If we consider two points on a turning disc at distance $r_{1}$ and $r_{2}$ from the center of the disc we can make a number of observations.

First the distance traveled by the points is quite different, points that are far from the center go through a greater distance.

However, the angular displacement is the same for both points, suggesting that angular displacement is a good variable to describe the motion of the whole rotating object.

We would like to be able to have a straight forward relationship between the actual distance traveled and the angular displacement, which we can have if we express our angular displacement in radians.

As the relationship between the arc length on a circle and the angle which it subtends in radians is

$l=r\theta$

a change in the angular displacement of $\Delta\theta$ results in a change in the displacement

$\Delta l=r\Delta\theta$

For change in angular displacement $\Delta \theta$ in a time interval $\Delta t$ we can define an average angular velocity

$\bar{\omega}=\frac{\Delta \theta}{\Delta t}$

As we are now accustomed, we can also define an instantaneous velocity

$\omega=\lim_{\Delta t\to 0}\frac{\Delta \theta}{\Delta t}=\frac{d\theta}{dt}$

The units of angular velocity can be expressed as either $\mathrm{rad\,s^{-1}}$ or $\mathrm{s^{-1}}$

Previously we only considered circular motion in which the angular velocity $\omega$ remained constant with time, but of course we can also define an angular acceleration

$\bar{\alpha}=\frac{\Delta \omega}{\Delta t}$

$\alpha=\lim_{\Delta t\to 0}\frac{\Delta \omega}{\Delta t}=\frac{d\omega}{dt}$

The units of angular acceleration can be expressed as either $\mathrm{rad\,s^{-2}}$ or $\mathrm{s^{-2}}$

A point at distance $r$ from the center of rotation will have at any time a tangential velocity of magnitude

$v=\omega r$

and a tangential acceleration (not to be confused with the centripetal acceleration) of

$a=\alpha r$

Rotational motion, when not accelerated, can be considered to be a form of periodic motion, and so relationships between the angular velocity, frequency and period are useful.

$T=\frac{2\pi}{\omega}$

$f=\frac{\omega}{2\pi}$

$\omega=2\pi f$

We'd also like to be able to express the centripetal acceleration in terms of $\omega$

$a_{R}=\frac{v^{2}}{r}=\frac{(\omega r)^{2}}{r}=\omega^{2}r$

The total acceleration of an object in accelerated rotational motion will be the vector sum of two perpendicular vectors, the tangential acceleration $\vec{a}_{tang}$, and the radial acceleration $\vec{a}_{R}$.

The magnitude of the total acceleration is

$\vec{a}=\sqrt{a_{tang}^2+a_{R}^2}=\sqrt{\alpha^2r^2+\omega^{4}r^2}=r\sqrt{\alpha^2+\omega^4}$

and it is directed an angle

$\arctan\frac{\alpha}{\omega^{2}}$ away from the radial direction.

As we know, translational velocity and acceleration are vector quantities. While we have defined angular velocity and acceleration we can see that they can represent points that are moving in directions that change with time, which would make them difficult to represent with vectors defined in cartesian coordinates.

We can however consider an axial vector, ie. a vector that points in the direction about which rotation occurs as a good way of representing the direction of these rotational quantities.

We give the direction of rotation around the axes according to a right-hand rule.

At the beginning of the course we derived, using calculus, a set of equations for motion under constant acceleration.

$v= v_{0}+at$

$x= x_{0}+v_{0}t+\frac{1}{2}at^2$

$v^{2}=v_{0}^2+2a(x-x_{0})$

We can equally derive similar equations for our rotational quantities. Indeed as we can see that the relationships between the new rotational quantities we have now are exactly the same as those between the translational quantities we can simply rewrite the translational motion equations in terms of rotational variables.

$\omega= \omega_{0}+\alpha t$

$\theta= \theta_{0}+\omega_{0}t+\frac{1}{2}\alpha t^2$

$\omega^{2}=\omega_{0}^2+2\alpha(\theta-\theta_{0})$

Let's consider a wheel that rolls without slipping. For an object to roll without slipping there is a frictional requirement; it is actually not possible for an object to roll on a frictionless surface, it would instead slide.

The point around which the wheel turns, which is also it's center of mass, is seen to execute translational motion with velocity $\vec{v}$

In the reference frame of the ground a point on the edge of the wheel comes to rest when it is contact with the ground. However from the perspective of the center of the wheel, the motion of the point of the edge of the wheel is purely rotational with $\omega=\frac{v}{r}$ or $v=\omega r$.

We can describe the displacement of a point on the edge of the wheel in the reference frame of the ground by adding the rotational motion of the point to the translational motion of the center of the wheel. This gives the equation

$\vec{r}=vt\hat{i}+r\cos(\theta(t))\hat{i}+(r\sin(\theta(t))+r)\hat{j}$

$\to \vec{r}=(r\omega t+r\cos(\omega t))\hat{i}+(r\sin(\omega t)+r)\hat{j}$

A path of this kind is called a cycloid

Differentiating with respect to time give us the velocity

$\vec{v}=(r\omega-r\omega(\sin(\omega t)))\hat{i}+r\omega\cos(\omega t)\hat{j}$

More on wheels and rolling at Physclips

As we've seen the angular velocity of a wheel is $\omega=\frac{v}{r}$ where $v$ is the tangential velocity of the wheel and $r$ is it's radius. The total velocity of a point on the wheels radius is the sum of the tangential motion around the center of the wheel and the tangential velocity of the center of the wheel and is a cycloid.

For an object to roll without slipping there is a frictional requirement; it is actually not possible for an object to roll on a frictionless surface, it would instead slide. Note that the point in contact with the ground during rolling is actually motionless, so the relevant coefficient of friction is the static coefficient of friction, whereas for sliding the kinetic coefficient of friction is relevant. There is no work done by the frictional force as it is directed perpendicular to the displacement (shown by the small black arrows in the diagram).

In the same way as a force causes linear acceleration, $\vec{F}=m\vec{a}$, there must be an analogous quantity and equation related to angular acceleration. From experience, for example, when we use a wrench on a nut, we can expect that the angular equivalent of Force, which is called Torque, will depend on the amount of Force applied, the distance at which it is applied and the direction of its application.

We define the torque due to a force $F$ applied at a distance $R$ from a pivot point as either

$\tau\equiv RF_{\perp}$ or $\tau\equiv R_{\perp}F$

and we can calculate the magnitude of the torque using

$\tau=RF\sin\theta$

The units of torque are $\mathrm{Nm}$