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# Centripetal expedience

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 Rs and Rt, and θs and θt, respectively. Then S = 2πRs and T = 2πRt. Also, Δs = Rs Δθs, Δt = Rt Δθt, Rs = Rtv, and Rt = Rsu, with velocity, v, and legerity, u.

What is the 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, Rs, which changes its angle but not its magnitude. The distance traversed in one cycle is S = 2πRs. The period or duration of one cycle is T = 2πRt = 2πRs/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 2πv. The magnitude of the acceleration is the change in velocity divided by the duration:

a = 2πv / T = 2πv / (2πRs/v) = / Rs.

Another derivation uses a diagram such as this:

Substituting Rs for r and θs for θ, the derivation is as follows:

Δs = Rs Δθs

v = |Δs| / |Δt|, and

a = |a| = |Δv| / |Δt| = v θs| / |Δt| = v s| / (Rst|) = t| / (Rst|) = / Rs.

a = / Rs = / (Rtv) = v / Rt = Rs / Rt².

What is the expedience that occurs in uniform circular motion? 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, Rt, which changes its angle but not its magnitude. The period or duration of one cycle is T = 2πRt. The distance traversed by one cycle is S = 2πRs = 2πRt/u.

The legerity vector of this object can also be represented by a vector that changes its angle but not its magnitude. The accumulated change in legerity is 2πu. The magnitude of the expedience is the change in legerity divided by the distance:

b = 2πu / S = 2πu / (2πRt/u) = / Rt.

A second derivation uses a diagram similar to the one above with these substitutions: rRt, θ → θt, st, and vu. Then,

Δt = Rt Δθt,

u = |Δt| / |Δs|, and

b = |b| = |Δu| / |Δs| = uθt| / |Δs| = ut| / (Rts|) = u² |Δs| / (Rts|) = / Rt.