Lesson 2 · 30 min
Position in Polar and Cylindrical Coordinates
Before you can find a velocity or an acceleration you need the position. In polar coordinates it is a distance \(\colR{r}\) from a fixed point and an angle \(\colT{\theta}\); in space, add a height \(\colZ{z}\). A moving particle is then three functions of time.
Learning objectives
- Locate a point by its polar coordinates \((r, \theta)\), and convert to and from \((x, y)\) in every quadrant.
- Extend polar coordinates to cylindrical coordinates \((r, \theta, z)\) for motion in space.
- Write the position vector as \(\rvec = r\,\er + z\,\ez\) and explain why it has no \(\et\) term.
- Evaluate the position of a particle whose motion is given as \(r(t)\), \(\theta(t)\) and \(z(t)\).
Polar coordinates in a plane
Choose a fixed point \(O\), the origin (or pole), and a fixed reference line through it, usually the \(+x\) axis. A point \(P\) in the plane is then located by two numbers:
\(r\): the radial coordinate
The distance from \(O\) to \(P\), in meters. It is never negative.
\(\theta\): the angular coordinate
The angle from the reference line to the line \(OP\), positive counter-clockwise, in radians.
The right triangle in Figure 2.1 gives the conversions. They hold in every quadrant, because \(\cos\theta\) and \(\sin\theta\) carry the signs of \(x\) and \(y\):
Polar ↔ rectangular
\[ x = r\cos\theta, \qquad y = r\sin\theta \] \[ r = \sqrt{x^2 + y^2}, \qquad \theta = \atantwo(y, x) \]Example 2.1 — Radar to map coordinates
A radar at \(O\) detects an aircraft at a horizontal range \(r = 8\ \text{km}\), at \(\theta = 135^\circ\) from the \(x\)-axis (east), measured counter-clockwise. Find its \(x\) and \(y\) coordinates.
Show solution
The aircraft is northwest of the radar: \(x \lt 0\) and \(y \gt 0\), as expected for an angle between \(90^\circ\) and \(180^\circ\).
Example 2.2 — Rectangular to polar, in quadrant III
A particle is at \((x, y) = (-3, -4)\ \text{m}\). Find \(r\) and \(\theta\).
Show solution
A calculator gives \(\tan^{-1}\!\big((-4)/(-3)\big) = \tan^{-1}(1.333) = 53.13^\circ\), which points into quadrant I. The particle is in quadrant III (both coordinates negative), so add \(180^\circ\):
\[ \theta = 53.13^\circ + 180^\circ = 233.13^\circ \approx 4.069\ \text{rad} \]\(\atantwo(-4, -3)\) gives \(-126.87^\circ\), the same direction measured clockwise; add \(360^\circ\) to get \(233.13^\circ\).
Adding the height: cylindrical coordinates
For motion in space, make the origin's reference line part of a right-handed \(x\), \(y\), \(z\) frame and add the height:
\(r\)
Distance from the \(z\)-axis (not from \(O\)), measured horizontally.
\(\theta\)
Angle around the \(z\)-axis from the \(+x\) axis, counter-clockwise as seen from above (\(+z\)).
\(z\)
Height along the axis, the same \(z\) as in rectangular coordinates.
So \(x = r\cos\theta\), \(y = r\sin\theta\) and \(z = z\). The point \(P'\) directly below \(P\) on the floor \(z = 0\) has the polar coordinates \((r, \theta)\) of \(P\).
The position vector
Define the radial unit vector \(\er\) at \(P\): the horizontal unit vector pointing from the \(z\)-axis toward \(P\). Then the vector from \(O\) to \(P\) is a horizontal step of length \(r\) along \(\er\) followed by a vertical step \(z\) along \(\ez\):
Position vector
\[ \rvec_P = r\,\colR{\er} + z\,\colZ{\ez} \qquad \text{(in a plane: } \rvec_P = r\,\er\text{)} \]Its magnitude is the distance from \(O\): \(|\rvec_P| = \sqrt{r^2 + z^2}\).
There is no \(\theta\) term: the angle is hidden in the direction of \(\er\), which depends on \(\theta\). That is the source of every new term in the velocity and acceleration. When the particle moves round, \(\er\) turns, and Lesson 3 shows how fast.
A moving particle: \(r(t)\), \(\theta(t)\), \(z(t)\)
A motion is described by giving the three coordinates as functions of time. Often each one comes straight from a part of the machine: the arm angle \(\theta(t)\) from the motor, the extension \(r(t)\) from a slider or a cord, the height \(z(t)\) from a lift. To find where the particle is at a given instant, evaluate the three functions and, if needed, convert to \(x\), \(y\), \(z\).
Example 2.3 — Where is the gripper?
The gripper of a cylindrical robot moves with \(r = (0.5 + 0.5t)\ \text{m}\), \(\theta = 2t\ \text{rad}\) and \(z = (1.5 + 0.3t)\ \text{m}\), where \(t\) is in seconds. Find its cylindrical and rectangular coordinates at \(t = 1\ \text{s}\) and at \(t = 3\ \text{s}\).
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At \(t = 1\ \text{s}\): \(r = 1\ \text{m}\), \(\theta = 2\ \text{rad}\ (114.6^\circ)\), \(z = 1.8\ \text{m}\). Then
\[ \begin{aligned} x &= 1\cos 2 \approx -0.4161\ \text{m} \\ y &= 1\sin 2 \approx 0.9093\ \text{m} \end{aligned} \]At \(t = 3\ \text{s}\): \(r = 2\ \text{m}\), \(\theta = 6\ \text{rad}\ (343.8^\circ)\), \(z = 2.4\ \text{m}\). Then
\[ \begin{aligned} x &= 2\cos 6 \approx 1.920\ \text{m} \\ y &= 2\sin 6 \approx -0.5588\ \text{m} \end{aligned} \]Between the two instants the arm has turned through \(4\ \text{rad}\), about \(229^\circ\). Set your calculator to radians: \(\cos 2\) in degree mode gives \(0.9994\), which is wrong here.
Check your understanding
Key takeaways
- Polar coordinates: \(r \ge 0\) is the distance from \(O\), and \(\theta\) the angle from the \(+x\) axis, counter-clockwise, in radians.
- \(x = r\cos\theta\), \(y = r\sin\theta\); \(r = \sqrt{x^2 + y^2}\) and \(\theta = \atantwo(y, x)\). Check the quadrant whenever \(x \lt 0\).
- Cylindrical coordinates add the height \(z\); \(r\) is then measured from the \(z\)-axis.
- The position vector is \(\rvec = r\,\er + z\,\ez\). The angle is hidden in the direction of \(\er\).
- A motion is \(r(t)\), \(\theta(t)\), \(z(t)\). Next, Lesson 3 shows how the unit vectors turn as \(\theta\) changes.