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🍏Principles of Physics I Unit 9 Review

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9.1 Rotational Motion and Angular Quantities

9.1 Rotational Motion and Angular Quantities

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
🍏Principles of Physics I
Unit & Topic Study Guides

Angular Quantities and Rotational Motion

Rotational motion describes objects spinning around an axis. The quantities used to describe it mirror linear motion: instead of position, velocity, and acceleration along a line, you track angle, angular velocity, and angular acceleration around a circle. Mastering the parallels between linear and angular quantities makes this topic much more manageable.

Angular vs. Linear Quantities

Every linear quantity you already know has a rotational counterpart. The table below lays out the parallels:

Linear QuantitySymbol / UnitsAngular QuantitySymbol / Units
Displacementxx (m)Angular displacementθ\theta (rad)
Velocityvv (m/s)Angular velocityω\omega (rad/s)
Accelerationaa (m/s²)Angular accelerationα\alpha (rad/s²)

Angular displacement (θ\theta) measures how far an object has rotated, in radians. One full revolution equals 2π2\pi radians (360°).

Angular velocity (ω\omega) is how fast the angle is changing:

ωavg=ΔθΔt\omega_{avg} = \frac{\Delta \theta}{\Delta t} and instantaneously ω=dθdt\omega = \frac{d\theta}{dt}

A fan blade spinning at a steady rate has constant ω\omega. The sign of ω\omega tells you the direction of rotation (counterclockwise is typically positive).

Angular acceleration (α\alpha) is how fast the angular velocity is changing:

αavg=ΔωΔt\alpha_{avg} = \frac{\Delta \omega}{\Delta t} and instantaneously α=dωdt\alpha = \frac{d\omega}{dt}

A merry-go-round speeding up from rest has a positive α\alpha; one slowing to a stop has a negative α\alpha.

Relationships in Circular Motion

The bridge between linear and angular quantities depends on the radius rr (the distance from the rotation axis to the point of interest):

  • Arc length: s=rθs = r\theta — A point on the rim of a Ferris wheel with radius 10 m that rotates through π\pi rad travels s=10π31.4s = 10 \cdot \pi \approx 31.4 m along its circular path.
  • Tangential velocity: v=rωv = r\omega — Two seats on the same merry-go-round spin with the same ω\omega, but the outer seat has a larger vv because its rr is bigger.
  • Tangential acceleration: at=rαa_t = r\alpha — This is the component of linear acceleration along the direction of motion (tangent to the circle). It exists only when α0\alpha \neq 0.

A few useful conversions:

  • 1 revolution=360°=2π rad1 \text{ revolution} = 360° = 2\pi \text{ rad}
  • ω=2πf\omega = 2\pi f, where ff is frequency in Hz (revolutions per second)
Angular vs linear quantities, 10.1 Angular Acceleration – College Physics

Rotational Kinematics Equations

These are the same kinematic equations from linear motion, with angular variables swapped in. They apply whenever angular acceleration α\alpha is constant.

EquationWhat it's useful for
ω=ω0+αt\omega = \omega_0 + \alpha tFinding final angular velocity when you know time
θ=θ0+ω0t+12αt2\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2Finding angular displacement when you know time
ω2=ω02+2α(θθ0)\omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0)Relating angular velocity and displacement (no time)
θ=θ0+12(ω+ω0)t\theta = \theta_0 + \frac{1}{2}(\omega + \omega_0)tUsing average angular velocity

The substitution pattern is straightforward: replace xx with θ\theta, vv with ω\omega, and aa with α\alpha.

Solving rotational kinematics problems:

  1. List your knowns (typically three of ω0\omega_0, ω\omega, α\alpha, θθ0\theta - \theta_0, tt).

  2. Identify the unknown you need.

  3. Pick the equation that contains your three knowns and the one unknown.

  4. Solve algebraically, then substitute values. Keep angles in radians throughout.

Centripetal Acceleration

Centripetal Acceleration Concept

Any object moving in a circle is constantly changing direction, which means its velocity vector is changing even if its speed stays the same. That change in velocity requires an acceleration directed toward the center of the circle. This is centripetal acceleration.

ac=v2r=rω2a_c = \frac{v^2}{r} = r\omega^2

A few things to note:

  • Centripetal acceleration doesn't change the object's speed; it only redirects the velocity vector inward.
  • In uniform circular motion (constant speed), aca_c is the only acceleration present. If the object is also speeding up or slowing down, there's a tangential component ata_t as well, and the total acceleration is the vector sum of aca_c and ata_t.

Centripetal force is just Newton's second law applied to circular motion:

Fc=mac=mv2r=mrω2F_c = ma_c = m\frac{v^2}{r} = mr\omega^2

This isn't a new type of force. Something real must supply it: tension in a string, gravity for an orbiting planet, friction on a car rounding a curve, or the normal force on a banked road. When solving problems, always identify what is providing the centripetal force rather than just labeling it FcF_c.