Centripetal force

(PD) Image: John R. Brews
Centripetal force F_C upon an object held in circular motion by a string of length R. The string is under tension F_T, as shown separately to the left.

Centripetal force (from Latin centrum "center" and petere "to seek") is a force that makes a body follow a curved path. According to Newton's laws of motion, any body that is not in uniform motion in a straight line is subject to a force. Thus, a body following a curved path is subject to a force, and the component of that force directed perpendicular to the velocity of the body, toward the instantaneous center of curvature of the path is called centripetal force.

An important point is that it is deviation from a straight path by itself that requires centripetal force, and the body does not have to accelerate along the path. According to Newton's laws, velocity is a vector that points in the direction of motion, so a change of direction implies a change of velocity. Any change of velocity v, including just a change of direction, constitutes an acceleration a, and therefore requires a force according to the famous rule: F = ma.

The figure shows the radially inward centripetal force F_C provided by the string that is necessary to force the object to travel the circular path of radius R at a constant speed v. The body in turn provides an equal but oppositely directed tensile force F_T on the string as shown to the left (a consequence of Newton's law that to every force there is an equal and opposite reaction force), and the string also is subject to the same tensile force at its other end, in opposite direction, as provided by the fixed centerpost.

Formula

(PD) Image: John R. Brews
Local coordinate system for planar motion on a curve. Two different positions are shown for distances s and s + ds along the curve. At each position s, unit vector u_n points along the outward normal to the curve and unit vector u_t is tangential to the path. The radius of curvature of the path is ρ as found from the rate of rotation of the tangent to the curve with respect to arc length, and is the radius of the osculating circle at position s. The unit circle on the left shows the rotation of the unit vectors with s.

An analysis of centripetal force provides a formula relating the required force to the properties of the path, namely the path's curvature, and to the speed of travel and mass of an object moving along the path. The result determines the centripetal force as:

$\mathbf {F_{C}} =-m{\frac {v^{2}}{\rho }}\mathbf {u} _{\mathrm {n} }(s),$

where m is the mass of the moving body, v is the speed of travel along the path, u_n is a unit vector normal to the path and pointing radially outward from the center of curvature and ρ is the radius of curvature of the path.

Quantitative treatment

The analysis behind the above formula for F_C is described below. It is an example of kinematics, that is, the analysis of the forces required to maintain a certain path, as opposed to dynamics, the analysis of what path results from known forces. The analysis can be provided by introducing a local coordinate system.

By local coordinates is meant a set of coordinates that travel with the particle. Suppose distance along the path of the particle is the arc length s, considered to be a known function of time.

$s=s(t)\ .$

At a point corresponding to s, an infinitesimal increase in s by ds is tangent to the path, and provides the direction of a unit vector u_t tangent to the path. At the same point a vector u_n is defined perpendicular to u_t and pointing outward, away from the center of curvature. Thus, the local unit vectors have orientation determined by the path of the particle. The unit vectors are shown at two points in the figure, corresponding to positions s and s+ds.

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 u_n (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.

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 a 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:

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

Because unit vectors cannot change magnitude, their rate of change is always perpendicular to their direction (see the left-hand insert in the figure):

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

Consequently, the velocity is:

$\mathbf {v} (t)=v\mathbf {u} _{\mathrm {t} }(s)\ ;$

and using the chain-rule of differentiation, the acceleration is:

$\mathbf {a} (t)={\frac {\mathrm {d} v}{\mathrm {d} t}}\mathbf {u} _{\mathrm {t} }(s)-{\frac {v^{2}}{\rho }}\mathbf {u} _{\mathrm {n} }(s)\ ;$

with the tangential acceleration of magnitude:

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

The centripetal acceleration is the normal component:

$\mathbf {a_{C}} =-{\frac {v^{2}}{\rho }}\mathbf {u} _{\mathrm {n} }(s),$

and the centripetal force is this acceleration multiplied by the mass of the moving body. Thus, in this local coordinate system the acceleration resembles the expression for nonuniform circular motion with the local radius ρ(s).

Extension of this approach to three dimensional space curves (curves that cannot be contained in a single plane) leads to the Frenet-Serret formulas.

Alternative approach

Looking at the figure, 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.

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:

$\left[x'(s)^{2}+y'(s)^{2}\right]=1\ .\ .\ .\ .\ .\ \mathrm {(Eq.1)}$

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:

$\sin \theta ={\frac {y'(s)}{\sqrt {x'(s)^{2}+y'(s)^{2}}}}=y'(s)\ ;\ \ \mathrm {and} \ \ \cos \theta ={\frac {x'(s)}{\sqrt {x'(s)^{2}+y'(s)^{2}}}}=x'(s)\ .$

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:

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

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)\$ $={\frac {y''(s)}{x'(s)}}=-{\frac {x''(s)}{y'(s)}}\ ,$

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:

$\mathbf {a} (s)={\frac {\mathrm {d} }{\mathrm {d} t}}\mathbf {v} (s)$  $={\frac {\mathrm {d} }{\mathrm {d} t}}\left[{\frac {\mathrm {d} s}{\mathrm {d} t}}\left(x'(s),\ y'(s)\right)\right]\$

$=\left({\frac {\mathrm {d} ^{2}s}{\mathrm {d} t^{2}}}\right)\mathbf {u} _{\mathrm {t} }(s)+\left({\frac {\mathrm {d} s}{\mathrm {d} t}}\right)^{2}\left(x''(s),\ y''(s)\right)$

$=\left({\frac {\mathrm {d} ^{2}s}{\mathrm {d} t^{2}}}\right)\mathbf {u} _{\mathrm {t} }(s)-\left({\frac {\mathrm {d} s}{\mathrm {d} t}}\right)^{2}{\frac {1}{\rho }}\mathbf {u} _{\mathrm {n} }(s)\ ,$

as can be verified by taking the dot product with the unit vectors u_t(s) and u_n(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 the figure might suggest about neglecting the variation in ρ.

Example: nonuniform circular motion

To illustrate the above formulas, let x, y be given as:

$x=R\cos {\frac {s}{R}}\ ;\ y=R\sin {\frac {s}{R}}\ .$

Then:

$x^{2}+y^{2}=R^{2}\ ,$

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

$y^{\prime }(s)=\cos {\frac {s}{R}}\ ;\ x^{\prime }(s)=-\sin {\frac {s}{R}}\ ,$

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

With these results one can verify that the radius of curvature of the path ρ is the same as the radius of the circle R:

$x^{\prime }(s)^{2}+y^{\prime }(s)^{2}=1\ ;\ {\frac {1}{\rho }}=y^{\prime \prime }(s)x^{\prime }(s)-y^{\prime }(s)x^{\prime \prime }(s)={\frac {1}{R}}\ .$

The unit vectors also can be found:

$\mathbf {u} _{\mathrm {t} }(s)=\left[-\sin {\frac {s}{R}}\ ,\ \cos {\frac {s}{R}}\right]\ ;\ \mathbf {u} _{\mathrm {n} }(s)=\left[\cos {\frac {s}{R}}\ ,\ \sin {\frac {s}{R}}\right]\ ,$

which serve to show that s = 0 is located at position [R, 0] and s = πR/2 at [0, R], 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:

${\frac {\mathrm {d} }{\mathrm {d} s}}\mathbf {u} _{\mathrm {t} }(s)=-{\frac {1}{R}}\left[\cos {\frac {s}{R}}\ ,\ \sin {\frac {s}{R}}\right]=-{\frac {1}{R}}\mathbf {u} _{\mathrm {n} }(s)\ ;$

$\ {\frac {\mathrm {d} }{\mathrm {d} s}}\mathbf {u} _{\mathrm {n} }(s)={\frac {1}{R}}\left[-\sin {\frac {s}{R}}\ ,\ \cos {\frac {s}{R}}\right]={\frac {1}{R}}\mathbf {u} _{\mathrm {t} }(s)\ .$

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:

$\mathbf {v} =v(t)\mathbf {u} _{\mathrm {t} }(s)\ ,$

$\mathbf {a} ={\frac {\mathrm {d} v}{\mathrm {d} t}}\mathbf {u} _{\mathrm {t} }(s)+v{\frac {\mathrm {d} }{\mathrm {d} t}}\mathbf {u} _{\mathrm {t} }(s)={\frac {\mathrm {d} v}{\mathrm {d} t}}\mathbf {u} _{\mathrm {t} }(s)-v{\frac {1}{R}}\mathbf {u} _{\mathrm {n} }(s){\frac {\mathrm {d} s}{\mathrm {d} t}}$

$\mathbf {a} ={\frac {\mathrm {d} v}{\mathrm {d} t}}\mathbf {u} _{\mathrm {t} }(s)-{\frac {v^{2}}{R}}\mathbf {u} _{\mathrm {n} }(s)\ ,$

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

Polar coordinates

(PD) Image: John R. Brews
Polar unit vectors at two times t and t + dt for a particle with trajectory r ( t ); on the left the unit vectors u_ρ and u_θ at the two times are moved so their tails all meet, and are shown to trace an arc of a unit radius circle. Their rotation in time dt is dθ, just the same angle as the rotation of the trajectory r ( t ).

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 u_ρ and an angular unit vector u_θ, as shown in the figure. 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 u_ρ travels with the particle and always points in the same direction as r(t). Unit vector u_θ also travels with the particle and stays orthogonal to u_ρ. Thus, u_ρ and u_θ 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 the figure, it is seen that u_ρ and u_θ 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 u_ρ is needed. Because u_ρ is a unit vector, its magnitude is fixed, and it can change only in direction, that is, its change du_ρ has a component only perpendicular to u_ρ. When the trajectory r(t) rotates an amount dθ, u_ρ, which points in the same direction as r(t), also rotates by dθ. See the figure. Therefore the change in u_ρ 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 u_θ is found. As with u_ρ, u_θ is a unit vector and can only rotate without changing size. To remain orthogonal to u_ρ while the trajectory r(t) rotates an amount dθ, u_θ, which is orthogonal to r(t), also rotates by dθ. See the figure. Therefore, the change du_θ is orthogonal to u_θ and proportional to dθ (see the figure):

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

The figure shows the sign to be negative: to maintain orthogonality, if du_ρ is positive with dθ, then du_θ must decrease.

Substituting the derivative of u_ρ 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 u_ρ and u_θ, 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 = v_θ, 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. If this acceleration is multiplied by the particle mass, the leading term is the centripetal force.

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

Applications

The notion that curved paths require forces is non-intuitive in some cases, so some examples may assist the interpretation of the role of centripetal force. For example, one may consider the simple case of spinning a weight at the end of a string. In this case, one is quite aware that one must pull on the string to effect the curved motion, and as soon as the string is released the weight flies off, in what careful observation shows to be a straight line (apart from the effects of the Earth's gravity).

A less obvious case is spinning a bucket of water, which causes the spinning water to mount the sides of the bucket to form a parabolic surface. In this case, an element of water travels in a circle about the center of rotation of the bucket, and so each element of water requires a centripetal force to create the curved path. For the water next to the bucket, the walls of the bucket provide this inward force, and these peripheral elements of water transmit some of this force to the water interior to them that are further away from the walls of the bucket. For an interior element of the water, the inward force is transmitted by the water at a larger radius to the inner element, and it in turn transmits a force to the elements interior to itself. By forming a parabolic surface increasing in depth toward the walls of the bucket, the force exerted upon an element of water by its neighboring elements is tilted to provide a radial centripetal component just sufficient to keep that element in a circular path, while the vertical component of this force exactly counters the downward force of gravity, which itself has no radial component and so cannot contribute to the centripetal force.

Banked turn

(PD) Image: John R. Brews
Upper panel: Ball on a banked circular track moving with constant speed v; Lower panel: Forces on the ball.

As another example from common experience, consider the banked turn, as found on major highways. The upper panel in the figure 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 track. 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.

Assuming no acceleration in the direction of the path, so the ball maintains a constant speed v, the lower panel 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 mg 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 ma_n. The centripetal force demanded by the curved motion also is shown. This centripetal force is not a third force applied to the ball, but rather is the net force on the ball resulting from vector addition of the normal force and the force of gravity. The resultant or net force on the ball found by vector addition of the normal force exerted by the road and vertical force due to gravity must equal the centripetal force dictated by the need to travel a circular path. The curved motion is maintained so long as this net force provides the centripetal force requisite to the motion.

The horizontal net force on the ball is pointed toward the center of curvature of its path, and is the horizontal component of the normal force from the road, which has magnitude |F_h| = m|a_n|sinθ. The vertical component of the force from the road must counteract the gravitational force: |F_v| = m|a_n|cosθ = m|g|, which implies |a_n|=|g| / cosθ. Substituting into the above formula for |F_h| yields a horizontal force to be:

$|\mathbf {F} _{\mathrm {h} }|=m|\mathbf {g} |{\frac {\mathrm {sin} \ \theta }{\mathrm {cos} \ \theta }}=m|\mathbf {g} |\mathrm {tan} \ \theta \ .$

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 F_C of magnitude:

$|\mathbf {F} _{\mathrm {C} }|=m|\mathbf {a} _{\mathrm {C} }|={\frac {m|\mathbf {v} |^{2}}{R}}\ .$

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

$m|\mathbf {g} |\mathrm {tan} \ \theta ={\frac {m|\mathbf {v} |^{2}}{R}}\ ,$

or,

$\mathrm {tan} \ \theta ={\frac {|\mathbf {v} |^{2}}{|\mathbf {g} |R}}\ .$

As the angle of bank θ approaches 90°, the tangent function approaches infinity, allowing larger values for |v|²/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.

Historical remarks

Historically, the notion that a curved path requires a centripetal force evoked some controversy. Initially, philosophers such as Descartes and Bishop Berkeley took the very reasonable view that in a universe with only one body in it, it would be impossible to tell if the body were moving at all, never mind in a curved path. That suggested that it was the relative motions of bodies that mattered, as more than one body was needed just to provide a reference point. Unfortunately, it is not possible to experience a universe containing only one body, so this logically appealing viewpoint cannot be tested. Because the universe contains many bodies, a version of this proposal due to Ernst Mach was that curved motion was curved relative to the fixed stars. The "fixed stars" were a simplified reference to the universe as a whole. Today, the notion is that the curved path should be observed in an "inertial frame of reference", which is to say, in one of a privileged set of frames that move uniformly with respect to each other, and in which the laws of physics take on their "simplest" form. To within experimental errors of observation, a reference frame attached to the universe is one of these inertial frames.