Illustration 10.2: Motion about a Fixed Axis

Please wait for the animation to completely load.

Many objects rotate (spin) about a fixed axis. Shown is a wheel of radius 5 cm rotating about a fixed axis at a constant rate (position is given in centimeters and time is given in seconds). Restart.

Consider the various points on the surface of the rotating wheel. By watching the line rotate, we can see that the wheel is rotating at a constant rate. In fact, if we watch a point on the surface of the wheel (a radius of 5 cm), we would say it has a constant speed. What about a point halfway out (a radius of 2.5 cm)? It also has a constant speed. But how does this speed compare to the speed of a point on the surface of the wheel?

We can determine this by first considering a quantity that is not related to the radius, the angular speed, ω. The angular speed is the angular displacement divided by the time interval (in this Illustration, since there is no acceleration, the average speed and the instantaneous speed are the same). So what is the wheel's angular speed? From one revolution of the wheel, the angular displacement is 2π and the time interval (called the period, T) is 5 seconds. Therefore the angular speed is the angular displacement over the time interval (ω = 2 π/T) 0.4π radians/s = 1.256 radians/s.

How do we relate the angular speed to the linear (tangential) speed of a point on the wheel? First consider the speed of a point on the surface of the wheel. It is again easiest to measure the speed by considering one rotation of the wheel. In this case, the distance traveled by that point is 2πr, and therefore the average (and in this case instantaneous) tangential speed is 2πr/T = 2π cm/s = 6.28 cm/s. The relationship between the angular speed and the tangential speed must be ω = v/r. (Recall that we found above that ω = 2π/T.)

This works because there is a relationship between the angular displacement and the tangential displacement (the arc length s), namely that Δθ = Δs/r. This also must be the case for an angular displacement of one revolution: 2π = 2πr/r.

Since linear velocity is a vector we might expect that angular velocity is a vector as well. This is indeed the case. So in which direction does the angular velocity point for the rotating wheel? We use the right-hand rule (RHR) to determine the direction of the angular velocity. Using your right hand, curl your fingers in the direction of the rotation of the wheel: the direction your thumb points is the direction of the wheel's angular velocity. Here, ω is into the page (computer screen). This may seem weird; after all you might say that the wheel is rotating clockwise. Clockwise is not a good description, as it does not imply a vector-like quantity and the description is not unique. Why is it not unique? If you were on the other side of the page or computer screen, you would say that the wheel is rotating counterclockwise instead!

Can you guess what the relationship between the angular acceleration and the tangential acceleration is? Well, given that acceleration is change in velocity for a given time, Δv/Δt, you have probably guessed that angular acceleration, called α, is the change in angular velocity for a given time, Δω/Δt. Given how v and ω are related to each other, it must be that a = α r.

Problem authored by Mario Belloni and Wolfgang Christian.
HTML updated for JavaScript by Ricky Davidson.
Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.