Circular motion

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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.

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 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 object's velocity vector is constantly changing direction.

1 Constant speed
2 Variable speed
3 See also
4 External links

[edit] Constant speed

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

Consider a body of one kilogram, moving in a circle of radius one meter, with an angular velocity of one radian per second.

The speed is one meter per second
The inward acceleration of one meter per second per second.
It is subject to a centripetal force of one kilogram meter per second per second, which is one newton.
The momentum of the body is one kg·m·s⁻¹.
The moment of inertia is one kg·m².
The angular momentum is one kg·m²·s⁻¹.
The kinetic energy is 1/2 joule.
The circumference of the orbit is 2π (~ 6.283) meters.
The period of the motion is 2π second per turn.
The frequency is (2π)⁻¹ hertz.
From the point of view of quantum mechanics, the system is in an excited state having quantum number ~ 9.48×10³⁵.

Then consider a body of mass m, moving in a circle of radius r, with an angular velocity of ω.

The speed is v = r·ω.
The inward acceleration is a = r·ω ² = r ⁻¹·v ².
The centripetal force is F = m·a = r·m·ω ² = r⁻¹·m·v ².
The momentum of the body is p = m·v = r·m·ω.
The moment of inertia is I = r ²·m.
The angular momentum is madharchod L = r·m·v = r ²·m·ω = I·ω.
The kinetic energy is E = 2⁻¹·m·v ² = 2⁻¹·r ²·m·ω ² = (2·m)⁻¹·p ² = 2⁻¹·I·ω ² = (2·I)⁻¹·L ² .
The circumference of the orbit is 2·π·r.
The period of the motion is T = 2·π·r·ω ⁻¹.
The frequency is f = T ⁻¹ . (Instead of letter f, the frequency is often denoted by the Greek letter ν, which however is almost indistinguishable from the letter v used here for velocity).
The quantum number is J = 2·π·L h⁻¹.

[edit] Mathematical description

[edit] Using real numbers

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

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

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

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

The derivatives of these equations are

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

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

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

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

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

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

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

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

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

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

[edit] Using complex numbers

The complex number z=x+iy satisfies the second order differential equation of the circular motion

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

The first order differential equation of the circular motion is

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

The solution to this equation is

$\ z = R e^{i \omega t}$

where R is a complex constant.

So the exponential function of iωt describes the circular motion. This leads to simpler calculations than the real number representation, which involves the trigonometric functions.

[edit] Deriving the centripetal force

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

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

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

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

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

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

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

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

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

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

Newton's second law of motion is usually expressed as

$F = m a \,$

which together with equation (11) implies that

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

(QED).

[edit] 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

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

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

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

which simplifies to

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

Combining equations (14) and (10) then yields

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

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

[edit] Variable speed

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.