires have radius 0.30 m. What is the angular velocity of the tires?
v = rω, so ω = v/r = 20/0.30 = 66.7 rad/s
That's about 10.6 revolutions per second. Faster cars have faster-spinning tires for the same tire size.
Topic 5.3
Torque
The rotational equivalent of force
The Big Idea
A torque is what makes things spin. It depends on THREE things: how much force you apply (F), how far from the pivot you apply it (r), and the angle between the force and the lever arm (θ). The formula: τ = rF·sin(θ). Units: N·m. Only the component of force perpendicular to the lever arm produces torque — a force directly toward the pivot does nothing.
Key Properties
Formula
τ = rF·sin(θ). Or "lever arm" × force.
Direction
+ counterclockwise; − clockwise (convention).
Zero Torque
When force points along the lever arm (sin 0° = 0).
Units
N·m. NOT joules — torque ≠ energy.
Example
Opening a door
You push on a door with 20 N of force perpendicular to its surface, 0.80 m from the hinges. What torque do you exert?
If you push at the same force but only 0.20 m from the hinges (close to the hinge), τ = (0.20)(20) = 4 N·m. Much less torque! That's why door handles are far from hinges.
Topic 5.4
Rotational Inertia
Mass distribution determines how hard an object is to spin
The Big Idea
Rotational inertia (I) is the rotational analog of mass. It tells you how hard it is to angularly accelerate an object. For a point mass: I = mr². For systems of point masses, add them up: I = Σm_i·r_i². The crucial thing: r is SQUARED, so mass far from the axis matters way more than mass close to the axis. Units: kg·m².
What Matters
Total Mass
More mass = more I (linear).
Mass Distribution
Mass far from axis matters much more (r is squared).
Choice of Axis
Same object has different I depending on where the axis is.
Hoop vs Disk
A hoop has higher I than a solid disk (same m, same r).
Example
Two point masses on a rod
Two 2 kg masses are on a massless rod, each 0.5 m from the center pivot. Find I about the center.
If we move the masses out to 1.0 m each instead: I = (2)(1)² + (2)(1)² = 4.0 kg·m². Doubling the distance QUADRUPLES the rotational inertia — even though the mass is unchanged.
Topic 5.5
Rotational Equilibrium
When torques cancel: Newton's first law in rotational form
The Big Idea
An object is in rotational equilibrium when the net torque is zero (Στ = 0). Its angular velocity stays constant — usually zero (not spinning), but it could be spinning at a steady rate. Static equilibrium requires BOTH the force condition (ΣF = 0) AND the torque condition (Στ = 0). Classic test problem: a uniform beam supported by a cable.
Recipe for Equilibrium Problems
Draw a force diagram showing all forces and where each acts.
Pick a smart pivot. Choose one through an unknown force to eliminate it.
A 40 kg kid sits 2.0 m to the left of a seesaw pivot. How far to the right should a 25 kg kid sit to balance it?
Equilibrium: τ_left = τ_right (in magnitude). The torques are (m·g·d) on each side.
(40)(10)(2.0) = (25)(10)(d)
800 = 250d → d = 3.2 m
The lighter kid sits farther from the pivot — more lever arm to balance less weight.
Topic 5.6
Newton's Second Law in Rotational Form
τ_net = Iα — solve it just like F = ma
The Big Idea
Just as F = ma describes how forces accelerate masses linearly, τ_net = Iα describes how torques angularly accelerate rotating objects. Bigger net torque → bigger angular acceleration. Bigger rotational inertia → smaller angular acceleration for the same torque. Solve these the same way you solve F = ma problems — just with rotational variables throughout.
The Direct Analogy
Linear
F = ma
Rotational
τ_net = Iα
Translation
F → τ, m → I, a → α
Solve The Same Way
Find net torque, divide by I, get α.
Example
Angular acceleration of a wheel
A wheel has rotational inertia I = 0.50 kg·m². A net torque of 4.0 N·m is applied to it. What is its angular acceleration?
Same problem-solving as F = ma: rearrange to get the unknown by itself, then plug in numbers. If you doubled the rotational inertia (to 1.0 kg·m²), α would halve to 4.0 rad/s² for the same torque.
1 / 6
How to use this visual review
Take 1–2 minutes per slide. Read the Big Idea first, then the definitions, then the worked example. Six slides total — one per topic.
Use the topic pills to jump to any topic, or use the arrow keys to step through them in order.
The key insight to internalize is the linear ↔ rotational analogy. Once you have it, every Unit 5 problem becomes a translation of a familiar Unit 1 or Unit 2 problem.