With circular motion there is a radius and circumference that may be measured as distance or duration. Call the spatial circumference *S*, and the temporal circumference *T*, which is known as the period. Distinguish the spatial and temporal versions of the radius, *R*, and the angle of motion, *θ*, by using *R _{s}* and

*R*, and

_{t}*θ*and

_{s}*θ*, respectively. Then

_{t}*S*= 2π

*R*and

_{s}*T*= 2π

*R*. Also, Δ

_{t}*s*=

*R*Δ

_{s}*θ*, Δ

_{s}*t*=

*R*Δ

_{t}*θ*,

_{t}*R*=

_{s}*R*, and

_{t}v*R*=

_{t}*R*, with velocity,

_{s}u*v*, and lenticity,

*w*.

The acceleration is directed from the center, and so is centripetal. What is the magnitude of acceleration that occurs in uniform circular motion? *Christiaan Huygens* was the first to answer this in 1658. Here is a simple derivation:

An object in uniform circular motion traverses a circle at constant speed, *v*. Its spatial position can be represented by a vector, *R _{s}*, which changes its angle but not its magnitude. The distance traversed in one cycle is

*S*= 2π

*R*. The period or duration of one cycle is

_{s}*T*= 2π

*R*= 2πR

_{t}_{s}/

*v*.

The velocity vector of this object can also be represented by a vector that changes its angle but not its magnitude. The accumulated change in velocity is 2π*v*. The magnitude of the acceleration is the change in velocity divided by the duration:

*a* = 2π*v* / *T* = 2π*v* / (2π*R _{s}*/

*v*) =

*v²*/

*R*.

_{s}Another derivation uses a diagram such as this:

Substituting *R _{s}* for

*r*and

*θ*for

_{s}*θ*, the derivation is as follows:

Δ*s** =* *R _{s}* Δ

*θ*

_{s}*v* = |Δ*s*| / |Δ*t*|, and

*a* = |**a**| = |Δ**v**| / |Δ*t*| = *v *|Δ*θ _{s}*| / |Δ

*t*| =

*v*|Δ

*s*| / (

*R*|Δ

_{s}*t*|) =

*v²*|Δ

*t*| / (

*R*|Δ

_{s}*t*|) =

*v²*/

*R*.

_{s}Additionally,

*a* = *v²* / *R _{s}* =

*v²*/ (

*R*) =

_{t}v*v*/

*R*=

_{t}*R*/

_{s}*R*².

_{t}What is the *relentation* that occurs in uniform circular motion? It is directed toward the periphery, and so is centrifugal. A simple derivation follows the first method above:

An object in uniform circular motion traverses a circle at constant pace, *u*. Its time position can be represented by a vector, *R _{t}*, which changes its angle but not its magnitude. The period or duration of one cycle is

*T*= 2π

*R*. The distance traversed by one cycle is

_{t}*S*= 2π

*R*= 2π

_{s}*R*/

_{t}*w*.

The lenticity vector of this object can also be represented by a vector that changes its angle but not its magnitude. The accumulated change in lenticity is 2π*w*. The magnitude of the relentation is the change in lenticity divided by the distance:

*b* = 2π*w* / S = 2π*w* / (2π*R _{t}*/

*w*) =

*w²*/

*R*.

_{t}A second derivation uses a diagram similar to the one above with these substitutions: *r* → *R _{t}*,

*θ*→

*θ*,

_{t}*s*↔

*t*, and

*v*→

*w*. Then,

*Δt = R _{t} Δθ_{t},*

*w* = |Δ*t*| / |Δ*s*|, and

*b* = |**b**| = |Δ**w**| / |Δ*s*| = *w* |Δ*θ _{t}*| / |Δ

*s*| =

*w*|Δ

*t*| / (

*R*|Δ

_{t}*s*|) =

*w*² |Δ

*s*| / (

*R*|Δ

_{t}*s*|) =

*w²*/

*R*.

_{t}Additionally,

*b* = *w²* / *R _{t}* =

*w²*/ (

*R*) =

_{s}w*w*/

*R*=

_{s}*R*/

_{t}*R*².

_{s}