University Physics I · 2D & 3D Kinematics · 3.5
Tangential and Radial Acceleration
On a curved path, acceleration splits into a tangential part that changes speed and a radial part that changes direction. The parts are perpendicular, so neither can do the other's job.
Build the model
Connect the measurement to the mechanism.
Write velocity as a speed times a unit vector along the path, v⃗ = v T̂. Differentiating with the product rule gives two terms and only two: one from the speed changing, one from the direction changing. The first, aₜ = dv/dt, lies along the path and is the only thing that changes speed.
The second, aᵣ = v²/ρ, points perpendicular to the path toward the centre of curvature and is the only thing that turns the velocity. Because the two are perpendicular they add by Pythagoras, |a⃗| = √(aₜ² + aᵣ²), and the total leans off the inward radius by an angle set by their ratio — forward of the radius when speeding up, behind it when slowing. Uniform circular motion is the case aₜ = 0 with ρ fixed; straight-line motion is the case aᵣ = 0.
- Simple definition
- Tangential acceleration is the part of the acceleration along the path, which changes speed; radial acceleration is the perpendicular part pointing toward the centre of curvature, which changes direction.
- Example
- A car accelerating out of a roundabout carries both parts at once: it gains speed along the road while the road is still turning its velocity toward the centre of the curve.
Any acceleration on a curve is exactly two pieces: one changes the speed, the other turns the velocity.
T̂ tangent to the path, N̂ inward normal
Rate of change of speed: positive when speeding up, negative when braking, zero at steady speed.
m s⁻²; signed along the motion
The turning part: it grows with speed squared and shrinks as the curve straightens out.
m s⁻²; toward the centre of curvature, with ω = v/ρ
The parts never cancel, so the total is always at least as big as the larger of them.
Perpendicular parts add by Pythagoras
How far the acceleration arrow leans off the radius: forward when speeding up, back when slowing.
φ measured from the inward radius
On a circle the curvature is fixed, so angular acceleration supplies the whole tangential part.
ρ = r; α in rad s⁻²
Two terms, and only two
Write the velocity as a speed times a unit vector along the path: v⃗ = v T̂. The product rule gives a⃗ = (dv/dt) T̂ + v (dT̂/dt). The first term is the speed changing with the direction held fixed. The second is the direction changing with the speed held fixed. Because T̂ has fixed length, dT̂/dt is perpendicular to it, and dT̂/ds = N̂/ρ defines the radius of curvature ρ, so dT̂/dt = (v/ρ) N̂ with N̂ the unit normal pointing toward the centre of the curve. That gives a⃗ = (dv/dt) T̂ + (v²/ρ) N̂. There is no third term: a particle on a path can speed up, turn, or do both.
Only the parallel part changes speed
Speed is the length of the velocity, so differentiate v² = v⃗ · v⃗ to get 2v (dv/dt) = 2 v⃗ · a⃗, which rearranges to dv/dt = a⃗ · v̂. Any part of the acceleration perpendicular to the velocity drops out of that dot product and cannot change the speed, however large it is. That is why uniform circular motion holds its speed: the acceleration is entirely perpendicular to the velocity. Read backwards, an object whose speed is momentarily steady has no tangential part at that instant, so whatever acceleration it does have is purely radial.
The radius of curvature is local
ρ is the radius of the circle that best fits the path where the object is now, not the size of the whole trajectory, and radial here means toward that local centre, not away from some fixed origin. On a circular track ρ is the track radius; on a parabola it changes from point to point. It is often easier to get it from the motion than from geometry, since ρ = v²/aᵣ. Take a ball launched at 25 m s⁻¹ at 60° above the horizontal. At the top of the flight the velocity is horizontal at 25 cos 60° = 12.5 m s⁻¹ and gravity is entirely perpendicular to it, so aᵣ = 9.81 m s⁻² and ρ = 12.5²/9.81 = 16 m. The path is locally a 16 m circle even though nothing is going round in circles.
Adding the parts
Because the parts are perpendicular, combine them with Pythagoras and take the direction from their ratio. A car takes a bend of radius 50 m at 15 m s⁻¹ while braking at 2.0 m s⁻². The radial part is aᵣ = 15²/50 = 4.5 m s⁻² inward, the tangential part is 2.0 m s⁻² backwards along the road, and the total is √(4.5² + 2.0²) = 4.9 m s⁻². It still points into the bend, but rotated back from the radius by arctan(2.0/4.5) = 24°, leaning against the travel direction because the car is slowing. Accelerating at the same rate would give the same magnitude with the 24° lean forwards.
Reading the tilt
The acceleration arrow on a curve always points to the concave side, because aᵣ = v²/ρ is never negative. The lean off the radius carries the rest of the story: forward of the radius means speeding up, behind it means slowing, and exactly along the radius means dv/dt = 0, so the speed is momentarily stationary — either steady, as in uniform circular motion, or at a turning point. On a straight run ρ is infinite, aᵣ vanishes, and the arrow lies along the line of travel. Sketching v⃗ and a⃗ at a few points therefore shows where the speed is rising and where it is falling before any number is computed.
Two familiar limits, and the dynamics
Uniform circular motion is this decomposition with aₜ = 0 and ρ fixed, and straight-line kinematics is the same decomposition with aᵣ = 0. Non-uniform circular motion needs both, and Newton's second law splits the same way into ΣFᵣ = mv²/r toward the centre and ΣFₜ = maₜ along the motion. The parts also grow at different rates: a car pulling away from rest around a 50 m circle at a steady aₜ = 2.0 m s⁻² has v² = 2aₜs, so the radial part climbs in step with the distance covered, and with t², while the tangential part holds still. The two match when v = √(raₜ) = √(50 × 2.0) = 10 m s⁻¹, five seconds in. After that, turning takes the larger share of the grip.
Change one variable at a time
Make the relationship visible.
Set aₜ to zero and the total arrow drops onto the radius — uniform circular motion; drag it negative and the total leans back against the travel direction while the radial arrow never moves.
RADIAL v²/ρ2.25 m s⁻²
TANGENTIAL dv/dt2.0 m s⁻²
TOTAL |a|3.01 m s⁻²
TILT OFF RADIUS42 °
Live interpretationRADIAL v²/ρ: 2.25 m s⁻². TANGENTIAL dv/dt: 2.0 m s⁻². TOTAL |a|: 3.01 m s⁻². TILT OFF RADIUS: 42 °
Catch the common trap
Explain before calculating.
A motorcycle rounds a bend of radius 100 m at 20 m s⁻¹. At that instant it is speeding up at 3.0 m s⁻². What is the magnitude of its total acceleration?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA cyclist rides round a circular bend of radius 25.0 m at a steady 8.0 m s⁻¹. Find the tangential and radial parts of the acceleration, and the magnitude of the total.
- The speed is steady, so the tangential part is aₜ = dv/dt = 0 — nothing along the path is changing the speed.
- The bend still turns the velocity: aᵣ = v²/ρ = (8.0 m s⁻¹)² / 25.0 m = 64/25.0 = 2.56 m s⁻², pointing at the centre.
- Perpendicular parts add by Pythagoras: |a⃗| = √(0² + 2.56²) = 2.56 m s⁻², so the whole acceleration lies along the inward radius — steady speed is not zero acceleration.
Answeraₜ = 0; aᵣ = |a⃗| = 2.56 m s⁻², directed toward the centre of the bend.
MediumA car rounds a bend of radius 120 m while braking at 3.5 m s⁻². At the instant its speed is 24 m s⁻¹, find the radial and tangential parts, the magnitude of the total acceleration, and the angle the total makes with the inward radius.
- Radial part, from the speed and the local curvature: aᵣ = v²/ρ = (24 m s⁻¹)² / 120 m = 576/120 = 4.80 m s⁻², pointing at the centre of the bend.
- Tangential part is the signed rate of change of speed: braking gives aₜ = −3.5 m s⁻², i.e. 3.5 m s⁻² backwards along the road.
- The parts are perpendicular, so |a⃗| = √(4.80² + 3.50²) = √(23.04 + 12.25) = √35.29 = 5.94 m s⁻².
- Direction: tan φ = |aₜ|/aᵣ = 3.50/4.80 = 0.729, so φ = 36°, measured behind the inward radius because the car is slowing.
Answeraᵣ = 4.80 m s⁻², aₜ = −3.5 m s⁻²; |a⃗| = 5.94 m s⁻², tilted 36° behind the inward radius.
HardA ball is thrown at 20 m s⁻¹, 55° above the horizontal; ignore air resistance. At the instant its velocity points 30° above the horizontal, on the way up, find its speed, the tangential and radial parts of its acceleration, and the radius of curvature of its path there.
- Gravity is the whole acceleration, |a⃗| = g = 9.81 m s⁻² downward; only the angle between g and the velocity decides the split.
- The horizontal velocity never changes: vₓ = 20 cos 55° = 11.472 m s⁻¹, so at 30° the speed is v = vₓ/cos 30° = 11.472/0.8660 = 13.25 m s⁻¹ (carry the extra digit).
- Resolve g along the velocity: aₜ = −g sin 30° = −9.81 × 0.5000 = −4.91 m s⁻², negative because the ball is still climbing and gravity opposes the motion.
- Resolve g across the velocity: aᵣ = g cos 30° = 9.81 × 0.8660 = 8.50 m s⁻², perpendicular to the path and pointing to the concave side. Check: √(4.905² + 8.496²) = 9.81 m s⁻², so the parts rebuild g.
- The local curvature is whatever makes that radial part work: ρ = v²/aᵣ = (13.25)²/8.50 = 175.6/8.50 = 20.7 m — the path is locally a 20.7 m circle, with nothing going round in circles.
Answerv = 13.2 m s⁻¹; aₜ = −4.91 m s⁻², aᵣ = 8.50 m s⁻²; ρ = 20.7 m.