# Circular motion

In physics, circular motion is rotation with constant speed around in a circle: a circular path or a circular orbit. It is one of the simplest cases of accelerated motion. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. We can talk about circular motion of an object if we ignore its size, so that we have the motion of a point mass in a plane.

Circular motion involves acceleration of the moving object by a centripetal force which pulls the moving object towards the center of the circular orbit. Without this acceleration, the object would move inertially in a straight line, according to Newton's first law of motion. Circular motion is accelerated even though the speed is constant, because the velocity of the moving object is constantly changing.

Examples of circular motion are: an artificial satellite orbiting the Earth in geosynchronous orbit, a stone which is tied to a rope and is being swung in circles (cf. hammer throw), a racecar turning through a curve in a racetrack, an electron moving perpendicular to a uniform magnetic field, a gear turning inside a mechanism.

A special kind of circular motion is when an object rotates around its own center of mass. This can be called spinning motion, or rotational motion.

Circular motion is characterized by an orbital radius r, a speed v, the mass m of the object which moves in a circle, and the magnitude F of the centripetal force. These quantities all relate to each other through the equation

${\displaystyle F={mv^{2} \over r}}$

which is always true for circular motion.

## Mathematical description

Circular motion can be described by means of parametric equations, viz.

${\displaystyle x(t)=R\,\cos \,\omega t,\qquad \qquad (1)}$
${\displaystyle y(t)=R\,\sin \,\omega t,\qquad \qquad (2)}$

where R and ω are coefficients. Equations (1) and (2) describe motion around a circle centered at the origin with radius R. The derivatives of these equations are

${\displaystyle {\dot {x}}(t)=-R\omega \,\sin \,\omega t,\qquad \qquad (3)}$
${\displaystyle {\dot {y}}(t)=R\omega \,\cos \,\omega t.\qquad \qquad (4)}$

The vector (x,y) is the position vector of the object undergoing the circular motion. The vector ${\displaystyle ({\dot {x}},{\dot {y}})}$, given by equations (3) and (4), is the velocity vector of the moving object. This velocity vector is perpendicular to the position vector, and it is tangent to the circular path of the moving object. The velocity vector must be considered to have its tail located at the head of the position vector. The tail of the position vector is located at the origin.

The derivatives of equations (3) and (4) are

${\displaystyle {\ddot {x}}(t)=-R\omega ^{2}\,\cos \,\omega t,\qquad \qquad (5)}$
${\displaystyle {\ddot {y}}(t)=-R\omega ^{2}\,\sin \,\omega t.\qquad \qquad (6)}$

The vector ${\displaystyle ({\ddot {x}},{\ddot {y}})}$, called the acceleration vector, is given by equations (5) and (6). It has its tail at the head of the position vector, but it points in the direction opposite to the position vector. This means that circular motion can be described by differential equations, thus

${\displaystyle {\ddot {x}}=-\omega ^{2}x,}$
${\displaystyle {\ddot {y}}=-\omega ^{2}y,}$

or letting x denote the position vector, then circular motion can be described by a single vector differential equation

${\displaystyle {\ddot {\mathbf {x} }}=-\omega ^{2}\mathbf {x} .}$

The quantity ω is the angular velocity.

## Deriving the centripetal force

From equations (5) and (6) it is evident that the magnitude of the acceleration is

${\displaystyle a=\omega ^{2}R.\qquad \qquad (7)}$

The angular frequency ω is expressed in terms of the period T as

${\displaystyle \omega ={2\pi \over T}.\qquad \qquad (8)}$

The speed v around the orbit is given by the circumference divided by the period:

${\displaystyle v={2\pi R \over T}.\qquad \qquad (9)}$

Comparing equations (8) and (9), we deduce that

${\displaystyle v=\omega R.\qquad \qquad (10)}$

Solving equation (10) for ω and substituting into equation (7) yields

${\displaystyle a={v^{2} \over R}.\qquad \qquad (11)}$

Newton's second law of motion is usually expressed as

${\displaystyle F=ma\,}$

which together with equation (11) implies that

${\displaystyle F={mv^{2} \over R},\qquad \qquad (12)}$

## Kepler's third law

For satellites tethered to a body of mass M at the origin by means of a gravitational force, the centripetal force is also equal to

${\displaystyle F={GMm \over R^{2}}\qquad \qquad (13)}$

where G is the gravitational constant, 6.67 × 10−11 N-m2/kg2. Combining equations (12) and (13) yields

${\displaystyle {GMm \over R^{2}}={mv^{2} \over R}}$

which simplifies to

${\displaystyle GM=Rv^{2}.\qquad \qquad (14)}$

Combining equations (14) and (10) then yields

${\displaystyle \omega ^{2}R^{3}=GM\ }$

which is a form of Kepler's harmonic law of planetary motion.