Uniform Circular Motion (UCM)

 

If you start reading the title of this topic, it won’t be difficult for you to realize what is this about. We will describe the characteristics of any movement which path is a circle and, at the same time, its speed remains constant. Notice that its written “constant speed”, not velocity!.You are now probably thinking about different examples of UCM. A carousel, the earth rotation, and thousand more.

It is a key concept to differenciate SPEED and VELOCITY, because in UCM the speed is constant BUT the velocity is NOT! Why? Because the direction changes continously while the object is moving in the circle.

Look at the scheme, in differents positions, the VELOCITY VECTOR points in differents directions, so, the velocity changes at every movement


 

 

MEASURING THE UCM

Time. The time that it takes a whole lap is called PERIOD. Its symbol is T. For example, the period of the earth rotation is 24 hours.

 

Speed. In a whole lap, the distance covered is the perimeter of the circle (\(2 \pi r\)) so, the speed is very easy to find,

\(v = {{2 \pi r} \over T}\)

Acceleration. What??? It is an UNIFORM MOTION and yet has acceleration?? YES. Please, remember that the SPEED is constant BUT NOT THE VELOCITY!. The acceleration is due to the change of the direction of the velocity. This acceleration doesn’t make it go faster or slower. It is only in charge of changing the direction of the velocity vector.

This kind of acceleration points to the center of the circle, so it is called CENTRIPETAL acceleration ac . Centripetal is a word that cames from greek language (centri = center, petare = go). This centripetal acceleration keeps the circular path. If this acelleration dissapears suddenly, the path becomes a straight line.

You can calculate the centripetal acceleration by

\(a_c = {{v^2} \over r}\)

 

Look at something else: in each position, the aceleration vector has different direction, so, even thoug the magnitude of acceleration is constant, the acceleration is not.

 

FORCES IN A UCM


When we start talking about forces in UCM, another word shows up to describe a force: centrifugal. But, it is a misleading word, because it describes a force which points outwards. This description can be used if we see the circular motion from the center of the circle. (Watch out! If we describe a UCM from the center of the circle, it is a non inertial frame of reference, so it isn’t convenient at all)

Perhaps, the word “centrifugal” reminds you the washing machines or ohter gadgets like that, gadgets which rotates. That is why the word “centrifugal” is a misleading word.

So, which forces are involved in a UCM? Let’s apply the Newton’s 2nd Law.

According to Newton 2nd Law, the net force is

 

\(F_n = ma\)

 

So, in a UCM, the net force will be a centripetal force, and if we remember how to calculate the centripetal acceleration, we can conclude that the net force is a centripetal force as follows

\(F_c = {{mv^2} \over r}\)

 

Let’s highlight some concepts

1- An Uniform Circular Motion have a constant speed but not a constant velocity

2- The acceleration in a UCM is a centripetal acceleration, and the net force, is called centripetal force. Both vectors points to the center of the circle.

3- The speed in a UCM can be determined by \(v = {2 \pi r \over T}\) and the centripetal acceleration by \(a_c ={ v^2 \over r}\)

 

Exercises

1- Imagine that you turn a stone tied to a 75 cm lenght rope. The stone takes a full turn in 0,98 seg

a- Calculate the speed of the stone.

b- Outline this situation, including velocity and aceleration vectors

c- Is it the stone in equilibrium?

a- The speed in the UCM can be determined by \(v = {2 \pi r \over T}\)

The lenght of the rope is the radius of the circle, and 0,98 s is the time that the stone takes a full turn, so the period T is 0,98 s. Let’s calculate

 

\(v = {{2*3.14*0.75m} \over 0.98s} = 4.80 m/s\)

Remember using the International Sistem of units

b-

 

c- The stone isn’t in equilibrium. Equilibrium means that the net force is zero and in a UCM the net force is the centripetal force. Eventhoug the magnitude of the velocity of the stone is constant, the direction is not.

 


 

2- A car is travelling on a road at 15 m/s.  It takes a curve that has a radius of 200 m.

a- Calculate the centripetal acceleration

b- The driver of the car feels that is being thrown outward from the center. Is it caused by a centrifugal force?


a- The centripetal acceleration can be calculated by

\(a_c = {{v^2} \over r}\)


 

\(a_c = {15^2 \over 200} = 112.5 m/s²\)

 

b- What the driver feels is described in non inertial frame of reference (NIFR). The driver is an observer sit into the car, a car which accelerates, so is an observer in a NIFR. It isn’t convenient to describe the forces in a NIFR.

 

Let’s change the frame of reference. We can describe the whole situation looking at it as if we are standing on the road, outside the car. It is an inertial frame of reference. In this case, we will say that the inertia causes the the driver tends to keep the straight path, but the car is taking the curve.