Angular Motion and Fixed-Axis Kinematics

Angular velocity and angular acceleration

One angular motion, different linear speeds

For rotation about a fixed axis, choose an angular coordinate θ measured in radians from a reference direction. Over a finite time, average angular velocity is Δθ/Δt. Its instantaneous value is dθ/dt. Every point of a rigid body sweeps through the same angular displacement during a given interval. A point farther from the axis follows a larger circular arc and therefore has a greater linear speed, even though the angular speed is shared.

The speed relation follows by differentiating arc length s = r⊥θ for a fixed particle radius. Treat angular velocity as a vector along the axis using the right-hand rule. Anticlockwise rotation as viewed from the positive end of the axis gives a vector toward that end. Reversing the spin reverses the vector. For a fixed axis, the signed scalar ω describes the same information more simply.

Deriving the vector relation

Take an origin O on the rotation axis. Split the particle position OP into OC, along the axis, and CP, perpendicular from the axis to the particle. The cross product ω × OC vanishes because those vectors are parallel. The remaining product ω × CP is tangential, has magnitude ωr⊥ and points along the particle’s motion. It is therefore precisely the linear velocity.

Angular acceleration and changing direction

Angular acceleration is dω/dt. For a fixed axis, its direction is along or opposite that axis according to the sign of angular-speed change. If the axis moves, angular acceleration can also reflect a changing direction of ω. A constant angular speed about a fixed axis does not imply zero linear acceleration of a rim point: its velocity direction keeps changing.

Kinematics with constant angular acceleration

One coordinate describes the fixed-axis rotation

With an axis fixed in space and body, angular position θ specifies orientation relative to a chosen reference direction. A full turn adds 2π radians. θ can accumulate through multiple revolutions even though the body eventually repeats its orientation. Choose a positive rotational sense, usually anticlockwise as viewed from the positive axis end. A negative angular velocity means the opposite sense, not a negative angular speed.

Deriving the constant-acceleration equations

Assume α is constant over the entire interval and let ω₀ and θ₀ be the values at t = 0. Integrate dω = α dt from the initial to the final state: ω − ω₀ = αt. This gives the first equation. Integrate dθ = (ω₀ + αt) dt to obtain θ − θ₀ = ω₀t + αt²/2. The initial angle must be retained when the requested quantity is final angular position.

The ω–t graph is a straight line, so its signed trapezium area gives angular displacement Δθ = (ω₀ + ω)t/2. Substitute t = (ω − ω₀)/α, for nonzero α, and use (ω + ω₀)(ω − ω₀) = ω² − ω₀². Rearrangement produces the third relation. The calculus route is α = ω dω/dθ, followed by integrating ω dω = α dθ. Both methods give the same result.

Use ω = ω₀ + αt if angle is absent; use Δθ = ω₀t + αt²/2 when final velocity is absent; use the squared equation if time is absent. A wheel slowing in the positive direction has negative α. If that acceleration continues after it stops, it reverses, so angular distance and signed angular displacement differ. Split at the stopping instant when calculating total turns travelled. A constant braking model normally ends when the wheel comes to rest.

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