Rotational Kinematics
- Define angular velocity and acceleration as derivatives ω = dθ/dt and α = dω/dt
- Apply the constant-angular-acceleration equations
- Relate linear and angular quantities with v = rω and a_t = rα
Angles have their own kinematics
Rotational motion mirrors translation exactly, with angular position θ replacing x. The angular velocity is the derivative ω = dθ/dt, and the angular acceleration is α = dω/dt = d²θ/dt². Everything you learned for x, v, a carries over — you differentiate to go down the chain and integrate to go back up, just with angular variables measured in radians.
Linking angular to linear
A point at radius r on a rotating body traces an arc, so its linear speed is v = rω and its tangential acceleration is a_t = rα. It also has a centripetal acceleration a_c = ω²r = v²/r directed inward, present whenever it moves in a circle. When α is constant, the rotational kinematic equations ω = ω₀ + αt and θ = θ₀ + ω₀t + ½αt² hold — the exact analogs of the linear ones.
A wheel rotates so that its angular position is θ(t) = 2t³ − t (radians, seconds). Find ω(t) and α(t), and evaluate both at t = 1 s.
- 1.Differentiate θ to get angular velocity: ω(t) = dθ/dt = 6t² − 1.
- 2.Differentiate again for angular acceleration: α(t) = dω/dt = 12t.
- 3.At t = 1: ω(1) = 6(1) − 1 = 5 rad/s.
- 4.At t = 1: α(1) = 12(1) = 12 rad/s².
A wheel rotates with angular position θ(t) = 2t³ radians. What is its angular velocity at t = 2 s?
A point on the rim of a wheel of radius 0.5 m moves as the wheel spins at ω = 8 rad/s. What is the point's linear (tangential) speed?
Set up a two-column analogy sheet: x↔θ, v↔ω, a↔α, m↔I, F↔τ, p↔L. Every translational relationship has a rotational twin, so once you know one side you can write the other by swapping symbols.
Answer the 2 checkpoints as you read.
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