Editing Centripetal force (section)

==== Derivation using vectors ====

[[File:Circular motion vectors.svg|right|thumb|Vector relationships for uniform circular motion; vector '''Ω''' representing the rotation is normal to the plane of the orbit with polarity determined by the [[right-hand rule]] and magnitude ''dθ'' /''dt''.]]
The image at right shows the vector relationships for uniform circular motion. The rotation itself is represented by the angular velocity vector '''Ω''', which is normal to the plane of the orbit (using the [[right-hand rule]]) and has magnitude given by:
: <math> |\mathbf{\Omega}| = \frac {\mathrm{d} \theta } {\mathrm{d}t} = \omega \ , </math>

with ''θ'' the angular position at time ''t''. In this subsection, d''θ''/d''t'' is assumed constant, independent of time. The distance traveled '''dℓ''' of the particle in time d''t'' along the circular path is
: <math> \mathrm{d}\boldsymbol{\ell} = \mathbf {\Omega} \times \mathbf{r}(t) \mathrm{d}t \ , </math>

which, by properties of the [[vector cross product]], has magnitude ''r''d''θ'' and is in the direction tangent to the circular path.

Consequently,
: <math>\frac {\mathrm{d} \mathbf{r}}{\mathrm{d}t} = \lim_{{\Delta}t \to 0} \frac {\mathbf{r}(t + {\Delta}t)-\mathbf{r}(t)}{{\Delta}t} = \frac{\mathrm{d} \boldsymbol{\ell}}{\mathrm{d}t} \ .</math>

In other words,
: <math> \mathbf{v}\ \stackrel{\mathrm{def}}{ = }\ \frac {\mathrm{d} \mathbf{r}}{\mathrm{d}t} = \frac {\mathrm{d}\mathbf{\boldsymbol{\ell}}}{\mathrm{d}t} = \mathbf {\Omega} \times \mathbf{r}(t)\ . </math>

Differentiating with respect to time,
<math display="block"> \mathbf{a}\ \stackrel{\mathrm{def}}{ = }\ \frac {\mathrm{d} \mathbf{v}} {d\mathrm{t}} = \mathbf {\Omega} \times \frac{\mathrm{d} \mathbf{r}(t)}{\mathrm{d}t} = \mathbf{\Omega} \times \left[ \mathbf {\Omega} \times \mathbf{r}(t)\right] \ .</math>

[[Triple product#Vector triple product|Lagrange's formula]] states:
<math display="block"> \mathbf{a} \times \left ( \mathbf{b} \times \mathbf{c} \right ) = \mathbf{b} \left ( \mathbf{a} \cdot \mathbf{c} \right ) - \mathbf{c} \left ( \mathbf{a} \cdot \mathbf{b} \right ) \ .</math>

Applying Lagrange's formula with the observation that '''Ω • r'''(''t'') = 0 at all times,
<math display="block"> \mathbf{a} = - {|\mathbf{\Omega|}}^2 \mathbf{r}(t) \ .</math>

In words, the acceleration is pointing directly opposite to the radial displacement '''r''' at all times, and has a magnitude:
<math display="block"> |\mathbf{a}| = |\mathbf{r}(t)| \left ( \frac {\mathrm{d} \theta}{\mathrm{d}t} \right) ^2 = r {\omega}^2 </math>
where vertical bars |...| denote the vector magnitude, which in the case of '''r'''(''t'') is simply the radius ''r'' of the path. This result agrees with the previous section, though the notation is slightly different.

When the rate of rotation is made constant in the analysis of [[#Nonuniform circular motion|nonuniform circular motion]], that analysis agrees with this one.

A merit of the vector approach is that it is manifestly independent of any coordinate system.