Formula Sheet · Cylindrical Coordinates

Coordinates and position

Figure F.1 Walk out \(r\) at angle \(\theta\) to reach \(P'\), then rise \(z\) to \(P\).

\(\colR{r}\): distance from the \(z\)-axis (in a plane: from \(O\)); \(r \ge 0\)
\(\colT{\theta}\): angle from the \(+x\) axis, counter-clockwise seen from \(+z\), in radians
\(\colZ{z}\): height along the axis

To rectangular

\[ x = r\cos\theta, \quad y = r\sin\theta \]

From rectangular

\[ r = \sqrt{x^2 + y^2}, \quad \theta = \atantwo(y, x) \]

Position vector: \(\rvec = r\,\er + z\,\ez\) (no \(\et\) term), with \(|\rvec| = \sqrt{r^2 + z^2}\). A motion is \(r(t)\), \(\theta(t)\), \(z(t)\); \(N\) rpm \(= 2\pi N/60\) rad/s.

Finding \(\theta\): the quadrant table

The angle theta from x and y, by the location of the point
\((x, y)\) lies in	\(\theta\) in \([0, 2\pi)\)
I: \(x > 0,\ y \ge 0\)	\(\arctan(y/x)\)
II or III: \(x \lt 0\)	\(\arctan(y/x) + \pi\)
IV: \(x > 0,\ y \lt 0\)	\(\arctan(y/x) + 2\pi\)
\(x = 0\)	\(\tfrac{\pi}{2}\) if \(y > 0\), \(\tfrac{3\pi}{2}\) if \(y \lt 0\)

A calculator's \(\tan^{-1}\) key returns only \(-\tfrac{\pi}{2}\) to \(\tfrac{\pi}{2}\): sketch the point first, then pick the row.

Unit vectors that turn

\[ \begin{aligned} \colR{\er} &= \cos\theta\,\ihat + \sin\theta\,\jhat \\ \colT{\et} &= -\sin\theta\,\ihat + \cos\theta\,\jhat \\ \colZ{\ez} &= \khat \end{aligned} \]

Time derivatives

\[ \begin{aligned} \dot{\mathbf{u}}_r &= \dot\theta\,\et \\ \dot{\mathbf{u}}_\theta &= -\dot\theta\,\er \\ \dot{\mathbf{u}}_z &= \mathbf{0} \end{aligned} \]

Components of a vector (\(\theta\) = angle of the particle's position)

\[ \begin{aligned} F_r &= F_x\cos\theta + F_y\sin\theta \\ F_\theta &= -F_x\sin\theta + F_y\cos\theta \end{aligned} \]

\[ \begin{aligned} F_x &= F_r\cos\theta - F_\theta\sin\theta \\ F_y &= F_r\sin\theta + F_\theta\cos\theta \end{aligned} \]

Weight with \(\theta\) from the horizontal in a vertical plane (\(y\) up): \(W_r = -mg\sin\theta\), \(W_\theta = -mg\cos\theta\).

Velocity

\[ \vvec = \dot r\,\colR{\er} + r\dot\theta\,\colT{\et} + \dot z\,\colZ{\ez} \]

\(v_r = \dot r\) (moving out), \(v_\theta = r\dot\theta\) (going round), \(v_z = \dot z\) (climbing). Speed \(v = \sqrt{\dot r^2 + (r\dot\theta)^2 + \dot z^2}\).

Direction in the \(r\)–\(\theta\) plane: \(\tan\psi = v_\theta/v_r\), measured from the radial line. A tracker reads \(\dot r = v_r\) and \(\dot\theta = v_\theta/r\).

Acceleration

\[ \avec = \left(\ddot r - r\dot\theta^2\right)\colR{\er} + \left(r\ddot\theta + 2\dot r\dot\theta\right)\colT{\et} + \ddot z\,\colZ{\ez} \]

The terms of the acceleration
Term	Name	Zero when
\(\ddot r\)	radial (sliding)	\(\dot r\) constant
\(-r\dot\theta^2\)	centripetal, toward the axis	\(\dot\theta = 0\)
\(r\ddot\theta\)	angular acceleration	\(\dot\theta\) constant
\(2\dot r\dot\theta\)	Coriolis	\(\dot r = 0\) or \(\dot\theta = 0\)
\(\ddot z\)	axial	\(\dot z\) constant

Circle about \(O\) (\(r\) constant): \(a_r = -r\dot\theta^2 = -v^2/r\), \(a_\theta = r\ddot\theta = \dot v\). Given \(r(t)\), \(\theta(t)\): differentiate twice, evaluate, substitute.

Paths given as \(r = f(\theta)\)

Chain rule

\[ \dot r = \frac{dr}{d\theta}\,\dot\theta, \qquad \ddot r = \frac{d^2r}{d\theta^2}\,\dot\theta^2 + \frac{dr}{d\theta}\,\ddot\theta \]

Tangent angle

\[ \tan\psi = \frac{r}{dr/d\theta} \]

Common paths and their derivatives
Path	\(f'\)	\(f''\)
\(r = b\theta\) (spiral)	\(b\)	\(0\)
\(r = a(1 + \cos\theta)\)	\(-a\sin\theta\)	\(-a\cos\theta\)
\(r = d\cos\theta\) (circle through \(O\))	\(-d\sin\theta\)	\(-d\cos\theta\)
\(r = ae^{k\theta}\) (\(\psi\) constant)	\(kr\)	\(k^2 r\)

\(\psi\) is measured from the extended radial line to the tangent, positive toward \(\et\).

Equations of motion

\[ \textstyle\sum F_r = m\left(\ddot r - r\dot\theta^2\right), \] \[ \textstyle\sum F_\theta = m\left(r\ddot\theta + 2\dot r\dot\theta\right), \] \[ \textstyle\sum F_z = m\ddot z \]

Procedure: free-body and kinetic diagrams with \(\er\), \(\et\) at the particle; kinematics (\(r, \dot r, \ddot r, \dot\theta, \ddot\theta\)); one equation per direction; solve and check signs.

Smooth rod or slot turning with the arm: force \(N\,\et\) only (no \(\er\) part).
Spring along the arm from \(O\): \(-k(r - r_0)\,\er\). Cord through \(O\): \(-T\,\er\) with \(T \ge 0\).
Collar free on a driven rod: \(\ddot r = r\dot\theta^2\) and \(N = 2m\dot r\dot\theta\).

Guides and central forces

Pin in a smooth fixed guide, pushed by a smooth arm (horizontal plane)

\[ \begin{gathered} \mathbf{N} = N(-\sin\psi\,\er + \cos\psi\,\et) \\ N = -\frac{m a_r}{\sin\psi}, \qquad F = m a_\theta - N\cos\psi \end{gathered} \]

Central force (\(\sum F_\theta = 0\): cord through \(O\), gravity of a planet)

\[ r^2\dot\theta = h = \text{constant}, \] \[ v_\theta = \frac{h}{r}, \] \[ \textstyle\sum F_r = m\left(\ddot r - \dfrac{h^2}{r^3}\right) \]

Orbits: \(r_p v_p = r_a v_a\) at perigee and apogee. Pulled cord: \(\dot\theta_2 = \dot\theta_1 (r_1/r_2)^2\).

Choosing a coordinate system

Which components make the equations simplest
Use	When the problem has
Rectangular \(x, y, z\)	forces in fixed directions: projectiles, blocks on fixed inclines
Path \(t, n\)	a known path and its radius of curvature: cars on curves, roller coasters
Polar \(r, \theta\)	a rotating arm, rod or slot; radar data \(r, \theta\); a path \(r = f(\theta)\); a cord through \(O\)
Cylindrical \(r, \theta, z\)	polar motion plus a height: helix, spiral chute, robot arm that lifts

All systems give the same \(\vvec\) and \(\avec\), so \(v\) and \(a\) agree. Circle centered on \(O\): \(a_n = -a_r = v^2/r\), \(a_t = a_\theta\).

Common mistakes

Trusting \(\arctan(y/x)\). It misses the quadrant whenever \(x \lt 0\).
A \(\theta\)-term in \(\rvec\). \(\rvec = r\,\er + z\,\ez\), never \(r\,\er + \theta\,\et\).
\(v_\theta = \dot\theta\). It is \(r\dot\theta\): check the units (m/s, not rad/s).
\(a_r = \ddot r\). It is \(\ddot r - r\dot\theta^2\); constant rates do not mean zero acceleration.
Forgetting the Coriolis term \(2\dot r\dot\theta\) when the particle moves in or out while turning.
\(\ddot r = f''\ddot\theta\) on a path. It is \(f''\dot\theta^2 + f'\ddot\theta\).
Degrees in a formula. \(r\dot\theta\) and \(r\dot\theta^2\) need rad/s.
Assuming \(r^2\dot\theta\) is constant when an arm or rod pushes sideways (\(\sum F_\theta \ne 0\)).

Check values

Test your calculator, spreadsheet or code on these before trusting it with a new problem.

Example 5.3: \(r = 1 + 0.2t^2\ \text{m}\), \(\theta = 0.5t^2\ \text{rad}\), at \(t = 2\ \text{s}\)

Values at t = 2 s for Example 5.3
\(r,\ \dot r,\ \ddot r\)	\(1.8\) m	\(0.8\) m/s	\(0.4\) m/s²
\(\theta,\ \dot\theta,\ \ddot\theta\)	\(2\) rad	\(2\) rad/s	\(1\) rad/s²
\(v_r,\ v_\theta,\ v\)	\(0.8\)	\(3.6\)	\(3.688\) m/s
\(a_r,\ a_\theta,\ a\)	\(-6.8\)	\(5.0\)	\(8.440\) m/s²
\(\sum F_r,\ \sum F_\theta\) (\(m = 2\) kg)	\(-13.6\) N	\(10.0\) N

Conversions

\((x, y) = (-3, 4)\): \(r = 5\), \(\theta = 2.214\) rad \((126.87^\circ)\). \((-3, -4)\): \(\theta = 4.069\) rad \((233.13^\circ)\); code returns \(-2.214\), so add \(2\pi\). \(60\) rpm \(= 2\pi\) rad/s \(= 6.283\) rad/s.

Worked in Lesson 5 and Lesson 2

Finding \(\theta\) with a calculator or code

These return \(\theta\) from \(-\pi\) to \(\pi\) (\(-180^\circ\) to \(180^\circ\) in DEG mode). For \(0 \le \theta \lt 2\pi\), add \(2\pi\) to a negative result, or use a mod form.

Calculator: x first

Use RAD mode for anything with \(\dot\theta\) or \(\ddot\theta\). \(\tan^{-1}\) alone needs the quadrant table; Pol(x, y) (Casio) and R►Pθ(x, y) (TI) get the quadrant right.

Excel / Google Sheets: x first

=ATAN2(x, y)\(-\pi\) to \(\pi\) =MOD(ATAN2(x, y), 2*PI())\(0\) to \(2\pi\)

ATAN2(0, 0) returns #DIV/0! (the particle is at \(O\)).

Python / NumPy: y first

math.atan2(y, x) % (2*math.pi)\(0\) to \(2\pi\) math.hypot(x, y)\(r\) np.arctan2(y, x) % (2*np.pi)arrays

C, Java, JavaScript: also \((y, x)\), but % and fmod keep the sign; add \(2\pi\) if negative.

MATLAB: atan2 y first, cart2pol x first

mod(atan2(y, x), 2*pi)\(0\) to \(2\pi\) [theta, rho] = cart2pol(x, y)\(\theta\) first, \(-\pi\) to \(\pi\)