# Uniform circular motion (MCU): formulas, characteristics

A particle has **uniformÂ ****circularÂ ****motion**Â (MCU) when its path is a circumference and it also travels it with constant speed.Â Many objects such as parts of machinery and motors, for example, have this kind of movement, among which computer hard drives, fan blades, shafts and many other things stand out.

Uniform circular motion is also a good approximation for the motion of some celestial bodies such as the Earth.Â Actually the Earth’s orbit is elliptical, as indicated by Kepler’s laws.Â However, the eccentricity of the orbit is small and as aÂ first approximation can be considered circular which simplifies some calculations, such as finding the speed of the Earth when it moves around theÂ sunÂ .

In describing uniform circular motion, the same parameters are used as in rectilinear motion, namely: position, displacement, time, velocity, and acceleration.

Acceleration?Â Yes, indeed, uniform circular motion is accelerated, even though its speedÂ *v*Â is constant.Â This is because the velocityÂ ** v**Â , which is a vector and therefore is bold, continually changes its direction as the object or particle rotates.Â Any change inÂ

**is produced by an acceleration, which, as will be seen, is directed towards the center of the circular path.**

*v*Uniform circular motion is motion in theÂ *xy*Â planeÂ , therefore it is motion in two dimensions.Â However, it is possible to express it more conveniently by the angle Î¸ that the particle sweeps, measured with respect to the horizontal axis or another suitable reference axis.

Even if it is an extended object, its particles always sweep the same angle, even if they have different coordinatesÂ *(x, y)* .

**Characteristics of uniform circular motion**

The characteristics of uniform circular motion can be summarized as follows:

-The trajectory is a circumference, therefore it is a movement in the plane.

-The speedÂ *v*Â is constant, but the speedÂ ** v is**Â not, because it continually changes direction and direction to accommodate the rotation of the mobile.

-The velocity vectorÂ ** v**Â is always tangential to the circumference and perpendicular to the radial direction.

-The angular velocity Ï‰ is constant.

-Despite being uniform, there is an acceleration to explain these changes in the direction of speed.Â This acceleration is the centripetal acceleration.

-The centripetal acceleration and velocity are perpendicular to each other.

-It is a periodic or repetitive movement, therefore the period and frequency magnitudes are defined for it.

**Uniform circular motion formulas**

In this scheme there is a particle P spinning counterclockwise with MCU, according to the direction and the sense of the velocity vectorÂ **v**Â drawn.

To specify the position vector it is necessary to have a reference point and the ideal point is the center of the circumference O that coincides with the center of the Cartesian coordinate system in the xy plane.

**Position vector**

It is denoted as r (t) and is directed from the origin to the point P where the particle is located.Â At a given instant t, in Cartesian coordinates, it is written as:

*r**Â (t) = x (t)Â iÂ + y (t)Â j*

WhereÂ **i**Â andÂ **j**Â are the perpendicular unit vectors in theÂ *x*Â andÂ *y*Â directionsÂ respectively.Â From the graph it can be seen that the modulus of the vectorÂ **r**Â (t) always equalsÂ *R*Â , the radius of the circumference.Â If Î¸ is the angle thatÂ **r makes**Â with the horizontal axis, the position is also equal to:

*r**Â (t) = [RcosÂ **Î¸Â **(t)]Â iÂ + [RsenÂ *

*Î¸Â*

*(t)]Â*

**j**The angle thatÂ *r**Â (t) makes*Â with the horizontal axis is a central angle and its value is:

*Î¸ = s / R*

Where s is the arc of circumference traveled and R the radius.Â This angleÂ *Î¸*Â is a function of time, so it can be writtenÂ *Î¸ = Î¸Â **(t),*Â called theÂ *angular position*Â .

Since the speed is constant, the particle describes equal angles in equal times and in analogy with the uniform rectilinear motion, it is written:

*Î¸ = Î¸ (t) =Â **Î¸Â _{or}Â + Ï‰t*

HereÂ Â *Î¸Â _{o}*Â is the initial angle measured in radians with respect to the reference axis, it can be 0 or any value and Ï‰ is the angular speed.

**Angular velocity and linear velocity**

Angular velocity is the first derivative of angular position and is denoted as Ï‰.Â Its value is constant for uniform circular motion, since equal angles are swept in equal times.Â In other words:

The units of linear speed in uniform circular motion are the same as for linear motion: m / s (in the SI International System), km / h, cm / s, and others.

**Centripetal acceleration**

In the figure below there is a particle moving clockwise around the circumference with constant speed.Â This means that the velocity vector always has the same modulus, but it changes direction to accommodate the circumference.

Any change in velocity results in acceleration, which by definition i

The triangle formed byÂ **vÂ **_{2}Â ,Â **vÂ **_{1}Â and Î”Â **v**Â is similar to the triangle with sidesÂ **rÂ **_{2}Â ,Â **rÂ **_{1}Â and Î”Â **l**Â , where Î”Ï† is the central angle.Â The magnitudes ofÂ **rÂ **_{2}Â andÂ **rÂ **_{1}Â are equal, so:

rÂ _{2}Â = rÂ _{1}Â = r

Then, of both triangles we have these relations for the angle:

Î”Ï† = Î”r / r;Â Î”Ï† = Î”v / v

The bold type is not necessary, since the measure of the angle depends on the magnitudes of these vectors.Â Matching the previous expressions it follows that:

**Period and frequency**

As circular motion is repetitive, its periodÂ *T*Â is definedÂ as the time it takes for the mobile to make one complete revolution.Â Since the length of the circumference of radius R is 2Ï€R, the angle swept in radians on the complete turn is 2Ï€ radians and it takes time T, the angular velocity is:

Ï‰ = 2Ï€ / T

T = 2Ï€ / Ï‰

The period of uniform circular motion is measured in seconds in the International System.

For its part, the frequencyÂ *f*Â is the number of turns per unit of time and is the reciprocal or inverse of the period:

f = n / t = 1 / T

The unit of frequency in the International System is sÂ ^{-1}Â .

**Examples of uniform circular motion**

Many objects rotate to produce various effects: wheels, discs, and turbines.Â Once the operating speed is reached, the rotation is usually carried out at a constant speed.Â Circular motion is so common in everyday life that you hardly ever think about it, so here are some close examples that illustrate it very well:

**The movement of the Earth**

The Earth and the other planets of the Solar System move in elliptical paths of small eccentricity, except forÂ MercuryÂ , which means that in first approximation, it can be assumed that their movement is uniform circular.

With this, you have a good idea of â€‹â€‹the speed of translation around the Sun, since in the case of the Earth, the period of the movement is known: one year or 365 days.

**Particles on the edge of a disk**

The spinning particles on the edge of an old record player or fan blade follow a uniform circular motion once the device reaches its playback speed.

**Hubble space telescope**

The Hubble Space Telescope circles the Earth at about 7550 m / s.

**Centrifuges**

The washing machines carry out a spinning process to squeeze the clothes, which consists of rotating the container drum at high speed.Â The dryers also rotate for a period of time in a uniform circular motion.

Centrifugation is also used in laboratories to separate compounds, for example, and thus separate their constituents by difference in densities.Â Whenever we talk about centrifugation, there is a circular motion that is uniform, at least for a time.

**Garden watering cans**

Many garden sprinklers rotate at a constant speed so that the ground is watered evenly.

**sports**

In the hammer throw for example, which is an Olympic discipline, the athlete forcefully spins a metal ball using a steel cable attached to the grip.Â The objective is to send the ball as far as possible, but without leaving a certain area.

**Exercise resolved**

A particle moves in a circle of radius 2m with a constant speed v = 8 m / s, counterclockwise.Â Initially the particle was atÂ **r**Â = +2Â **j**Â m.Â Calculate:

a) The angular velocity Ï‰

b) Its angular position Î¸ (t)

c) The period of movement

d) Centripetal acceleration.

e) Particle position after passing t = Ï€ / 4 s

**Solution to**

From the formula v = RÏ‰ it follows that:

Ï‰ = v / R = (8 m / s) / 2m = 4rad âˆ™ sÂ ^{-1}

**Solution b**

Taking the positive x axis as the reference axis, the particle is initially at 90Âº = Ï€ / 2 radians with respect to said axis, since theÂ statementÂ says that the initial position is +2Â **j**Â m, that is, the particle is in y = 2m when the movement begins to follow.

Î¸ = Î¸ (t) = Î¸Â _{or}Â + Ï‰t = Ï€ / 2 + 4t

**Solution c**

T = 2Ï€ / Ï‰ = 2Ï€ / 4 s = 0.5 Ï€ s

**Solution d**

a = vÂ ^{2}Â / R = (8 m / s)Â ^{2}Â /2 m = 32 m / sÂ ^{2}

**Solution e**

Î¸ (t) = Ï€ / 2 + 4t â†’ Î¸ (Ï€ / 4) = Ï€ / 2 + 4 âˆ™ (Ï€ / 4) = 3Ï€ / 2 radians

This means that after that time, the particle is in the position y = -2mÂ **j.Â **It makes sense because t = Ï€ / 4 s is half the period, therefore the particle traveled a 180Âº angle counterclockwise from its initial position and it has to be just in the opposite position.