###Introduction###
The dynamics of circular motion refer to the theoretical aspects of forces in circular motion. Circular motion dynamics revolve around a few key formulas pertaining to forces and acceleration. First, we have the formula for acceleration in uniform circular motion:
$$a_R = \frac{v^2}{R}$$
However, we can multiple both sides of the equation by the mass of the object:
$$ma_R = \frac{mv^2}{R}$$
And by Newton's Second Law:
$$F_R = \frac{mv^2}{R}$$
Finally, sometimes an object will have an additional force being applied to it beyond the centerpointing force. Usually this second force will be friction or gravity.
###Calculating Forces in Circular Motion###
The equation
$$F_R = \frac{mv^2}{R}$$
is central to calculating forces pertaining to circular motion.
Example 1: A spaceship of mass 5000 kilograms is floating in outer space in a uniform circle of radius 1000 meters at a constant speed of 2 meters per second. What is the centerpointing force on the spaceship?
Solution: Use the equation for a centripetal force:
$$F_R = \frac{mv^2}{R} = \frac{5000 \cdot 2^2}{1000} = \frac{20000}{1000} = 20 \; N$$
Example 2: A 0.05-kilogram frisbee is flying in a circular arc where the radius is 12 meters and the constant speed is 30 meters per second. What is the centerpointing force on the frisbee?
Solution: Again just plug-and-chug:
$$F_R = \frac{0.05 \cdot 30^2}{12} = 3.75 \; N$$
###Finding Other Values###
Sometimes it is instead necessary to find other quantities besides the force, such as the radius of the circle formed or the speed of the object. This section explores problem-solving in uniform circular motion with these goals in mind.
Example 3: A block of mass \(2\) kilograms is sliding around in a circle at a constant speed of \(4\) meters per second. If the coefficient of kinetic friction between the block and the ground is \(0.3\) and the constant acceleration of the block is \(10\) meters per second squared, find the radius of the circle formed by the block.
Solution: The fact that the speed, radius, and acceleration are constant mean that the block is undergoing uniform circular motion. There are two forces acting on the block on the plane of the ground: the centerpointing applied force on the block, and friction. The sum of the forces on this plane is
$$\sum{F} = F_R - F_{fr}$$
By Newton's Second Law:
$$ma = F_R - F_{fr}$$
Now we know the formula for the applied force:
$$ma = \frac{mv^2}{R} - F_{fr}$$
We also know the formula for the frictional force:
$$ma = \frac{mv^2}{R} - mg\mu_k$$
Divide away \(m\):
$$a = \frac{v^2}{R} - g\mu_k$$
Now solve for \(R\):
$$aR = v^2 - Rg\mu_k \Rightarrow$$
$$aR + Rg\mu_k = v^2 \Rightarrow$$
$$R(a + g\mu_k) = v^2 \Rightarrow$$
$$R = \frac{v^2}{a + g\mu_k}$$
Feel free to check the units on the right-hand side to ensure they are units of length. Now plug in the known values:
$$R = \frac{4^2}{10 + 9.8(0.3)} = 1.236 \; m$$
Example 4: A stone is rolling around in a uniform circular motion. The stone has a mass of \(1\) kilogram, and a radius of \(12\) meters. The constant speed of the stone is \(6\) meters per second. The acceleration of the box is \(0.2\) meters per second squared. What is the coefficient of kinetic friction on the stone?
Solution: According to Newton's Second Law:
$$ma = \frac{mv^2}{R} - mg\mu_k$$
We need to solve for \(\mu_k\). Divide away all of the \(m\)s:
$$a = \frac{v^2}{R} - g\mu_k$$
Isolate the term with \(\mu_k\):
$$a + g\mu_k = \frac{v^2}{R} \Rightarrow$$
$$g\mu_k = \frac{v^2}{R} - a \Rightarrow$$
$$\mu_k = \frac{v^2}{gR} - \frac{a}{g}$$
Plug in the known values:
$$\mu_k = \frac{6^2}{9.8 \cdot 12} - \frac{0.2}{9.8} = .286$$
Example 5: You are playing with a yo-yo with a mass of \(.68\) kilograms (ignore the mass of the string). The length of the string is \(.2\) meters. If the yo-yo has a constant acceleration of \(4.5\) meter per second squared, what is its speed at the botton of the yo-yo?
Solution: There are two forces acting on the yo-yo: the tension of the string and gravity. At the bottom of the yo-yo, the force of tension points upward and gravity points downward. By Newton's Second Law:
$$\sum_{F} = \frac{mv^2}{R} - mg \Rightarrow$$
$$ma = \frac{mv^2}{R} - mg \Rightarrow$$
$$a = \frac{v^2}{R} - g$$
Now that the equation is simplified we solve for \(v\):
$$aR = v^2 - gR \Rightarrow$$
$$v^2 = aR + gR \Rightarrow$$
$$v = \sqrt{R(a + g)} = \sqrt{.2(4.5 + 9.8)} = 1.691 \; \frac{m}{s}$$
As you can see, combining equations of uniform circular motion allows for a lot of problem-solving opportunities. It is important to realize how circular motion dynamics relates to Newton's Second Law since all of these problems pertain to forces.