Editing Centripetal force (section)

===== Example: circular motion =====

To illustrate the above formulas, let ''x'', ''y'' be given as:
: <math>x = \alpha \cos \frac{s}{\alpha} \ ; \ y = \alpha \sin\frac{s}{\alpha} \ .</math>

Then:
: <math>x^2 + y^2 = \alpha^2 \ , </math>

which can be recognized as a circular path around the origin with radius ''α''. The position ''s'' = 0 corresponds to [''α'', 0], or 3 o'clock. To use the above formalism, the derivatives are needed:

: <math>y^{\prime}(s) = \cos \frac{s}{\alpha} \ ; \ x^{\prime}(s) = -\sin \frac{s}{\alpha} \ , </math>
: <math>y^{\prime\prime}(s) = -\frac{1}{\alpha}\sin\frac{s}{\alpha} \ ; \ x^{\prime\prime}(s) = -\frac{1}{\alpha}\cos \frac{s}{\alpha} \ . </math>

With these results, one can verify that:
: <math> x^{\prime}(s)^2 + y^{\prime}(s)^2 = 1 \ ; \ \frac{1}{\rho} = y^{\prime\prime}(s)x^{\prime}(s)-y^{\prime}(s)x^{\prime\prime}(s) = \frac{1}{\alpha} \ . </math>

The unit vectors can also be found:
: <math>\mathbf{u}_\mathrm{t}(s) = \left[-\sin\frac{s}{\alpha} \ , \ \cos\frac{s}{\alpha} \right] \ ; \ \mathbf{u}_\mathrm{n}(s) = \left[\cos\frac{s}{\alpha} \ , \ \sin\frac{s}{\alpha} \right] \ , </math>

which serve to show that ''s'' = 0 is located at position [''ρ'', 0] and ''s'' = ''ρ''π/2 at [0, ''ρ''], which agrees with the original expressions for ''x'' and ''y''. In other words, ''s'' is measured counterclockwise around the circle from 3 o'clock. Also, the derivatives of these vectors can be found:
: <math>\frac{\mathrm{d}}{\mathrm{d}s}\mathbf{u}_\mathrm{t}(s) = -\frac{1}{\alpha} \left[\cos\frac{s}{\alpha} \ , \ \sin\frac{s}{\alpha} \right] = -\frac{1}{\alpha}\mathbf{u}_\mathrm{n}(s) \ ; </math>
: <math> \ \frac{\mathrm{d}}{\mathrm{d}s}\mathbf{u}_\mathrm{n}(s) = \frac{1}{\alpha} \left[-\sin\frac{s}{\alpha} \ , \ \cos\frac{s}{\alpha} \right] = \frac{1}{\alpha}\mathbf{u}_\mathrm{t}(s) \ . </math>

To obtain velocity and acceleration, a time-dependence for ''s'' is necessary. For counterclockwise motion at variable speed ''v''(''t''):
: <math>s(t) = \int_0^t \ dt^{\prime} \ v(t^{\prime}) \ , </math>
where ''v''(''t'') is the speed and ''t'' is time, and ''s''(''t'' = 0) = 0. Then:
: <math>\mathbf{v} = v(t)\mathbf{u}_\mathrm{t}(s) \ ,</math>
: <math>\mathbf{a} = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s) + v\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s) = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s)-v\frac{1}{\alpha}\mathbf{u}_\mathrm{n}(s)\frac{\mathrm{d}s}{\mathrm{d}t} </math>
: <math>\mathbf{a} = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s)-\frac{v^2}{\alpha}\mathbf{u}_\mathrm{n}(s) \ , </math>
where it already is established that α = ρ. This acceleration is the standard result for [[non-uniform circular motion]].