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Centripetal force

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From Wikipedia, the free encyclopedia

Force directed to the center of rotation

Not to be confused withCentrifugal force. For other meanings of "centripetal", seeCentripetal (disambiguation).

A particle is disturbed from its uniform linear motion by a series of short kicks (1, 2, ...), giving its trajectory a nearly circular shape. The force is referred to as a centripetal force in the limit of a continuously acting force directed towards the center of curvature of the path.

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Centripetal force (fromLatincentrum, "center" andpetere, "to seek"^[1]) is theforce that makes a body follow a curvedpath. The direction of the centripetal force is alwaysorthogonal to the motion of the body and towards the fixed point of the instantaneouscenter of curvature of the path.Isaac Newton coined the term,^[2] describing it as "a force by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre".^[3] InNewtonian mechanics, gravity provides the centripetal force causing astronomicalorbits.

One common example involving centripetal force is the case in which a body moves with uniform speed along a circular path. The centripetal force is directed at right angles to the motion and also along the radius towards the centre of the circular path.^[4]^[5] The mathematical description was derived in 1659 by the Dutch physicistChristiaan Huygens.^[6]^[7]

Formula

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From thekinematics of curved motion it is known that an object moving attangential speedv along a path withradius of curvaturer accelerates toward the center of curvature at a rate ${\textbf {a}}_{c}=\lim _{\Delta t\to 0}{\frac {\Delta {\textbf {v}}}{\Delta t}},\quad a_{c}={\frac {v^{2}}{r}}$ Here, $a_{c}$ is thecentripetal acceleration and $\Delta {\textbf {v}}$ is thedifference between the velocity vectors at $t+\Delta {t}$ and $t {\displaystyle t}$ .

ByNewton's second law, the cause of acceleration is a net force acting on the object, which is proportional to its massm and its acceleration. The force, usually referred to as acentripetal force, has a magnitude^[8] $F_{c}=ma_{c}=m{\frac {v^{2}}{r}}$ and is, like centripetal acceleration, directed toward the center of curvature of the object's trajectory.

Derivation

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The centripetal acceleration can be inferred from the diagram of the velocity vectors at two instances. In the case of uniform circular motion the velocities have constant magnitude. Because each one is perpendicular to its respective position vector, simple vector subtraction implies two similar isosceles triangles with congruent angles – one comprising abase of $\Delta {\textbf {v}}$ and aleg length of $v {\displaystyle v}$ , and the other abase of $\Delta {\textbf {r}}$ (position vectordifference) and aleg length of $r {\displaystyle r}$ :^[9] ${\frac {|\Delta {\textbf {v}}|}{v}}={\frac {|\Delta {\textbf {r}}|}{r}}$ $|\Delta {\textbf {v}}|={\frac {v}{r}}|\Delta {\textbf {r}}|$ Therefore, $|\Delta {\textbf {v}}|$ can be substituted with ${\frac {v}{r}}|\Delta {\textbf {r}}|$ :^[9] $a_{c}=\lim _{\Delta t\to 0}{\frac {|\Delta {\textbf {v}}|}{\Delta t}}={\frac {v}{r}}\lim _{\Delta t\to 0}{\frac {|\Delta {\textbf {r}}|}{\Delta t}}={\frac {v^{2}}{r}}$ The direction of the force is toward the center of the circle in which the object is moving, or theosculating circle (the circle that best fits the local path of the object, if the path is not circular).^[10]The speed in the formula is squared, so twice the speed needs four times the force, at a given radius.

This force is also sometimes written in terms of theangular velocityω of the object about the center of the circle, related to the tangential velocity by the formula $v=\omega r$ so that $F_{c}=mr\omega ^{2}\,.$

Expressed using theorbital periodT for one revolution of the circle, $\omega ={\frac {2\pi }{T}}$ the equation becomes^[11] $F_{c}=mr\left({\frac {2\pi }{T}}\right)^{2}.$

In particle accelerators, velocity can be very high (close to thespeed of light in vacuum) so the same rest mass now exerts greater inertia (relativistic mass) thereby requiring greater force for the same centripetal acceleration, so the equation becomes:^[12] $F_{c}={\frac {\gamma mv^{2}}{r}}$ where $\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$ is theLorentz factor.

Thus the centripetal force is given by: $F_{c}=\gamma mv\omega$ which is the rate of change ofrelativistic momentum $\gamma mv$ .

Sources

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A body experiencinguniform circular motion requires a centripetal force, towards the axis as shown, to maintain its circular path.

In the case of an object that is swinging around on the end of a rope in a horizontal plane, the centripetal force on the object is supplied by the tension of the rope. The rope example is an example involving a 'pull' force. The centripetal force can also be supplied as a 'push' force, such as in the case where the normal reaction of a wall supplies the centripetal force for awall of death or aRotor rider.

Newton's idea of a centripetal force corresponds to what is nowadays referred to as acentral force. When asatellite is inorbit around aplanet, gravity is considered to be a centripetal force even though in the case of eccentric orbits, the gravitational force is directed towards the focus, and not towards the instantaneous center of curvature.^[13]

Another example of centripetal force arises in the helix that is traced out when a charged particle moves in a uniformmagnetic field in the absence of other external forces. In this case, the magnetic force is the centripetal force that acts towards the helix axis.

Analysis of several cases

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

Uniform circular motion

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Calculus derivation

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In two dimensions, the position vector ${\textbf {r}}$ , which has magnitude (length) $r {\displaystyle r}$ and directed at an angle $\theta$ above the x-axis, can be expressed inCartesian coordinates using theunit vectors ${\hat {\mathbf {x} }}$ and ${\hat {\mathbf {y} }}$ :^[14] ${\textbf {r}}=r\cos(\theta ){\hat {\mathbf {x} }}+r\sin(\theta ){\hat {\mathbf {y} }}.$

The assumption ofuniform circular motion requires three things:

The object moves only on a circle.
The radius of the circle $r {\displaystyle r}$ does not change in time.
The object moves with constantangular velocity $\omega$ around the circle. Therefore, $\theta =\omega t$ where $t {\displaystyle t}$ is time.

Thevelocity ${\textbf {v}}$ andacceleration ${\textbf {a}}$ of the motion are the first and second derivatives of position with respect to time:

${\textbf {r}}=r\cos(\omega t){\hat {\mathbf {x} }}+r\sin(\omega t){\hat {\mathbf {y} }},$ ${\textbf {v}}={\dot {\textbf {r}}}=-r\omega \sin(\omega t){\hat {\mathbf {x} }}+r\omega \cos(\omega t){\hat {\mathbf {y} }},$ ${\textbf {a}}={\ddot {\textbf {r}}}=-\omega ^{2}(r\cos(\omega t){\hat {\mathbf {x} }}+r\sin(\omega t){\hat {\mathbf {y} }}).$

The term in parentheses is the original expression of ${\textbf {r}}$ inCartesian coordinates. Consequently, ${\textbf {a}}=-\omega ^{2}{\textbf {r}}.$ The negative sign shows that the acceleration is pointed towards the center of the circle (opposite the radius), hence it is called "centripetal" (i.e. "center-seeking"). While objects naturally follow a straight path (due toinertia), this centripetal acceleration describes the circular motion path caused by a centripetal force.

Derivation using vectors

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Vector relationships for uniform circular motion; vectorΩ representing the rotation is normal to the plane of the orbit with polarity determined by theright-hand rule and magnitudedθ /dt.

The image at right shows the vector relationships for uniform circular motion. The rotation itself is represented by the angular velocity vectorΩ, which is normal to the plane of the orbit (using theright-hand rule) and has magnitude given by:

|\mathbf {\Omega } |={\frac {\mathrm {d} \theta }{\mathrm {d} t}}=\omega \ ,

withθ the angular position at timet. In this subsection, dθ/dt is assumed constant, independent of time. The distance traveleddℓ of the particle in time dt along the circular path is

\mathrm {d} {\boldsymbol {\ell }}=\mathbf {\Omega } \times \mathbf {r} (t)\mathrm {d} t\ ,

which, by properties of thevector cross product, has magnituderdθ and is in the direction tangent to the circular path.

Consequently,

{\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}=\lim _{{\Delta }t\to 0}{\frac {\mathbf {r} (t+{\Delta }t)-\mathbf {r} (t)}{{\Delta }t}}={\frac {\mathrm {d} {\boldsymbol {\ell }}}{\mathrm {d} t}}\ .

In other words,

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

Differentiating with respect to time, $\mathbf {a} \ {\stackrel {\mathrm {def} }{=}}\ {\frac {\mathrm {d} \mathbf {v} }{d\mathrm {t} }}=\mathbf {\Omega } \times {\frac {\mathrm {d} \mathbf {r} (t)}{\mathrm {d} t}}=\mathbf {\Omega } \times \left[\mathbf {\Omega } \times \mathbf {r} (t)\right]\ .$

Lagrange's formula states: $\mathbf {a} \times \left(\mathbf {b} \times \mathbf {c} \right)=\mathbf {b} \left(\mathbf {a} \cdot \mathbf {c} \right)-\mathbf {c} \left(\mathbf {a} \cdot \mathbf {b} \right)\ .$

Applying Lagrange's formula with the observation thatΩ • r(t) = 0 at all times, $\mathbf {a} =-{|\mathbf {\Omega |} }^{2}\mathbf {r} (t)\ .$

In words, the acceleration is pointing directly opposite to the radial displacementr at all times, and has a magnitude: $|\mathbf {a} |=|\mathbf {r} (t)|\left({\frac {\mathrm {d} \theta }{\mathrm {d} t}}\right)^{2}=r{\omega }^{2}$ where vertical bars |...| denote the vector magnitude, which in the case ofr(t) is simply the radiusr of the path. This result agrees with the previous section, though the notation is slightly different.

When the rate of rotation is made constant in the analysis ofnonuniform circular motion, that analysis agrees with this one.

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

Example: The banked turn

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Main article:Banked turn

Nonuniform circular motion

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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 the image at right. This case is used to demonstrate a derivation strategy based on apolar coordinate system.

Letr(t) be a vector that describes the position of apoint mass as a function of time. Since we are assumingcircular motion, letr(t) =R·u_r, whereR is a constant (the radius of the circle) andu_r is theunit vector pointing from the origin to the point mass. The direction ofu_r is described byθ, the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates,u_θ is perpendicular tou_r and points in the direction of increasingθ. These polar unit vectors can be expressed in terms ofCartesian unit vectors in thex andy directions, denoted ${\hat {\mathbf {i} }}$ and ${\hat {\mathbf {j} }}$ respectively:^[19] $\mathbf {u} _{r}=\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}$ and $\mathbf {u} _{\theta }=-\sin \theta \ {\hat {\mathbf {i} }}+\cos \theta \ {\hat {\mathbf {j} }}.$

One can differentiate to find velocity: ${\begin{aligned}\mathbf {v} &=r{\frac {d\mathbf {u} _{r}}{dt}}\\&=r{\frac {d}{dt}}\left(\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}{\frac {d}{d\theta }}\left(\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}\left(-\sin \theta \ {\hat {\mathbf {i} }}+\cos \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}\mathbf {u} _{\theta }\\&=\omega r\mathbf {u} _{\theta }\end{aligned}}$ whereω is the angular velocitydθ/dt.

This result for the velocity matches expectations that the velocity should be directed tangentially to the circle, and that the magnitude of the velocity should berω. Differentiating again, and noting that ${\frac {d\mathbf {u} _{\theta }}{dt}}=-{\frac {d\theta }{dt}}\mathbf {u} _{r}=-\omega \mathbf {u} _{r}\ ,$ we find that the acceleration,a is: $\mathbf {a} =r\left({\frac {d\omega }{dt}}\mathbf {u} _{\theta }-\omega ^{2}\mathbf {u} _{r}\right)\ .$

Thus, the radial and tangential components of the acceleration are: $\mathbf {a} _{r}=-\omega ^{2}r\ \mathbf {u} _{r}=-{\frac {|\mathbf {v} |^{2}}{r}}\ \mathbf {u} _{r}$ and $\mathbf {a} _{\theta }=r\ {\frac {d\omega }{dt}}\ \mathbf {u} _{\theta }={\frac {d|\mathbf {v} |}{dt}}\ \mathbf {u} _{\theta }\ ,$ 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 aperpendicular component that changes the direction of motion (the centripetal acceleration), and a parallel, ortangential component, that changes the speed.

General planar motion

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Position vectorr, always points radially from the origin.

Velocity vectorv, always tangent to the path of motion.

Acceleration vectora, not parallel to the radial motion but offset by the angular and Coriolis accelerations, nor tangent to the path but offset by the centripetal and radial accelerations.

Kinematic vectors in plane polar coordinates. Notice the setup is not restricted to 2d space, but a plane in any higher dimension.

Polar unit vectors at two timest andt +dt for a particle with trajectoryr (t ); on the left the unit vectorsu_ρ andu_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 timedt isdθ, just the same angle as the rotation of the trajectoryr (t ).

Polar coordinates

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The above results can be derived perhaps more simply inpolar 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 vectoru_ρ and an angular unit vectoru_θ, as shown above.^[20] A particle at positionr is described by:

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

where the notationρ is used to describe the distance of the path from the origin instead ofR to emphasize that this distance is not fixed, but varies with time. The unit vectoru_ρ travels with the particle and always points in the same direction asr(t). Unit vectoru_θ also travels with the particle and stays orthogonal tou_ρ. Thus,u_ρ andu_θ form a local Cartesian coordinate system attached to the particle, and tied to the path travelled by the particle.^[21] By moving the unit vectors so their tails coincide, as seen in the circle at the left of the image above, it is seen thatu_ρ andu_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 angleθ(t) asr(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 vectoru_ρ is needed. Becauseu_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 tou_ρ. When the trajectoryr(t) rotates an amount dθ,u_ρ, which points in the same direction asr(t), also rotates by dθ. See image above. Therefore, the change inu_ρ is

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

{\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 ofu_θ is found. As withu_ρ,u_θ is a unit vector and can only rotate without changing size. To remain orthogonal tou_ρ while the trajectoryr(t) rotates an amount dθ,u_θ, which is orthogonal tor(t), also rotates by dθ. See image above. Therefore, the change du_θ is orthogonal tou_θ and proportional to dθ (see image above):

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

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

Substituting the derivative ofu_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}}=v_{\rho }\mathbf {u} _{\rho }+v_{\theta }\mathbf {u} _{\theta }=\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 ofu_ρ andu_θ, the acceleration of the particle is:^[22]

{\begin{aligned}\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} v_{\rho }}{\mathrm {d} t}}-{\frac {v_{\theta }^{2}}{\rho }}\right]+\mathbf {u} _{\theta }\left[{\frac {2}{\rho }}v_{\rho }v_{\theta }+\rho {\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {v_{\theta }}{\rho }}\right]\,.\end{aligned}}

As a particular example, if the particle moves in a circle of constant radiusR, 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 {v^{2}}{r}}\right]+\mathbf {u} _{\theta }\left[{\frac {\mathrm {d} v}{\mathrm {d} t}}\right]\$

where $v=v_{\theta }.$

These results agree with those above fornonuniform circular motion. See also the article onnon-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 toangular acceleration is sometimes called theEuler force.^[23]

For trajectories other than circular motion, for example, the more general trajectory envisioned in the image above, the instantaneous center of rotation and radius of curvature of the trajectory are related only indirectly to the coordinate system defined byu_ρ andu_θ 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.^[24]^[25] To deal directly with this issue, local coordinates are preferable, as discussed next.

Local coordinates

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Local coordinates mean a set of coordinates that travel with the particle,^[26] and have orientation determined by the path of the particle.^[27] Unit vectors are formed as shown in the image at right, both tangential and normal to the path. This coordinate system sometimes is referred to asintrinsic orpath coordinates^[28]^[29] ornt-coordinates, fornormal-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.^[30]

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

s=s(t)\ .

A center of curvature is defined at each positions located a distanceρ (theradius of curvature) from the curve on a line along the normalu_n (s). The required distanceρ(s) at arc lengths 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 isθ(s), thenρ(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 thecurvatureκ is a signed quantity.

A geometric approach to finding the center of curvature and the radius of curvature uses a limiting process leading to theosculating circle.^[31]^[32] See image above.

Using these coordinates, the motion along the path is viewed as a succession of circular paths of ever-changing center, and at each positions constitutesnon-uniform circular motion at that position with radiusρ. 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 speedv 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 the image above):^[33]

{\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 and acceleration are:^[32]^[34]^[35]

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

and using thechain-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 fornonuniform circular motion with the local radiusρ(s), and the centripetal acceleration is identified as the second term.^[36]

Extending this approach to three dimensional space curves leads to theFrenet–Serret formulas.^[37]^[38]

Alternative approach

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Looking at the image above, one might wonder whether adequate account has been taken of the difference in curvature betweenρ(s) andρ(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 oncurvature.

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 lengths, let the path be described as:^[39] $\mathbf {r} (s)=\left[x(s),\ y(s)\right].$

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 tos. The magnitude of this displacement is ds, showing that:^[40]

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

(Eq. 1)

This displacement is necessarily a tangent to the curve ats, 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 vectordot product is zero. The unit magnitude of these vectors is a consequence ofEq. 1. Using the tangent vector, the angleθ of the tangent to the curve is given by: $\sin \theta ={\frac {y'(s)}{\sqrt {x'(s)^{2}+y'(s)^{2}}}}=y'(s)\ ;$ 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 ofθ 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 ofEq. 1: $x'(s)x''(s)+y'(s)y''(s)=0\ .$ With these results, the acceleration can be found: ${\begin{aligned}\mathbf {a} (s)&={\frac {\mathrm {d} }{\mathrm {d} t}}\mathbf {v} (s)={\frac {\mathrm {d} }{\mathrm {d} t}}\left[{\frac {\mathrm {d} s}{\mathrm {d} t}}\left(x'(s),\ y'(s)\right)\right]\\&=\left({\frac {\mathrm {d} ^{2}s}{\mathrm {d} t^{2}}}\right)\mathbf {u} _{\mathrm {t} }(s)+\left({\frac {\mathrm {d} s}{\mathrm {d} t}}\right)^{2}\left(x''(s),\ y''(s)\right)\\&=\left({\frac {\mathrm {d} ^{2}s}{\mathrm {d} t^{2}}}\right)\mathbf {u} _{\mathrm {t} }(s)-\left({\frac {\mathrm {d} s}{\mathrm {d} t}}\right)^{2}{\frac {1}{\rho }}\mathbf {u} _{\mathrm {n} }(s)\end{aligned}}$ as can be verified by taking the dot product with the unit vectorsu_t(s) andu_n(s). This result for acceleration is the same as that for circular motion based on the radiusρ. 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 withs is handled completely formally, consistent with a geometric interpretation, but not relying upon it, thereby avoiding any questions the image above might suggest about neglecting the variation inρ.

Example: circular motion

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To illustrate the above formulas, letx,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 radiusα. The positions = 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 can also 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 thats = 0 is located at position [ρ, 0] ands =ρπ/2 at [0,ρ], which agrees with the original expressions forx andy. 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 fors is necessary. For counterclockwise motion at variable speedv(t):

s(t)=\int _{0}^{t}\ dt^{\prime }\ v(t^{\prime })\ ,

wherev(t) is the speed andt is time, ands(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 fornon-uniform circular motion.

Notes and references

[edit]

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