Uniform Circular Motion

On a circle of fixed radius, one number locates the particle: the angle θ from a reference axis, counted counterclockwise. Uniform motion means θ grows at a steady rate, θ(t) = θ₀ + ωt, so the angular speed ω = dθ/dt is a constant in rad s⁻¹. Radians are what make this work: arc length is s = Rθ only when θ is in radians, so the distance travelled is Rωt and the speed is ds/dt = ωR with no conversion factor. The period T is the time for θ to grow by 2π, so T = 2π/ω, and the frequency f = 1/T counts revolutions per second. A washing-machine drum at 1200 rpm makes f = 1200/60 = 20.0 revolutions per second, so T = 0.0500 s and ω = 2πf = 125.7 rad s⁻¹. Writing that 20.0 s⁻¹ down as ω is the commonest slip in the topic, and it leaves ω too small by a factor of 2π.

Put the origin at the centre and start the particle on the +x axis. Its position is r(t) = R cos(ωt) î + R sin(ωt) ĵ, and differentiating each component gives v(t) = −ωR sin(ωt) î + ωR cos(ωt) ĵ. Two facts follow at once. The magnitude is |v| = ωR √(sin²ωt + cos²ωt) = ωR, the same at every instant. And the dot product r ⋅ v = −ωR² cos(ωt) sin(ωt) + ωR² sin(ωt) cos(ωt) = 0, so the velocity is perpendicular to the radius: it lies along the tangent, pointing the way the particle is going round. At t = 0 the particle sits at (R, 0) moving in the +ĵ direction, which is counterclockwise for ω > 0. For a 0.25 m-radius drum at 125.7 rad s⁻¹ the speed is ωR = 31.4 m s⁻¹, about 113 km h⁻¹, for a point that never leaves a circle half a metre across.

Differentiating v(t) gives a(t) = −ω²R cos(ωt) î − ω²R sin(ωt) ĵ, which is exactly −ω² times r(t). So a = −ω²r: the acceleration has magnitude a_c = ω²R and points opposite r, from the particle toward the centre, at every instant. Substituting ω = v/R gives the same magnitude as v²/R. Because a ⋅ v = 0, the acceleration never has a component along the motion, which is why the speed stays fixed: d(v ⋅ v)/dt = 2a ⋅ v = 0. Notice what is and is not constant. The magnitude ω²R is fixed, but the direction rotates with the particle, so a is not a constant vector, and v = v₀ + at or any other constant-acceleration equation gives nonsense here. The drum at 125.7 rad s⁻¹ and R = 0.25 m has a_c = (125.7 rad s⁻¹)² × 0.25 m = 3.95 × 10³ m s⁻², roughly 400 times g, with its speed never changing at all.

The result can be seen without components. Redraw each velocity vector with its tail at one fixed point. Every one has length v and each is 90° ahead of r, so while the particle goes round once, the tip of v also goes round once, on a circle of radius v, at the same angular speed ω. The acceleration is the velocity of that tip, and the rule just derived for any point moving round a circle, speed = radius × ω, gives |a| = ωv = v²/R. Its direction is tangent to the velocity circle, 90° ahead of v and so 180° from r: toward the centre. The same picture shows the limit at work. Over a finite turn Δθ the chord joining the two velocity tips has length 2v sin(Δθ/2), so the average acceleration is (v²/R) × sin(Δθ/2)/(Δθ/2). A quarter turn gives 0.900 of v²/R, a 10° turn gives 0.9987 of it, and the instantaneous value is the limit as Δθ → 0.

v²/R, ω²R and 4π²R/T² are one quantity, but they seem to predict opposite trends with R until you ask what stays fixed. On a rigid rotating object every point shares ω and T, so a_c = ω²R grows in proportion to R. On a playground roundabout with T = 6.0 s, a child 1.5 m from the axis accelerates at 4π² × 1.5 m/(6.0 s)² = 1.64 m s⁻², and one 3.0 m out at 3.29 m s⁻², twice as much, because the outer child also moves twice as fast. A car holding 25 m s⁻¹ is the opposite case: v is fixed, so a_c = v²/R falls as 1/R, from 6.25 m s⁻² on a 100 m bend to 1.56 m s⁻² on a 400 m one. The period form carries the steepest dependence of all: halve T at fixed R and a_c rises fourfold. Decide which variable is fixed before choosing the formula, and the trend cannot come out backwards.

The standard form quietly assumed three conveniences, and real problems rarely grant all of them. A starting angle adds a phase, r = R cos(ωt + θ₀) î + R sin(ωt + θ₀) ĵ, which moves where the particle is at t = 0 but changes no magnitude. Clockwise motion is the same formula with ω negative: the velocity reverses, while a = −ω²r does not, because ω enters squared. And when the centre sits at c rather than at the origin, every result holds with r replaced by r − c, the vector from the centre to the particle, so a = −ω²(r − c). Take a particle going clockwise at 3.0 rad s⁻¹ round a 0.50 m circle centred at c = (2.0 î + 1.0 ĵ) m. At the top of the circle, (2.0 î + 1.5 ĵ) m, r − c = 0.50 ĵ m, so a = −(3.0 s⁻¹)² × 0.50 ĵ m = −4.5 ĵ m s⁻², straight down at the centre, while v = 1.5 î m s⁻¹, to the right, as clockwise motion across the top requires. A wrong direction almost always means one of these three was forgotten, not that the physics changed.