A topic in the Open Knowledge Graph — a free, open map of 15,290 topics and the order to learn them in.
Fixed-axis rotation describes the motion of a rigid body that rotates about a stationary axis. Every point in the body moves in a circular arc centered on the axis, so the kinematics of any point can be expressed in terms of the angular quantities: angular position theta, angular velocity omega = d(theta)/dt, and angular acceleration alpha = d(omega)/dt. The relationships mirror rectilinear particle kinematics: alpha = d(omega)/dt, omega = d(theta)/dt, and alpha*d(theta) = omega*d(omega). For constant angular acceleration, the familiar constant-acceleration equations apply with theta, omega, and alpha replacing s, v, and a. The velocity and acceleration of any point P at radial distance r from the axis are v = omega*r (tangential), a_t = alpha*r (tangential acceleration), and a_n = omega2*r (centripetal acceleration directed toward the axis).
Draw the analogy to rectilinear kinematics explicitly: theta <-> s, omega <-> v, alpha <-> a. Solve constant angular acceleration problems using the rotational kinematic equations first, then find the linear velocity and acceleration of specific points using the r-omega and r-alpha relationships. Work problems that combine angular kinematics with gear or belt connections between rotating bodies.
In particle kinematics you described motion along a straight line using position s, velocity v = ds/dt, and acceleration a = dv/dt. Fixed-axis rotation is the direct rotational analogue: replace the linear coordinates with angular coordinates. Angular position θ (radians) locates the body, angular velocity ω = dθ/dt describes how fast it spins, and angular acceleration α = dω/dt describes how that spin rate changes. Every kinematic equation from rectilinear motion has an identical twin in rotation — just swap s → θ, v → ω, a → α. If α is constant, the constant-acceleration equations apply: ω = ω₀ + αt, θ = θ₀ + ω₀t + ½αt², and ω² = ω₀² + 2α(θ − θ₀). This one-to-one correspondence means you already know half of rotational kinematics — you just need to translate.
The connection between the rotation of the body and the motion of any specific point P on that body is where radial distance r enters. Every point traces a circular arc, so its speed is tangential: v = ωr. This is not an approximation — for a rigid body, the entire body rotates as one unit, so a point twice as far from the axis moves twice as fast. The acceleration of point P has two components. The tangential acceleration a_t = αr points along the arc direction and changes the speed. The centripetal (normal) acceleration a_n = ω²r points radially inward toward the rotation axis and arises purely from the changing direction of the velocity vector. Crucially, a_n is present whenever ω ≠ 0, even if α = 0 — a spinning body at constant speed still requires centripetal acceleration at every point on it.
When dealing with connected rotating parts — a motor shaft driving a gear, which drives a belt, which drives another shaft — you translate between components using the constraint that belt speed or contact speed must match at the interface. If gear A (radius r_A) meshes with gear B (radius r_B), then their contact speeds are equal: ω_A·r_A = ω_B·r_B. This relationship governs all gear trains, belt-pulley systems, and chain drives. Write this constraint first, use it to express all angular velocities in terms of one unknown, then apply whatever kinematics equation the problem requires.
When α is not constant, you cannot use the constant-α formulas. Instead, you must integrate. If α is given as a function of time, integrate once to get ω(t) and again for θ(t). If α is given as a function of θ, use the identity α·dθ = ω·dω (obtained by writing α = dω/dt = (dω/dθ)(dθ/dt) = ω·dω/dθ) and integrate to find ω as a function of θ directly. Recognizing which form of α you have — time-dependent or position-dependent — determines which integration strategy to apply, and choosing the wrong one is a common source of algebraic dead-ends.