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In physics, circular motion is rotation along a circle: a circular path or a circular orbit. It can be uniform, that is, with constant angular rate of rotation, or non-uniform, that is, with a changing rate of rotation. 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. For example, the center of mass of a body can undergo circular motion.## Formulas for uniform circular motion

## Constant speed

In the simplest case the speed, mass and radius are constant.## 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.## Description of circular motion using polar coordinates

## Description of circular motion using complex numbers

Circular motion can be described using complex numbers. Let the $x$ axis be the real axis and the $y$ axis be the imaginary axis. The position of the body can then be given as $z$, a complex "vector":
## See also

## External links

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 race track, an electron moving perpendicular to a uniform magnetic field, a gear turning inside a mechanism.

Circular motion is accelerated even if the angular rate of rotation is constant, because the object's velocity vector is constantly changing direction. Such change in direction of velocity 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 in a straight line, according to Newton's laws of motion.

For motion in a circle of radius R, the circumference of the circle is C = 2π R. If the period for one rotation is T, the angular rate of rotation ω is:

- $omega\; =\; frac\; \{2\; pi\}\{T\}\; .$

The speed of the object traveling the circle is

- $v,\; =\; frac\; \{2\; pi\; R\; \}\; \{T\}\; =\; omega\; R$

The angle θ swept out in a time t is:

- $theta\; =\; 2\; pi\; frac\{t\}\{T\}\; =\; omega\; t,$

The acceleration due to change in the direction of the velocity is found by noticing that the velocity completely rotates direction in the same time T the object takes for one rotation. Thus, the velocity vector sweeps out a path of length 2π v every T seconds, or:

- $a,\; =\; frac\; \{2\; pi\; v\; \}\{T\}\; =\; omega^2\; R\; ,$

and is directed radially inward.

The vector relationships are shown in Figure 1. The axis of rotation is shown as a vector Ω perpendicular to the plane of the orbit and with a magnitude ω = dθ / dt. The direction of Ω is chosen using the right-hand rule. With this convention for depicting rotation, the velocity is given by a vector cross product as

- $mathbf\{v\}\; =\; boldsymbol\; Omega\; times\; mathbf\; r\; ,$

which is a vector perpendicular to both Ω and r (t ), tangential to the orbit, and of magnitude ω R. Likewise, the acceleration is given by

- $mathbf\{a\}\; =\; boldsymbol\; Omega\; times\; mathbf\; v\; ,$

which is a vector perpendicular to both Ω and v (t ) of magnitude ω |v| = ω^{2} R and directed exactly opposite to r (t ).

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

- The speed is one metre per second
- The inward acceleration is one metre per second per second.
- It is subject to a centripetal force of one kilogram metre 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·m
^{2}. - The angular momentum is one kg·m
^{2}·s^{−1}. - The kinetic energy is 1/2 joule.
- The circumference of the orbit is 2π (~ 6.283) metres.
- The period of the motion is 2π seconds 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 ~ 9.48×10
^{35}.

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 centripetal (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 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·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}

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.

During circular motion the body moves on a curve that can be described in polar coordinate system as a fixed distance R from the center of the orbit taken as origin, oriented at an angle θ (t) from some reference direction. See Figure 2. The displacement vector $stackrel\{vec\; r\}\{\}$ is the radial vector from the origin to the particle location:

- $vec\; r=R\; hat\; u\_R\; (t)\; ,$

where $hat\; u\_R\; (t)$ is the unit vector parallel to the radius vector at time t and pointing away from the origin. It is handy to introduce the unit vector orthogonal to $hat\; u\_R$ as well, namely $hat\; u\_theta$. It is customary to orient $hat\; u\_theta$ to point in the direction of travel along the orbit.

The velocity is the time derivative of the displacement:

- $vec\; v\; =\; frac\; \{d\}\{dt\}\; vec\; r(t)\; =\; frac\; \{d\; R\}\{dt\}\; hat\; u\_R\; +\; Rfrac\; \{d\; hat\; u\_R\; \}\; \{dt\}\; .$

Because the radius of the circle is constant, the radial component of the velocity is zero. The unit vector $hat\; u\_R$ has a time-invariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angle θ the same as the angle of $vec\; r\; (t)$. If the particle displacement rotates through an angle dθ in time dt, so does $hat\; u\_R$, describing an arc on the unit circle of magnitude dθ. See the unit circle at the left of Figure 2. Hence:

- $frac\; \{d\; hat\; u\_R\; \}\; \{dt\}\; =\; frac\; \{d\; theta\; \}\; \{dt\}\; hat\; u\_theta\; ,$

where the direction of the change must be perpendicular to $hat\; u\_R$ (or, in other words, along $hat\; u\_theta$) because any change d$hat\; u\_R$ in the direction of $hat\; u\_R$ would change the size of $hat\; u\_R$. The sign is positive, because an increase in dθ implies the object and $hat\; u\_R$ have moved in the direction of $hat\; u\_theta$. Hence the velocity becomes:

- $vec\; v\; =\; frac\; \{d\}\{dt\}\; vec\; r(t)\; =\; Rfrac\; \{d\; hat\; u\_R\; \}\; \{dt\}\; =\; R\; frac\; \{d\; theta\; \}\; \{dt\}\; hat\; u\_theta\; =\; R\; omega\; hat\; u\_theta\; .$

The acceleration of the body can also be broken into radial and tangential components. The acceleration is the time derivative of the velocity:

- $vec\; a\; =\; frac\; \{d\}\{dt\}\; vec\; v\; =\; frac\; \{d\}\{dt\}\; left(R\; omega\; hat\; u\_theta\; right)\; .$

- $=R\; left(frac\; \{d\; omega\}\{dt\}\; hat\; u\_theta\; +\; omega\; frac\; \{d\; hat\; u\_theta\}\{dt\}\; right)\; .$

The time derivative of $hat\; u\_theta$ is found the same way as for $hat\; u\_R$. Again, $hat\; u\_theta$ is a unit vector and its tip traces a unit circle with an angle that is π/2 + θ. Hence, an increase in angle dθ by $vec\; r\; (t)$ implies $hat\; u\_theta$ traces an arc of magnitude dθ, and as $hat\; u\_theta$ is orthogonal to $hat\; u\_R$, we have:

- $frac\; \{d\; hat\; u\_theta\; \}\; \{dt\}\; =\; -frac\; \{d\; theta\; \}\; \{dt\}\; hat\; u\_R\; =\; -omega\; hat\; u\_R\; ,$

where a negative sign is necessary to keep $hat\; u\_theta$ orthogonal to $hat\; u\_R$. (Otherwise, the angle between $hat\; u\_theta$ and $hat\; u\_R$ would decrease with increase in dθ.) See the unit circle at the left of Figure 2. Consequently the acceleration is:

- $vec\; a\; =\; R\; left(frac\; \{d\; omega\}\{dt\}\; hat\; u\_theta\; +\; omega\; frac\; \{d\; hat\; u\_theta\}\{dt\}\; right)$

- $=R\; frac\; \{d\; omega\}\{dt\}\; hat\; u\_theta\; -\; omega^2\; R\; hat\; u\_R\; .$

The centripetal acceleration is the radial component, which is directed radially inward:

- $vec\; a\_R=\; -omega\; ^2R\; hat\; u\_R\; ,$

- $vec\; a\_\{theta\}=\; R\; frac\; \{d\; omega\}\{dt\}\; hat\; u\_theta\; =\; frac\; \{d\; R\; omega\}\{dt\}\; hat\; u\_theta\; =frac\; \{d\; |vec\; v|\}\{dt\}\; hat\; u\_theta\; .$

- $z=x+iy=R(cos\; theta\; +i\; sin\; theta)=Re^\{itheta\}\; ,$

- $theta\; =theta\; (t)\; ,$

- $dot\; R\; =ddot\; R\; =0\; ,$

- $v=dot\; z\; =\; R\; frac\; \{d\}\{d\; theta\}left(i\; thetaright)\; e^\{i\; theta\}\; =\; iRdot\; theta\; e^\{itheta\}\; =\; iomega\; cdot\; Re^\{itheta\}=\; iomega\; z$

- $a=dot\; v\; =idot\; omega\; z\; +i\; omega\; dot\; z\; =(idot\; omega\; z\; -omega^2)z$

- $=\; left(idot\; omega-omega^2\; right)\; R\; e^\{itheta\}$

- $=-omega^2\; R\; e^\{itheta\}\; +\; dot\; omega\; e^\{ifrac\{pi\}\{2\}\}R\; e^\{itheta\}\; .$

The first term is opposite to the direction of the displacement vector and the second is perpendicular to it, just like the earlier results.

- Circular Motion - a chapter from an online textbook

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Last updated on Saturday October 11, 2008 at 10:48:32 PDT (GMT -0700)

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