Uniform Circular Motion

In this section I derive a simple and very useful equation for the magnitude of the acceleration of an object undergoing constant acceleration. The law of sines is involved, so I've recapped it in figure f.

f / The law of sines.

The derivation is brief, but the method requires some explanation and justification. The idea is to calculate a Δv vector describing the change in the velocity vector as the object passes through an angle θ. We then calculate the acceleration, a = Δv/Δt. The astute reader will recall, however, that this equation is only valid for motion with constant acceleration. Although the magnitude of the acceleration is constant for uniform circular motion, the acceleration vector changes its direction, so it is not a constant vector, and the equation a = Δv/Δt does not apply. The justification for using it is that we will then examine its behavior when we make the time interval very short, which means making the angle θ very small. For smaller and smaller time intervals, the Δv/Δt expression becomes a better and better approximation, so that the final result of the derivation is exact.

g / Deriving |a| = |v|2/r for uniform circular motion.

In figure g/1, the object sweeps out an angle θ. Its direction of motion also twists around by an angle θ, from the vertical dashed line to the tilted one. Figure g/2 shows the initial and final velocity vectors, which have equal magnitude, but directions differing by θ. In g/3, I've reassembled the vectors in the proper positions for vector subtraction. They form an isosceles triangle with interior angles θ ,η, and η. (Eta, η, is my favorite Greek letter.) The law of sines gives

This tells us the magnitude of Δv, which is one of the two ingredients we need for calculating the magnitude of a = Δv/Δt. The other ingredient is Δt. The time required for the object to move through the angle θ is

Now if we measure our angles in radians we can use the definition of radian measure, which is (angle) = (length of arc)/(radius), giving Δt = θr/|v|. Combining this with the first expression involving |Δv| gives

When θ becomes very small, the small-angle approximation sin θ ≈ θ applies, and also η becomes close to 90 °, so sin η ≈ 1, and we have an equation for |a|:

Force required to turn on a bike.

Don't hug the center line on a curve!

Acceleration related to radius and period of rotation.

A clothes dryer.

More about clothes dryers!.

→ Solved problem: The tilt-a-whirl page 231, problem 6

→ Solved problem: An off-ramp page 231, problem 7

Discussion Questions

A

A certain amount of force is needed to provide the acceleration of circular motion. What if were are exerting a force perpendicular to the direction of motion in an attempt to make an object trace a circle of radius r , but the force isn't as big as m|v|2/r?

B

i / Discussion question B. An artist's conception of a rotating space colony in the form of a giant wheel. A person living in this noninertial frame of reference has an illusion of a force pulling her outward, toward the deck, for the same reason that a person in the pickup truck has the illusion of a force pulling the bowling ball. By adjusting the speed of rotation, the designers can make an acceleration |v|2/r equal to the usual acceleration of gravity on earth. On earth, your acceleration standing on the ground is zero, and a falling rock heads for your feet with an acceleration of 9.8 m/s2. A person standing on the deck of the space colony has an upward acceleration of 9.8 m/s2, and when she lets go of a rock, her feet head up at the nonaccelerating rock. To her, it seems the same as true gravity. Suppose a rotating space station, as in figure i on page 227, is built. It gives its occupants the illusion of ordinary gravity. What happens when a person in the station lets go of a ball? What happens when she throws a ball straight "up" in the air (i.e., towards the center)?