Lesson 3 · 30 min

Unit Vectors That Turn

\(\er\) points away from the axis and \(\et\) points around it, both from wherever the particle is now. As the particle moves round, they turn with it. How fast they turn is the key to every velocity and acceleration formula that follows.

Learning objectives

Draw the unit vectors \(\er\), \(\et\) and \(\ez\) at a point, and check that they form a right-handed set.
Write \(\er\) and \(\et\) in terms of \(\ihat\) and \(\jhat\).
Show that \(d\er/dt = \dot\theta\,\et\) and \(d\et/dt = -\dot\theta\,\er\), and explain them geometrically.
Resolve a force (or any vector) into radial and transverse components, and convert back to \(x\) and \(y\).

Three unit vectors at the particle

At the particle \(P\), with coordinates \((r, \theta, z)\), set up three perpendicular unit vectors, each pointing in the direction in which one coordinate increases:

\(\er\): radial

Horizontal, from the \(z\)-axis out through \(P\): the direction of increasing \(r\).

\(\et\): transverse

Horizontal, perpendicular to \(\er\), pointing the way \(\theta\) increases (counter-clockwise seen from above).

\(\ez\): axial

Along the \(z\)-axis, the same as \(\khat\). It never changes.

They are a right-handed set, like \(\ihat\), \(\jhat\), \(\khat\): \(\er \times \et = \ez\). Unlike \(\ihat\) and \(\jhat\), the first two depend on where the particle is. Resolving them along \(x\) and \(y\) (Figure 3.1) gives

Unit vectors in terms of \(\ihat\) and \(\jhat\)

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

Check one case: at \(\theta = 90^\circ\) the particle is on the \(+y\) axis, so "away from the axis" is \(\jhat\) and "counter-clockwise" is \(-\ihat\). The formulas give \(\er = \jhat\) and \(\et = -\ihat\).

Figure 3.1 Top view. Drag \(P\) (or use the slider). The unit vectors \(\er\) (orange) and \(\et\) (violet) at \(P\) are one grid ring long; the fixed \(\ihat\) and \(\jhat\) sit at the origin. Turn on the components to see \(\er = \cos\theta\,\ihat + \sin\theta\,\jhat\) or \(\et = -\sin\theta\,\ihat + \cos\theta\,\jhat\) as dashed legs. Move \(P\) out along the same line: \(\er\) and \(\et\) do not change. Move it round: they turn.

Figure 3.2 The same unit vectors in space, with \(\ez\) pointing up. Press Play orbit to carry \(P\) around the \(z\)-axis: the local arrows \(\er\), \(\et\), \(\ez\) travel with \(P\), and \(\er\) and \(\et\) turn, while \(\ihat\), \(\jhat\), \(\khat\) at the origin never move. Drag \(r\) or \(z\) instead: the local arrows slide but do not turn.

How fast do the unit vectors turn?

A vector of constant length can still change: its direction can turn. As the particle moves, \(\theta\) is a function of time, so \(\er\) and \(\et\) are functions of time too. Differentiate the \(\ihat\), \(\jhat\) forms with the chain rule (\(\ihat\) and \(\jhat\) are constant):

\[ \begin{aligned} \frac{d\er}{dt} &= \frac{d}{dt}\left(\cos\theta\,\ihat + \sin\theta\,\jhat\right) = \left(-\sin\theta\,\ihat + \cos\theta\,\jhat\right)\dot\theta = \dot\theta\,\et \\[4pt] \frac{d\et}{dt} &= \frac{d}{dt}\left(-\sin\theta\,\ihat + \cos\theta\,\jhat\right) = \left(-\cos\theta\,\ihat - \sin\theta\,\jhat\right)\dot\theta = -\dot\theta\,\er \end{aligned} \]

Time derivatives of the unit vectors

\[ \frac{d\colR{\er}}{dt} = \dot\theta\,\colT{\et}, \qquad \frac{d\colT{\et}}{dt} = -\dot\theta\,\colR{\er}, \qquad \frac{d\colZ{\ez}}{dt} = \mathbf{0} \]

The unit vectors turn only when \(\theta\) changes. Moving in \(r\) or \(z\) does not turn them.

Hibbeler writes these as \(\dot{\mathbf{u}}_r = \dot\theta\,\mathbf{u}_\theta\) and \(\dot{\mathbf{u}}_\theta = -\dot\theta\,\mathbf{u}_r\). The geometry says the same thing. Turn by a small angle \(\Delta\theta\): the tip of \(\er\) slides a distance of about \(\Delta\theta\) around the unit circle, in the direction of \(\et\). The tip of \(\et\) slides the same distance, toward \(-\er\).

Figure 3.3 Both unit vectors drawn from the origin at \(\theta\) (solid) and at \(\theta + \Delta\theta\) (dashed). The arrows joining the tips are the changes \(\Delta\er\) and \(\Delta\et\). Shrink \(\Delta\theta\): \(\Delta\er\) lines up with \(\et\) and \(|\Delta\er|/\Delta\theta\) approaches 1. Likewise \(\Delta\et\) lines up with \(-\er\).

So for a small \(\Delta\theta\), \(\Delta\er \approx \Delta\theta\,\et\). Divide by the time \(\Delta t\) the turn takes and let \(\Delta t \to 0\): \(\Delta\theta/\Delta t\) becomes \(\dot\theta\), and \(d\er/dt = \dot\theta\,\et\), exactly the chain-rule result.

Radial and transverse components of a vector

Any vector at \(P\), such as a force \(\Fvec\), can be written either way:

\[ \Fvec = F_x\,\ihat + F_y\,\jhat + F_z\,\khat = F_r\,\er + F_\theta\,\et + F_z\,\ez \]

Its radial component \(F_r\) and transverse component \(F_\theta\) are its projections onto \(\er\) and \(\et\). The \(z\) component is the same in both. Use the dot product with the unit vectors above:

Rectangular ↔ radial and transverse

\[ \begin{aligned} F_r &= \Fvec\cdot\er = F_x\cos\theta + F_y\sin\theta & \qquad F_x &= F_r\cos\theta - F_\theta\sin\theta \\ F_\theta &= \Fvec\cdot\et = -F_x\sin\theta + F_y\cos\theta & \qquad F_y &= F_r\sin\theta + F_\theta\cos\theta \end{aligned} \]

Here \(\theta\) is the angle of the particle's position, not of the force.

Figure 3.4 A force \(\Fvec\) (teal) acts on the particle at \(P\). Set its rectangular components and the particle's angle \(\theta\). The dashed orange and violet arrows are \(F_r\,\er\) and \(F_\theta\,\et\); they add up to \(\Fvec\). Keep \(\Fvec\) fixed and drag \(\theta\): the force does not change, but its radial and transverse components trade off as the axes \(\er\), \(\et\) turn.

Example 3.1 — Weight on a rotating arm

A \(2\ \text{kg}\) collar rides on an arm that turns in a vertical plane about \(O\). The arm is at \(\theta = 30^\circ\) above the horizontal \(x\)-axis, with \(y\) pointing up. Find the radial and transverse components of the collar's weight.

Show solution

The weight is \(\mathbf{W} = -mg\,\jhat\), so \(W_x = 0\) and \(W_y = -mg = -(2)(9.81) = -19.62\ \text{N}\). Then

\[ \begin{aligned} W_r &= W_x\cos\theta + W_y\sin\theta = -mg\sin\theta = -19.62\sin 30^\circ = -9.81\ \text{N} \\ W_\theta &= -W_x\sin\theta + W_y\cos\theta = -mg\cos\theta = -19.62\cos 30^\circ \approx -16.99\ \text{N} \end{aligned} \]

Both are negative: the weight pulls the collar back toward \(O\) along the arm (\(-\er\)) and against increasing \(\theta\) (\(-\et\)). Check: \(\sqrt{9.81^2 + 16.99^2} \approx 19.62\ \text{N}\). You will use \(W_r = -mg\sin\theta\) and \(W_\theta = -mg\cos\theta\) for arms in vertical planes in Lesson 7.

Example 3.2 — A force in rectangular components

A force \(\Fvec = (30\,\ihat + 40\,\jhat)\ \text{N}\) acts on a particle at \(\theta = 60^\circ\). Find \(F_r\) and \(F_\theta\).

Show solution

\[ \begin{aligned} F_r &= 30\cos 60^\circ + 40\sin 60^\circ = 15 + 34.64 = 49.64\ \text{N} \\ F_\theta &= -30\sin 60^\circ + 40\cos 60^\circ = -25.98 + 20 = -5.98\ \text{N} \end{aligned} \]

Check: \(\sqrt{49.64^2 + 5.98^2} \approx 50.0\ \text{N} = \sqrt{30^2 + 40^2}\). Resolving changes the components, never the magnitude. The force points almost straight out from the axis (\(F_r\) is large) and slightly clockwise (\(F_\theta \lt 0\)). Try it in Figure 3.4.

Check your understanding

Key takeaways

\(\er\) points away from the axis, \(\et\) around it (the way \(\theta\) increases), and \(\ez\) along it; \(\er \times \et = \ez\).
\(\er = \cos\theta\,\ihat + \sin\theta\,\jhat\) and \(\et = -\sin\theta\,\ihat + \cos\theta\,\jhat\) depend on \(\theta\), so they turn as the particle moves round.
\(d\er/dt = \dot\theta\,\et\), \(d\et/dt = -\dot\theta\,\er\) and \(d\ez/dt = \mathbf{0}\).
\(F_r = F_x\cos\theta + F_y\sin\theta\) and \(F_\theta = -F_x\sin\theta + F_y\cos\theta\), with \(\theta\) the angle of the particle's position.
Next, Lesson 4 differentiates \(\rvec = r\,\er + z\,\ez\) to get the velocity.