In physics, circular motion is rotation along a circle: a circular path or a circular orbit. 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.

Constant speed

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In the simplest case the speed is constant. It is one of the simplest cases of accelerated motion.

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

$F\; =\; \{m\; v^2\; \backslash over\; r\}$

which is always true for circular motion.

Since $v\; =\; r\; \backslash omega\backslash$, the above equation can be expressed as $F\; =\; m\; r\; \backslash omega\backslash \; ^2$

Mathematical description

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

$x(t)\; =\; R\; \backslash ,\; \backslash cos\; \backslash ,\; \backslash omega\; t,\; \backslash qquad\; \backslash qquad\; (1)$

$y(t)\; =\; R\; \backslash ,\; \backslash sin\; \backslash ,\; \backslash omega\; t,\; \backslash qquad\; \backslash qquad\; (2)$

where R and ω are coefficients. Equations (1) and (2) describe motion around a circle centered at the origin with radius R. The quantity ω is the angular velocity, and t is the time.

The derivatives of these equations are

$\backslash dot\{x\}(t)\; =\; -\; R\; \backslash omega\; \backslash ,\; \backslash sin\; \backslash ,\; \backslash omega\; t,\; \backslash qquad\; \backslash qquad\; (3)$

$\backslash dot\{y\}(t)\; =\; R\; \backslash omega\; \backslash ,\; \backslash cos\; \backslash ,\; \backslash omega\; t.\; \backslash qquad\; \backslash qquad\; (4)$

The vector $\backslash \; (x,y)$ is the position vector of the object undergoing the circular motion. The vector $(\backslash dot\{x\},\backslash dot\{y\})$, 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

$\backslash ddot\{x\}(t)\; =\; -\; R\; \backslash omega^2\; \backslash ,\; \backslash cos\; \backslash ,\; \backslash omega\; t,\; \backslash qquad\; \backslash qquad\; (5)$

$\backslash ddot\{y\}(t)\; =\; -\; R\; \backslash omega^2\; \backslash ,\; \backslash sin\; \backslash ,\; \backslash omega\; t.\; \backslash qquad\; \backslash qquad\; (6)$

The vector $(\backslash ddot\{x\},\backslash ddot\{y\})$, 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

$\backslash ddot\{x\}\; =\; -\; \backslash omega^2\; x,$

$\backslash ddot\{y\}\; =\; -\; \backslash omega^2\; y,$

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

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

The second order differential equation of the circular motion is

$\backslash ddot\{z\}\; =\; -\; \backslash omega^2\; z$

The first order differential equation of the circular motion is

$\backslash dot\{z\}\; =\; i\; \backslash omega\; z$

Deriving the centripetal force

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

$a\; =\; \backslash omega^2\; R.\; \backslash qquad\; \backslash qquad\; (7)$

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

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

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

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

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

$v\; =\; \backslash omega\; R.\; \backslash qquad\; \backslash qquad\; (10)$

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

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

Newton's second law of motion is usually expressed as

$F\; =\; m\; a\; \backslash ,$

which together with equation (11) implies that

$F\; =\; \{m\; v^2\; \backslash over\; R\},\backslash qquad\; \backslash qquad\; (12)$

(QED).

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

$F\; =\; \{G\; M\; m\; \backslash over\; R^2\}\; \backslash qquad\; \backslash qquad\; (13)$

where G is the gravitational constant, 6.67·10⁻¹¹ N·m²·kg⁻². Combining equations (12) and (13) yields

$\{G\; M\; m\; \backslash over\; R^2\}\; =\; \{m\; v^2\; \backslash over\; R\}$

which simplifies to

$G\; M\; =\; R\; v^2.\; \backslash qquad\; \backslash qquad\; (14)$

Combining equations (14) and (10) then yields

$\backslash omega^2\; R^3\; =\; G\; M\; \backslash$

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

Variable speed

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In the general case, circular motion requires that the total force can be decomposed into the centripetal force required to keep the orbit circular, and a force tangent to the circle, causing a change of speed.

The magnitude of the centripetal force depends on the instantaneous speed.

In the case of an object at the end of a rope, subjected to a force, we can decompose the force into a radial and a lateral component. The radial component is either outward or inward.

See also

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• angular momentum
• simple harmonic motion

Classical mechanics

תנועה מעגלית | Pohyb po kružnici | 등속원운동 | Moto circolare | Ruch obrotowy | Movimento circular | Kroženje | Pyörimisliike than the required centripetal force, then the stress in the rope provides the difference, provided that the rope is strong enough. If the radial component is inward and more than the required centripetal force, then circular motion is not maintained.

An example is a rotation of an object at the end of a rope, in a vertical plane. If the speed is large enough, circular motion is maintained.

In the case of a rigid body with a hinge, the motion is circular anyway, because the stress can both pull and push.