Definitions

# Centripetal force

The centripetal force is the external force required to make a body follow a curved path. Hence centripetal force is a kinematic force requirement, not a particular kind of force, like gravity or electromagnetic force. Isaac Newton's description is found in the Principia. Many different examples are examined, one being the following: Any force (gravitational, electromagnetic, etc.) (or combination of forces) can act to provide a centripetal force. An example for the case of uniform circular motion is shown in Figure 1.

Centripetal force is directed inward, toward the center of curvature of the path. The term centripetal force comes from the Latin words centrum ("center") and petere ("tend towards", "aim at."), and can also be derived from Newton's original definitions described in the Principia

Centripetal force should not be confused with centrifugal force (see the Common misunderstandings section below).

## Simple example: uniform circular motion

The velocity vector is defined by the speed and also by the direction of motion. Objects experiencing no net force do not accelerate and, hence, move in a straight line with constant speed: they have a constant velocity. However, even an object moving in a circle at constant speed has a changing direction of motion. The rate of change of the object's velocity vector in this case is the centripetal acceleration. See Figure 1.

The centripetal acceleration varies with the radius $R$ of the path and speed $v$ of the object, becoming larger for greater speed (at constant radius) and smaller radius (at constant speed). If an object is traveling in a circle with a varying speed, its acceleration can be divided into two components, a radial acceleration (the centripetal acceleration that changes the direction of the velocity) and a tangential acceleration that changes the magnitude of the velocity.

Below a number of examples of increasing complexity are discussed, and formulas for the motion are derived.

## How is centripetal force provided?

Centripetal force is inferred from the trajectory of the object, without regard for how the path was arrived at (regardless of the origin of the forces involved). The studies of trajectories and the forces they imply is kinematics, while the study of which motions result from given physical forces is kinetics, the other branch of dynamics.

Supposing the analysis of a trajectory (kinematics) has concluded that for an object to follow the observed path a centripetal force is required, one might reasonably ask the (kinetic) question: "Where is the centripetal force coming from?"

For a satellite in orbit around a planet, the centripetal force is supplied by the gravitational attraction between the satellite and the planet, and acts toward the center of mass of the two objects. For an object at the end of a rope rotating about a vertical axis, the centripetal force is the horizontal component of the tension of the rope, which acts towards the center of mass between the axis of rotation and the rotating object. For a spinning object, internal tensile stress is the centripetal force that holds the object together in one piece.

## Common misunderstandings

Centripetal force should not be confused with centrifugal force. Centripetal force is a kinematic force requirement deduced from an observed trajectory, not a kinetic force like gravity or electrical forces. Centripetal force requirements may be deduced from a trajectory in any frame of reference (although the trajectory of an object and the deduced centripetal force will vary from one frame to another). Because centripetal force is a kinematic force requirement inferred from an established trajectory, it is not used to deduce a trajectory from a physical situation, and centripetal force is not included in the inventory of forces that are used in applying Newton's laws F = m a to calculate a trajectory.

Centrifugal force, on the other hand, is treated in a rotating frame as a kinetic force, that is, as part of the inventory of forces used in Newton's laws to predict motion. Centrifugal force is a fictitious force, however, that arises only when motion is described or experienced in a rotating reference frame, and it does not exist in an inertial frame of reference.

Centripetal force should not be confused with central force, either. To reiterate, centripetal force is a force requirement necessary for a curved trajectory to be possible: it is not a type of force, such as a nuclear or gravitational force. In contrast, central forces does refer to a type of force, more exactly to a class of physical forces between two objects that meet two conditions: (1) their magnitude depends only on the distance between the two objects and (2) their direction points along the line connecting the centers of these two objects. Examples of central forces include the gravitational force between two masses and the electrostatic force between two charges.

As an example relating these terms, the centripetal force implied by the circular motion of an object often is provided by a central force.

## Analysis of several cases

Below are three examples of increasing complexity, with derivations of the formulas governing velocity and acceleration.

### Uniform circular motion

Uniform circular motion refers to the case of constant rate of rotation. Here are two approaches to describing this case.

#### Geometric derivation

The circle on the left in Figure 2 shows an object moving on a circle at constant speed at two different times in its orbit. Its position is given by R and its velocity is v.

The velocity vector v is always perpendicular to the position vector (since the velocity vector is always tangent to the R circle); thus, since R moves in a circle, so does v. The circular motion of the velocity is shown in the circle on the right of Figure 2, along with its acceleration a. Just as velocity is the rate of change of position, acceleration is the rate of change of velocity.

Since the position and velocity vectors move in tandem, they go around their circles in the same time T. That time equals the distance traveled divided by the velocity


T = frac{2pi R}{v} and, by analogy,


T = frac{2pi v}{a}

Setting these two equations equal and solving for $a$, we get


a = frac{v^{2}}{R}

The angular rate of rotation in radians / s is:

$omega = frac \left\{2 pi\right\} \left\{T\right\} .$

Comparing the two circles in Figure 2 also shows that the acceleration points toward the center of the R circle. For example, in the left circle in Figure 2, the position vector R pointing at 12 o'clock has a velocity vector v pointing at 9 o'clock, which (switching to the circle on the right) has an acceleration vector a pointing at 6 o'clock. So the acceleration vector is opposite to R and toward the center of the R circle.

#### Derivation using vectors

Figure 3 shows the vector relationships for uniform circular motion. The rotation itself is represented by the vector Ω, which is normal to the plane of the orbit (using the right-hand rule) and has magnitude given by:

$left| boldsymbol \left\{Omega\right\}right| = frac \left\{d theta \right\} \left\{dt\right\} = omega ,$

with θ the angular position at time t. In this subsection, dθ / dt is assumed constant, independent of time t. The displacement of the particle in time dt along the circular path is

$dboldsymbol\left\{ell\right\} = boldsymbol \left\{Omega\right\} times mathbf\left\{r\right\} \left(t \right) dt ,$

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

Consequently,

$frac \left\{d mathbf\left\{r\right\}\right\}\left\{dt\right\} = frac \left\{mathbf\left\{r\right\}\left(t + dt\right)-mathbf\left\{r\right\}\left(t \right)\right\}\left\{dt\right\} = frac\left\{d boldsymbol\left\{ell\right\}\right\}\left\{dt\right\} .$

In other words,

$mathbf\left\{v\right\} stackrel\left\{mathrm\left\{def\right\}\right\}\left\{=\right\} frac \left\{d mathbf\left\{r\right\}\right\}\left\{dt\right\} = frac \left\{dboldsymbol\left\{ell\right\}\right\}\left\{dt\right\} = boldsymbol \left\{Omega\right\} times mathbf\left\{r\right\} \left(t \right) .$

Differentiating with respect to time,

$mathbf\left\{a\right\} stackrel\left\{mathrm\left\{def\right\}\right\}\left\{=\right\} frac \left\{d mathbf\left\{v\right\}\right\} \left\{dt\right\} = boldsymbol \left\{Omega\right\} times frac\left\{d mathbf\left\{r\right\} \left(t \right)\right\}\left\{dt\right\} = boldsymbol \left\{Omega\right\} times left\left[boldsymbol \left\{Omega\right\} times mathbf\left\{r\right\} \left(t\right)right\right] .$

Lagrange's formula states:

a × (b × c) = b(a · c) − c(a · b).

Applying Lagrange's formula with the observation that Ω • r (t) = 0 at all times,

$mathbf\left\{a\right\} = - \left\{Omega\right\}^2 mathbf\left\{r\right\} \left(t\right) .$

In words, the acceleration is pointing directly opposite to the radial displacement r at all times, and has a magnitude:

$|mathbf\left\{a\right\}| = |mathbf\left\{r\right\}| left \left(frac \left\{d theta\right\}\left\{dt\right\} right\right) ^2 = R \left\{omega\right\}^2$

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 if the substitution is made for rate of rotation in terms of the period of rotation T:

$frac \left\{d theta \right\}\left\{dt\right\} = omega = frac \left\{2 pi \right\} \left\{T\right\} = frac \left\{v\right\}\left\{R\right\} .$

Not surprisingly, this result also agrees with that of the nonuniform circular motion when the rate of rotation is made constant.

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

#### Example: The banked turn

Figure 4 shows a ball in circular motion on a banked curve. The curve is banked at an angle θ from the horizontal, and the surface of the road is considered to be slippery. The object is to find what angle θ the bank must have so the ball does not slide off the road (the so-called "angle of bank"). Intuition tells us that on a flat curve with no banking at all, the ball will simply slide off the road; while with a very steep banking, the ball will slide to the center unless it travels the curve rapidly.

The right side of Figure 4 indicates the forces on the ball. There are two forces; one is the force of gravity vertically downward through the center of mass of the ball m g where m is the mass of the ball and g is the gravitational acceleration; the second is the upward normal force exerted by the road perpendicular to the road surface m an. The horizontal net force on the ball is the horizontal component of the force from the road, which has magnitude Fh = m an sin θ. The vertical component of the force from the road must counteract the gravitational force, that is Fv = m an cos θ = m g, so m an = m g / cos θ. Accordingly one finds the net horizontal force to be:

$F_h = frac \left\{m g \right\} \left\{ mathrm\left\{cos\right\} theta\right\} mathrm \left\{sin\right\} theta = mg mathrm\left\{tan\right\} theta .$

On the other hand, at velocity v on a circular path of radius R, kinematics says that the force needed to turn the ball continuously into the turn is the radially inward centripetal force Fc of magnitude:

$F_c = m a _c = mfrac\left\{v^2\right\}\left\{R\right\} .$

Consequently the ball is in a stable path when the angle of the road is set to satisfy the condition:

$mg mathrm\left\{tan\right\} theta = mfrac\left\{v^2\right\}\left\{R\right\} ,$

or,

$mathrm\left\{tan\right\} theta = frac \left\{v^2\right\} \left\{gR\right\} .$

As the angle of bank θ approaches 90°, the tangent function approaches infinity, allowing larger values for v2 / R. In words, this equation states that for faster speeds (bigger v) the road must be banked more steeply (a larger value for θ), and for sharper turns (smaller R) the road also must be banked more steeply, which accords with intuition. When the angle θ does not satisfy the above condition, the horizontal component of force exerted by the road does not provide the correct centripetal force, and an additional frictional force tangential to the road surface is called upon to provide the difference. If friction cannot do this (that is, the coefficient of friction is exceeded), the ball slides to a different radius where the balance can be realized.

These ideas apply to air flight as well. See the FAA pilot's manual.

### Nonuniform circular motion

As a generalization of the uniform circular motion case, suppose the angular rate of rotation is not constant. The acceleration now has a tangential component, as shown in Figure 5. This case is used to demonstrate a derivation strategy based upon a polar coordinate system.

Let r(t) be a vector that describes the position of a point mass as a function of time. Since we are assuming circular motion, let r(t) = R·ur, where R is a constant (the radius of the circle) and ur is the unit vector pointing from the origin to the point mass. The direction of ur is described by θ, the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates, uθ is perpendicular to ur and points in the direction of increasing θ. These polar unit vectors can be expressed in terms of Cartesian unit vectors in the x and y directions, denoted i and j respectively:

ur = cos(θ) i + sin(θ) j

and

uθ = sin(θ) i + cos(θ) j.

Note: unlike the Cartesian unit vectors i and j, which are constant, in polar coordinates the direction of the unit vectors ur and uθ depend on θ, and so in general have non-zero time derivatives.

We differentiate to find velocity:

$mathbf\left\{v\right\} = R frac \left\{d mathbf\left\{u_r\right\}\right\}\left\{dt\right\} = R frac \left\{d\right\}\left\{dt\right\} left\left(mathrm\left\{cos\right\} theta mathbf\left\{i\right\} + mathrm\left\{sin\right\} theta mathbf\left\{j\right\}right\right)$
$= R frac \left\{d theta\right\} \left\{dt\right\} left\left(-mathrm\left\{sin\right\} theta mathbf\left\{i\right\} + mathrm\left\{cos\right\} theta mathbf\left\{j\right\}right\right),$

$= R frac\left\{dtheta\right\}\left\{dt\right\} mathbf\left\{u_theta\right\} ,$

$= R omega mathbf\left\{u_theta\right\} ,$

where ω is the angular velocity dθ/dt.

This result for the velocity matches expectations that the velocity should be directed tangential to the circle, and that the magnitude of the velocity should be ωR. Differentiating again, and noting that

$\left\{frac \left\{dmathbf\left\{u_theta\right\}\right\}\left\{dt\right\}\right\} = -frac\left\{dtheta\right\}\left\{dt\right\} mathbf\left\{u_r\right\} = - omega mathbf\left\{u_r\right\} ,$

we find that the acceleration, a is:

$mathbf\left\{a\right\} = R left\left(frac \left\{domega\right\}\left\{dt\right\} mathbf\left\{u_theta\right\} - omega^2 mathbf\left\{u_r\right\} right\right) .$

Thus, the radial and tangential components of the acceleration are:

$mathbf\left\{a\right\}_\left\{mathrm\left\{r\right\}\right\} = - omega^\left\{2\right\} R mathbf\left\{u_r\right\}=- frac$   and   $mathbf\left\{a\right\}_\left\{mathrm\left\{theta\right\}\right\}=R frac \left\{domega\right\}\left\{dt\right\} mathbf\left\{u_theta\right\} = frac \left\{d | mathbf\left\{v\right\} | \right\}\left\{dt\right\} mathbf\left\{u_theta\right\} ,$

where |v| = Rω is the magnitude of the velocity (the speed).

These equations express mathematically that, in the case of an object that moves along a circular path with a changing speed, the acceleration of the body may be decomposed into a perpendicular component that changes the direction of motion (the centripetal acceleration), and a parallel, or tangential component, that changes the speed.

### General planar motion

#### 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:

$mathbf\left\{r\right\} = rho mathbf\left\{u\right\}_\left\{rho\right\} ,$

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

$mathbf\left\{v\right\} = frac \left\{d rho \right\}\left\{dt\right\} mathbf\left\{u\right\}_\left\{rho\right\} + rho frac \left\{d mathbf\left\{u\right\}_\left\{rho\right\}\right\}\left\{dt\right\} .$

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

$d mathbf\left\{u\right\}_\left\{rho\right\} = mathbf\left\{u\right\}_\left\{theta\right\} dtheta ,$

or

$frac \left\{d mathbf\left\{u\right\}_\left\{rho\right\}\right\}\left\{dt\right\} = mathbf\left\{u\right\}_\left\{theta\right\} frac \left\{dtheta\right\}\left\{dt\right\} .$

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):

$frac\left\{d mathbf\left\{u\right\}_\left\{theta\right\}\right\}\left\{dt\right\} = -frac \left\{d theta\right\} \left\{dt\right\} mathbf\left\{u\right\}_\left\{rho\right\} .$

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

Substituting the derivative of uρ into the expression for velocity:

$mathbf\left\{v\right\} = frac \left\{d rho \right\}\left\{dt\right\} mathbf\left\{u\right\}_\left\{rho\right\} + rho mathbf\left\{u\right\}_\left\{theta\right\} frac \left\{d theta\right\} \left\{dt\right\} =mathbf\left\{v\right\}_\left\{rho\right\} + mathbf\left\{v\right\}_\left\{theta\right\} .$

To obtain the acceleration, another time differentiation is done:

$mathbf\left\{a\right\} = frac \left\{d^2 rho \right\}\left\{dt^2\right\} mathbf\left\{u\right\}_\left\{rho\right\}+ frac \left\{d rho \right\}\left\{dt\right\} frac\left\{d mathbf\left\{u\right\}_\left\{rho\right\}\right\}\left\{dt\right\} +frac \left\{d rho\right\}\left\{dt\right\} mathbf\left\{u\right\}_\left\{theta\right\} frac \left\{d theta\right\} \left\{dt\right\}+rho frac\left\{d mathbf\left\{u\right\}_\left\{theta\right\}\right\}\left\{dt\right\} frac \left\{d theta\right\} \left\{dt\right\}+rho mathbf\left\{u\right\}_\left\{theta\right\} frac \left\{d^2 theta\right\} \left\{dt^2\right\} .$

Substituting the derivatives of uρ and uθ, the acceleration of the particle is:

$mathbf\left\{a\right\} = frac \left\{d^2 rho \right\}\left\{dt^2\right\} mathbf\left\{u\right\}_\left\{rho\right\}+ 2frac \left\{d rho\right\}\left\{dt\right\} mathbf\left\{u\right\}_\left\{theta\right\} frac \left\{d theta\right\} \left\{dt\right\}-rho mathbf\left\{u\right\}_\left\{rho\right\}left\left(frac \left\{d theta\right\} \left\{dt\right\}right\right)^2+rho mathbf\left\{u\right\}_\left\{theta\right\} frac \left\{d^2 theta\right\} \left\{dt^2\right\} ,$
$=mathbf\left\{u\right\}_\left\{rho\right\} left\left[frac \left\{d^2 rho \right\}\left\{dt^2\right\}-rholeft\left(frac \left\{d theta\right\} \left\{dt\right\}right\right)^2 right\right] + mathbf\left\{u\right\}_\left\{theta\right\}left\left[2frac \left\{d rho\right\}\left\{dt\right\} frac \left\{d theta\right\} \left\{dt\right\}+rho frac \left\{d^2 theta\right\} \left\{dt^2\right\}right\right]$
$=mathbf\left\{u\right\}_\left\{rho\right\} left\left[frac \left\{d|mathbf\left\{v\right\}_\left\{rho\right\}$
{dt}-frac As a particular example, if the particle moves in a circle of constant radius R, then dρ / dt = 0, v = vθ, and:

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

$s = s\left(t\right) .$

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:

$frac\left\{1\right\} \left\{rho \left(s\right)\right\} =kappa \left(s\right) = frac \left\{dtheta\right\}\left\{ds\right\} .$

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:

$omega\left(s\right) = frac\left\{dtheta\right\}\left\{dt\right\} = frac\left\{dtheta\right\}\left\{ds\right\} frac \left\{ds\right\}\left\{dt\right\} =frac\left\{1\right\}\left\{rho \left(s\right)\right\} frac \left\{ds\right\}\left\{dt\right\} =frac\left\{v\left(s\right)\right\}\left\{rho \left(s\right)\right\} ,$
with the local speed v given by:

$v\left(s\right) = frac \left\{ds\right\}\left\{dt\right\} .$

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):

$frac\left\{dmathbf\left\{u_n\right\}\left(s\right)\right\}\left\{ds\right\} = mathbf\left\{u_t\right\}\left(s\right)frac\left\{dtheta\right\}\left\{ds\right\} = mathbf\left\{u_t\right\}\left(s\right)frac\left\{1\right\}\left\{rho\right\} ;$  $frac\left\{dmathbf\left\{u_t\right\}\left(s\right)\right\}\left\{ds\right\} = -mathbf\left\{u_n\right\}\left(s\right)frac\left\{dtheta\right\}\left\{ds\right\} =- mathbf\left\{u_n\right\}\left(s\right)frac\left\{1\right\}\left\{rho\right\} .$
Consequently, the velocity and acceleration are:
$mathbf\left\{v\right\}\left(t\right) = v mathbf\left\{u_t\right\}\left(s\right) ;$
and using the chain-rule of differentiation:

$mathbf\left\{a\right\}\left(t\right) = frac\left\{dv\right\}\left\{dt\right\} mathbf\left\{u_t\right\}\left(s\right) - frac\left\{v^2\right\}\left\{rho\right\}mathbf\left\{u_n\right\}\left(s\right) ;$   with the tangential acceleration $frac\left\{dv\right\}\left\{dt\right\} = frac\left\{dv\right\}\left\{ds\right\} frac\left\{ds\right\}\left\{dt\right\} =frac\left\{dv\right\}\left\{ds\right\} v .$

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). 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:

$mathbf\left\{r\right\}\left(s\right) = left\left[x\left(s\right), y\left(s\right) right\right] .$
Then an incremental displacement along the path ds is described by:
$dmathbf\left\{r\right\}\left(s\right) = left\left[dx\left(s\right), dy\left(s\right) right\right]=left\left[x\text{'}\left(s\right), y\text{'}\left(s\right) right\right] ds ,$
where primes are introduced to denote derivatives with respect to s. The magnitude of this displacement is ds, showing that:
$left\left[x\text{'}\left(s\right)^2 + y\text{'}\left(s\right)^2 right\right] = 1 .$    (Eq. 1)
This displacement is necessarily tangent to the curve at s, showing that the unit vector tangent to the curve is:
$mathbf\left\{u\right\}_t\left(s\right) = left\left[x\text{'}\left(s\right), y\text{'}\left(s\right) right\right] ,$
while the outward unit vector normal to the curve is
$mathbf\left\{u\right\}_n\left(s\right) = left\left[y\text{'}\left(s\right), -x\text{'}\left(s\right) right\right] ,$
Orthogonality can be verified by showing the vector dot product is zero. The unit magnitude of these vectors is a consequence of Eq. 1. Using the tangent vector, the angle of the tangent to the curve, say θ, is given by:
$sin theta =frac\left\{y\text{'}\left(s\right)\right\}\left\{sqrt\left\{x\text{'}\left(s\right)^2+y\text{'}\left(s\right)^2\right\}\right\} = y\text{'}\left(s\right) ;$   and   $cos theta =frac\left\{x\text{'}\left(s\right)\right\}\left\{sqrt\left\{x\text{'}\left(s\right)^2+y\text{'}\left(s\right)^2\right\}\right\} = x\text{'}\left(s\right) .$
The radius of curvature is introduced completely formally (without need for geometric interpretation) as:
$frac\left\{1\right\}\left\{rho\right\} = frac\left\{dtheta\right\}\left\{ds\right\} .$
The derivative of θ can be found from that for sin θ:
$frac\left\{d sintheta\right\}\left\{ds\right\} = cos theta frac \left\{dtheta\right\}\left\{ds\right\} = frac\left\{1\right\}\left\{rho\right\} cos theta$
$= frac\left\{1\right\}\left\{rho\right\} x\text{'}\left(s\right) .$
Now:
$frac\left\{d sin theta \right\}\left\{ds\right\} = frac\left\{d\right\}\left\{ds\right\} frac\left\{y\text{'}\left(s\right)\right\}\left\{sqrt\left\{x\text{'}\left(s\right)^2+y\text{'}\left(s\right)^2\right\}\right\}$  $= frac\left\{y$(s)x'(s)^2-y'(s)x'(s)x(s)} {left(x'(s)^2+y'(s)^2right)^{3/2}} ,
in which the denominator is unity. With this formula for the derivative of the sine, the radius of curvature becomes:
$frac \left\{dtheta\right\}\left\{ds\right\} = frac\left\{1\right\}\left\{rho\right\} = y$(s)x'(s) - y'(s)x(s)  $=frac\left\{y$(s)}{x'(s)} = -frac{x(s)}{y'(s)} ,
where the equivalence of the forms stems from differentiation of Eq. 1:
$x\text{'}\left(s\right)x$(s) + y'(s)y(s) = 0 .
With these results, the acceleration can be found:
$mathbf\left\{a\right\}\left(s\right) = frac\left\{d\right\}\left\{dt\right\}mathbf\left\{v\right\}\left(s\right)$  $= frac\left\{d\right\}\left\{dt\right\}left\left[frac\left\{ds\right\}\left\{dt\right\} left\left(x\text{'}\left(s\right), y\text{'}\left(s\right) right\right) right\right]$
$= left\left(frac\left\{d^2s\right\}\left\{dt^2\right\}right\right)mathbf\left\{u\right\}_t\left(s\right) +left\left(frac\left\{ds\right\}\left\{dt\right\}right\right) ^2 left\left(x$(s), y(s) right)
$= left\left(frac\left\{d^2s\right\}\left\{dt^2\right\}right\right)mathbf\left\{u\right\}_t\left(s\right) - left\left(frac\left\{ds\right\}\left\{dt\right\}right\right) ^2 frac\left\{1\right\}\left\{rho\right\} mathbf\left\{u\right\}_n\left(s\right) ,$
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\left\{s\right\}\left\{alpha\right\} ;$$y=alpha sinfrac\left\{s\right\}\left\{alpha\right\} .$
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:
$y\text{'}\left(s\right) = cos frac\left\{s\right\}\left\{alpha\right\} ; x\text{'}\left(s\right) = -sin frac\left\{s\right\}\left\{alpha\right\}$
$y$(s) = -frac{1}{alpha}sinfrac{s}{alpha} ; x(s) = -frac{1}{alpha}cos frac{s}{alpha} .
With these results one can verify that:
$x\text{'}\left(s\right)^2 + y\text{'}\left(s\right)^2 = 1$ ; $frac\left\{1\right\}\left\{rho\right\} = y$(s)x'(s)-y(s)x''(s) = frac{1}{alpha} .
The unit vectors also can be found:
$mathbf\left\{u\right\}_t\left(s\right) = left\left[-sinfrac\left\{s\right\}\left\{alpha\right\}, cosfrac\left\{s\right\}\left\{alpha\right\} right\right]$ ; $mathbf\left\{u\right\}_n\left(s\right) = left\left[cosfrac\left\{s\right\}\left\{alpha\right\}, sinfrac\left\{s\right\}\left\{alpha\right\} right\right] ,$
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:
$frac\left\{d\right\}\left\{ds\right\}mathbf\left\{u\right\}_t\left(s\right) = -frac\left\{1\right\}\left\{alpha\right\} left\left[cosfrac\left\{s\right\}\left\{alpha\right\}, sinfrac\left\{s\right\}\left\{alpha\right\} right\right]$  $= -frac\left\{1\right\}\left\{alpha\right\}mathbf\left\{u\right\}_n\left(s\right) ;$

$frac\left\{d\right\}\left\{ds\right\}mathbf\left\{u\right\}_n\left(s\right)$  $= frac\left\{1\right\}\left\{alpha\right\} left\left[-sinfrac\left\{s\right\}\left\{alpha\right\}, cosfrac\left\{s\right\}\left\{alpha\right\} right\right]$  $= frac\left\{1\right\}\left\{alpha\right\}mathbf\left\{u\right\}_t\left(s\right) .$

To obtain velocity and acceleration, a time-dependence for s is necessary. For counterclockwise motion at variable speed v(t):

$s\left(t\right) = int_0^t dt\text{'} v\left(t\text{'}\right) ,$
where v(t) is the speed and t is time, and s(t=0) = 0. Then:
$mathbf\left\{v\right\} = v\left(t\right)mathbf\left\{u\right\}_t\left(s\right) ,$

$mathbf\left\{a\right\} = frac\left\{dv\right\}\left\{dt\right\}mathbf\left\{u\right\}_t\left(s\right)+vfrac\left\{d\right\}\left\{dt\right\}mathbf\left\{u\right\}_t\left(s\right)$  $= frac\left\{dv\right\}\left\{dt\right\}mathbf\left\{u\right\}_t\left(s\right)-vfrac\left\{1\right\}\left\{alpha\right\}mathbf\left\{u\right\}_n\left(s\right)frac\left\{ds\right\}\left\{dt\right\}$

$=frac\left\{dv\right\}\left\{dt\right\}mathbf\left\{u\right\}_t\left(s\right)-frac\left\{v^2\right\}\left\{alpha\right\}mathbf\left\{u\right\}_n\left(s\right) ,$
where it already is established that α = ρ. This acceleration is the standard result for non-uniform circular motion.