Unit 5.1 | Rotational Kinematics
- 17 minutes
- 1 formulas
- 28 questions
- 2 videos
Every Unit 1 equation has a rotational twin. This lesson shows how to describe spin, link it to the speed of a point, and solve the four rotational kinematics problem types.
Unit 5 Breakdown
In this lesson:
- Describe rotation with angle, angular velocity and angular acceleration
- Use the rotational kinematic equations, the same ones you already know from Unit 1
- Connect rotation to the motion of a point: [katex] s = r\theta [/katex], [katex] v = r\omega [/katex] and [katex] a_t = r\alpha [/katex]
- The 4 problem types: spin-ups and slow-downs, points at different radii, rolling and unwinding, and angular motion graphs
Introduction
This lesson covers rotational kinematics: how to describe spin before asking what causes it. It starts Unit 5, torque and rotation, and ends with the four rotational kinematics problem types on the AP exam.
Good news: you already know the math. Every rotational quantity has a linear twin, and every kinematic equation from Unit 1 has a rotational version. Swap the letters and the problems solve the same way.
Angle, Angular Velocity and Angular Acceleration
- Angular position [katex] \theta [/katex]: how far the object has turned, in radians. One full turn is [katex] 2\pi [/katex] rad.
- Angular velocity [katex] \omega = \frac{\Delta\theta}{\Delta t} [/katex]: how fast it turns, in rad/s.
- Angular acceleration [katex] \alpha = \frac{\Delta\omega}{\Delta t} [/katex]: how fast [katex] \omega [/katex] changes, in rad/s².
Every point on a rigid object turns through the same angle in the same time. So the whole object has one [katex] \theta [/katex], one [katex] \omega [/katex] and one [katex] \alpha [/katex]. That is what makes rotation simpler than it looks.
Rotation has a direction
Like velocity, angular velocity has a sign. Counterclockwise is positive and clockwise is negative, unless the problem says otherwise. Speeding up means [katex] \omega [/katex] and [katex] \alpha [/katex] have the same sign. Slowing down means opposite signs.
Convert first. [katex] 1 \text{ rev} = 2\pi \text{ rad} [/katex], and rpm means revolutions per minute, so [katex] 1 \text{ rpm} = \frac{2\pi}{60} \text{ rad/s} [/katex]. Many wrong answers in this unit come from skipping this step.
The Rotational Kinematic Equations
When the angular acceleration is constant, the Unit 1 equations carry straight over. Replace [katex] x [/katex] with [katex] \theta [/katex], [katex] v [/katex] with [katex] \omega [/katex] and [katex] a [/katex] with [katex] \alpha [/katex]:
| Linear | Rotational |
| [katex] v = v_0 + at [/katex] | [katex] \omega = \omega_0 + \alpha t [/katex] |
| [katex] \Delta x = v_0 t + \tfrac{1}{2}at^2 [/katex] | [katex] \Delta\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 [/katex] |
| [katex] v^2 = v_0^2 + 2a\Delta x [/katex] | [katex] \omega^2 = \omega_0^2 + 2\alpha\Delta\theta [/katex] |
| [katex] \Delta x = \tfrac{v_0 + v}{2}\,t [/katex] | [katex] \Delta\theta = \tfrac{\omega_0 + \omega}{2}\,t [/katex] |
The strategy is the same as Unit 1: list what you know, then pick the one equation that holds your unknown and skips the variable you do not have.
Linking Rotation to a Point on the Object
The whole object shares [katex] \omega [/katex], but its points do not share a speed. A point far from the axis travels a bigger circle in the same time, so it moves faster. Here is where the link comes from:
- Arc length is radius times angle, with the angle in radians: [katex] s = r\theta [/katex].
- Divide both sides by time: [katex] \frac{\Delta s}{\Delta t} = r\frac{\Delta\theta}{\Delta t} [/katex], so [katex] v = r\omega [/katex].
- Do the same with the change in speed: [katex] \frac{\Delta v}{\Delta t} = r\frac{\Delta\omega}{\Delta t} [/katex], so [katex] a_t = r\alpha [/katex].
[katex] v = r\omega \qquad a_t = r\alpha [/katex]
Here [katex] a_t [/katex] is the tangential acceleration, the part that changes the point’s speed. The point also has the centripetal acceleration from circular motion, [katex] a_c = \frac{v^2}{r} = \omega^2 r [/katex], which points to the center and only changes the direction.
These links only work with angles in radians. Degrees and revolutions break them.
Every rotational kinematics question on the AP exam is one of the four types below.
| Problem type | The clue | The setup |
| 1. Spin-ups and slow-downs | “rpm”, “revolutions”, “comes to rest” | Convert to rad, then a rotational kinematic equation |
| 2. Points at different radii | two points on one disk, “on the rim”, “halfway out” | Same [katex] \omega [/katex] for both, then [katex] v = r\omega [/katex], [katex] a_c = \omega^2 r [/katex] |
| 3. Rolling, unwinding and belts | a wheel rolls, a rope unwinds, a belt joins two wheels | The rim and what it touches share one speed: [katex] v = r\omega [/katex] |
| 4. Angular motion graphs | a graph of [katex] \theta [/katex], [katex] \omega [/katex] or [katex] \alpha [/katex] against [katex] t [/katex] | Slope and area, exactly like Unit 1 graphs |
Problem Type 1: Spin-ups and slow-downs
How to spot it: Turning rates in rpm or rev/s, “how many revolutions”, or something that “comes to rest” with a constant angular acceleration.
Setup: Convert to radians and seconds. List [katex] \omega_0 [/katex], [katex] \omega [/katex], [katex] \alpha [/katex], [katex] t [/katex] and [katex] \Delta\theta [/katex], then pick the equation that skips the one you do not know.
A ceiling fan spinning at [katex] 180\,\text{rpm} [/katex] is switched off and coasts to a stop in [katex] 12\,\text{s} [/katex] with a constant angular acceleration. Find its angular acceleration and how many revolutions it makes while stopping.
Convert: [katex] \omega_0 = 180 \cdot \frac{2\pi}{60} = 6\pi \approx 18.8 \text{ rad/s} [/katex]. Then [katex] \alpha = \frac{0 – 6\pi}{12} = -\frac{\pi}{2} \approx -1.57 \text{ rad/s}^2 [/katex]. The angle is [katex] \Delta\theta = \frac{\omega_0 + \omega}{2}t = \frac{6\pi}{2}(12) = 36\pi \text{ rad} [/katex], and [katex] \frac{36\pi}{2\pi} = 18 [/katex] revolutions.
AP trap: Plugging rpm or revolutions straight into the equations. Convert to rad/s and rad first. Divide by [katex] 2\pi [/katex] at the end only if the question asks for revolutions.
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Problem Type 2: Points at different distances from the axis
How to spot it: Two points on the same rotating object, or a point “on the rim”, “halfway out”, “at radius r”.
Setup: Every point shares [katex] \omega [/katex] and [katex] \alpha [/katex]. Only the linear quantities scale with radius: [katex] v = r\omega [/katex], [katex] a_t = r\alpha [/katex], [katex] a_c = \omega^2 r [/katex].
A merry-go-round turns at [katex] 1.5\,\text{rad/s} [/katex]. Child A sits [katex] 1.0\,\text{m} [/katex] from the center and child B sits [katex] 2.0\,\text{m} [/katex] from the center. Compare their angular speeds, linear speeds and centripetal accelerations.
Both turn at [katex] 1.5\,\text{rad/s} [/katex]. [katex] v_A = (1.0)(1.5) = 1.5 \text{ m/s} [/katex] and [katex] v_B = (2.0)(1.5) = 3.0 \text{ m/s} [/katex]. Centripetal: [katex] a_{c,A} = (1.5)^2(1.0) = 2.25 \text{ m/s}^2 [/katex] and [katex] a_{c,B} = 4.5 \text{ m/s}^2 [/katex]. Twice the radius gives twice the speed and twice the centripetal acceleration.
If the object is speeding up, a point on it has both accelerations. [katex] a_t = r\alpha [/katex] stays constant when [katex] \alpha [/katex] is constant, but [katex] a_c = \omega^2 r [/katex] grows as [katex] \omega [/katex] grows.
AP trap: Saying the outer point has a bigger angular velocity. It covers more distance, but the same angle, in the same time.
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Problem Type 3: Rolling, unwinding and belts
How to spot it: A wheel rolls without slipping, a rope or cable winds onto a spool, or a belt or chain connects two wheels.
Setup: Whatever touches the rim without slipping moves with the rim: [katex] \Delta x = r\Delta\theta [/katex], [katex] v = r\omega [/katex], [katex] a = r\alpha [/katex]. Two wheels joined by a belt share the rim speed, not [katex] \omega [/katex].
A bucket hangs from a rope wound on a drum of radius [katex] 0.10\,\text{m} [/katex]. The drum turns at [katex] 4.0\,\text{rad/s} [/katex]. How fast does the bucket rise, and how far does it rise in [katex] 5.0\,\text{s} [/katex]?
[katex] v = r\omega = (0.10)(4.0) = 0.40 \text{ m/s} [/katex]. In [katex] 5.0\,\text{s} [/katex] the drum turns [katex] \Delta\theta = (4.0)(5.0) = 20 \text{ rad} [/katex], so the rope winds up [katex] \Delta x = r\Delta\theta = (0.10)(20) = 2.0 \text{ m} [/katex].
AP trap: With two wheels on one belt, giving them the same [katex] \omega [/katex]. The belt speed is the same at both rims, so [katex] \omega_1 r_1 = \omega_2 r_2 [/katex] and the small wheel spins faster.
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Problem Type 4: Angular motion graphs
How to spot it: A graph of [katex] \theta [/katex], [katex] \omega [/katex] or [katex] \alpha [/katex] against time.
Setup: Read it like a Unit 1 graph. Slope of [katex] \theta [/katex] vs. [katex] t [/katex] is [katex] \omega [/katex]. Slope of [katex] \omega [/katex] vs. [katex] t [/katex] is [katex] \alpha [/katex]. Area under [katex] \omega [/katex] vs. [katex] t [/katex] is [katex] \Delta\theta [/katex]. Area under [katex] \alpha [/katex] vs. [katex] t [/katex] is [katex] \Delta\omega [/katex].
A wheel speeds up, cruises, then stops
The graph above shows a wheel’s angular velocity. Find its angular acceleration from 0 to [katex] 4\,\text{s} [/katex], and the total angle it turns in [katex] 10\,\text{s} [/katex].
Slope: [katex] \alpha = \frac{12 – 0}{4} = 3 \text{ rad/s}^2 [/katex]. Area: a triangle [katex] \tfrac{1}{2}(4)(12) = 24 [/katex], a rectangle [katex] (3)(12) = 36 [/katex] and a triangle [katex] \tfrac{1}{2}(3)(12) = 18 [/katex], so [katex] \Delta\theta = 78 \text{ rad} [/katex], about 12.4 revolutions.
AP trap: Reading the height of the graph instead of its slope. On an [katex] \omega [/katex] vs. [katex] t [/katex] graph, a high flat line means fast and steady, so [katex] \alpha = 0 [/katex].
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Lesson 5.2 Preview
In the next lesson we ask what changes rotation. In Unit 2 a net force changes linear motion. For rotation, that job belongs to torque.
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This article was written by Jason