Cylindrical / polar coordinates

A particle sliding along a rotating slot, a radar dish tracking a moving aircraft, a cam-follower tracing a curved groove — none of these paths are naturally described by fixed xx-yy axes. What matters is the distance rr from a fixed point and the angle θ\theta swept out, and both can change with time at once. Polar coordinates (r,θ)(r,\theta) track exactly that, and cylindrical coordinates simply bolt on an independent zz-axis for out-of-plane motion. The twist that makes this coordinate system trickier than rectangular is that its own unit vectors, e^r\hat e_r and e^θ\hat e_\theta, rotate with the particle — so differentiating position to get velocity and acceleration is not just "differentiate the components." This article derives v⃗\vec v and a⃗\vec a in polar form from first principles, works two fully numerical examples, and flags the single term (the Coriolis-type 2r˙θ˙2\dot r\dot\theta) that costs the most marks under exam pressure.

The core insight

Because the radial and transverse unit vectors e^r,e^θ\hat e_r,\hat e_\theta rotate with the particle, differentiating r⃗=re^r\vec r = r\hat e_r twice produces extra centripetal (−rθ˙2-r\dot\theta^2) and Coriolis-type (2r˙θ˙2\dot r\dot\theta) terms that have no counterpart in rectangular coordinates.

The core idea

Two rotating unit vectors. Define e^r\hat e_r pointing radially outward from the origin to the particle, and e^θ\hat e_\theta perpendicular to it, in the direction of increasing θ\theta. Unlike the fixed i^,j^\hat i, \hat j of rectangular coordinates, these two rotate as θ\theta changes, so they carry their own time derivatives:

e^˙r=θ˙ e^θ,e^˙θ=−θ˙ e^r,\dot{\hat e}_r = \dot\theta\,\hat e_\theta, \qquad \dot{\hat e}_\theta = -\dot\theta\,\hat e_r,

where: θ˙\dot\theta is the angular velocity of the position vector (rad/s). This single fact — that the "constant" unit vectors of polar coordinates are not actually constant — is the entire source of the extra terms below.

Position and velocity. Position is simply r⃗=r e^r\vec r = r\,\hat e_r (all the direction information sits in e^r\hat e_r, all the magnitude in rr). Differentiating once, and using the product rule together with e^˙r=θ˙ e^θ\dot{\hat e}_r = \dot\theta\,\hat e_\theta, gives velocity:

v⃗=r˙ e^r+rθ˙ e^θ,\vec v = \dot r\,\hat e_r + r\dot\theta\,\hat e_\theta,

where: vr=r˙v_r = \dot r is the radial component (how fast the particle moves outward or inward) and vθ=rθ˙v_\theta = r\dot\theta is the transverse component (how fast it sweeps sideways, which depends on both how far out it is and how fast it is turning).

Acceleration. Differentiating v⃗\vec v again, carrying the product rule through both rotating unit vectors, gives

a⃗=(r¨−rθ˙2) e^r+(rθ¨+2r˙θ˙) e^θ,\vec a = (\ddot r - r\dot\theta^2)\,\hat e_r + (r\ddot\theta + 2\dot r\dot\theta)\,\hat e_\theta,

so that

ar=r¨−rθ˙2,aθ=rθ¨+2r˙θ˙.a_r = \ddot r - r\dot\theta^2, \qquad a_\theta = r\ddot\theta + 2\dot r\dot\theta.

where: r¨\ddot r is the radial acceleration a passenger moving only along rr would feel, −rθ˙2-r\dot\theta^2 is the familiar centripetal term (present even when rr is constant), rθ¨r\ddot\theta is ordinary angular acceleration times radius, and 2r˙θ˙2\dot r\dot\theta is a Coriolis-type coupling term that only appears when rr and θ\theta change simultaneously — it has no rectangular-coordinate counterpart and is the term students most often drop.

Two limiting checks. For pure circular motion (r=Rr = R constant, so r˙=r¨=0\dot r = \ddot r = 0), the formulas collapse to the familiar ar=−Rθ˙2a_r = -R\dot\theta^2 (centripetal, magnitude v2/Rv^2/R) and aθ=Rθ¨a_\theta = R\ddot\theta (tangential) — exactly the circular-motion results from earlier in the course. For cylindrical (3-D) motion, add an independent z(t)z(t) axis exactly as in rectangular kinematics: v⃗=r˙ e^r+rθ˙ e^θ+z˙ k^\vec v = \dot r\,\hat e_r + r\dot\theta\,\hat e_\theta + \dot z\,\hat k, with az=z¨a_z = \ddot z — no new coupling terms appear because zz is orthogonal to the rotating plane.

Figure
Radial and transverse unit vectors at a particle position, rotating as theta increases, with r and theta labelled
Polar unit vectors rotate with the particle O theta r e_r e_theta e_r points outward along r; e_theta is perpendicular, in the direction of increasing theta
Derivation AAcceleration in polar coordinates from a rotating position vector

The goal is to derive the full acceleration a⃗=(r¨−rθ˙2)e^r+(rθ¨+2r˙θ˙)e^θ\vec a = (\ddot r - r\dot\theta^2)\hat e_r + (r\ddot\theta + 2\dot r\dot\theta)\hat e_\theta starting only from r⃗=re^r\vec r = r\hat e_r and the two unit-vector derivatives e^˙r=θ˙e^θ\dot{\hat e}_r = \dot\theta\hat e_\theta, e^˙θ=−θ˙e^r\dot{\hat e}_\theta = -\dot\theta\hat e_r, using nothing beyond the ordinary product rule applied twice — the entire difficulty is bookkeeping, not new physics.

Step 1. Differentiate the position vector r⃗=re^r\vec r = r\hat e_r once, treating both rr and e^r\hat e_r as functions of time. The product rule gives v⃗=r˙ e^r+r e^˙r\vec v = \dot r\,\hat e_r + r\,\dot{\hat e}_r. Substituting e^˙r=θ˙ e^θ\dot{\hat e}_r = \dot\theta\,\hat e_\theta, this becomes v⃗=r˙ e^r+rθ˙ e^θ\vec v = \dot r\,\hat e_r + r\dot\theta\,\hat e_\theta, matching the velocity result stated above.

Step 2. Differentiate v⃗\vec v term by term. Write v⃗=r˙ e^r+rθ˙ e^θ\vec v = \dot r\,\hat e_r + r\dot\theta\,\hat e_\theta as a sum of two products, each needing its own product rule: ddt(r˙ e^r)=r¨ e^r+r˙ e^˙r\frac{d}{dt}(\dot r\,\hat e_r) = \ddot r\,\hat e_r + \dot r\,\dot{\hat e}_r, and ddt(rθ˙ e^θ)=(r˙θ˙+rθ¨) e^θ+rθ˙ e^˙θ\frac{d}{dt}(r\dot\theta\,\hat e_\theta) = (\dot r\dot\theta + r\ddot\theta)\,\hat e_\theta + r\dot\theta\,\dot{\hat e}_\theta.

Step 3. Substitute both unit-vector derivatives. In the first term, r˙ e^˙r=r˙θ˙ e^θ\dot r\,\dot{\hat e}_r = \dot r\dot\theta\,\hat e_\theta. In the second term, rθ˙ e^˙θ=rθ˙(−θ˙ e^r)=−rθ˙2 e^rr\dot\theta\,\dot{\hat e}_\theta = r\dot\theta(-\dot\theta\,\hat e_r) = -r\dot\theta^2\,\hat e_r. Collecting every piece: a⃗=r¨ e^r+r˙θ˙ e^θ+r˙θ˙ e^θ+rθ¨ e^θ−rθ˙2 e^r\vec a = \ddot r\,\hat e_r + \dot r\dot\theta\,\hat e_\theta + \dot r\dot\theta\,\hat e_\theta + r\ddot\theta\,\hat e_\theta - r\dot\theta^2\,\hat e_r.

Step 4. Group by direction. The e^r\hat e_r terms are r¨ e^r−rθ˙2 e^r=(r¨−rθ˙2)e^r\ddot r\,\hat e_r - r\dot\theta^2\,\hat e_r = (\ddot r - r\dot\theta^2)\hat e_r. The e^θ\hat e_\theta terms are r˙θ˙ e^θ+r˙θ˙ e^θ+rθ¨ e^θ=(rθ¨+2r˙θ˙)e^θ\dot r\dot\theta\,\hat e_\theta + \dot r\dot\theta\,\hat e_\theta + r\ddot\theta\,\hat e_\theta = (r\ddot\theta + 2\dot r\dot\theta)\hat e_\theta — note the factor of two arises because two separate places in the differentiation each contributed a r˙θ˙\dot r\dot\theta term, one from differentiating rr in the rθ˙e^θr\dot\theta\hat e_\theta product, one from differentiating e^θ\hat e_\theta itself. This is exactly why the coupling term is so easy to under-count: skip either source and you lose a factor of r˙θ˙\dot r\dot\theta, not the whole term.

Step 5. Read off the final result: ar=r¨−rθ˙2a_r = \ddot r - r\dot\theta^2 and aθ=rθ¨+2r˙θ˙a_\theta = r\ddot\theta + 2\dot r\dot\theta, matching the boxed formulas above and reducing correctly to ar=−Rθ˙2a_r = -R\dot\theta^2, aθ=Rθ¨a_\theta = R\ddot\theta when r=Rr = R is held constant.

Worked example

A particle moves so that r=2tr = 2t (m) and θ=t2\theta = t^2 (rad), with tt in seconds. Find the speed and the magnitude of acceleration at t=1 st = 1\ \text{s}.

Step 1. Differentiate the given motion. From r=2tr = 2t: r˙=2 m/s\dot r = 2\ \text{m/s}, r¨=0\ddot r = 0. From θ=t2\theta = t^2: θ˙=2t rad/s\dot\theta = 2t\ \text{rad/s}, θ¨=2 rad/s2\ddot\theta = 2\ \text{rad/s}^2. At t=1 st = 1\ \text{s} these give r=2 mr = 2\ \text{m}, r˙=2 m/s\dot r = 2\ \text{m/s}, r¨=0\ddot r = 0, θ˙=2 rad/s\dot\theta = 2\ \text{rad/s}, θ¨=2 rad/s2\ddot\theta = 2\ \text{rad/s}^2.

Step 2. Compute the velocity components: vr=r˙=2 m/sv_r = \dot r = 2\ \text{m/s} and vθ=rθ˙=2(2)=4 m/sv_\theta = r\dot\theta = 2(2) = 4\ \text{m/s}. The speed is v=vr2+vθ2=4+16=20=4.47 m/sv = \sqrt{v_r^2 + v_\theta^2} = \sqrt{4 + 16} = \sqrt{20} = 4.47\ \text{m/s}.

Step 3. Compute the acceleration components: ar=r¨−rθ˙2=0−2(2)2=−8 m/s2a_r = \ddot r - r\dot\theta^2 = 0 - 2(2)^2 = -8\ \text{m/s}^2, and aθ=rθ¨+2r˙θ˙=2(2)+2(2)(2)=4+8=12 m/s2a_\theta = r\ddot\theta + 2\dot r\dot\theta = 2(2) + 2(2)(2) = 4 + 8 = 12\ \text{m/s}^2.

Step 4. Combine to get the magnitude: a=ar2+aθ2=64+144=208=14.42 m/s2a = \sqrt{a_r^2 + a_\theta^2} = \sqrt{64 + 144} = \sqrt{208} = 14.42\ \text{m/s}^2. Answer: v=4.47 m/sv = 4.47\ \text{m/s}, a=14.42 m/s2a = 14.42\ \text{m/s}^2 at t=1 st = 1\ \text{s}, with the negative ara_r showing the centripetal term already dominates the small r¨=0\ddot r = 0 contribution.

Competitive stretch

Now let r=0.5t2r = 0.5t^2 (m) with a rotating slot holding θ˙=3 rad/s\dot\theta = 3\ \text{rad/s} constant (θ¨=0\ddot\theta = 0). At t=2 st = 2\ \text{s}: r=2 mr = 2\ \text{m}, r˙=t=2 m/s\dot r = t = 2\ \text{m/s}, r¨=1 m/s2\ddot r = 1\ \text{m/s}^2. Then vr=2 m/sv_r = 2\ \text{m/s}, vθ=2(3)=6 m/sv_\theta = 2(3) = 6\ \text{m/s}, v=4+36=6.32 m/sv = \sqrt{4+36} = 6.32\ \text{m/s}; and ar=1−2(3)2=−17 m/s2a_r = 1 - 2(3)^2 = -17\ \text{m/s}^2, aθ=0+2(2)(3)=12 m/s2a_\theta = 0 + 2(2)(3) = 12\ \text{m/s}^2, so a=289+144=20.8 m/s2a = \sqrt{289+144} = 20.8\ \text{m/s}^2. Despite θ¨=0\ddot\theta = 0, aθ≠0a_\theta \ne 0 — it comes entirely from the Coriolis-type term 2r˙θ˙2\dot r\dot\theta.

Common traps

  • Differentiating r⃗=re^r\vec r = r\hat e_r with the ordinary product rule but forgetting e^˙r=θ˙e^θ\dot{\hat e}_r = \dot\theta\hat e_\theta — this drops the entire rθ˙e^θr\dot\theta\hat e_\theta term from the velocity.
  • Dropping one of the two sources of 2r˙θ˙2\dot r\dot\theta in aθa_\theta — the coupling term gets a factor of two precisely because it arises independently from differentiating rr and from differentiating e^θ\hat e_\theta; many students write only r˙θ˙\dot r\dot\theta.
  • Confusing rθ˙r\dot\theta (transverse velocity, m/s) with θ˙\dot\theta (angular velocity, rad/s) — they carry different units and are never interchangeable.
  • Applying the simple circular-motion shortcut a=v2/ra = v^2/r (which assumes aθ=0a_\theta = 0) to a problem where rr is actually changing with time, silently discarding a nonzero aθa_\theta.

Ask the tutor: Want to see the 2r˙θ˙2\dot r\dot\theta term derived on a whiteboard, one line at a time, or try a fresh r(t)r(t), θ(t)\theta(t) pair yourself and check your vr,vθ,ar,aθv_r, v_\theta, a_r, a_\theta against the tutor.

Why do we even need e_r and e_theta instead of just using x and y?

Because for paths that are naturally circular, spiralling, or defined by a changing radius and angle — a rotating slot, radar tracking, a cam follower — resolving motion along the radial and transverse directions is far simpler than working out messy, coupled x(t)x(t) and y(t)y(t) expressions. The price is that e^r\hat e_r and e^θ\hat e_\theta themselves rotate, so differentiation picks up the extra centripetal and Coriolis-type terms. You choose the coordinate system that matches the geometry of the problem, and polar coordinates are the natural match whenever rr or θ\theta (or both) is the quantity actually varying.

Is the 2 r-dot theta-dot term the same as the Coriolis force from rotating reference frames?

It has the same mathematical origin — a rotating unit vector picking up an extra term under differentiation — and the factor of two arises the same way, but here it is simply a term in the acceleration of a particle described in polar coordinates, not a force in a separately rotating frame. In this course you will meet a true Coriolis force later when analyzing motion relative to rotating frames; for now, treat 2r˙θ˙2\dot r\dot\theta purely as a kinematic term that must be included whenever both rr and θ\theta vary together.

How does cylindrical differ from polar if the formulas look almost the same?

Cylindrical coordinates are just planar polar coordinates with an independent third axis z(t)z(t) bolted on, exactly as in rectangular kinematics: v⃗=r˙e^r+rθ˙e^θ+z˙k^\vec v = \dot r\hat e_r + r\dot\theta\hat e_\theta + \dot z\hat k and az=z¨a_z = \ddot z. Because zz is perpendicular to the rotating rr-θ\theta plane, it contributes no coupling terms of its own — you simply solve the zz-motion the same way you would in rectangular coordinates and add it on separately from the polar-plane result.

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