I’ll now attempt to relate everything I have said specifically to being
on a merry-go-round. Any results I give came from a hypothetical situation I
have created and the formulas and answers can be found in the “Sample Problem” section of the web page.

If we were in a perfect, frictionless, world I could be on a merry-go-round and achieve uniform circular motion. That is I would be traveling in a perfect circle at a constant speed, NOT a constant
velocity, because speed is scalar and velocity is a vector. Scalar measurements
have magnitude only, while a vector measurement has both magnitude and direction. Since
I would be traveling in a circle my direction would constantly be changing, so even though my magnitude would not change my
direction would so that would change my velocity.

To determine my velocity while riding the merry-go-round you can use a modified velocity formula. Normally for linear velocity you take the distance over the time, but for circular motion instead of distance
traveled in a line you take the distance around the circle (circumference aka. (2 pi)(radius)) over the time.

Lets say when the merry-go-round begins I start at the origin (center) of the circle. While standing there I would have have virtually no radius so my linear speed would be negligible. But as I traveled further from the center of the circle the distance from the center
(radius) would increase and increase my linear speed as I continue to travel out.

Anyone playing on a merry-go-round knows that by letting go of the bars on a merry-go-round you would fly off. But few people realize the physics behind falling off.
Every single time I fall off would result in the same thing, and by looking at me falling aerially you would notice
that I always fall off tangent to the circle. This is easy to explain using Newton’s
First Law of Motion. It states that objects tend to stay at a constant velocity
unless an outside net force acts upon it. The reason you always fall off tangent
to the merry-go-round is because in circular motion velocity is always tangent to the circle.
Since your body wants to stay at a constant velocity (inertia) it will continue forward when you let go, rather than
continually going around in a circle.

You can see that my linear speed depends on where I am standing on the merry-go-round. Another way to measure how fast you are going would be to see how many times I do
a rotation in one minute, or how many radians I cover in one second. You will
commonly see this in you car, on your tachometer. This measurement is known
as the angular speed and it is designated with the lowercase Greek letter omega (ω).
There are three versions of the angular speed formula depending on the information you have available to you. The first formula is for when you know the period of an object (time it takes for
an object to complete a full revolution). You take 2 pi over the period (usually
represented by T). The next formula involves the linear velocity and the radius
of the merry-go-round, you compute it by taking the linear velocity over the radius.
Finally the third formula is for when you only know the frequency (the amount of revolutions in one second denoted
with the unit Hertz or Hz) of the merry-go-round. The angular velocity can be
determined by multiplying 2 pi by the frequency. By using the second variation
of the angular speed formula you can determine the linear speed of an object at the edge of the circle by multiplying the
angular speed and the radius together.

The next basic topic after speed is acceleration.
Most people would say that I have no acceleration while on a merry-go round, but I do.
The next question to come up would be where am I accelerating? The easiest
way to determine the direction of my acceleration is to draw a vector diagram. By
using adding the vectors together and finding the resultant you can see the acceleration while under uniform acceleration
is always towards the center of the circle. To determine the acceleration another
formula can be used. The acceleration formula has two versions. The first utilizes the linear velocity and the radius; by taking the square of the linear velocity over
the radius. The second version again utilizes the radius but this time also uses
the period; take 4 times pi squared and that times the radius and finally all that over the period squared.

Next Thing when it comes to centripetal motion involves a little English knowledge. One of the most common mistakes of circular motion is to call it centrifugal motion
instead of centripetal motion, and any physicist will want to pull their hair out at the sound of someone saying centrifugal
motion. The reason why is because when you look at the roots of the words you
can plainly see the difference. Centrifugal means center fleeing while centripetal
means center seeking. Since all acceleration is directed towards the center it
is said that it is center seeking. You can see that centrifugal force doesn’t
exist and you can also see why some people may get mad why you say it. That is
why some physicist refer to centrifugal as the F-Word.

The final question that needs to be answered is weather or not any work is being
done on me as I ride my frictionless merry-go-round. This is a very simple task
because of course as in everything in physics, there is a formula. Work is equal
to force times displacement time cosine of theta. To make things easy I’ll
tell you now that the cosine theta will equal 0 and thus making the amount of work equal 0 joules. This means that the mechanical energy(potential energy added to the kinetic energy) this will also mean
that the kinetic energy remains constant and when that happens the speed does not change in the system unless something like
friction is added into the system. Sounds familiar? Just as stated in Newton’s first law of motion.