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5.1 - Rotational Kinematics

Introduction to Rotation

We’re shifting from linear kinematics (motion along a line) to rotational kinematics (motion around an axis). Many of the concepts directly mirror what we covered in Unit 1, just with new variables and a new perspective. Instead of asking “how far?” we now ask “how much did it rotate?” Instead of “how fast is it moving?” we ask “how fast is it spinning?”

Rigid Systems vs. Objects

So far we’ve modeled almost everything as an object: a single point with mass and no size, so every part of it moves the same way. That can’t describe spinning. As a wheel turns, the top moves one way, the bottom moves the opposite way, and points near the axle barely move at all.
For rotation we use a rigid system instead: a system that holds its shape (it doesn’t stretch, bend, or squish), but whose points move in different directions, and at different speeds, as it rotates. A spinning wheel, a swinging door, and a seesaw are all rigid systems. A rigid system can’t be modeled as an object, because an object has no size, so the model can’t tell the top of the wheel from the bottom or say where a force is applied. For a rigid system, where a force acts matters, which is the idea behind torque (5.3).
When you can ignore the spin: if the motion you care about is well described by the motion of the system’s center of mass, you can still treat the system as a single object. For example, Earth spins on its axis once a day, but when you study Earth’s yearly revolution around the center of mass of the Earth-Sun system, the spin doesn’t matter, so Earth is treated as an object. The same thing can be an object in one problem and a rigid system in another: a car is an object when you find its stopping distance, but each wheel is a rigid system when you ask how fast it spins.

Angular Displacement

Angular displacement (Δθ\Delta\theta) measures the angle through which a point or rigid system rotates about a specified axis.
Δθ=θf−θi\Delta \theta = \theta_f - \theta_i
where Δθ\Delta\theta is angular displacement, θf\theta_f is the final angular position, and θi\theta_i is the initial angular position.
Angular measurements in physics use radians, not degrees.
1 revolution=2π radians=360°1 radian≈57.3°π radians=180°1 \text{ revolution} = 2\pi \text{ radians} = 360° \qquad 1 \text{ radian} \approx 57.3° \qquad \pi \text{ radians} = 180°
Radians create a natural connection between angular and linear quantities (coming up next), and they simplify equations for calculus-based physics. One important note for calculators: treat radians as a unit, like meters, when doing these calculations. You don’t need to switch your calculator to radian mode for this, we’re using radians as a measurement unit here, not for trig functions.
5.1 - Rotational Kinematics
For example, a half rotation is Δθ=π\Delta\theta = \pi rad, a quarter rotation (90°) is Δθ=π2\Delta\theta = \frac{\pi}{2} rad, and three complete rotations is Δθ=3(2π)=6π\Delta\theta = 3(2\pi) = 6\pi rad.

Angular Velocity

Angular velocity (ω\omega) is the rate at which angular position changes over time.
ωavg=ΔθΔt\omega_{\text{avg}} = \frac{\Delta \theta}{\Delta t}
with units of radians per second (rad/s). The lowercase Greek letter omega represents angular velocity, just as vv represents linear velocity.
Angular velocity tells you how fast something’s spinning. A car wheel might have ω=50\omega = 50 rad/s (spinning rapidly), a clock’s second hand has ω=2π60≈0.105\omega = \frac{2\pi}{60} \approx 0.105 rad/s (one revolution per minute), and Earth’s rotation is ω=2π86400≈7.3×10−5\omega = \frac{2\pi}{86400} \approx 7.3 \times 10^{-5} rad/s (one revolution per day).
For calculus-based physics, instantaneous angular velocity at a specific moment is ω=lim⁡Δt→0ΔθΔt=dθdt\omega = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt}, but AP Physics 1 mainly works with average angular velocity.

Angular Acceleration

Angular acceleration (α\alpha) is the rate at which angular velocity changes over time.
α=ΔωΔt\alpha = \frac{\Delta \omega}{\Delta t}
with units of radians per second squared (rad/s²). It tells you how quickly the spin speeds up or slows down. A ceiling fan starting up has positive angular acceleration (speeding up), a braking bicycle wheel has negative angular acceleration (slowing down), and a constant-speed turntable has zero angular acceleration (steady rotation).
By convention, positive α\alpha means increasing angular velocity in the positive direction, and negative α\alpha means decreasing angular velocity.

Linear and Rotational Quantities Side by Side

Rotational kinematics directly parallels linear kinematics, every linear quantity has a rotational counterpart.
Linear Quantity
Symbol
Rotational Analog
Symbol
Position
xx
Angular position
θ\theta
Displacement
Δx\Delta x
Angular displacement
Δθ\Delta \theta
Velocity
vv
Angular velocity
ω\omega
Acceleration
aa
Angular acceleration
α\alpha
Time
tt
Time
tt
If you understand linear motion, you already understand rotational motion, just by substitution.

Rotational Kinematic Equations (Constant Angular Acceleration)

For constant angular acceleration, we have equations directly analogous to the linear CAKE equations. Recall the linear versions:
v=v0+atΔx=v0t+12at2v2=v02+2aΔxv = v_0 + at \qquad \Delta x = v_0 t + \frac{1}{2}at^2 \qquad v^2 = v_0^2 + 2a\Delta x
Substituting angular quantities gives the rotational versions:
Final angular velocity (missing Δθ\Delta\theta):
ω=ω0+αt\omega = \omega_0 + \alpha t
Angular displacement (missing ω\omega, quadratic in time):
Δθ=ω0t+12αt2orθ=θ0+ω0t+12αt2\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \qquad \text{or} \qquad \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2
The “timeless” equation (missing tt):
ω2=ω02+2αΔθ\omega^2 = \omega_0^2 + 2\alpha \Delta \theta
5.1 - Rotational Kinematics

Problem-Solving Strategy

The approach is identical to linear kinematics: list known values with their symbols (initial angular velocity ω0\omega_0, final angular velocity ω\omega, angular acceleration α\alpha, time tt, angular displacement Δθ\Delta\theta), identify the unknown, choose the equation that contains your unknown and three knowns, solve algebraically, then check units (rad/s, rad/s²) and whether the answer makes physical sense.

Worked Example: Ceiling Fan Starting Up

A ceiling fan starts from rest and accelerates uniformly at 3.5 rad/s² for 4.0 seconds. Find the final angular velocity, and the number of revolutions it completes.
Final angular velocity: given ω0=0\omega_0 = 0, α=3.5\alpha = 3.5 rad/s², t=4.0t = 4.0 s. Since Δθ\Delta\theta is missing, use ω=ω0+αt\omega = \omega_0 + \alpha t:
ω=0+(3.5)(4.0)=14 rad/s\omega = 0 + (3.5)(4.0) = 14 \text{ rad/s}
After 4 seconds, the fan spins at 14 radians per second.
Number of revolutions: using Δθ=ω0t+12αt2\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2, and since ω0=0\omega_0 = 0:
Δθ=12αt2=12(3.5)(4.0)2=12(3.5)(16)=28 radians\Delta \theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(3.5)(4.0)^2 = \frac{1}{2}(3.5)(16) = 28 \text{ radians}
Converting to revolutions using 1 revolution=2π1 \text{ revolution} = 2\pi radians:
revolutions=282π=286.283≈4.46\text{revolutions} = \frac{28}{2\pi} = \frac{28}{6.283} \approx 4.46
The fan completes about 4.5 rotations during the 4-second startup, and it’s important to report the numerical value (4.46), not leave the answer in terms of π\pi.

Unit Conversions for Angular Measurements

Radians to revolutions: revolutions=radians2π\text{revolutions} = \frac{\text{radians}}{2\pi}. Revolutions to radians: radians=revolutions×2π\text{radians} = \text{revolutions} \times 2\pi.
Degrees to radians (if needed): radians=degrees×π180\text{radians} = \text{degrees} \times \frac{\pi}{180}. For example, 90° = π2\frac{\pi}{2} rad.
RPM to rad/s: RPM (revolutions per minute) is common in the real world but needs converting for physics calculations.
ω (rad/s)=RPM×2π60=RPM×π30\omega \text{ (rad/s)} = \text{RPM} \times \frac{2\pi}{60} = \text{RPM} \times \frac{\pi}{30}
For example, an engine at 3000 RPM: ω=3000×π30=100π≈314\omega = 3000 \times \frac{\pi}{30} = 100\pi \approx 314 rad/s.

Special Case: Starting from Rest

Many problems have ω0=0\omega_0 = 0, which simplifies the equations to ω=αt\omega = \alpha t, Δθ=12αt2\Delta \theta = \frac{1}{2}\alpha t^2, and ω2=2αΔθ\omega^2 = 2\alpha \Delta \theta, cutting down on algebra.

Sign Convention

Rotation directions are described as clockwise or counterclockwise about a given axis, as seen from one viewpoint. One direction is chosen as mathematically positive and the other becomes negative. The standard choice is counterclockwise positive and clockwise negative. If nearly everything in a problem turns clockwise, you can call clockwise positive instead, as long as every angular quantity in that problem uses the same choice.
A positive angular velocity means spinning counterclockwise viewed from above, and negative means clockwise. Positive angular acceleration means speeding up counterclockwise, or slowing down clockwise, while negative angular acceleration means slowing down counterclockwise, or speeding up clockwise.

Graphs of Rotational Motion

Graphs of angular position, angular velocity, and angular acceleration versus time follow the same rules as the xx, vv, and aa graphs from Unit 1, with the angular variables swapped in.
Angular position vs. time (θ\theta vs. tt): the slope at any moment is the angular velocity ω\omega. A straight line means constant ω\omega, a curve that gets steeper means the spin is speeding up, a flat section means the system isn’t turning, and a downward slope means it’s turning in the negative (clockwise) direction.
Angular velocity vs. time (ω\omega vs. tt): the slope is the angular acceleration α\alpha, and the area between the graph and the time axis is the angular displacement Δθ\Delta\theta. Area above the axis counts as positive (counterclockwise) and area below counts as negative (clockwise).
Angular acceleration vs. time (α\alpha vs. tt): the area between the graph and the time axis is the change in angular velocity Δω\Delta\omega.

Worked Example: Washing Machine Drum

A washing machine drum turns one way and then the other to tumble the clothes. Its angular velocity for the first 9.0 s is graphed below, with counterclockwise as positive. Find (a) the angular acceleration in each stage, (b) when the drum reverses direction and its angular acceleration at that moment, (c) its angular displacement from 0 to 9.0 s, and (d) the total angle it turns through, in revolutions.
5.1 - Rotational Kinematics
(a) Angular acceleration: α\alpha is the slope of each straight segment.
α1=ΔωΔt=6.0 rad/s−02.0 s=3.0 rad/s2(0 to 2.0 s)\alpha_1 = \frac{\Delta\omega}{\Delta t} = \frac{6.0\text{ rad/s} - 0}{2.0\text{ s}} = 3.0\text{ rad/s}^2 \quad (0 \text{ to } 2.0\text{ s})α3=−6.0 rad/s−6.0 rad/s9.0 s−5.0 s=−3.0 rad/s2(5.0 to 9.0 s)\alpha_3 = \frac{-6.0\text{ rad/s} - 6.0\text{ rad/s}}{9.0\text{ s} - 5.0\text{ s}} = -3.0\text{ rad/s}^2 \quad (5.0 \text{ to } 9.0\text{ s})
From 2.0 s to 5.0 s the graph is flat, so α2=0\alpha_2 = 0 and the drum spins at a steady 6.0 rad/s.
(b) Reversing: the drum reverses when ω\omega changes sign, where the graph crosses the time axis at t=7.0t = 7.0 s. At that instant ω=0\omega = 0, but α\alpha is still −3.0-3.0 rad/s², not zero, because the slope of the line hasn’t changed. From 5.0 s to 7.0 s, ω\omega is positive and α\alpha is negative (opposite signs), so the drum slows down while turning counterclockwise. From 7.0 s to 9.0 s, both are negative (same signs), so it speeds up turning clockwise.
(c) Angular displacement: add the signed areas. The area above the axis (a triangle, a rectangle, and a triangle) is positive, and the triangle below the axis is negative:
Δθ=12(2.0)(6.0)+(3.0)(6.0)+12(2.0)(6.0)−12(2.0)(6.0)=6.0+18+6.0−6.0=+24 rad\Delta\theta = \tfrac{1}{2}(2.0)(6.0) + (3.0)(6.0) + \tfrac{1}{2}(2.0)(6.0) - \tfrac{1}{2}(2.0)(6.0) = 6.0 + 18 + 6.0 - 6.0 = +24\text{ rad}
The drum ends up 24 rad (about 3.8 revolutions) counterclockwise from where it started. Its average angular velocity over the 9.0 s is ωavg=ΔθΔt=24 rad9.0 s=2.7\omega_{\text{avg}} = \frac{\Delta\theta}{\Delta t} = \frac{24\text{ rad}}{9.0\text{ s}} = 2.7 rad/s.
(d) Total angle turned: count every area as positive, the way distance works in Unit 1:
6.0+18+6.0+6.0=36 rad=36 rad2π rad/rev≈5.7 revolutions6.0 + 18 + 6.0 + 6.0 = 36\text{ rad} = \frac{36\text{ rad}}{2\pi\text{ rad/rev}} \approx 5.7\text{ revolutions}
Check with a kinematic equation: from 5.0 s to 9.0 s the angular acceleration is constant, so Δθ=ω0t+12αt2=(6.0)(4.0)+12(−3.0)(4.0)2=24−24=0\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = (6.0)(4.0) + \frac{1}{2}(-3.0)(4.0)^2 = 24 - 24 = 0, which matches the +6.0+6.0 rad and −6.0-6.0 rad triangles canceling.
Here is the same motion on all three graphs. The θ\theta vs. tt graph curves upward while the drum speeds up, is a straight line while ω\omega is constant, peaks at t=7.0t = 7.0 s where ω=0\omega = 0, and then falls as the drum turns clockwise. The α\alpha vs. tt graph is a set of flat steps, and its areas (+6.0+6.0 rad/s from 0 to 2.0 s and −12-12 rad/s from 5.0 to 9.0 s) add up to −6.0-6.0 rad/s, the drum’s total change in angular velocity (from 0 to −6.0-6.0 rad/s). Spread over 9.0 s, that’s an average angular acceleration of αavg=ΔωΔt=−6.0 rad/s9.0 s=−0.67\alpha_{\text{avg}} = \frac{\Delta\omega}{\Delta t} = \frac{-6.0\text{ rad/s}}{9.0\text{ s}} = -0.67 rad/s², even though α\alpha was never actually −0.67-0.67 rad/s² at any instant.
5.1 - Rotational Kinematics

Real-World Applications

Automotive: wheels have positive angular acceleration while a car speeds up, spin at roughly 90 rad/s at highway speed (about 30 m/s on a tire of radius 0.33 m, using v=rωv = r\omega from 5.2), and anti-lock braking systems produce rapid angular deceleration with pulsing.
Machinery: motors and engines are rated in RPM, startup involves angular acceleration, and governors regulate angular velocity.
Sports: figure skating spins speed up when arms are pulled in (conservation of angular momentum, covered later), with a typical spin running 3 to 6 revolutions per second, or about 19 to 38 rad/s.
Astronomy: Earth’s rotation is ω=2π24 hours≈7.3×10−5\omega = \frac{2\pi}{24 \text{ hours}} \approx 7.3 \times 10^{-5} rad/s, essentially constant, with no meaningful angular acceleration.

Common Mistakes

Using degrees instead of radians: don’t plug Δθ=90°\Delta\theta = 90° directly into the rotational kinematic equations, convert to radians first, Δθ=π2\Delta\theta = \frac{\pi}{2} rad.
Wrong calculator mode: don’t switch your calculator to radian mode for these problems, keep it in whatever you normally use, radians here are a unit, not a trig function setting.
Forgetting to convert revolutions: don’t leave your answer in radians if the problem asks for revolutions, divide by 2π2\pi to convert.
Confusing linear and angular equations: don’t use v=v0+atv = v_0 + at for rotational motion, use the angular version, ω=ω0+αt\omega = \omega_0 + \alpha t.

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