Centripetal Acceleration

Constant speed, changing direction: the acceleration points at the centre.

12 minUpdated 21 October 2026centripetal accelerationradial unit vectortangential unit vectornon-uniform circular motion

Lesson at a glanceShow the key points

Velocity on a circle is always tangent, with magnitude rω.
Uniform circular motion: acceleration v²/r = ω²r directed toward the centre (centripetal acceleration).
Non-uniform motion adds a tangential component dv/dt; |a| = √[(v²/r)² + (dv/dt)²].
The resultant acceleration makes angle α with the radius where tan α = (dv/dt)/(v²/r).

A car taking a roundabout at a steady 30 km/h is accelerating the whole time. The speed never changes; the direction never stops changing.

Two unit vectors that ride along

Put the centre of the circle at the origin, and let the particle be at angular position θ. Two unit vectors are handy:

\hat e_r = \hat\imath\cos\theta + \hat\jmath\sin\theta \ \text{(outward radius)}, \qquad \hat e_t = -\hat\imath\sin\theta + \hat\jmath\cos\theta \ \text{(tangent, direction of increasing }\theta) \quad (7.6, 7.7)

The position vector is $\vec r = r(\hat\imath\cos\theta + \hat\jmath\sin\theta)$ . Differentiate with r fixed and θ changing:

\vec v = \frac{d\vec r}{dt} = r\omega(-\hat\imath\sin\theta + \hat\jmath\cos\theta) = r\omega\,\hat e_t

So the velocity is always along the tangent, with size $r\omega$ . Differentiate again (now ω may also change):

Acceleration on a circle(7.8)

\vec a = -\omega^2 r\,\hat e_r + \frac{dv}{dt}\,\hat e_t

Uniform circular motion

If the speed is constant, $dv/dt = 0$ and only the first term survives. It points along $-\hat e_r$ , straight at the centre:

Centripetal acceleration(7.9)

a_c = \omega^2 r = \frac{v^2}{r} \quad \text{toward the centre}

You saw the same thing from the velocity arrows in chapter 3: turn the velocity through a small angle and the change $\Delta\vec v$ points inward.

Interactive

Same speed, new direction

A ball moves round a circle at a constant 2 m/s. Its velocity arrow turns, so the change Δv = v₂ − v₁ is not zero. Notice Δv points toward the centre of the circle.

|\vec v_1| = |\vec v_2|

2.00m/s

|\Delta\vec v|

1.37m/s

Angle turned40°

Worked example

A steady lap

A bead moves at constant speed round a circle of radius 15 cm, completing each lap in 3.0 s. Find its acceleration.

When the speed changes too

If the speed changes, the acceleration has both parts:

a_r = \frac{v^2}{r}\ (\text{inward}), \qquad a_t = \frac{dv}{dt}\ (\text{along the motion}) \qquad (7.10)

|\vec a| = \sqrt{\left(\frac{v^2}{r}\right)^2 + \left(\frac{dv}{dt}\right)^2}, \qquad \tan\alpha = \frac{dv/dt}{v^2/r}

where α is the angle between the acceleration and the inward radius.

Interactive

Where does the acceleration point?

The particle is passing the 60° mark. With steady speed the acceleration points straight at the centre. Make the speed change (dv/dt ≠ 0) and a tangential part tilts the arrow.

a_r = v^2/r

4.50m/s²

a_t = dv/dt

0.00m/s²

|\vec a|

4.50m/s²

angle from the radius0.0°

Speed

v

3.0m/s

Rate of change of speed

dv/dt

0.0m/s²

Radius

r

2.0m

Worked example

Speeding round a bend

A particle moves on a circle of radius 50 cm with speed v = 3t (v in m/s, t in s). Find its radial and tangential acceleration at t = 2 s.

What to remember

1On a circle the velocity is tangent, with magnitude rω.
2Uniform motion: $a = v^2/r = \omega^2 r$ toward the centre.
3Changing speed adds a tangential component dv/dt.
4Total: $\sqrt{(v^2/r)^2 + (dv/dt)^2}$ .

Check yourself

If a car doubles its speed on the same circular track, its centripetal acceleration…

For uniform circular motion, which is constant?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchCentripetal force and acceleration intuitionKhan Academy WatchUniform Circular MotionCrash Course Physics #7 ReadAcceleration in circular motionThe Physics Classroom ReadUniform circular motionOpenStax University Physics §4.4 ReadCentripetal forceHyperPhysics

PreviousAngles, Angular Velocity & Angular Acceleration Next The Force Behind Circular Motion

Centripetal Acceleration

Constant speed, changing direction: the acceleration points at the centre.

12 minUpdated 21 October 2026centripetal accelerationradial unit vectortangential unit vectornon-uniform circular motion

Lesson at a glanceShow the key points

Velocity on a circle is always tangent, with magnitude rω.
Uniform circular motion: acceleration v²/r = ω²r directed toward the centre (centripetal acceleration).
Non-uniform motion adds a tangential component dv/dt; |a| = √[(v²/r)² + (dv/dt)²].
The resultant acceleration makes angle α with the radius where tan α = (dv/dt)/(v²/r).

A car taking a roundabout at a steady 30 km/h is accelerating the whole time. The speed never changes; the direction never stops changing.

Two unit vectors that ride along

Put the centre of the circle at the origin, and let the particle be at angular position θ. Two unit vectors are handy:

\hat e_r = \hat\imath\cos\theta + \hat\jmath\sin\theta \ \text{(outward radius)}, \qquad \hat e_t = -\hat\imath\sin\theta + \hat\jmath\cos\theta \ \text{(tangent, direction of increasing }\theta) \quad (7.6, 7.7)

The position vector is $\vec r = r(\hat\imath\cos\theta + \hat\jmath\sin\theta)$ . Differentiate with r fixed and θ changing:

\vec v = \frac{d\vec r}{dt} = r\omega(-\hat\imath\sin\theta + \hat\jmath\cos\theta) = r\omega\,\hat e_t

So the velocity is always along the tangent, with size $r\omega$ . Differentiate again (now ω may also change):

Acceleration on a circle(7.8)

\vec a = -\omega^2 r\,\hat e_r + \frac{dv}{dt}\,\hat e_t

Uniform circular motion

If the speed is constant, $dv/dt = 0$ and only the first term survives. It points along $-\hat e_r$ , straight at the centre:

Centripetal acceleration(7.9)

a_c = \omega^2 r = \frac{v^2}{r} \quad \text{toward the centre}

You saw the same thing from the velocity arrows in chapter 3: turn the velocity through a small angle and the change $\Delta\vec v$ points inward.

Interactive

Same speed, new direction

A ball moves round a circle at a constant 2 m/s. Its velocity arrow turns, so the change Δv = v₂ − v₁ is not zero. Notice Δv points toward the centre of the circle.

|\vec v_1| = |\vec v_2|

2.00m/s

|\Delta\vec v|

1.37m/s

Angle turned40°

Worked example

A steady lap

A bead moves at constant speed round a circle of radius 15 cm, completing each lap in 3.0 s. Find its acceleration.

When the speed changes too

If the speed changes, the acceleration has both parts:

a_r = \frac{v^2}{r}\ (\text{inward}), \qquad a_t = \frac{dv}{dt}\ (\text{along the motion}) \qquad (7.10)

|\vec a| = \sqrt{\left(\frac{v^2}{r}\right)^2 + \left(\frac{dv}{dt}\right)^2}, \qquad \tan\alpha = \frac{dv/dt}{v^2/r}

where α is the angle between the acceleration and the inward radius.

Interactive

Where does the acceleration point?

The particle is passing the 60° mark. With steady speed the acceleration points straight at the centre. Make the speed change (dv/dt ≠ 0) and a tangential part tilts the arrow.

a_r = v^2/r

4.50m/s²

a_t = dv/dt

0.00m/s²

|\vec a|

4.50m/s²

angle from the radius0.0°

Speed

v

3.0m/s

Rate of change of speed

dv/dt

0.0m/s²

Radius

r

2.0m

Worked example

Speeding round a bend

A particle moves on a circle of radius 50 cm with speed v = 3t (v in m/s, t in s). Find its radial and tangential acceleration at t = 2 s.

What to remember

1On a circle the velocity is tangent, with magnitude rω.
2Uniform motion: $a = v^2/r = \omega^2 r$ toward the centre.
3Changing speed adds a tangential component dv/dt.
4Total: $\sqrt{(v^2/r)^2 + (dv/dt)^2}$ .

Check yourself

If a car doubles its speed on the same circular track, its centripetal acceleration…

For uniform circular motion, which is constant?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PreviousAngles, Angular Velocity & Angular Acceleration Next The Force Behind Circular Motion

Two unit vectors that ride along#

Uniform circular motion#

When the speed changes too#

Check yourself

Keep exploring

Two unit vectors that ride along#

Uniform circular motion#

When the speed changes too#

Check yourself

Keep exploring

Two unit vectors that ride along

Uniform circular motion

When the speed changes too

Two unit vectors that ride along

Uniform circular motion

When the speed changes too