Lesson 4 · 30 min

Velocity

Differentiate the position \(\rvec = r\,\er + z\,\ez\) and the turning unit vector adds a second term. A particle moves out from the axis at the rate \(\dot r\) and round it at \(r\dot\theta\): those are the radial and transverse components of its velocity.

Learning objectives

Derive \(\vvec = \dot r\,\er + r\dot\theta\,\et + \dot z\,\ez\) from the position vector.
Explain what the radial component \(v_r = \dot r\) and the transverse component \(v_\theta = r\dot\theta\) measure.
Find the speed and the direction of the velocity from its components.
Find \(\dot r\) and \(\dot\theta\) from a known velocity, as a radar or a tracking camera does.

Differentiating the position

The velocity is the time derivative of the position vector. Use the product rule on \(r\,\er\), because both the length \(r\) and the direction \(\er\) change, and use \(d\er/dt = \dot\theta\,\et\) from Lesson 3. The unit vector \(\ez\) is constant.

\[ \begin{aligned} \vvec = \frac{d\rvec}{dt} &= \frac{d}{dt}\left(r\,\er\right) + \frac{d}{dt}\left(z\,\ez\right) \\ &= \dot r\,\er + r\,\frac{d\er}{dt} + \dot z\,\ez \\ &= \dot r\,\er + r\dot\theta\,\et + \dot z\,\ez \end{aligned} \]

Velocity in cylindrical coordinates

\[ \vvec = \dot r\,\colR{\er} + r\dot\theta\,\colT{\et} + \dot z\,\colZ{\ez} \] \[ v_r = \dot r, \qquad v_\theta = r\dot\theta, \qquad v_z = \dot z, \qquad v = |\vvec| = \sqrt{\dot r^2 + (r\dot\theta)^2 + \dot z^2} \]

For motion in a plane, drop the \(z\) term: \(\vvec = \dot r\,\er + r\dot\theta\,\et\).

\(v_r = \dot r\), the radial component, is how fast the distance from the axis grows: an arm telescoping in or out, a collar sliding along a rod, a cord being reeled in. It is negative when \(r\) decreases.
\(v_\theta = r\dot\theta\), the transverse component, is the speed of going round. \(\dot\theta\) is the angular velocity of the line \(OP\) (in rad/s), and the same turning rate carries a point farther out faster: this is the familiar \(v = \omega r\).
\(v_z = \dot z\) is the climbing speed, exactly as in rectangular coordinates.

The three components are perpendicular, so the speed follows from Pythagoras. The velocity is always tangent to the path; its components describe how that tangent direction splits into "out", "round" and "up".

See the two components

Figure 4.1 Choose a motion and press Play, or drag \(P\) along its path. The green arrow is the velocity \(\vvec\), always tangent to the path. Its components \(v_r\,\er\) (dashed orange, along \(OP\)) and \(v_\theta\,\et\) (dashed violet, across \(OP\)) are drawn tip to tail and add up to \(\vvec\). In Cyclist on a straight road, \(\vvec\) never changes, yet \(v_r\) and \(v_\theta\) do, because \(\er\) and \(\et\) turn as the cyclist passes.

Speed and direction

In the \(r\)–\(\theta\) plane the velocity makes an angle \(\psi\) with the radial line \(OP\) (measured from \(\er\) toward \(\et\)), where

\[ \tan\psi = \frac{v_\theta}{v_r} = \frac{r\dot\theta}{\dot r} \]

A purely radial motion (\(\dot\theta = 0\)) has \(\psi = 0\); motion on a circle about \(O\) (\(\dot r = 0\)) has \(\psi = 90^\circ\). Lesson 6 shows that \(\psi\) depends only on the shape of the path. If the motion is in space, the angle of \(\vvec\) above the horizontal is \(\tan^{-1}\!\big(v_z/\sqrt{v_r^2 + v_\theta^2}\big)\).

Example 4.1 — A robot arm that extends while it slews

A cylindrical robot rotates its arm at a constant \(\dot\theta = 2\ \text{rad/s}\), extends it at \(\dot r = 0.5\ \text{m/s}\) and raises it at \(\dot z = 0.3\ \text{m/s}\). Find the velocity of the gripper and its speed at the instant when \(r = 1\ \text{m}\).

Show solution

\[ \begin{aligned} v_r &= \dot r = 0.5\ \text{m/s} \\ v_\theta &= r\dot\theta = (1)(2) = 2\ \text{m/s} \\ v_z &= \dot z = 0.3\ \text{m/s} \end{aligned} \] \[ \vvec = (0.5\,\er + 2\,\et + 0.3\,\ez)\ \text{m/s}, \qquad v = \sqrt{0.5^2 + 2^2 + 0.3^2} = \sqrt{4.34} \approx 2.083\ \text{m/s} \]

Most of the speed comes from swinging round. In the horizontal plane the velocity makes \(\psi = \tan^{-1}(2/0.5) \approx 75.96^\circ\) with the arm, and it climbs at \(\tan^{-1}\!\big(0.3/\sqrt{0.5^2 + 2^2}\big) \approx 8.27^\circ\) above the horizontal. Figure 4.2 shows this instant.

Figure 4.2 The robot of Example 4.1 in space: \(r = (0.5 + 0.5t)\ \text{m}\), \(\theta = 2t\ \text{rad}\), \(z = (1.5 + 0.3t)\ \text{m}\). The green arrow is \(\vvec\); the thin arrows are its components along \(\er\), \(\et\) and \(\ez\), drawn tip to tail. Press Play, or Jump to Example 4.1 to see the instant \(t = 1\ \text{s}\). Drag the view to rotate it.

Tracking: from velocity to \(\dot r\) and \(\dot\theta\)

A radar or a tracking camera at \(O\) measures \(r\), \(\theta\) and their rates. The relations work both ways: given \(\dot r\) and \(\dot\theta\) you get the velocity, and given the velocity you get the rates the tracker must follow, \(\dot r = v_r\) and \(\dot\theta = v_\theta/r\).

Example 4.2 — Radar tracking an aircraft

A radar measures an aircraft at a range \(r = 8\ \text{km}\), with the range decreasing at \(150\ \text{m/s}\) and the line of sight turning at \(\dot\theta = 0.02\ \text{rad/s}\) (assume level flight, so the motion is in a plane). Find the aircraft's speed and the angle between its velocity and the line from the aircraft toward the radar.

Show solution

Work in meters: \(r = 8000\ \text{m}\), \(\dot r = -150\ \text{m/s}\) (the range decreases).

\[ \begin{aligned} v_r &= \dot r = -150\ \text{m/s} \\ v_\theta &= r\dot\theta = (8000)(0.02) = 160\ \text{m/s} \\ v &= \sqrt{(-150)^2 + 160^2} \approx 219.3\ \text{m/s} \end{aligned} \]

The line toward the radar is \(-\er\). The velocity has \(150\ \text{m/s}\) along \(-\er\) and \(160\ \text{m/s}\) along \(\et\), so it makes \(\tan^{-1}(160/150) \approx 46.85^\circ\) with that line, turned toward \(\et\). The aircraft is not flying straight at the radar, which is why the line of sight turns.

Example 4.3 — Following a cyclist

A camera at \(O\) films a cyclist riding at a constant \(5\ \text{m/s}\) along a straight road. Take the road as the line \(y = 10\ \text{m}\), traveled in the \(+x\) direction. How fast is the distance to the camera changing, and how fast must the camera turn, when the cyclist is at \(x = 10\ \text{m}\)? How fast must it turn as the cyclist passes closest?

Show solution

At \((10, 10)\ \text{m}\): \(r = \sqrt{200} \approx 14.14\ \text{m}\) and \(\theta = 45^\circ\). The velocity is \(\vvec = 5\,\ihat\ \text{m/s}\). Resolve it along \(\er\) and \(\et\) (Lesson 3):

\[ \begin{aligned} v_r &= v_x\cos\theta + v_y\sin\theta = 5\cos 45^\circ \approx 3.536\ \text{m/s} \\ v_\theta &= -v_x\sin\theta + v_y\cos\theta = -5\sin 45^\circ \approx -3.536\ \text{m/s} \end{aligned} \]

So the distance grows at \(\dot r = 3.536\ \text{m/s}\), and the camera turns at

\[ \dot\theta = \frac{v_\theta}{r} = \frac{-3.536}{14.14} = -0.25\ \text{rad/s} \]

(clockwise, since \(\theta\) decreases as the cyclist rides on). At the closest point, \((0, 10)\ \text{m}\), the velocity is all transverse: \(v_\theta = -5\ \text{m/s}\) at \(r = 10\ \text{m}\), so \(\dot\theta = -0.5\ \text{rad/s}\), the fastest turn. In Figure 4.1, choose Cyclist on a straight road to watch \(v_r\) and \(v_\theta\) trade off while \(\vvec\) stays constant.

Check your understanding

Key takeaways

\(\vvec = \dot r\,\er + r\dot\theta\,\et + \dot z\,\ez\). The \(r\dot\theta\) term comes from differentiating the turning unit vector \(\er\).
\(v_r = \dot r\) measures moving out from the axis, \(v_\theta = r\dot\theta\) going round it, and \(v_z = \dot z\) climbing.
\(v = \sqrt{\dot r^2 + (r\dot\theta)^2 + \dot z^2}\), and in the plane \(\tan\psi = v_\theta/v_r\) gives the direction relative to the radial line.
A constant velocity can still have changing \(v_r\) and \(v_\theta\), because the unit vectors turn.
Next, Lesson 5 differentiates \(\vvec\) to get the acceleration, with its centripetal and Coriolis terms.