User:John R. Brews/Centrifugal force

Polar coordinates
The above results can be derived perhaps more simply in polar coordinates, and at the same time extended to general motion within a plane, as shown next. Polar coordinates in the plane employ a radial unit vector uρ and an angular unit vector uθ, as shown in Figure 6. A particle at position r is described by:


 * $$\mathbf{r} = \rho \mathbf{u}_{\rho} \, $$

where the notation ρ is used to describe the distance of the path from the origin instead of R to emphasize that this distance is not fixed, but varies with time. The unit vector uρ travels with the particle and always points in the same direction as r(t). Unit vector uθ also travels with the particle and stays orthogonal to uρ. Thus, uρ and uθ form a local Cartesian coordinate system attached to the particle, and tied to the path traveled by the particle. By moving the unit vectors so their tails coincide, as seen in the circle at the left of Figure 6, it is seen that uρ and uθ form a right-angled pair with tips on the unit circle that trace back and forth on the perimeter of this circle with the same angle θ(t) as r(t).

When the particle moves, its velocity is


 * $$ \mathbf{v} = \frac {\mathrm{d} \rho }{\mathrm{d}t} \mathbf{u}_{\rho} + \rho \frac {\mathrm{d} \mathbf{u}_{\rho}}{\mathrm{d}t} \ . $$

To evaluate the velocity, the derivative of the unit vector uρ is needed. Because uρ is a unit vector, its magnitude is fixed, and it can change only in direction, that is, its change duρ has a component only perpendicular to uρ. When the trajectory r(t) rotates an amount dθ, uρ, which points in the same direction as r(t), also rotates by dθ. See Figure 6. Therefore the change in uρ is


 * $$ \mathrm{d} \mathbf{u}_{\rho} = \mathbf{u}_{\theta} \mathrm{d}\theta \, $$

or


 * $$ \frac {\mathrm{d} \mathbf{u}_{\rho}}{\mathrm{d}t} = \mathbf{u}_{\theta} \frac {\mathrm{d}\theta}{\mathrm{d}t} \ . $$

In a similar fashion, the rate of change of uθ is found. As with uρ, uθ is a unit vector and can only rotate without changing size. To remain orthogonal to uρ while the trajectory r(t) rotates an amount dθ, uθ, which is orthogonal to r(t), also rotates by dθ. See Figure 6. Therefore, the change duθ is orthogonal to uθ and proportional to dθ (see Figure 6):


 * $$ \frac{\mathrm{d} \mathbf{u}_{\theta}}{\mathrm{d}t} = -\frac {\mathrm{d} \theta} {\mathrm{d}t} \mathbf{u}_{\rho} \ . $$

Figure 6 shows the sign to be negative: to maintain orthogonality, if duρ is positive with dθ, then duθ must decrease.

Substituting the derivative of uρ into the expression for velocity:


 * $$ \mathbf{v} = \frac {\mathrm{d} \rho }{\mathrm{d}t} \mathbf{u}_{\rho} + \rho \mathbf{u}_{\theta} \frac {\mathrm{d} \theta} {\mathrm{d}t} = \mathbf{v}_{\rho} + \mathbf{v}_{\theta} \ . $$

To obtain the acceleration, another time differentiation is done:
 * $$ \mathbf{a} = \frac {\mathrm{d}^2 \rho }{\mathrm{d}t^2} \mathbf{u}_{\rho} + \frac {\mathrm{d} \rho }{\mathrm{d}t} \frac{\mathrm{d} \mathbf{u}_{\rho}}{\mathrm{d}t} + \frac {\mathrm{d} \rho}{\mathrm{d}t} \mathbf{u}_{\theta} \frac {\mathrm{d} \theta} {\mathrm{d}t} + \rho \frac{\mathrm{d} \mathbf{u}_{\theta}}{\mathrm{d}t} \frac {\mathrm{d} \theta} {\mathrm{d}t} + \rho \mathbf{u}_{\theta} \frac {\mathrm{d}^2 \theta} {\mathrm{d}t^2} \ . $$

Substituting the derivatives of uρ and uθ, the acceleration of the particle is:
 * $$ \mathbf{a} = \frac {\mathrm{d}^2 \rho }{\mathrm{d}t^2} \mathbf{u}_{\rho} + 2\frac {\mathrm{d} \rho}{\mathrm{d}t} \mathbf{u}_{\theta} \frac {\mathrm{d} \theta} {\mathrm{d}t}-\rho \mathbf{u}_{\rho}\left( \frac {\mathrm{d} \theta} {\mathrm{d}t}\right)^2 + \rho \mathbf{u}_{\theta} \frac {\mathrm{d}^2 \theta} {\mathrm{d}t^2} \, $$
 * $$ = \mathbf{u}_{\rho} \left[ \frac {\mathrm{d}^2 \rho }{\mathrm{d}t^2}-\rho\left( \frac {\mathrm{d} \theta} {\mathrm{d}t}\right)^2 \right] + \mathbf{u}_{\theta}\left[ 2\frac {\mathrm{d} \rho}{\mathrm{d}t} \frac {\mathrm{d} \theta} {\mathrm{d}t} + \rho \frac {\mathrm{d}^2 \theta} {\mathrm{d}t^2}\right] \ $$
 * $$ = \mathbf{u}_{\rho} \left[ \frac {\mathrm{d}|\mathbf{v}_{\rho}|}{\mathrm{d}t}-\frac{|\mathbf{v}_{\theta}|^2}{\rho}\right] + \mathbf{u}_{\theta}\left[ \frac{2}{\rho}|\mathbf{v}_{\rho}||\mathbf{v}_{\theta}| + \rho\frac{\mathrm{d}}{\mathrm{d}t}\frac{|\mathbf{v}_{\theta}|}{\rho}\right] \ .$$

As a particular example, if the particle moves in a circle of constant radius R, then dρ/dt = 0, v = vθ, and:


 * $$\mathbf{a} = \mathbf{u}_{\rho} \left[ -\rho\left( \frac {\mathrm{d} \theta} {\mathrm{d}t}\right)^2 \right] + \mathbf{u}_{\theta}\left[ \rho \frac {\mathrm{d}^2 \theta} {\mathrm{d}t^2}\right] \ $$
 * $$ = \mathbf{u}_{\rho} \left[ -\frac{|\mathbf{v}|^2}{R}\right] + \mathbf{u}_{\theta}\left[ \frac {\mathrm{d} |\mathbf{v}|} {\mathrm{d}t}\right] \ . $$

These results agree with those above for nonuniform circular motion. See also the article on non-uniform circular motion. If this acceleration is multiplied by the particle mass, the leading term is the centripetal force and the negative of the second term related to angular acceleration is sometimes called the Euler force.

For trajectories other than circular motion, for example, the more general trajectory envisioned in Figure 6, the instantaneous center of rotation and radius of curvature of the trajectory are related only indirectly to the coordinate system defined by uρ and uθ and to the length |r(t)| = ρ. Consequently, in the general case, it is not straightforward to disentangle the centripetal and Euler terms from the above general acceleration equation. To deal directly with this issue, local coordinates are preferable, as discussed next.

Local coordinates
By local coordinates is meant a set of coordinates that travel with the particle, and have orientation determined by the path of the particle. Unit vectors are formed as shown in Figure 7, both tangential and normal to the path. This coordinate system sometimes is referred to as intrinsic or path coordinates  or nt-coordinates, for normal-tangential, referring to these unit vectors. These coordinates are a very special example of a more general concept of local coordinates from the theory of differential forms.

Distance along the path of the particle is the arc length s, considered to be a known function of time.


 * $$ s = s(t) \ . $$

A center of curvature is defined at each position s located a distance ρ (the radius of curvature) from the curve on a line along the normal un (s). The required distance ρ(s) at arc length s is defined in terms of the rate of rotation of the tangent to the curve, which in turn is determined by the path itself. If the orientation of the tangent relative to some starting position is θ(s), then ρ(s) is defined by the derivative dθ/ds:


 * $$\frac{1} {\rho (s)} = \kappa (s) = \frac {\mathrm{d}\theta}{\mathrm{d}s}\ . $$

The radius of curvature usually is taken as positive (that is, as an absolute value), while the curvature κ is a signed quantity.

A geometric approach to finding the center of curvature and the radius of curvature uses a limiting process leading to the osculating circle. See Figure 7.

Using these coordinates, the motion along the path is viewed as a succession of circular paths of ever-changing center, and at each position s constitutes non-uniform circular motion at that position with radius ρ. The local value of the angular rate of rotation then is given by:


 * $$ \omega(s) = \frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\mathrm{d}\theta}{\mathrm{d}s} \frac {\mathrm{d}s}{\mathrm{d}t} = \frac{1}{\rho(s)}\ \frac {\mathrm{d}s}{\mathrm{d}t} = \frac{v(s)}{\rho(s)}\ ,$$

with the local speed v given by:


 * $$ v(s) = \frac {\mathrm{d}s}{\mathrm{d}t}\ . $$

As for the other examples above, because unit vectors cannot change magnitude, their rate of change is always perpendicular to their direction (see the left-hand insert in Figure 7):


 * $$\frac{d\mathbf{u}_\mathrm{n}(s)}{ds} = \mathbf{u}_\mathrm{t}(s)\frac{d\theta}{ds} = \mathbf{u}_\mathrm{t}(s)\frac{1}{\rho} \ ; $$&ensp;&ensp;$$\frac{d\mathbf{u}_\mathrm{t}(s)}{\mathrm{d}s} = -\mathbf{u}_\mathrm{n}(s)\frac{\mathrm{d}\theta}{\mathrm{d}s} = - \mathbf{u}_\mathrm{n}(s)\frac{1}{\rho} \ . $$

Consequently, the velocity and acceleration are:
 * $$ \mathbf{v}(t) = v \mathbf{u}_\mathrm{t}(s)\ ; $$

and using the chain-rule of differentiation:


 * $$ \mathbf{a}(t) = \frac{\mathrm{d}v}{\mathrm{d}t} \mathbf{u}_\mathrm{t}(s) - \frac{v^2}{\rho}\mathbf{u}_\mathrm{n}(s) \ ; $$ with the tangential acceleration $$\frac{\mathrm{\mathrm{d}}v}{\mathrm{\mathrm{d}}t} = \frac{\mathrm{d}v}{\mathrm{d}s}\ \frac{\mathrm{d}s}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}s}\ v \ . $$

In this local coordinate system the acceleration resembles the expression for nonuniform circular motion with the local radius ρ(s), and the centripetal acceleration is identified as the second term.

Extension of this approach to three dimensional space curves leads to the Frenet-Serret formulas.

Alternative approach
Looking at Figure 7, one might wonder whether adequate account has been taken of the difference in curvature between ρ(s) and ρ(s + ds) in computing the arc length as ds = ρ(s)dθ. Reassurance on this point can be found using a more formal approach outlined below. This approach also makes connection with the article on curvature.

To introduce the unit vectors of the local coordinate system, one approach is to begin in Cartesian coordinates and describe the local coordinates in terms of these Cartesian coordinates. In terms of arc length s let the path be described as:
 * $$\mathbf{r}(s) = \left[ x(s),\ y(s) \right] \ . $$

Then an incremental displacement along the path ds is described by:
 * $$\mathrm{d}\mathbf{r}(s) = \left[ \mathrm{d}x(s),\ \mathrm{d}y(s) \right] = \left[ x'(s),\ y'(s) \right] \mathrm{d}s \, $$

where primes are introduced to denote derivatives with respect to s. The magnitude of this displacement is ds, showing that:
 * $$\left[ x'(s)^2 + y'(s)^2 \right] = 1 \ . $$&ensp;&ensp;&ensp;&ensp;(Eq. 1)

This displacement is necessarily tangent to the curve at s, showing that the unit vector tangent to the curve is:
 * $$\mathbf{u}_\mathrm{t}(s) = \left[ x'(s), \ y'(s) \right] \, $$

while the outward unit vector normal to the curve is
 * $$\mathbf{u}_\mathrm{n}(s) = \left[ y'(s),\ -x'(s) \right] \, $$

Orthogonality can be verified by showing that the vector dot product is zero. The unit magnitude of these vectors is a consequence of Eq. 1. Using the tangent vector, the angle θ of the tangent to the curve is given by:
 * $$\sin \theta = \frac{y'(s)}{\sqrt{x'(s)^2 + y'(s)^2}} = y'(s) \ ;$$ &ensp; and &ensp; $$\cos \theta = \frac{x'(s)}{\sqrt{x'(s)^2 + y'(s)^2}} = x'(s) \ .$$

The radius of curvature is introduced completely formally (without need for geometric interpretation) as:
 * $$\frac{1}{\rho} = \frac{\mathrm{d}\theta}{\mathrm{d}s}\ . $$

The derivative of θ can be found from that for sinθ:
 * $$\frac{\mathrm{d} \sin\theta}{\mathrm{d}s} = \cos \theta \frac {\mathrm{d}\theta}{\mathrm{d}s} = \frac{1}{\rho} \cos \theta \ = \frac{1}{\rho} x'(s)\ . $$

Now:
 * $$\frac{\mathrm{d} \sin \theta }{\mathrm{d}s} = \frac{\mathrm{d}}{\mathrm{d}s} \frac{y'(s)}{\sqrt{x'(s)^2 + y'(s)^2}}$$&ensp;&ensp;$$ = \frac{y(s)x'(s)^2-y'(s)x'(s)x(s)} {\left(x'(s)^2 + y'(s)^2\right)^{3/2}}\, $$

in which the denominator is unity. With this formula for the derivative of the sine, the radius of curvature becomes:
 * $$\frac {\mathrm{d}\theta}{\mathrm{d}s} = \frac{1}{\rho} = y(s)x'(s) - y'(s)x(s)\ $$&ensp;$$ = \frac{y(s)}{x'(s)} = -\frac{x(s)}{y'(s)} \ ,$$

where the equivalence of the forms stems from differentiation of Eq. 1:
 * $$x'(s)x(s) + y'(s)y(s) = 0 \ . $$

With these results, the acceleration can be found:
 * $$\mathbf{a}(s) = \frac{\mathrm{d}}{\mathrm{d}t}\mathbf{v}(s) $$&ensp;&ensp;$$ = \frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{\mathrm{d}s}{\mathrm{d}t} \left( x'(s), \ y'(s) \right) \right]\ $$
 * $$ = \left(\frac{\mathrm{d}^2s}{\mathrm{d}t^2}\right)\mathbf{u}_\mathrm{t}(s) + \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right) ^2 \left(x(s),\ y(s) \right) $$
 * $$ = \left(\frac{\mathrm{d}^2s}{\mathrm{d}t^2}\right)\mathbf{u}_\mathrm{t}(s) - \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right) ^2 \frac{1}{\rho} \mathbf{u}_\mathrm{n}(s) \, $$

as can be verified by taking the dot product with the unit vectors ut(s) and un(s). This result for acceleration is the same as that for circular motion based on the radius ρ. Using this coordinate system in the inertial frame, it is easy to identify the force normal to the trajectory as the centripetal force and that parallel to the trajectory as the tangential force. From a qualitative standpoint, the path can be approximated by an arc of a circle for a limited time, and for the limited time a particular radius of curvature applies, the centrifugal and Euler forces can be analyzed on the basis of circular motion with that radius.

This result for acceleration agrees with that found earlier. However, in this approach the question of the change in radius of curvature with s is handled completely formally, consistent with a geometric interpretation, but not relying upon it, thereby avoiding any questions Figure 7 might suggest about neglecting the variation in ρ.

Example: circular motion
To illustrate the above formulas, let x, y be given as:
 * $$x = \alpha \cos \frac{s}{\alpha} \ ; \ y = \alpha \sin\frac{s}{\alpha} \ .$$

Then:
 * $$x^2 + y^2 = \alpha^2 \, $$

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:


 * $$y^{\prime}(s) = \cos \frac{s}{\alpha} \ ; \ x^{\prime}(s) = -\sin \frac{s}{\alpha} \, $$
 * $$y^{\prime\prime}(s) = -\frac{1}{\alpha}\sin\frac{s}{\alpha} \ ; \ x^{\prime\prime}(s) = -\frac{1}{\alpha}\cos \frac{s}{\alpha} \ . $$

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

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

which serve to show that s = 0 is located at position [ρ, 0] and s = ρπ/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:
 * $$\frac{\mathrm{d}}{\mathrm{d}s}\mathbf{u}_\mathrm{t}(s) = -\frac{1}{\alpha} \left[\cos\frac{s}{\alpha} \, \ \sin\frac{s}{\alpha} \right] = -\frac{1}{\alpha}\mathbf{u}_\mathrm{n}(s) \ ; $$
 * $$ \ \frac{\mathrm{d}}{\mathrm{d}s}\mathbf{u}_\mathrm{n}(s) = \frac{1}{\alpha} \left[-\sin\frac{s}{\alpha} \, \ \cos\frac{s}{\alpha} \right] = \frac{1}{\alpha}\mathbf{u}_\mathrm{t}(s) \ . $$

To obtain velocity and acceleration, a time-dependence for s is necessary. For counterclockwise motion at variable speed v(t):
 * $$s(t) = \int_0^t \ dt^{\prime} \ v(t^{\prime}) \, $$

where v(t) is the speed and t is time, and s(t = 0) = 0. Then:
 * $$\mathbf{v} = v(t)\mathbf{u}_\mathrm{t}(s) \ ,$$
 * $$\mathbf{a} = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s) + v\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s) = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s)-v\frac{1}{\alpha}\mathbf{u}_\mathrm{n}(s)\frac{\mathrm{d}s}{\mathrm{d}t} $$
 * $$\mathbf{a} = \frac{\mathrm{d}v}{\mathrm{d}t}\mathbf{u}_\mathrm{t}(s)-\frac{v^2}{\alpha}\mathbf{u}_\mathrm{n}(s) \, $$

where it already is established that α = ρ. This acceleration is the standard result for non-uniform circular motion.