Editing Circular motion (section)

==== In polar coordinates ====
{{See also|Velocity#Polar coordinates}}
[[File:Vectors in polar coordinates.PNG|thumb|350px|Figure 4: Polar coordinates for circular trajectory. On the left is a unit circle showing the changes <math>\mathbf{d\hat\mathbf{u}_R} </math> and <math>\mathbf{d\hat\mathbf{u}_\theta}</math> in the unit vectors <math>\mathbf{\hat\mathbf{u}_R} </math> and <math>\mathbf{\hat\mathbf{u}_\theta}</math> for a small increment <math>d \theta</math> in angle <math>\theta</math>.]]
During circular motion, the body moves on a curve that can be described in the [[polar coordinate system]] as a fixed distance {{math|''R''}} from the center of the orbit taken as the origin, oriented at an angle {{math|''θ''(''t'')}} from some reference direction. See Figure 4. The displacement ''vector'' <math>\mathbf{r}</math> is the radial vector from the origin to the particle location:
<math display="block">\mathbf{r}(t) = R \hat\mathbf{u}_R(t)\,,</math>
where <math>\hat\mathbf{u}_R(t)</math> is the [[unit vector]] parallel to the radius vector at time {{mvar|t}} and pointing away from the origin. It is convenient to introduce the unit vector [[Orthogonality (mathematics)#Euclidean vector spaces|orthogonal]] to <math>\hat\mathbf{u}_R(t)</math> as well, namely <math>\hat\mathbf{u}_\theta(t)</math>. It is customary to orient <math>\hat\mathbf{u}_\theta(t)</math> to point in the direction of travel along the orbit.

The velocity is the time derivative of the displacement:
<math display="block">\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = \frac{d R}{dt} \hat\mathbf{u}_R(t) + R \frac{d \hat\mathbf{u}_R}{dt} \, .</math>

Because the radius of the circle is constant, the radial component of the velocity is zero. The unit vector <math>\hat\mathbf{u}_R(t)</math> has a time-invariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angle {{mvar|θ}} the same as the angle of <math>\mathbf{r}(t)</math>. If the particle displacement rotates through an angle {{math|''dθ''}} in time {{math|''dt''}}, so does <math>\hat\mathbf{u}_R(t)</math>, describing an arc on the unit circle of magnitude {{math|''dθ''}}. See the unit circle at the left of Figure 4. Hence: 
<math display="block">\frac{d \hat\mathbf{u}_R}{dt} = \frac{d \theta}{dt} \hat\mathbf{u}_\theta(t) \, ,</math>
where the direction of the change must be perpendicular to <math>\hat\mathbf{u}_R(t)</math> (or, in other words, along <math>\hat\mathbf{u}_\theta(t)</math>) because any change <math>d\hat\mathbf{u}_R(t)</math> in the direction of <math>\hat\mathbf{u}_R(t)</math> would change the size of <math>\hat\mathbf{u}_R(t)</math>. The sign is positive because an increase in {{math|''dθ''}} implies the object and <math>\hat\mathbf{u}_R(t)</math> have moved in the direction of <math>\hat\mathbf{u}_\theta(t)</math>.
Hence the velocity becomes:
<math display="block">\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = R\frac{d \hat\mathbf{u}_R}{dt} = R \frac{d \theta}{dt} \hat\mathbf{u}_\theta(t) = R \omega \hat\mathbf{u}_\theta(t) \, .</math>

The acceleration of the body can also be broken into radial and tangential components. The acceleration is the time derivative of the velocity:
<math display="block">\begin{align}
\mathbf{a}(t) &= \frac{d}{dt} \mathbf{v}(t) = \frac{d}{dt} \left(R \omega \hat\mathbf{u}_\theta(t) \right) \\
              &= R \left( \frac{d \omega}{dt} \hat\mathbf{u}_\theta(t) + \omega \frac{d \hat\mathbf{u}_\theta}{dt} \right) \, .
\end{align}</math>

The time derivative of <math>\hat\mathbf{u}_\theta(t)</math> is found the same way as for <math>\hat\mathbf{u}_R(t)</math>. Again, <math>\hat\mathbf{u}_\theta(t)</math> is a unit vector and its tip traces a unit circle with an angle that is {{math|''π''/2 + ''θ''}}. Hence, an increase in angle {{math|''dθ''}} by <math>\mathbf{r}(t)</math> implies <math>\hat\mathbf{u}_\theta(t)</math> traces an arc of magnitude {{math|''dθ''}}, and as <math>\hat\mathbf{u}_\theta(t)</math> is orthogonal to <math>\hat\mathbf{u}_R(t)</math>, we have:
<math display="block">\frac{d \hat\mathbf{u}_\theta}{dt} = -\frac{d \theta}{dt} \hat\mathbf{u}_R(t) = -\omega \hat\mathbf{u}_R(t) \, ,</math>
where a negative sign is necessary to keep <math>\hat\mathbf{u}_\theta(t)</math> orthogonal to <math>\hat\mathbf{u}_R(t)</math>. (Otherwise, the angle between <math>\hat\mathbf{u}_\theta(t)</math> and <math>\hat\mathbf{u}_R(t)</math> would ''decrease'' with an increase in {{math|''dθ''}}.) See the unit circle at the left of Figure 4. Consequently, the acceleration is:
<math display="block">\begin{align}
\mathbf{a}(t) &= R \left( \frac{d \omega}{dt} \hat\mathbf{u}_\theta(t) + \omega \frac{d \hat\mathbf{u}_\theta}{dt} \right) \\
              &= R \frac{d \omega}{dt} \hat\mathbf{u}_\theta(t) - \omega^2 R \hat\mathbf{u}_R(t) \,.
\end{align}</math>

The [[centripetal force|centripetal acceleration]] is the radial component, which is directed radially inward:
<math display="block">\mathbf{a}_R(t) = -\omega^2 R \hat\mathbf{u}_R(t) \, ,</math>
while the tangential component changes the [[Vector (geometry)#Length|magnitude]] of the velocity: 
<math display="block">\mathbf{a}_\theta(t) = R \frac{d \omega}{dt} \hat\mathbf{u}_\theta(t) = \frac{d R \omega}{dt} \hat\mathbf{u}_\theta(t) = \frac{d \left|\mathbf{v}(t)\right|}{dt} \hat\mathbf{u}_\theta(t) \, .</math>