### Velocity of an object moving in a circle

Velocity along a straight line is measured in units of length divided by time. In modern physics that's usually meters per second (m·s-1). By now you should have had plenty of experience working with linear velocities and the vectors that represent them.

Things are different on the rotational world. In the rotational world, we measure velocity in a circular motion in terms of angle traversed divided by time, usually in units of degrees per second or radians per second (rad·s-1).

We usually denote angular velocity with the Greek lower case letter omega, ω

$$\omega = \frac{angle}{time}$$

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#### The Greek alphabet

 alpha Α α beta Β β gamma Γ γ delta Δ δ epsilon Ε ε zeta Ζ ζ eta Η η theta Θ θ iota Ι ι kappa Κ κ lambda Λ λ mu Μ μ nu Ν ν xi Ξ ξ omicron Ο ο pi Π π rho Ρ ρ sigma Σ σ tau Τ τ upsilon Υ υ phi Φ φ chi Χ χ psi Ψ ψ omega Ω ω

### Average and instantaneous angular velocity

The average angular velocity of an object traveling rotating about an axis is

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

where Δθ is the change in angle over the change in time, Δt. Recall that a bar over a quantity in this context means "mean" or "average." In such a calculation, we have no details about acceleration during the time period Δt, so, just like we saw for linear velocity, this is an average between two endpoints:

$$\bar{\omega} = \frac{\theta_f - \theta_i}{\Delta t}$$

where θf and θi are the final and initial angle values, and Δt = tf - ti.

We can also write the average value of two known instantaneous angular velocities, ω1 and ω2 as

$$\bar{\omega} = \frac{\omega_1 + \omega_2}{2}$$

The instantaneous angular velocity is found when Δt becomes infinitesimally small, or

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

This is read "the limit of Δθ/Δt as Δt approaches zero."

In calculus (which you don't need to know to work through the rest of this section), the angular velocity is the first derivative of the angle with respect to time:

$$\omega = \frac{d\theta}{dt}$$

We'll need to be clear about when we're working with average and instantaneous angular velocities.

### Relationship between angular and linear velocity

You might have watched a bicycle tire rotate and noticed that the tire moves much faster than the hub in the center. Your observation was correct. While the angular velocity of the hub and tire were the same, their linear velocities were different.

Consider the car tire show here. Let's think about two points on the tire. One on the surface, at a radius of r1 from the center, and one at a shorter radius, r2.

Now if the tire is rotating at radians per second (that would be 60 rpm), then both points on the more-or-less rigid tire would be going around at the same rate; they would both have the same angular velocity.

But we could also ask how far each point travels in the same time. The point at r1 travels the full circumference of the tire, C = 2πr1. The point at r2 travels a shorter distance, C2 = 2πr2. It's shorter because r2 < r1.

The linear velocity of the point at r1 is 2πr1/1 s, a distance divided by a time, and the linear velocity of the point at r2 is 2πr2/1 s.

For an object rotating around an axis like the wheel, every point on the object has the same angular velocity, but not necessarily the same linear velocity.

I've left off the units of the radii here, but that's not important to the discussion. Points farther from the center of the rotating body have the same angular velocity as any other point, but a greater linear velocity.

#### Relationship

Consider the figure below, describing the rotation of an object about an axis.

The arc S can be determined very simply if we work in radians. We simply set up the proportion: "The angle θ is to a full circle () as the arc S is to the full circumference (2πr):

$$\frac{\theta}{2 \pi} = \frac{S}{2 \pi r}$$

The 's cancel, giving the angle in terms of the arc length and radius.

$$\theta = \frac{S}{r}$$

Now the average angular velocity is:

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

In the last step we substituted S/r for θ. Now we showed that

$$\frac{\Delta S}{\Delta t} = \bar{v}$$

Therefore the angular velocity is related to the velocity (average or instantaneous) by:

$$\omega = \frac{v}{r}$$

or

$$v = r \cdot \omega$$

The two linear velocity vectors, tangents to the circle in the last figure, are given by v = r·ω.

For an object rotating around an axis, every point on the object has the same angular velocity, but not necessarily the same linear velocity.

### The angular velocity vector

Like linear velocity, angular velocity is a vector quantity. But because a rotating body is in continuous motion, how do we decide the direction of the angular velocity vector?

We overcome that problem by choosing to have the angular velocity vector to be aligned with the axis of rotation, as shown.

One problem remains, however: The axis of rotation has two ends. Which way does the vector point?

We decide to use what is called the right-hand rule to decide. The right hand rule works as shown in the picture.

If I imagine wrapping the fingers of my right hand around the axis of rotation, fingers pointing in the direction of the rotation, then my thumb can only point one way. That's the direction of the angular velocity vector.

### Example 1

Calculate the angular velocity and the linear velocity of a child revolving at the edge of a playground merry-go-round that is 3.6 m in diameter, if the child completes one revolution every 2.1 seconds.

Solution: The angular velocity is easy to calculate. One revolution of 2π rad takes 2.1 s, so

\begin{align} \bar{\omega} &= \frac{\Delta \theta}{\Delta t} = \frac{2\pi \; rad}{2.1 \; s} \\ \\ &= 3.- \; rad/s \end{align}

(That's about 171˚ per second.) Now the linear velocity should just be v = r·ω. That's

\begin{align} v = r \cdot \omega \\ &= (1.8 \; m)(3.0 \; rad/s) \\ &= 5.4 \; m/s \end{align}

The circumference of the merry-go-round is

\begin{align} C &= 2\pi r \\ &= 2\pi (1.8 \; m) = 11.3 \; m \end{align}

So 5.4 m/s seems about right for 171˚/s.

### Example 2

Calculate the angular velocity of Earth (a) as it rotates about its axis, and (b) in its orbit around the sun.

Solution: For part (a), we know that Earth makes one complete rotation (2π rad) in 24 hours. To put that in radians per second is just a simple unit conversion:

\begin{align} \omega &= \left( \frac{2\pi \; rad}{24 \; h} \right) \left( \frac{1 \; h}{3600 \; s} \right)\\ \\ &= 0.0000727 \; rad/s \\ &= 73 \; \mu rad/s \end{align}

In the last step, an awkward number of radians has been converted to a nicer number om micro-radians. It's good to know your metric prefixes and be able to do that when you want.

For part (b), we know that Earth completes one revolution about the sun (in a nearly circular orbit) every 365.25 days, so it's a similar calculation:

\begin{align} \omega &= \left( \frac{2\pi \; rad}{365.25 \; d} \right) \left( \frac{1 \; d}{24 \; h} \right) \left( \frac{1 \; h}{3600 \; s} \right)\\ \\ &= 0.000000119 \; rad/s \\ &= 0.2 \; \mu rad/s \end{align}

It's interesting to go a little further here and calculate the linear velocities of the surface of Earth as it spins and of Earth as it revolves about the sun.

For the first, the radius of Earth is about 6.4 x 106 m, so

\begin{align} v &= r \cdot \omega \\ &= (6.4 \times 10^6 \; m)(7.27 \times 10^{-5} \; rad/s) \\ &= 465 \; m/s \end{align}

That's pretty fast. Let's put it in terms of miles per hour to see just how fast.

$$\left( \frac{465 \; m}{s} \right) \left( \frac{1 \; mi}{1609 \; m} \right) \left( \frac{3600 \; s}{1 \; h} \right) \\ = 1041 \; mi/h$$

Sot that's surprising. If you're standing on the surface of Earth at the equator, "not moving," you're already moving at more than 1000 mi./h !

We can repeat the calculation for the linear velocity of Earth as it orbits the sun. The mean Earth-Sun distance is 1.496 x 1011 m:

\begin{align} v &= r \cdot \omega \\ &= (1.496 \times 10^{11} \; m)(1.99 \times 10^{-7} \; rad/s) \\ &= 29,770 \; m/s \end{align}

... and 29,770 m/s in miles per hour is:

$$\left( \frac{29,770 \; m}{s} \right) \left( \frac{1 \; mi}{1609 \; m} \right) \left( \frac{3600 \; s}{1 \; h} \right) \\ = 66,609 \; mi/h$$

### Practice problems

 1 Let's say you are designing a wind turbine to generate electricity. the tower height is 200 ft to safely accommodate a 105 ft. rotor diameter. If the shaft of the rotor is connected directly to a generator that needs to spin at 1000 revolutions per minute (rpm), how fast would the ends of the turbine blades be turning in order to acheive this? Is this feasible? If not, what's your solution? Solution Solution 2 A typical road bicycle has a wheel diameter of 700 mm. How many times does the wheel rotate in covering 25 miles on the road? Calculate the angular velocity of such a wheel when the bike is rolling at 18 mi/h. (1 mi = 1609 m) Solution Solution 3 Problem Solution Solution 4 Problem Solution Solution 5 Problem Solution Solution

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