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Section 4.5 Application: Circular Motion: Velocity and Angular Velocity

A second physical application of the cross product describes the motion of an object travelling in a circle: a rider on a circular track, a ball whirled on a string, a point on a spinning wheel, or the moon on its (nearly circular) orbit. In each case the object turns about a fixed centre, and the natural way to keep track of where it is at time \(t\) is not by its coordinates \(x\) and \(y\) but by the single angle \(\theta(t)\) it has swung through. Our goal is to study how the velocity and the rate of change of that angle are related.
Throughout this section we write unit vectors with hats, so that \(\hat i\text{,}\) \(\hat j\) and \(\hat k\) are the usual fixed unit vectors \(\mathbf i\text{,}\) \(\mathbf j\) and \(\mathbf k\) along the coordinate axes.

Subsection 4.5.1 A Frame that Turns with the Particle

Let an object move on a circle of radius \(r\) centred at the origin of the \(xy\)-plane, and let \(\theta(t)\) be the angle its position makes with the positive \(x\)-axis at time \(t\text{,}\) as in Figure 4.23. At the point \(P\) where the particle sits we attach two unit vectors:
  • \(\hat r(t)\text{,}\) pointing radially outward, away from the centre, and
  • \(\hat\theta(t)\text{,}\) perpendicular to \(\hat r(t)\) and tangent to the circle, pointing in the direction of increasing \(\theta\text{.}\)
Reading their components off the figure,
\begin{align} \hat r(t) \amp = \cos\theta(t)\,\hat i + \sin\theta(t)\,\hat j,\tag{4.11}\\ \hat\theta(t) \amp = -\sin\theta(t)\,\hat i + \cos\theta(t)\,\hat j.\tag{4.12} \end{align}
The position of the particle is then simply
\begin{equation} \vec r(t) = r\,\hat r(t).\tag{4.13} \end{equation}
The pair \(\hat i\text{,}\) \(\hat j\) is fixed once and for all, while the pair \(\hat r(t)\text{,}\) \(\hat\theta(t)\) rides along with the particle and turns as it goes; this is exactly what makes the second pair convenient. Note that together with \(\hat k\) the moving pair forms a right-handed frame, since
\begin{align} \hat r \times \hat\theta \amp = \begin{vmatrix} \hat i \amp \hat j \amp \hat k \\ \cos\theta \amp \sin\theta \amp 0 \\ -\sin\theta \amp \cos\theta \amp 0 \end{vmatrix}\notag\\ \amp = \left(\cos^2\theta + \sin^2\theta\right)\hat k = \hat k.\tag{4.14} \end{align}
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A circle of radius r centred at the origin. A particle sits at a point P on the circle, its position vector making an angle theta of t with the positive x-axis. At P the unit vector r hat points radially outward along the position vector, and the unit vector theta hat is perpendicular to it, tangent to the circle in the counterclockwise direction. The fixed unit vectors i and j are also drawn at P for comparison.
Figure 4.23. The moving frame at the point \(P\text{:}\) \(\hat r(t)\) points radially outward and \(\hat\theta(t)\) is tangent to the circle. The fixed vectors \(\hat i\) and \(\hat j\) are drawn at \(P\) for comparison; unlike \(\hat r\) and \(\hat\theta\text{,}\) they do not turn as the particle moves.

Subsection 4.5.2 The Velocity in Circular Motion

Because the particle moves, the angle \(\theta\) depends on \(t\text{,}\) and so do the two moving unit vectors. Differentiating (4.11) and (4.12) with the chain rule gives
\begin{align} \frac{d\hat r}{dt} \amp = \left(-\sin\theta\,\hat i + \cos\theta\,\hat j\right)\frac{d\theta}{dt} = \frac{d\theta}{dt}\,\hat\theta,\tag{4.15}\\ \frac{d\hat\theta}{dt} \amp = \left(-\cos\theta\,\hat i - \sin\theta\,\hat j\right)\frac{d\theta}{dt} = -\frac{d\theta}{dt}\,\hat r.\tag{4.16} \end{align}
Each unit vector has constant length, so it can only turn; and turning a unit vector moves its tip perpendicular to itself, which is exactly what (4.15) and (4.16) say.
Now differentiate the position (4.13). The radius \(r\) is constant on a circle, so only the unit vector needs differentiating:
\begin{equation} \vec v(t) = \frac{d\vec r}{dt} = r\,\frac{d\hat r}{dt} = r\,\frac{d\theta}{dt}\,\hat\theta(t).\tag{4.17} \end{equation}
The velocity has no \(\hat r\)-component at all: it points purely along \(\hat\theta\text{,}\) tangent to the circle. Its \(\hat\theta\)-component
\begin{equation} v_\theta = r\,\frac{d\theta}{dt}\tag{4.18} \end{equation}
is called the tangential component of the velocity. It is positive when the particle turns counterclockwise and negative when it turns clockwise.

Definition 4.24 Angular Speed.

The angular speed of the particle is the magnitude of the rate of change of its angle,
\begin{equation} \omega = \left|\frac{d\theta}{dt}\right|,\tag{4.19} \end{equation}
measured in radians per second. By (4.17), the speed of the particle and its angular speed are related by
\begin{equation} v = \left\|\vec v\right\| = r\,\omega.\tag{4.20} \end{equation}
It is worth seeing where (4.20) comes from geometrically, without any differentiation of unit vectors. In a short time interval \(\Delta t\) the particle moves from \(\vec r(t)\) to \(\vec r(t+\Delta t)\text{,}\) turning through a small angle \(\Delta\theta\text{,}\) as in Figure 4.25. The two position vectors have the same length \(r\text{,}\) so they form an isosceles triangle whose third side is the displacement \(\Delta\vec r = \vec r(t+\Delta t) - \vec r(t)\text{.}\) Splitting that triangle down the middle gives
\begin{equation} \left\|\Delta\vec r\right\| = 2r\sin\frac{\Delta\theta}{2}.\tag{4.21} \end{equation}
For a small angle the power series \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\) is dominated by its first term, so \(\sin(\Delta\theta/2)\approx \Delta\theta/2\) and the chord is very nearly the arc it cuts off: \(\left\|\Delta\vec r\right\| \approx r\,\Delta\theta\text{.}\) Dividing by \(\Delta t\) and letting \(\Delta t \to 0\text{,}\)
\begin{align*} v \amp = \lim_{\Delta t\to 0}\frac{\left\|\Delta\vec r\right\|}{\Delta t} = r\lim_{\Delta t\to 0}\frac{\left|\Delta\theta\right|}{\Delta t}\\ \amp = r\left|\frac{d\theta}{dt}\right| = r\,\omega, \end{align*}
which is (4.20) again. The direction comes out of the same picture: as \(\Delta t \to 0\) the chord \(\Delta\vec r\) lines up with the tangent line to the circle, so \(\vec v\) is perpendicular to \(\vec r\text{,}\) in agreement with (4.17).
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Two position vectors of equal length r are drawn from the centre of a dashed circle, separated by a small angle delta theta. The chord delta r joins the tip of the first to the tip of the second. A dashed tangent line touches the circle at the first point, and the velocity arrow at that point lies along it.
Figure 4.25. The displacement \(\Delta\vec r\) is the chord subtending the angle \(\Delta\theta\text{.}\) For small \(\Delta\theta\) the chord is nearly the arc, \(\left\|\Delta\vec r\right\| \approx r\,\Delta\theta\text{,}\) and its direction approaches that of the tangent line, along which \(\vec v\) points.

Subsection 4.5.3 Angular Velocity as a Vector

So far the rate of turning has been a number, \(d\theta/dt\text{.}\) But a rotation has a direction as well as a rate: the plane of the circle can be tilted any which way, and the particle can go around it either way. Both pieces of information can be carried by a single vector, provided we point it along the axis of rotation rather than along the motion.
We always choose a right-handed coordinate system, with \(\theta\) increasing in the counterclockwise direction when viewed from the positive \(z\)-axis.

Definition 4.26 Angular Velocity Vector.

For an object moving in a circle in the \(xy\)-plane about the origin, the angular velocity vector is
\begin{equation} \vec\omega = \frac{d\theta}{dt}\,\hat k = \omega_z\,\hat k,\tag{4.22} \end{equation}
directed along the axis of rotation. Its SI units are \(\text{rad}\cdot\text{s}^{-1}\text{,}\) and the angular speed of (4.19) is its magnitude, \(\omega = \left\|\vec\omega\right\| = \left|\omega_z\right|\text{.}\)
The sign of \(\omega_z\) records which way the object goes around, and the right-hand rule turns that sign into a direction in space: curl the fingers of your right hand in the direction of the motion and your thumb points along \(\vec\omega\text{.}\) If the particle circulates counterclockwise as seen from above, then \(\omega_z = d\theta/dt > 0\) and \(\vec\omega\) points in the \(+\hat k\)-direction; if it circulates clockwise, then \(\omega_z < 0\) and \(\vec\omega\) points in the \(-\hat k\)-direction. The two cases are shown in Figure 4.27.
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Two panels showing a particle on a circular orbit in the horizontal plane, drawn in an oblique view with the z-axis vertical. In the left panel the particle circulates counterclockwise as seen from above and the angular velocity vector points up along the positive z-direction. In the right panel the particle circulates clockwise and the angular velocity vector points down along the negative z-direction. In both panels the velocity is tangent to the orbit.
Figure 4.27. The angular velocity vector points along the axis of rotation, in the direction given by the right-hand rule: up the \(z\)-axis for counterclockwise motion (left), down it for clockwise motion (right). Note that \(\vec\omega\) is not tangent to the orbit — the velocity \(\vec v\) is.

Subsection 4.5.4 Velocity as a Cross Product

We now have two vectors attached to the motion: the angular velocity \(\vec\omega\text{,}\) along the axis, and the position \(\vec r\text{,}\) in the plane of the circle. The velocity is perpendicular to both, which is precisely the job of the cross product.
To verify this, first compute \(\hat k \times \hat r\) using (4.11):
\begin{align} \hat k \times \hat r \amp = \begin{vmatrix} \hat i \amp \hat j \amp \hat k \\ 0 \amp 0 \amp 1 \\ \cos\theta \amp \sin\theta \amp 0 \end{vmatrix}\notag\\ \amp = -\sin\theta\,\hat i + \cos\theta\,\hat j = \hat\theta.\tag{4.24} \end{align}
Therefore, using (4.22) and (4.13),
\begin{align*} \vec\omega \times \vec r \amp = \left(\frac{d\theta}{dt}\,\hat k\right)\times\left(r\,\hat r\right) = r\,\frac{d\theta}{dt}\left(\hat k \times \hat r\right)\\ \amp = r\,\frac{d\theta}{dt}\,\hat\theta = \vec v, \end{align*}
by (4.17). See Figure 4.29.
The geometry of the cross product delivers all three features of the velocity at once, with no computation:
  • Magnitude. Since \(\vec\omega\) is perpendicular to the plane of the circle, it is perpendicular to \(\vec r\text{,}\) so \(\left\|\vec v\right\| = \left\|\vec\omega\right\|\left\|\vec r\right\| \sin 90^\circ = \omega r\text{,}\) recovering (4.20).
  • Direction. The cross product is perpendicular to \(\vec r\text{,}\) so the velocity is tangent to the circle; and it is perpendicular to \(\vec\omega\text{,}\) so the motion stays in the plane of the orbit.
  • Sense. The right-hand rule applied to \(\vec\omega \times \vec r\) automatically sends the particle counterclockwise when \(\omega_z > 0\) and clockwise when \(\omega_z < 0\text{,}\) since reversing \(\vec\omega\) reverses the cross product.
There is a bonus. A piece of \(\vec r\) running along the axis is parallel to \(\vec\omega\) and so contributes nothing to the cross product, since \(\hat k \times \left(z\,\hat k\right) = \vec 0\text{.}\) Consequently (4.23) remains true when \(\vec r\) is measured from any point on the axis of rotation, not just from the centre of the circle.
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A circular orbit in the horizontal plane drawn in an oblique view, with the z-axis vertical. The angular velocity vector points straight up along the z-axis from the centre O. The position vector runs from O out to the particle on the orbit, and the velocity vector at the particle is tangent to the orbit, equal to the cross product of the angular velocity with the position vector.
Figure 4.29. The velocity of a particle in circular motion is \(\vec v = \vec\omega \times \vec r\text{.}\) Because \(\vec\omega\) and \(\vec r\) are perpendicular, the speed is \(\omega r\text{,}\) and the right-hand rule places \(\vec v\) tangent to the orbit in the direction of travel.

Subsection 4.5.5 Examples

Example 4.30 Angular velocity from an angle function.

A particle moves in a circle of radius \(R\) centred at the origin of the \(xy\)-plane; at \(t = 0\) it is on the positive \(x\)-axis. The angle it makes with the positive \(x\)-axis is \(\theta(t) = At - Bt^3\text{,}\) where \(A\) and \(B\) are positive constants.
  1. Find the angular velocity vector.
  2. Find the velocity vector, in polar coordinates.
  3. At what time \(t_1\) is the angular velocity zero?
  4. In which direction does the angular velocity point for \(t < t_1\text{,}\) and for \(t > t_1\text{?}\)
Solution.
The rate of change of the angle is
\begin{equation*} \frac{d\theta}{dt} = A - 3Bt^2. \end{equation*}
Part 1. By (4.22),
\begin{equation*} \vec\omega(t) = \frac{d\theta}{dt}\,\hat k = \left(A - 3Bt^2\right)\hat k. \end{equation*}
Part 2. By (4.17),
\begin{equation*} \vec v(t) = R\,\frac{d\theta}{dt}\,\hat\theta(t) = R\left(A - 3Bt^2\right)\hat\theta(t). \end{equation*}
Equivalently, one can obtain this from (4.23): \(\vec v = \vec\omega\times\vec r = \left(A - 3Bt^2\right)\hat k \times R\,\hat r = R\left(A-3Bt^2\right)\hat\theta\text{,}\) using (4.24).
Part 3. The angular velocity vanishes when
\begin{equation*} A - 3Bt_1^2 = 0 \quad\Longrightarrow\quad t_1 = \sqrt{\frac{A}{3B}}. \end{equation*}
At that instant the particle is momentarily at rest.
Part 4. For \(t < t_1\) we have \(A - 3Bt^2 > 0\text{,}\) so \(\vec\omega\) points in the \(+\hat k\)-direction and the particle travels counterclockwise. For \(t > t_1\) we have \(A - 3Bt^2 < 0\text{,}\) so \(\vec\omega\) points in the \(-\hat k\)-direction and the particle has reversed and travels clockwise.

Example 4.31 A turntable.

A record turntable spins at \(33\tfrac13\) revolutions per minute, counterclockwise when viewed from above. Take the axis of rotation to be the \(z\)-axis, with the turntable lying in the \(xy\)-plane, distances in meters.
  1. Find the angular velocity vector \(\vec\omega\text{.}\)
  2. Find the velocity of the point of the record at \(\vec r = \left\langle 0.12,\,0,\,0 \right\rangle\text{,}\) and its speed.
Solution.
Part 1. One revolution is \(2\pi\) radians, so the frequency \(f = 33\tfrac13\) rev/min \(= 0.5556\) rev/s corresponds to an angular speed
\begin{equation*} \omega = 2\pi f = 2\pi(0.5556) \approx 3.49\ \text{rad}\cdot\text{s}^{-1}. \end{equation*}
The rotation is counterclockwise seen from the \(+z\)-axis, so \(\omega_z > 0\) and
\begin{equation*} \vec\omega \approx \left\langle 0,\,0,\,3.49 \right\rangle \ \text{rad}\cdot\text{s}^{-1}. \end{equation*}
(For comparison, the period of one revolution is \(T = 2\pi/\omega = 1/f = 1.8\) s.)
Part 2. By (4.23),
\begin{align*} \vec v = \vec\omega\times\vec r \amp = \begin{vmatrix} \hat i \amp \hat j \amp \hat k \\ 0 \amp 0 \amp 3.49 \\ 0.12 \amp 0 \amp 0 \end{vmatrix}\\ \amp = \left(0 - 0\right)\hat i - \left(0 - (3.49)(0.12)\right)\hat j + \left(0 - 0\right)\hat k\\ \amp = \left\langle 0,\, 0.419,\, 0 \right\rangle\ \text{m}\cdot\text{s}^{-1}. \end{align*}
The point in question is on the positive \(x\)-axis, and its velocity points in the \(+y\)-direction — tangent to its circle and consistent with counterclockwise motion, as it should be. Its speed is \(0.419\ \text{m}\cdot\text{s}^{-1}\text{,}\) which agrees with (4.20): \(v = r\omega = (0.12)(3.49) \approx 0.419\ \text{m}\cdot\text{s}^{-1}\text{.}\)
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