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Split the Acceleration into a Radial and a Tangential Part

Circular motion has an angular description (θ, ω, α) and a linear one (arc, speed, acceleration). The two are linked by v = ω R, and the acceleration always splits into a part toward the centre and a part along the path.

NSEJS · Circular Motion · 9 linked questions · 100% coverage

Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction

Study deeper: NCERT Science · Circular Motion

Angular and linear quantities

ω = dθ/dt = 2π/T, so the second hand of a watch (T = 60 s) has ω = π/30 rad/s

If θ = 2t³ + 0.5, then ω = 6t² = 24 rad/s at t = 2 s

The linear velocity

v = ω × r as a vector: for ω = (3, -4, 1) and r = (5, -6, 6), v = (-18, -13, 2)

The velocity is perpendicular to both ω and r, ω is perpendicular to the plane of motion, and for motion in a plane ω and α are parallel (or antiparallel) along the axis, never perpendicular. For a particle moving at constant ω, the distance travelled is R ω t (the arc), but the magnitude of the displacement is the chord 2R sin(ωt/2).

Centripetal and tangential acceleration

The centripetal acceleration

a_c = v²/R = ω² R toward the centre, and the tangential acceleration aₜ = dv/dt is along the path

They are perpendicular, so

a = sqrt(a_c² + aₜ²)

A car slowing at a m/s² on a circle of radius R has

a = sqrt((v²/R)² + a²)

If the total acceleration is 7.5 m/s² at 30° to the radius on

R = 5 m, then v²/R = 7.5 cos 30°, so v = sqrt(7.5 x 5 x 0.866) = 5.7 m/s

With constant speed there is no tangential acceleration, so only a_c acts.

Using energy with tangential acceleration

A particle of mass 10 g on a circle of radius 6.4 cm that reaches kinetic energy 8 x 10^-4 J after two revolutions has gone

s = 4\pi R = 0.80 m

The work done by the tangential force is m aₜ

s = K, so aₜ = K / (m s) = 8 x 10⁻⁴ / (0.01 x 0.804) = 0.1 m/s², the constant tangential acceleration

OlympIQ exam lens

✓ω = dθ/dt = 2π/T; v = ω R; as vectors v = ω × r
✓a_c = v²/R = ω² R toward the centre; aₜ = dv/dt along the path; a = sqrt(a_c² + aₜ²)
✓Chord (displacement) = 2R sin(ωt/2); arc = R ω t
✓For planar circular motion ω and α are parallel to the axis; v is perpendicular to ω and to a_c
✓For constant tangential acceleration, use the work-energy theorem with the distance 2πR x revolutions

OlympIQ trap

Do not use the arc length as the displacement. In time t a particle on a circle travels an arc R ω t, but its displacement is the shorter straight-line chord 2R sin(ωt/2).

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