# Angular Kinematics

**Angular kinematics** is the study of rotational motion in the absence of forces. The equations of angular kinematics are extremely similar to the usual equations of kinematics, with quantities like displacements replaced by angular displacements and velocities replaced by angular velocities. Just as kinematics is routinely used to describe the trajectory of almost any physical system moving linearly, the equations of angular kinematics are relevant to most rotating physical systems.

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## Basic Equations of Angular Kinematics

In purely rotational (circular) motion, the equations of angular kinematics are:

\[v = r\omega, \qquad a_c = -r\omega^2, \qquad a = r\alpha\]

The tangential velocity \(v\) describes the velocity of an object tangent to its path in rotational motion at angular frequency \(\omega\) and radius \(r\). This is the velocity an object would follow if it suddenly broke free of rotational motion and traveled along a straight line. The rate of change of this velocity is the tangential acceleration \(a\). The centripetal acceleration \(a_c\) is a second acceleration experienced by rotating objects, because changing the direction of a velocity vector requires an acceleration. Since the direction of the velocity vector changes constantly in rotational motion, rotating objects must be continuously accelerated towards the axis of rotation by some force providing a centripetal acceleration.

From the above equations, the usual kinematic equations hold in angular form. If an object undergoes constant angular acceleration \(\alpha\), the total angular displacement is:

\[\theta - \theta_0 = \omega_0 t + \frac12 \alpha t^2\]

where \(\theta_0\) is the initial angle and \(\omega_0\) is the initial angular velocity. Similarly, the angular velocity changes according to:

\[\omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0)\]

in terms of the angular displacement, or

\[\omega = \omega_0 + \alpha t\]

in terms of time.

Though the above derivation gives the magnitudes of angular quantities correctly, it does not capture the fact that angular quantities are *also* vector quantities. The direction in which the angular velocity points can be found from the right-hand rule: curving the fingers of your right hand along the direction of rotation, your thumb points in the direction of the angular velocity vector, along the axis of rotation. This is true by definition; although it seems strange since the vector is perpendicular to the rotation, this definition turns out to be the only way to formulate a consistent vector theory of rotational forces.

A merry-go-round has a radius of about \(8 \text{ m}\). A child sitting on a horse on the outer edge sees his parents, standing still at the entrance, every \(20 \text{ s}\). If the child's horse could break free of its merry-go-round restraints and continue forward tangentially without accelerating, how fast would it be traveling in \(\text{m}/\text{s}\)?

## Derivation of Equations of Angular Kinematics

In a rotating frame of reference, it is often more convenient to use polar coordinates than Cartesian coordinates. Similarly, for vectors it is more convenient to use the **radial** and **tangential** vectors \(\hat{r}\) and \(\hat{\theta}\) in place of the usual Cartesian basis vectors \(\hat{x}\) and \(\hat{y}\). The radial vector to an object is defined so that it always points towards the object:

\[\hat{r} = \cos \theta\, \hat{x} + \sin \theta\, \hat{y}.\]

Note that the \(x\) and \(y\) components of the radial vector are just the usual polar coordinates, normalized to one. Similarly, the tangential vector \(\hat{\theta}\) is defined so that it is always orthogonal to the radial vector and tangent to the circle on which the radial vector lies:

\[\hat{\theta} = -\sin \theta\,\hat{x} + \cos \theta\, \hat{y}.\]

## Show that \(\frac{d\hat{r}}{dt} = \dot{\theta} \hat{\theta}\) and \(\frac{d\hat{\theta}}{dt} = -\dot{\theta} \hat{r}\), where dots indicate time derivatives.

Solution:

Note that in Cartesian coordinates, the derivatives of the basis vectors \(\hat{x}\), \(\hat{y}\), etc. always vanish, because these basis vectors are fixed. The polar basis vectors, however, rotate in time so that they are always pointing radially and tangentially along some trajectory. Computing the derivative from the definitions above using the chain rule:

\[ \begin{align} \frac{d\hat{r}}{dt} &= \frac{d}{dt} \left( \cos \theta\, \hat{x} + \sin \theta\, \hat{y}\right) = -\dot{\theta} \sin \theta\,\hat{x} + \dot{\theta} \cos \theta\,\hat{y} = \dot{\theta} \hat{\theta} \\ \frac{d\hat{\theta}}{dt} &= \frac{d}{dt} \left( -\sin \theta\, \hat{x} + \cos \theta\, \hat{y}\right) = -\dot{\theta} \cos \theta\,\hat{x} - \dot{\theta} \sin \theta\,\hat{y} = -\dot{\theta} \hat{r} \end{align} \] as claimed.

From the definitions above, all of the laws of angular kinematics are straightforward to derive. Suppose that the vector pointing to some rotating object is \(\vec{r} = r\hat{r}\) in polar coordinates, where \(r\) is the magnitude of the distance from the origin. The velocity of the object is then:

\[\frac{d\vec{r}}{dt} = \dot{r} \hat{r} +r\dot{\theta} \hat{\theta}.\]

The velocity has two components, as can be seen above. The first term, \(\dot{r} \hat{r}\), describes the radial velocity of the object away from the origin. The second term is the tangential velocity. Denoting \(\dot{\theta} = \omega\) as the **angular velocity**, the tangential velocity is just \(v = r\omega\).

Taking another derivative allows identification of the different terms contributing to the acceleration of the object:

\[ \begin{align} \frac{d^2 \vec{r}}{dt} &= \frac{d}{dt} \frac{d\vec{r}}{dt} = \frac{d}{dt} \left( \dot{r} \hat{r} +r\dot{\theta} \hat{\theta} \right) \\ &=(\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r} \dot{\theta}) \hat{\theta} \end{align} \]

The radial terms \(\ddot{r}\) and \(r\dot{\theta}^2 = r\omega^2\) describe the **radial acceleration** outward from the origin and the **centripetal acceleration** towards the origin, respectively. The tangential terms are the **tangential acceleration** \(a = r\ddot{\theta} = r\alpha\), where \(\alpha = \dot{\omega}\) is the **angular acceleration**, and \(2\dot{r} \dot{\theta} = 2\dot{r} \omega\), the **Coriolis acceleration**.

Remarkably, this derivation proves without reference to any forces at all that an object in circular motion with angular velocity \(\omega\) must accelerate radially inward with the centripetal acceleration \(a_c\) given above.

## Rotating Systems in Physics

Countless rotating systems in physics can be analyzed using the laws of angular kinematics; a few are explored in the following examples.

## As most kids learn, if you quickly rotate a tube containing a ball, the ball can be made to "slingshot" out the end at very high speeds. The same effect is visible in rotating a rod with a bead around it and many other physical scenarios. How fast does the velocity of the ball/bead increase if the tube/rod is rotated at constant angular velocity \(\omega\)?

Solution:

Since there is no radial force on the ball/bead, the total radial acceleration is zero according to Newton's second law. In polar coordinates this means that:

\[\ddot{r} - r\omega^2 = 0.\]

This differential equation in \(r\) is solved by:

\[r(t) = Ae^{\omega t} + Be^{-\omega t},\]

for some constants \(A\) and \(B\) depending on initial conditions. If the ball/bead starts from rest at radius \(r_0\), these constants are fixed to be:

\[A = B = \frac{r_0}{2},\]

so the solution for the velocity of the ball/bead is found by differentiating to be:

\[\dot{r}(t) = \frac{r_0 \omega}{2} e^{\omega t} - \frac{r_0 \omega}{2} e^{-\omega t}.\]

The velocity of the ball/bead grows exponentially, since the second term damps to zero quickly.

## A bead is lodged in a wheel that rolls without slipping with constant velocity \(V\). Show that the trajectory of the bead traces a cycloid in the lab frame.

Solution:

In the wheel reference frame, the bead is in uniform circular motion. If the wheel is of radius \(r\), the bead has coordinates and velocity with respect to the center of the wheel: \[ \begin{align} \vec{r} &= r\hat{r} \\ \vec{v} &= r \omega \hat{\theta} = V \hat{\theta} \end{align} \]

In the lab reference frame, the center of the wheel moves with velocity \(V\), so it is located at:

\[\vec{R} = Vt \hat{x} + r\hat{y} = r\omega t \hat{x} + r\hat{y},\]

keeping in mind that the center of the wheel is a height \(r\) above the ground. The position of the bead in the lab frame is therefore:

\[\vec{R} + \vec{r} = r\omega t \hat{x} + r\hat{y} + r\hat{r}\]

Now if the wheel rolls clockwise (i.e., to the right) without slipping, the angle the bead has traveled is:

\[\theta(t) = -\omega t\]

assuming the bead starts at \(\theta = 0\). So the position of the bead is:

\[ r\omega t \hat{x} + r\hat{y} + r\hat{r} = r(\omega t +\cos (\omega t)) \hat{x} + r(1-\sin (\omega t)) \hat{y}.\]

Below is a plot of the position above for \(r = \omega = 1\) to verify that it is indeed a cycloid:

:

## The fact that the Earth rotates means that everywhere on the surface of the earth is a rotating reference frame. This has observable effects from the Coriolis acceleration, most notably in the precession of the axis of rotation of a sufficiently large pendulum. This experimental apparatus is usually called a

Foucault pendulum.

Derive the precession of the Foucault pendulum assuming the Earth rotates with angular frequency \(\Omega\).

Solution:

The precession results from the fact that the plane in which the pendulum oscillates rotates with the rotation of the Earth. However, at higher latitudes \(\varphi\), this precession is slower than at the equator. In two dimensions, the Coriolis acceleration is then:

\[ \begin{align} \ddot{x} &= \Omega \sin \varphi \dot{y} \\ \ddot{y} &= \Omega \sin \varphi \dot{x} \end{align} \]

Small oscillations of a pendulum at frequency \(\omega\) obey Hooke's law. Using this fact and Newton's second law gives the following equations of motion:

\[ \begin{align} \ddot{x} &= -\omega^2 x + 2\Omega \sin \varphi \dot{y} \\ \ddot{y} &= -\omega^2 y - 2\Omega \sin \varphi \dot{x} \end{align} \]

The solution for the complex coordinate \(z= x+iy\) can be found by matrix ODE methods to be:

\[z = e^{-i\Omega \sin \varphi t} \left(A e^{i\omega t} + Be^{-i \omega t} \right).\]

for some constants \(A\) and \(B\) to be determined by initial conditions. The leading prefactor \(e^{-i\Omega \sin \varphi t}\) describes the \(z\) coordinate as rotating over time. Since \(z = x+iy\), the axis of the pendulum thus rotates in the \(x,y\) plane with frequency \(\Omega \sin \varphi\). Since \(\Omega = 2\pi \text{ rads}/\text{day}\), over the course of a single day, the pendulum oscillation precesses by an angle \(-2\pi \sin \varphi\).

## References

[1] D. Kleppner and R. Kolenkow, *An Introduction to Mechanics*. McGraw-Hill, 1973.

[2] Image from https://en.wikipedia.org/wiki/Coriolis_force#/media/File:Corioliskraftanimation.gif under Creative Commons licensing for reuse with modification.

[3] Image from https://en.wikipedia.org/wiki/Foucault_pendulum#/media/File:Foucault-rotz.gif under Creative Commons licensing for reuse with modification.

**Cite as:**Angular Kinematics.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/angular-kinematics-problem-solving/