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Circular Motion Kinematics

Level 1 - Physics topic page in Motion in Space.

Principle

Circular motion is constrained two-dimensional motion: the radius is fixed, but the useful direction axes move with the particle.

Earlier pages already introduced position, velocity, acceleration, and Cartesian components. This page adds the radial-tangential basis used for motion on a circle.

Notation

\(R\)
fixed circle radius
\(\mathrm{m}\)
\(\theta\)
angle measured from the positive x-axis
\(\mathrm{rad}\)
\(\omega=d\theta/dt\)
angular velocity
\(\mathrm{rad\,s^{-1}}\)
\(\alpha=d\omega/dt\)
angular acceleration
\(\mathrm{rad\,s^{-2}}\)
\(\hat e_r\)
unit vector from the center to the particle
1
\(\hat e_\theta\)
unit vector tangent to increasing \(\theta\)
1

Method

Derivation 1: Build the radial direction

Place the circle's center at the origin. A point on the circle is reached by moving a distance \(R\) at angle \(\\theta\). The x- and y-components are the two projections of that radius.

Project the radius
\[x=R\cos\theta,\qquad y=R\sin\theta\]
Write position
\[\vec r=x\hat{\imath}+y\hat{\jmath}=R\cos\theta\,\hat{\imath}+R\sin\theta\,\hat{\jmath}\]
Divide by the radius
\[\hat e_r=\frac{\vec r}{R}=\cos\theta\,\hat{\imath}+\sin\theta\,\hat{\jmath}\]
This is why \(\hat e_r\) points outward from the center.

Derivation 2: Build the tangent direction

The tangent direction is what the radial direction becomes when the angle increases a little. Differentiate the radial unit vector with respect to angle.

Differentiate with respect to angle
\[\frac{d\hat e_r}{d\theta}=-\sin\theta\,\hat{\imath}+\cos\theta\,\hat{\jmath}\]
Check its length
\[\left|\frac{d\hat e_r}{d\theta}\right|=\sqrt{\sin^2\theta+\cos^2\theta}=1\]
Name the tangent unit vector
\[\hat e_\theta=\frac{d\hat e_r}{d\theta}=-\sin\theta\,\hat{\imath}+\cos\theta\,\hat{\jmath}\]
It is perpendicular to \(\hat e_r\) and points in the direction of increasing angle.

Derivation 3: Differentiate position

Now use the chain rule. The radius \(R\) is constant, but the direction \(\\hat e_r\) changes because the angle changes.

Start with constrained position
\[\vec r=R\hat e_r\]
Apply the chain rule
\[\frac{d\hat e_r}{dt}=\frac{d\hat e_r}{d\theta}\frac{d\theta}{dt}=\omega\hat e_\theta\]
Velocity is tangent
\[\vec v=\frac{d\vec r}{dt}=R\omega\hat e_\theta\]

Derivation 4: Differentiate velocity

Acceleration has two pieces because both the angular speed and the tangent direction can change.

Use the product rule
\[\vec a=R\frac{d\omega}{dt}\hat e_\theta+R\omega\frac{d\hat e_\theta}{dt}\]
Differentiate the tangent basis
\[\frac{d\hat e_\theta}{dt}=-\omega\hat e_r\]
Collect radial and tangent parts
\[\vec a=-R\omega^2\hat e_r+R\alpha\hat e_\theta\]
The radial term is inward. The tangent term changes speed.

The sketch below is only an instantaneous direction map. It is not a graph of position against time.

3.20 deg90 deg180 deg270 degrvac
At one point on the circle, the radial and tangent directions are perpendicular.

Rules

These are the compact results from the derivations above.

Position
\[\vec r=R\hat e_r=R\cos\theta\,\hat{\imath}+R\sin\theta\,\hat{\jmath}\]
Radial direction
\[\hat e_r=\cos\theta\,\hat{\imath}+\sin\theta\,\hat{\jmath}\]
Tangent direction
\[\hat e_\theta=-\sin\theta\,\hat{\imath}+\cos\theta\,\hat{\jmath}\]
Velocity
\[\vec v=R\omega\hat e_\theta\]
Acceleration
\[\vec a=-R\omega^2\hat e_r+R\alpha\hat e_\theta\]
Uniform speed case
\[v=R|\omega|,\qquad a_c=R\omega^2=\frac{v^2}{R}\]

Examples

Question
A sensor moves on a circular track with
\[R=1.8\,\mathrm{m}\]
and constant angular speed
\[\omega=4.0\,\mathrm{rad\,s^{-1}}\]
Find the speed and acceleration magnitude.
Answer
The speed is tangential:
\[v=R\omega=1.8(4.0)=7.2\,\mathrm{m\,s^{-1}}\]
Since the speed is constant, acceleration is only centripetal:
\[a_c=R\omega^2=1.8(4.0)^2=28.8\,\mathrm{m\,s^{-2}}\]
directed inward.

Checks

  • Radians are required in angular equations.
  • Constant speed means zero tangent acceleration, not zero acceleration.
  • The inward term appears because the tangent direction itself turns as the particle moves.
  • If \(R\) is very large over a small arc, the inward acceleration becomes small and the motion locally resembles straight-line motion.

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Projectile Models

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Relative Motion