###Introduction###
Uniform circular motion refers to where an object moves around in a constant circle (i.e. the space which the circle encompasses never changes). The mechanics of circular motion are very interesting and largely revolve around one equation with some applications. Circular motion is largely dependant on the circle's radius. In fact,
$$a_R = \frac{v^2}{R}$$
In this case, \(a_R\) pertains to the acceleration of the object, \(v^2\) to its speed, and \(R\) to the circle's radius. Although some problems can be solved with this equation alone, some other formulas pertaining to circles are sometimes required.
###Finding the Acceleration###
As acceleration and force have a fundamental relationship, so oftentimes the acceleration of the object in circular motion is a target value that must be found.
Example 1: An object has a constant speed of \(7\) meters per second as it moves with a changing velocity in a uniform circle of radius \(2\) meters. What is the magnitude of the acceleration of the object?
Solution: We can use the equation \(a_R = \frac{v^2}{R}\). Plug in \(v\) and \(R\):
$$a_R = \frac{7^2}{2} = \frac{49}{2} = 24.5 \; \frac{m}{s^2}$$
Example 2: A pillow is being dragged around in a circle with a circumference of \(18\pi\). The pillow has a constant velocity of \(4\) meters per second.
a. What is the magnitude of acceleration of the pillow?
b. When the pillow is at the bottom of the circle, in what direction is the acceleration?
Solution:
a. We need the radius of the circle:
$$C = 2\pi R \Rightarrow$$
$$R = \frac{C}{2\pi} = \frac{18\pi}{2\pi} = 9$$
Now we can use the formula for the acceleration of an object in uniform circular motion:
$$a_R = \frac{v^2}{R} = \frac{4^2}{9} = 1.78 \; \frac{m}{s^2}$$
b. At the bottom of the circle, the object will begin moving upward. Since at the bottom of the circle it has no upward velocity, and it acquires a nonzero upward velocity, the acceleration is upward.
###Finding Other Quantities###
In some cases we may know the acceleration of the object and instead need to find the radius of the circle or the speed of the object. In this section we will show two examples of this type.
Example 3: Agh! You want to find the radius of a circle formed by a dog chasing its tail in uniform circular motion. But, your ruler broke! You observe that the dog's constant speed is \(1.5\) meters per second and its circular acceleration is \(4.5\) meters per second squared. What is the radius of the circle formed?
Solution: We know the equation for the acceleration of an object in uniform circular motion:
$$a_R = \frac{v^2}{R}$$
We can solve for the radius in this equation, as we are given the other two quantities:
$$Ra_R = v^2 \Rightarrow$$
$$R = \frac{v^2}{a_R}$$
Now we plug in \(v^2\) and \(a_R\):
$$R = \frac{1.5^2}{4.5} = 0.5 \; m$$
Example 4: A penguin slides around a circular ring of ice at a constant acceleration of \(12\) meters per second squared. The radius of the circle is \(3\) meters. What is the penguin's speed (given that this is uniform circular motion)?
Solution: In this case we want to solve for the speed of the object (the penguin):
$$a_R = \frac{v^2}{R} \Rightarrow$$
$$v^2 = Ra_R \Rightarrow v = \sqrt{Ra_R}$$
Plug and chug:
$$v = \sqrt{3 \cdot 12} = 6 \; \frac{m}{s}$$
This equation, while quite simple, is also very versatile. Introducing force (which will be done in another article) expands the usage of this equation even further.
Disclaimer on Symbols
Teachers and textbooks sometimes use a lowercase \(r\) to denote the radius of a circle. However it doesn't mean anything different in this context. Just be sure to be consistent with which symbol you use when working through a particular problem.