Editing Centripetal force (section)

=== Uniform circular motion ===
{{See also|Uniform circular motion}}

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

==== Calculus derivation ====
In two dimensions, the position vector <math>\textbf{r}</math>, which has magnitude (length) <math>r</math> and directed at an angle <math>\theta</math> above the x-axis, can be expressed in [[Cartesian coordinates]] using the [[unit vectors]] <math alt="x-hat">\hat\mathbf x</math> and <math alt="y-hat">\hat\mathbf y</math>:<ref>
{{cite book
| title = Vectors in physics and engineering
| author = A. V. Durrant
| publisher = CRC Press
| year = 1996
| isbn = 978-0-412-62710-1
| page = 103
| url = https://books.google.com/books?id=cuMLGAO-ii0C&pg=PA103
}}</ref>
<math display="block"> \textbf{r} = r \cos(\theta) \hat\mathbf x + r \sin(\theta) \hat\mathbf y. </math>

The assumption of [[uniform circular motion]] requires three things:
# The object moves only on a circle.
# The radius of the circle <math>r</math> does not change in time.
# The object moves with constant [[angular velocity]] <math>\omega</math> around the circle. Therefore, <math>\theta = \omega t</math> where <math>t</math> is time.

The [[velocity]] <math>\textbf{v}</math> and [[acceleration]] <math>\textbf{a}</math> of the motion are the first and second derivatives of position with respect to time:

<math display="block"> \textbf{r} = r \cos(\omega t) \hat\mathbf x + r \sin(\omega t) \hat\mathbf y, </math>
<math display="block" qid="Q11465"> \textbf{v} = \dot{\textbf{r}} = - r \omega \sin(\omega t) \hat\mathbf x + r \omega \cos(\omega t) \hat\mathbf y,  </math>
<math display="block" qid=Q11376> \textbf{a} = \ddot{\textbf{r}} = - \omega^2 (r \cos(\omega t) \hat\mathbf x + r \sin(\omega t) \hat\mathbf y). </math>

The term in parentheses is the original expression of <math>\textbf{r}</math> in [[Cartesian coordinates]]. Consequently,
<math display="block"> \textbf{a} = - \omega^2 \textbf{r}. </math>
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 to [[inertia]]), this centripetal acceleration describes the circular motion path caused by a centripetal force.

==== Derivation using vectors ====

[[File:Circular motion vectors.svg|right|thumb|Vector relationships for uniform circular motion; vector '''Ω''' representing the rotation is normal to the plane of the orbit with polarity determined by the [[right-hand rule]] and magnitude ''dθ'' /''dt''.]]
The image at right shows the vector relationships for uniform circular motion. The rotation itself is represented by the angular velocity vector '''Ω''', which is normal to the plane of the orbit (using the [[right-hand rule]]) and has magnitude given by:
: <math> |\mathbf{\Omega}| = \frac {\mathrm{d} \theta } {\mathrm{d}t} = \omega \ , </math>

with ''θ'' the angular position at time ''t''. In this subsection, d''θ''/d''t'' is assumed constant, independent of time. The distance traveled '''dℓ''' of the particle in time d''t'' along the circular path is
: <math> \mathrm{d}\boldsymbol{\ell} = \mathbf {\Omega} \times \mathbf{r}(t) \mathrm{d}t \ , </math>

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

Consequently,
: <math>\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} \ .</math>

In other words,
: <math> \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)\ . </math>

Differentiating with respect to time,
<math display="block"> \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] \ .</math>

[[Triple product#Vector triple product|Lagrange's formula]] states:
<math display="block"> \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 ) \ .</math>

Applying Lagrange's formula with the observation that '''Ω • r'''(''t'') = 0 at all times,
<math display="block"> \mathbf{a} = - {|\mathbf{\Omega|}}^2 \mathbf{r}(t) \ .</math>

In words, the acceleration is pointing directly opposite to the radial displacement '''r''' at all times, and has a magnitude:
<math display="block"> |\mathbf{a}| = |\mathbf{r}(t)| \left ( \frac {\mathrm{d} \theta}{\mathrm{d}t} \right) ^2 = r {\omega}^2 </math>
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, though the notation is slightly different.

When the rate of rotation is made constant in the analysis of [[#Nonuniform circular motion|nonuniform 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 ====
{{Main|Banked turn}}
{{See also|Reactive centrifugal force}}

[[File:Banked turn.svg|thumb|Upper panel: Ball on a banked circular track moving with constant speed ''v''; Lower panel: Forces on the ball]]

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.

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:
<math display="block" qid=Q11402> |\mathbf{F}_\mathrm{h}| = m |\mathbf{g}| \frac { \sin \theta}{ \cos \theta} = m|\mathbf{g}| \tan \theta \, . </math>

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'''<sub>c</sub> of magnitude:
<math display="block">|\mathbf{F}_\mathrm{c}| = m |\mathbf{a}_\mathrm{c}| = \frac{m|\mathbf{v}|^2}{r} \, . </math>

Consequently, the ball is in a stable path when the angle of the road is set to satisfy the condition:
<math display="block">m |\mathbf{g}| \tan \theta = \frac{m|\mathbf{v}|^2}{r} \, ,</math>
or,
<math display="block"> \tan \theta = \frac {|\mathbf{v}|^2} {|\mathbf{g}|r} \, .</math>

As the angle of bank ''θ'' approaches 90°, the [[tangent function]] approaches infinity, allowing larger values for |'''v'''|<sup>2</sup>/''r''. In words, this equation states that for greater 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.<ref name = Schaum>{{cite book |title = Schaum's Outline of Applied Physics |author = Arthur Beiser |page = 103 |url = https://books.google.com/books?id=soKguvJDgmsC&pg=PA103 |publisher = McGraw-Hill Professional |year = 2004 |location = New York |isbn = 978-0-07-142611-4 |access-date = 30 March 2021 |archive-date = 7 October 2024 |archive-url = https://web.archive.org/web/20241007055651/https://books.google.com/books?id=soKguvJDgmsC&pg=PA103 |url-status = live }}</ref><ref name = Darbyshire>{{cite book |title = Mechanical Engineering: BTEC National Option Units |author = Alan Darbyshire |page = 56 |url = https://books.google.com/books?id=fzfXLGpElZ0C&pg=PA57 |isbn = 978-0-7506-5761-7 |publisher = Newnes |year = 2003 |location = Oxford |access-date = 30 March 2021 |archive-date = 7 October 2024 |archive-url = https://web.archive.org/web/20241007055651/https://books.google.com/books?id=fzfXLGpElZ0C&pg=PA57 |url-status = live }}</ref>

These ideas apply to air flight as well. See the FAA pilot's manual.<ref name = FAA>{{cite book |title = Pilot's Encyclopedia of Aeronautical Knowledge |author = Federal Aviation Administration |page = Figure 3–21 |url = https://books.google.com/books?id=m5V04SXE4zQC&pg=PT33 |isbn = 978-1-60239-034-8 |year = 2007 |publisher = Skyhorse Publishing Inc. |location = Oklahoma City OK |no-pp = true |access-date = 30 March 2021 |archive-date = 7 October 2024 |archive-url = https://web.archive.org/web/20241007060154/https://books.google.co.id/books?id=m5V04SXE4zQC&pg=PT33&redir_esc=y |url-status = live }}</ref>