Centripetal force: Difference between revisions

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'''Centripetal force''' (~~from [[Latin]] ''~~centrum~~'', "~~center~~" and ''~~petere~~'', "~~to seek"<ref>{{cite book |title=A new universal etymological, technological and pronouncing dictionary of the English language: embracing all terms used in art, science, and literature, Volume 1 |first1=John |last1=Craig |publisher=Harvard University |year=1849 |page=291 |url=https://books.google.com/books?id=0nxBAAAAYAAJ}} [https://books.google.com/books?id=0nxBAAAAYAAJ&pg=PA291 Extract of page 291]</ref>) is the [[force]] that makes a body follow a curved [[trajectory|path]]. The direction of the centripetal force is always [[orthogonality|orthogonal]] to the motion of the body and towards the fixed point of the instantaneous [[osculating circle|center of curvature]] of the path. [[Isaac Newton]] coined the term,<ref>{{Cite book |last=Brackenridge |first=John Bruce |url=https://books.google.com/books?id=ovOTK7X_mMkC&pg=PA74 |title=The Key to Newton's Dynamics: The Kepler Problem and the Principia |date=1996 |publisher=University of California Press |isbn=978-0-520-91685-2 |location= |pages=74}}</ref> describing it as "a force by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre".<ref>{{cite book |last=Newton |first=Isaac |title=The Principia: Mathematical Principles of Natural Philosophy |publisher=Snowball Pub. |year=2010 |isbn=978-1-60796-240-3 |location=[S.l.] |pages=10}}</ref> In [[Newtonian mechanics]], gravity provides the centripetal force causing astronomical [[orbit]]s.

'''Centripetal force''' ({{etymology|la|{{lang|la|centrum}}|center||{{lang|la|petere}}|to seek}}<ref>{{cite book |title=A new universal etymological, technological and pronouncing dictionary of the English language: embracing all terms used in art, science, and literature, Volume 1 |first1=John |last1=Craig |publisher=Harvard University |year=1849 |page=291 |url=https://books.google.com/books?id=0nxBAAAAYAAJ}} [https://books.google.com/books?id=0nxBAAAAYAAJ&pg=PA291 Extract of page 291]</ref>) is the [[force]] that makes a body follow a curved [[trajectory|path]]. The direction of the centripetal force is always [[orthogonality|orthogonal]] to the motion of the body and towards the fixed point of the instantaneous [[osculating circle|center of curvature]] of the path. [[Isaac Newton]] coined the term,<ref>{{Cite book |last=Brackenridge |first=John Bruce |url=https://books.google.com/books?id=ovOTK7X_mMkC&pg=PA74 |title=The Key to Newton's Dynamics: The Kepler Problem and the Principia |date=1996 |publisher=University of California Press |isbn=978-0-520-91685-2 |location= |pages=74}}</ref> describing it as "a force by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre".<ref>{{cite book |last=Newton |first=Isaac |title=The Principia: Mathematical Principles of Natural Philosophy |publisher=Snowball Pub. |year=2010 |isbn=978-1-60796-240-3 |location=[S.l.] |pages=10}}</ref> In [[Newtonian mechanics]], gravity provides the centripetal force causing astronomical [[orbit]]s.

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.<ref name=Hibbeler>{{cite book |title=Engineering Mechanics: Dynamics |author=Russelkl C Hibbeler |chapter-url=https://books.google.com/books?id=tOFRjXB-XvMC&pg=PA131 |page=131 |chapter=Equations of Motion: Normal and tangential coordinates |isbn=978-0-13-607791-6 |year=2009 |edition=12 |publisher=Prentice Hall}}</ref><ref name=Tipler0>{{cite book |title=Physics for scientists and engineers |page=129 |author1=Paul Allen Tipler |author2=Gene Mosca |url=https://books.google.com/books?id=2HRFckqcBNoC&pg=PA129 |isbn=978-0-7167-8339-8 |edition=5th |publisher=Macmillan |year=2003 |access-date=4 November 2020 |archive-date=7 October 2024 |archive-url=https://web.archive.org/web/20241007055646/https://books.google.com/books?id=2HRFckqcBNoC&pg=PA129#v=onepage&q&f=false |url-status=live }}</ref> The mathematical description was derived in 1659 by the Dutch physicist [[Christiaan Huygens]].<ref>

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[[File:Velocity-acceleration.svg|left|frameless|upright=0.8]]

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 a [[base (geometry)|base]] of <math>\Delta \textbf{v}</math> and a [[isosceles triangle|leg]] length of <math>v</math>, and the other a [[base (geometry)|base]] of <math>\Delta \textbf{r}</math> (position vector [[Euclidean vector#Addition and subtraction|difference]]) and a [[isosceles triangle|leg]] length of <math>r</math>:<ref name="uniform_circular_motion">{{cite web |author=OpenStax CNX |title=Uniform Circular Motion |url=https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/4-4-uniform-circular-motion/ |access-date=25 December 2020 |archive-date=7 October 2024 |archive-url=https://web.archive.org/web/20241007055805/https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/4-4-uniform-circular-motion/ |url-status=live }}</ref>

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 a [[base (geometry)|base]] of <math>\Delta \textbf{v}</math> and a [[isosceles triangle|leg]] length of <math>v</math>, and the other a [[base (geometry)|base]] of <math>\Delta \textbf{r}</math> (position vector [[Euclidean vector#Addition and subtraction|difference]]) and a [[isosceles triangle|leg]] length of <math>r</math>

<math display="block">\frac{|\Delta \textbf{v}|}{v} = \frac{|\Delta \textbf{r}|}{r}</math>

<math display="block">|\Delta \textbf{v}| = \frac{v}{r}|\Delta \textbf{r}|</math>

Therefore, <math>|\Delta\textbf{v}|</math> can be substituted with <math>\frac{v}{r} |\Delta \textbf{r}|</math>:~~<ref name="uniform_circular_motion" />~~

Therefore, <math>|\Delta\textbf{v}|</math> can be substituted with <math>\frac{v}{r} |\Delta \textbf{r}|</math>:

<math display="block">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}</math>

The direction of the force is toward the center of the circle in which the object is moving, or the [[osculating circle]] (the circle that best fits the local path of the object, if the path is not circular).<ref>{{cite book

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The upper panel in the image at right 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 objective is to find what angle the bank must have so the ball does not slide off the road.<ref name = Lerner>{{cite book |title = Physics for Scientists and Engineers |author = Lawrence S. Lerner |page = 128 |url = https://books.google.com/books?id=kJOnAvimS44C&pg=PA129 |isbn = 978-0-86720-479-7 |year = 1997 |location = Boston |publisher = Jones & Bartlett Publishers |access-date = 30 March 2021 |archive-date = 7 October 2024 |archive-url = https://web.archive.org/web/20241007055705/https://books.google.com/books?id=kJOnAvimS44C&pg=PA129#v=onepage&q&f=false |url-status = live }}</ref> 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.

Apart from any acceleration that might occur in the direction of the path, the lower panel of the image above 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 at a right angle to the road surface ''m'''''a'''n. The centripetal force demanded by the curved motion is also shown above. This centripetal force is not a third force applied to the ball, but rather must be provided by 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.

Apart from any acceleration that might occur in the direction of the path, the lower panel of the image above 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 at a [[right angle]] to the road surface ''m'''''a'''n. The centripetal force demanded by the curved motion is also shown above. This centripetal force is not a third force applied to the ball, but rather must be provided by 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 the horizontal component of the force from the road, which has magnitude {{math|1={{!}}'''F'''h{{!}} = ''m''{{!}}'''a'''n{{!}} sin ''θ''}}. The vertical component of the force from the road must counteract the gravitational force: {{math|1={{!}}'''F'''v{{!}} = ''m''{{!}}'''a'''n{{!}} cos ''θ'' = ''m''{{!}}'''g'''{{!}}}}, which implies {{math|1={{!}}'''a'''n{{!}} = {{!}}'''g'''{{!}} / cos ''θ''}}. Substituting into the above formula for {{math|1={{!}}'''F'''h{{!}}}} yields a horizontal force to be:

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<math display="block">\mathbf{r} = \rho \mathbf{u}_{\rho} \ , </math>

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 travelled by the particle.<ref>Notice that this local coordinate system is not autonomous; for example, its rotation in time is dictated by the trajectory traced by the particle. The radial vector '''r'''(''t'') does not represent the [[Osculating circle|radius of curvature]] of the path.</ref> By moving the unit vectors so their tails coincide, as seen in the circle at the left of the image above, 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'').

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 travelled by the particle.<ref>Notice that this local coordinate system is not autonomous; for example, its rotation in time is dictated by the trajectory traced by the particle. The radial vector '''r'''(''t'') does not represent the [[Osculating circle|radius of curvature]] of the path.</ref> By moving the unit vectors so their tails coincide, as seen in the circle at the left of the image above, 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

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: <math>\frac{1} {\rho (s)} = \kappa (s) = \frac {\mathrm{d}\theta}{\mathrm{d}s}\ . </math>

The radius of curvature usually is taken as positive (that is, as an absolute value), while the ''[[Curvature#In terms of a general parametrization|curvature]]'' ''κ'' is a signed quantity.

The radius of curvature usually is taken as positive (that is, as an [[absolute value]]), while the ''[[Curvature#In terms of a general parametrization|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]].<ref name = osculating>The osculating circle at a given point ''P'' on a curve is the limiting circle of a sequence of circles that pass through ''P'' and two other points on the curve, ''Q'' and ''R'', on either side of ''P'', as ''Q'' and ''R'' approach ''P''. See the online text by Lamb: {{cite book |title = An Elementary Course of Infinitesimal Calculus|author = Horace Lamb |page = [https://archive.org/details/anelementarycou01lambgoog/page/n429 406] |url = https://archive.org/details/anelementarycou01lambgoog |quote = osculating circle.|publisher = University Press |year = 1897 |isbn = 978-1-108-00534-0 }}</ref><ref name = Chen0>{{cite book|title = An Introduction to Planar Dynamics|author1 = Guang Chen|author2 = Fook Fah Yap|edition = 3rd|url = https://books.google.com/books?id=xt09XiZBzPEC&pg=PA34|page = 34|isbn = 978-981-243-568-2|year = 2003|publisher = Central Learning Asia/Thomson Learning Asia|access-date = 30 March 2021|archive-date = 7 October 2024|archive-url = https://web.archive.org/web/20241007060232/https://books.google.com/books?id=xt09XiZBzPEC&pg=PA34|url-status = live}}</ref> See image above.

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 {{Classical mechanics|rotational}}
-'''Centripetal force''' (from [[Latin]] ''centrum'', "center" and ''petere'', "to seek"<ref>{{cite book |title=A new universal etymological, technological and pronouncing dictionary of the English language: embracing all terms used in art, science, and literature, Volume 1 |first1=John |last1=Craig |publisher=Harvard University |year=1849 |page=291 |url=https://books.google.com/books?id=0nxBAAAAYAAJ}} [https://books.google.com/books?id=0nxBAAAAYAAJ&pg=PA291 Extract of page 291]</ref>) is the [[force]] that makes a body follow a curved [[trajectory|path]]. The direction of the centripetal force is always [[orthogonality|orthogonal]] to the motion of the body and towards the fixed point of the instantaneous [[osculating circle|center of curvature]] of the path. [[Isaac Newton]] coined the term,<ref>{{Cite book |last=Brackenridge |first=John Bruce |url=https://books.google.com/books?id=ovOTK7X_mMkC&pg=PA74 |title=The Key to Newton's Dynamics: The Kepler Problem and the Principia |date=1996 |publisher=University of California Press |isbn=978-0-520-91685-2 |location= |pages=74}}</ref> describing it as "a force by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre".<ref>{{cite book |last=Newton |first=Isaac |title=The Principia: Mathematical Principles of Natural Philosophy |publisher=Snowball Pub. |year=2010 |isbn=978-1-60796-240-3 |location=[S.l.] |pages=10}}</ref> In [[Newtonian mechanics]], gravity provides the centripetal force causing astronomical [[orbit]]s.
+'''Centripetal force''' ({{etymology|la|{{lang|la|centrum}}|center||{{lang|la|petere}}|to seek}}<ref>{{cite book |title=A new universal etymological, technological and pronouncing dictionary of the English language: embracing all terms used in art, science, and literature, Volume 1 |first1=John |last1=Craig |publisher=Harvard University |year=1849 |page=291 |url=https://books.google.com/books?id=0nxBAAAAYAAJ}} [https://books.google.com/books?id=0nxBAAAAYAAJ&pg=PA291 Extract of page 291]</ref>) is the [[force]] that makes a body follow a curved [[trajectory|path]]. The direction of the centripetal force is always [[orthogonality|orthogonal]] to the motion of the body and towards the fixed point of the instantaneous [[osculating circle|center of curvature]] of the path. [[Isaac Newton]] coined the term,<ref>{{Cite book |last=Brackenridge |first=John Bruce |url=https://books.google.com/books?id=ovOTK7X_mMkC&pg=PA74 |title=The Key to Newton's Dynamics: The Kepler Problem and the Principia |date=1996 |publisher=University of California Press |isbn=978-0-520-91685-2 |location= |pages=74}}</ref> describing it as "a force by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre".<ref>{{cite book |last=Newton |first=Isaac |title=The Principia: Mathematical Principles of Natural Philosophy |publisher=Snowball Pub. |year=2010 |isbn=978-1-60796-240-3 |location=[S.l.] |pages=10}}</ref> In [[Newtonian mechanics]], gravity provides the centripetal force causing astronomical [[orbit]]s.
 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.<ref name=Hibbeler>{{cite book |title=Engineering Mechanics: Dynamics |author=Russelkl C Hibbeler |chapter-url=https://books.google.com/books?id=tOFRjXB-XvMC&pg=PA131 |page=131 |chapter=Equations of Motion: Normal and tangential coordinates |isbn=978-0-13-607791-6 |year=2009 |edition=12 |publisher=Prentice Hall}}</ref><ref name=Tipler0>{{cite book |title=Physics for scientists and engineers |page=129 |author1=Paul Allen Tipler |author2=Gene Mosca |url=https://books.google.com/books?id=2HRFckqcBNoC&pg=PA129 |isbn=978-0-7167-8339-8 |edition=5th |publisher=Macmillan |year=2003 |access-date=4 November 2020 |archive-date=7 October 2024 |archive-url=https://web.archive.org/web/20241007055646/https://books.google.com/books?id=2HRFckqcBNoC&pg=PA129#v=onepage&q&f=false |url-status=live }}</ref> The mathematical description was derived in 1659 by the Dutch physicist [[Christiaan Huygens]].<ref>
@@ Line 22: / Line 22: @@
 [[File:Velocity-acceleration.svg|left|frameless|upright=0.8]]
-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 a [[base (geometry)|base]] of <math>\Delta \textbf{v}</math> and a [[isosceles triangle|leg]] length of <math>v</math>, and the other a [[base (geometry)|base]] of <math>\Delta \textbf{r}</math> (position vector [[Euclidean vector#Addition and subtraction|difference]]) and a [[isosceles triangle|leg]] length of <math>r</math>:<ref name="uniform_circular_motion">{{cite web |author=OpenStax CNX |title=Uniform Circular Motion |url=https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/4-4-uniform-circular-motion/ |access-date=25 December 2020 |archive-date=7 October 2024 |archive-url=https://web.archive.org/web/20241007055805/https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/4-4-uniform-circular-motion/ |url-status=live }}</ref>
+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 a [[base (geometry)|base]] of <math>\Delta \textbf{v}</math> and a [[isosceles triangle|leg]] length of <math>v</math>, and the other a [[base (geometry)|base]] of <math>\Delta \textbf{r}</math> (position vector [[Euclidean vector#Addition and subtraction|difference]]) and a [[isosceles triangle|leg]] length of <math>r</math>
 <math display="block">\frac{|\Delta \textbf{v}|}{v} = \frac{|\Delta \textbf{r}|}{r}</math>
 <math display="block">|\Delta \textbf{v}| = \frac{v}{r}|\Delta \textbf{r}|</math>
-Therefore, <math>|\Delta\textbf{v}|</math> can be substituted with <math>\frac{v}{r} |\Delta \textbf{r}|</math>:<ref name="uniform_circular_motion" />
+Therefore, <math>|\Delta\textbf{v}|</math> can be substituted with <math>\frac{v}{r} |\Delta \textbf{r}|</math>:
 <math display="block">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}</math>
 The direction of the force is toward the center of the circle in which the object is moving, or the [[osculating circle]] (the circle that best fits the local path of the object, if the path is not circular).<ref>{{cite book
@@ Line 159: / Line 159: @@
 The upper panel in the image at right 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 objective is to find what angle the bank must have so the ball does not slide off the road.<ref name = Lerner>{{cite book |title = Physics for Scientists and Engineers |author = Lawrence S. Lerner |page = 128 |url = https://books.google.com/books?id=kJOnAvimS44C&pg=PA129 |isbn = 978-0-86720-479-7 |year = 1997 |location = Boston |publisher = Jones & Bartlett Publishers |access-date = 30 March 2021 |archive-date = 7 October 2024 |archive-url = https://web.archive.org/web/20241007055705/https://books.google.com/books?id=kJOnAvimS44C&pg=PA129#v=onepage&q&f=false |url-status = live }}</ref> 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.
-Apart from any acceleration that might occur in the direction of the path, the lower panel of the image above 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 at a right angle to the road surface ''m'''''a'''<sub>n</sub>. The centripetal force demanded by the curved motion is also shown above. This centripetal force is not a third force applied to the ball, but rather must be provided by 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.
+Apart from any acceleration that might occur in the direction of the path, the lower panel of the image above 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 at a [[right angle]] to the road surface ''m'''''a'''<sub>n</sub>. The centripetal force demanded by the curved motion is also shown above. This centripetal force is not a third force applied to the ball, but rather must be provided by 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 the horizontal component of the force from the road, which has magnitude {{math|1={{!}}'''F'''<sub>h</sub>{{!}} = ''m''{{!}}'''a'''<sub>n</sub>{{!}} sin ''θ''}}. The vertical component of the force from the road must counteract the gravitational force: {{math|1={{!}}'''F'''<sub>v</sub>{{!}} = ''m''{{!}}'''a'''<sub>n</sub>{{!}} cos ''θ'' = ''m''{{!}}'''g'''{{!}}}}, which implies {{math|1={{!}}'''a'''<sub>n</sub>{{!}} = {{!}}'''g'''{{!}} / cos ''θ''}}. Substituting into the above formula for {{math|1={{!}}'''F'''<sub>h</sub>{{!}}}} yields a horizontal force to be:
@@ Line 230: / Line 230: @@
 <math display="block">\mathbf{r} = \rho \mathbf{u}_{\rho} \ , </math>
-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'''<sub>ρ</sub> travels with the particle and always points in the same direction as '''r'''(''t''). Unit vector '''u'''<sub>θ</sub> also travels with the particle and stays orthogonal to '''u'''<sub>ρ</sub>. Thus, '''u'''<sub>ρ</sub> and '''u'''<sub>θ</sub> form a local Cartesian coordinate system attached to the particle, and tied to the path travelled by the particle.<ref>Notice that this local coordinate system is not autonomous; for example, its rotation in time is dictated by the trajectory traced by the particle. The radial vector '''r'''(''t'') does not represent the [[Osculating circle|radius of curvature]] of the path.</ref> By moving the unit vectors so their tails coincide, as seen in the circle at the left of the image above, it is seen that '''u'''<sub>ρ</sub> and '''u'''<sub>θ</sub> 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'').
+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'''<sub>ρ</sub> travels with the particle and always points in the same direction as '''r'''(''t''). Unit vector '''u'''<sub>θ</sub> also travels with the particle and stays orthogonal to '''u'''<sub>ρ</sub>. Thus, '''u'''<sub>ρ</sub> and '''u'''<sub>θ</sub> form a local Cartesian coordinate system attached to the particle, and tied to the path travelled by the particle.<ref>Notice that this local coordinate system is not autonomous; for example, its rotation in time is dictated by the trajectory traced by the particle. The radial vector '''r'''(''t'') does not represent the [[Osculating circle|radius of curvature]] of the path.</ref> By moving the unit vectors so their tails coincide, as seen in the circle at the left of the image above, it is seen that '''u'''<sub>ρ</sub> and '''u'''<sub>θ</sub> 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
@@ Line 288: / Line 288: @@
 : <math>\frac{1} {\rho (s)} = \kappa (s) = \frac {\mathrm{d}\theta}{\mathrm{d}s}\ . </math>
-The radius of curvature usually is taken as positive (that is, as an absolute value), while the ''[[Curvature#In terms of a general parametrization|curvature]]'' ''κ'' is a signed quantity.
+The radius of curvature usually is taken as positive (that is, as an [[absolute value]]), while the ''[[Curvature#In terms of a general parametrization|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]].<ref name = osculating>The osculating circle at a given point ''P'' on a curve is the limiting circle of a sequence of circles that pass through ''P'' and two other points on the curve, ''Q'' and ''R'', on either side of ''P'', as ''Q'' and ''R'' approach ''P''. See the online text by Lamb: {{cite book |title = An Elementary Course of Infinitesimal Calculus|author = Horace Lamb |page = [https://archive.org/details/anelementarycou01lambgoog/page/n429 406] |url = https://archive.org/details/anelementarycou01lambgoog |quote = osculating circle.|publisher = University Press |year = 1897 |isbn = 978-1-108-00534-0 }}</ref><ref name = Chen0>{{cite book|title = An Introduction to Planar Dynamics|author1 = Guang Chen|author2 = Fook Fah Yap|edition = 3rd|url = https://books.google.com/books?id=xt09XiZBzPEC&pg=PA34|page = 34|isbn = 978-981-243-568-2|year = 2003|publisher = Central Learning Asia/Thomson Learning Asia|access-date = 30 March 2021|archive-date = 7 October 2024|archive-url = https://web.archive.org/web/20241007060232/https://books.google.com/books?id=xt09XiZBzPEC&pg=PA34|url-status = live}}</ref> See image above.

Centripetal force: Difference between revisions

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