A Centripetal Force ( from Latin centrum , “centre” and peter , “to seek” [1] ) is a force that causes a body to follow a curved path . Its direction is always orthogonal to the motion of the body and of the instantaneous center of curvature of the path towards the fixed point . Isaac Newton described it as “a force by which bodies are pulled or propelled, or in any way directed towards a point as the centre”. In Newtonian mechanics , the centripetal force of gravity provides the astronomicalclasses .
A common example of a centripetal force is the case in which a body moves along a circular path with a uniform speed. The centripetal force is directed at right angles to the motion and along the radius towards the center of the circular path. [3] [4] The mathematical description was obtained in 1659 by the Dutch physicist Christiaan Huygens.
sources
The magnitude of the centripetal force on an object of mass m moving along a path with tangential speed v is radius r :
{\displaystyle F_{c}=ma_{c}={\frac {mv^{2}}{r}}}{\displaystyle a_{c}=\lim _{\Delta {t}\to 0}{\frac {|\Delta {\textbf {v}}|}{\Delta {t}}}}where is is the difference between centripetal acceleration and velocity vectors . Since the velocity vectors in the diagram above have a constant magnitude and since each one stands to their respective position vector, simple vector subtraction implies two similar isosceles triangles with favorable angles of the base and one in which a – The length of one leg , and the other of a base (position vector difference ) and the length of a leg :
a_c{\displaystyle \Delta {\textbf {v}}}{\displaystyle \Delta {\textbf {v}}}v{\displaystyle \Delta {\textbf {r}}}r{\displaystyle {\frac {|\Delta {\textbf {v}}|}{v}}={\frac {|\Delta {\textbf {r}}|}{r}}}{\displaystyle |\Delta {\textbf {v}}|={\frac {v}{r}}|\Delta {\textbf {r}}|}Therefore, can be replaced with :
{\displaystyle |\Delta {\textbf {v}}|}{\displaystyle {\frac {v}{r}}|\Delta {\textbf {r}}|}{\displaystyle 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}}}=\omega \lim _{{\Delta }t\to 0}{\frac {|\Delta {\textbf {r}}|}{\Delta {t}}}=v\omega ={\frac {v^{2}}{r}}}The direction of the force is toward the center of the circle in which the object is moving, or the circle of oscillation (the circle best suited for the object’s local path, if the path is not circular). [8] In the formula, speed is squared, so twice the speed requires four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force. This force is also sometimes written in terms of the angular velocity of the object about the center of the circle, related to the tangential velocity by the formula
v = \omega r
So That
{\displaystyle F_{c}=mr\omega ^{2}\,.}Expressed using the orbital period T for one revolution of the circle ,
{\displaystyle \omega ={\frac {2\pi }{T}}\,}the equation becomes
{\displaystyle F_{c}=mr\left({\frac {2\pi }{T}}\right)^{2}.}In particle accelerators, the velocity can be much higher (closer to the speed of light in vacuum) so the same rest mass now exerts more inertia (relative mass) requiring more force for the same centripetal acceleration, so the equation becomes :
{\displaystyle F_{c}={\frac {\gamma mv^{2}}{r}}}Where from
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}is the Lorentz factor .
Thus the centripetal force is given by:
{\displaystyle F_{c}=\gamma mv\omega }which is the rate of change of relative momentum
{\displaystyle \gamma mv}.sources say
In the case of an object moving at 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. Centripetal force can also be supplied as a ‘push’ force, as in the case where the normal reaction of the wall supplies the centripetal force to the wall of the die plunger .
This corresponds to Newton ‘s idea of a centripetal force, which is now called the centripetal force . When a satellite is in orbit around a planet , the instantaneous center of gravity is considered to be a centripetal force even though in the case of eccentric orbits, the gravitational force is directed toward the focus, and not toward the instantaneous center of curvature.
Another example of centripetal force arises in the helix which is detected when a charged particle moves in a uniform magnetic field in the absence of other external forces. In this case, the magnetic force is the centripetal force acting towards the helix axis.
Analysis of multiple cases
Below are three examples of increasing complexity, along with the derivation of the formulas governing velocity and acceleration.
Uniform circular motion
Uniform circular motion refers to the case of a constant rate of rotation. Here are two approaches to describe the matter.
calculus derivation
In two dimensions, the position vector , which has magnitude (length) and is directed at an angle above the x-axis, can be expressed in Cartesian coordinates using the unit vectors and :
\textbf{r}r\theta {\hat {x}}{\hat{y}}\textbf{r} = r \cos(\theta) \hat{x} + r \sin(\theta) \hat{y}.Assume uniform circular motion , which requires three things.
- The object moves only in a circle.
- The radius of the circle does not change in time.r
- The object moves with constant angular velocity around the circle. Therefore, where is the time.
\omega\theta = \omega tt
Now find the velocity and the acceleration of the motion by deriving the position with respect to time. va
\textbf{r} = r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y}
\dot{\textbf{r}} = \textbf{v} = - r \omega \sin(\omega t) \hat{x} + r \omega \cos(\omega t) \hat{y}\ddot{\textbf{r}} = \textbf{a} = - r \omega^2 \cos(\omega t) \hat{x} - r \omega^2 \sin(\omega t) \hat{y}\textbf{a} = - \omega^2 (r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y})Note that the root expression of the word in parentheses is in Cartesian coordinates. As a result, r
\textbf{a} = - \omega^2 \textbf{r}.The negative signifies that the acceleration is pointed toward the center of the circle (as opposed to the radius), so it is called “centripetal” (i.e. “centre-seeking”). While objects naturally follow a straight path (due to inertia), this centripetal acceleration describes a circular motion path due to a centripetal force.
Derivation using vectors
The image on the right shows the vector relation for uniform circular motion. The rotation itself is represented by the angular velocity vector , which is normal to the plane of the orbit (using the right-hand rule) and the magnitude is given by:
|\mathbf{\Omega}| = \frac {\mathrm{d} \theta } {\mathrm{d}t} = \omega \ ,with the angular position in time t . In this subsection , d / d t is assumed constant, independent of time. The distance d of the particle in time d t along the circular path is
\mathrm{d}\boldsymbol{\ell} = \mathbf {\Omega} \times \mathbf{r}(t) \mathrm{d}t \ ,which, by the properties of the vector cross product , has magnitude r d and is tangent to the direction for the circular path.
As a result,
\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)\ .differentiate with respect to time,
{\displaystyle \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:
{\displaystyle \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 that • r ( t ) = 0 at all times,
\mathbf{a} = - {|\mathbf{\Omega|}}^2 \mathbf{r}(t) \ .In words, the acceleration is always pointing directly opposite to the radial displacement r , and its magnitude is:
|\mathbf{a}| = |\mathbf{r}(t)| \left ( \frac {\mathrm{d} \theta}{\mathrm{d}t} \right) ^2 = r {\omega}^2where vertical bars |…| denote the vector magnitude, which in the case of r ( t ) is just the radius r of the path . This result agrees with the previous section, although the notation is slightly different.
When the rate of rotation is kept constant in the analysis of non-uniform circular motion, that analysis agrees.
A beauty of the vector approach is that it is clearly independent of any coordinate system.
Example: banked turn
The upper panel in the image to the right shows a ball in circular motion on an edge curve. The curve is bent at an angle to the horizontal , and the road surface is considered to be slippery. The objective is to find out what the angle of the bank should be so that the ball does not slip off the road. [13] Intuition tells us that, on a flat curve with no banking, the ball will simply slide off the road; Whereas with very fast banking, the ball will slide towards the center unless it travels a sharp curve.
The lower panel of the image above indicates the forces on the ball, in addition to any acceleration occurring in the direction of the path. There are two forces; The force of gravity of a ball vertically down through the center is 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 on the road surface m a nBut is placed at right angles. The centripetal force demanded by the rotational motion is also shown above. This centripetal force is not the third force exerted on the ball, but must be imparted by the net force on the ball as a result of the vector addition of the normal force and the gravitational force. The resultant or net force found on the ball by the vector addition must be equal to the normal force exerted by the road and the vertical force due to the centripetal force determined by the centripetal force needed to travel along a circular path. The rotational motion is maintained as long as this net force provides the centripetal force necessary for the motion.
The horizontal mesh force on the ball is the horizontal component of the force on the road, whose magnitude is F H | = M | a n | Sin . _ The vertical component of the force from the road must counteract the force of gravity: | F V | = M | a n | Because = m | _ G |, which means | a n | , g | / because . , , Substituting in the above formula for f h | Produces a horizontal force:
|\mathbf{F}_\mathrm{h}| = m |\mathbf{g}| \frac { \mathrm{sin}\ \theta}{ \mathrm {cos}\ \theta} = m|\mathbf{g}| \mathrm{tan}\ \theta \ .On the other hand, with velocity. V | On a circular path of radius r , the kinematics says that the force required to make the ball make continuous turns is an inward centripetal force of radius F c of magnitude:
|\mathbf{F}_\mathrm{c}| = m |\mathbf{a}_\mathrm{c}| = \frac{m|\mathbf{v}|^2}{r} \ .As a result, the ball is in a constant 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 approaches 90°, the tangent function approaches infinity, so that | , Larger values are obtained for V | 2 / r . In words, this equation states that for maximal speed (large | v |) the road is more steep (for a larger value ) , and the faster turns (for small r ) road too Should be trusted more rapidly, which accords with intuition. when angledoes not satisfy the above condition, then the horizontal component of the force exerted by the road does not provide the true centripetal force, and an additional frictional force called tangent is applied to the road surface 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 equilibrium can be felt. [14] [15]
These considerations also apply to air flight. See FAA Pilot’s Manual. [16]
non-uniform circular motion
As a generalization of the uniform circular motion case, suppose that the angular rate of rotation is not constant. The acceleration now has a tangent component, as shown in the image to the right. This case is used to demonstrate the derivation strategy based on the polar coordinate system.
Let r ( t ) be a vector that describes the position of the point mass as a function of time. Since we are assuming circular motion, let r ( t ) = R u r , where R is a constant (radius of the circle) and u r is the unit vector representing the point mass from the origin . The direction of u r is described by , the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates, U stands for U RAnd points in the direction of increasing . These polar unit vectors can be expressed as Cartesian unit vectors in the x and y directions, denoted i and j , respectively:
U r = cos i + sin j
AndU = -sin i + cos j . _
To find the velocity one can differentiate:
\mathbf{v} = r \frac {\mathrm{d} \mathbf{u}_\mathrm{r}}{\mathrm{d}t} = r \frac {\mathrm{d}}{\mathrm{d}t} \left( \mathrm{cos}\ \theta \ \mathbf{i} + \mathrm{sin}\ \theta \ \mathbf{j}\right)= r \frac {d \theta} {dt} \left( -\mathrm{sin}\ \theta \ \mathbf{i} + \mathrm{cos}\ \theta \ \mathbf{j}\right)\,= r \frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{u}_\mathrm{\theta} \,= \omega r \mathbf{u}_\mathrm{\theta} \,where is the angular velocity d / d t .
This result for velocity corresponds to the requirements that the velocity should be directed tangentially to the circle, and that the magnitude of the velocity should be rω . differentiate again, and pay attention to
{\frac {\mathrm{d}\mathbf{u}_\mathrm{\theta}}{\mathrm{d}t} = -\frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{u}_\mathrm{r} = - \omega \mathbf{u}_\mathrm{r}} \ ,We find that the acceleration, a is:
\mathbf{a} = r \left( \frac {\mathrm{d}\omega}{\mathrm{d}t} \mathbf{u}_\mathrm{\theta} - \omega^2 \mathbf{u}_\mathrm{r} \right) \ .Thus, the radial and tangential components of acceleration are:
\mathbf{a}_{\mathrm{r}} = - \omega^{2} r \ \mathbf{u}_\mathrm{r} = - \frac{|\mathbf{v}|^{2}}{r} \ \mathbf{u}_\mathrm{r} \ ~~And~~ \ \mathbf{a}_{\mathrm{\theta}} = r \ \frac {\mathrm{d}\omega}{\mathrm{d}t} \ \mathbf{u}_\mathrm{\theta} = \frac {\mathrm{d} | \mathbf{v} | }{\mathrm{d}t} \ \mathbf{u}_\mathrm{\theta} \ ,where | V | = r is the magnitude of velocity.
These equations express mathematically that, in the case of an object that moves along a circular path with a varying speed, the acceleration of the body can be decomposed into a perpendicular component that changes the direction of motion (central acceleration). , and a parallel, or tangent component, which varies speed.
normal planar motion
Kinetic vector in plane polar coordinates. Note that the setup is not restricted to 2d space, but a plane in any higher dimension.
polar coordinates
The above results can probably be more easily obtained in polar coordinates, and also extended to normal speeds within a plane, as shown next. The aircraft crew has polar coordinates consisting of a radial unit vector U and an angular unit vector U , as shown above. [18] A particle at position r is described by:
\mathbf{r} = \rho \mathbf{u}_{\rho} \ ,where the notation is used instead of r to describe the distance of the path from the origin to emphasize that this distance is not fixed, but varies with time. The particle in the same direction as the unit vector U and always travels along the point r ( t ). The unit vector U also travels with the particle and remains orthogonal to U . Thus, U and U form a local Cartesian coordinate system attached to the particle and bound by the path the particle travels. [19]If its tail coincides, as seen in the circle on the left of the image above, it is observed that the unit vectors U and U have a right angled pair with the tips of the unit circle that trace back and trace back to this circle. forward on the circumference with the same angle ( t ) as r ( t ).
When the particle is moving, 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 derivative of velocity, the unit vector u is needed. Because u is a unit vector , its magnitude is fixed, and it can only change in direction, that is, its change d u has a component only perpendicular to u . The trajectory when r ( t ) rotates by a sum d , u , which points in the same direction as r ( t ), also rotates by d . See picture above. Therefore, the change in u 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} \ .Similarly, the rate of change of U is found. As with U , U is a unit vector and can only rotate without resizing . To remain orthogonal to u while the trajectory r ( t ) rotates a sum d , u , which is orthogonal to r ( t ), also rotates by d . See picture above. Therefore, the change d is orthogonal to u u and proportional to d(see above image):
\frac{\mathrm{d} \mathbf{u}_{\theta}}{\mathrm{d}t} = -\frac {\mathrm{d} \theta} {\mathrm{d}t} \mathbf{u}_{\rho} \ .The image above shows the sign to be negative: to maintain orthogonality, if du is positive with d , then du must decrease.
In the expression for the u velocity of the derivative substitute :
\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} \ .In order to get the acceleration, the differentiation is done for the second time:
\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 derivatives u and u , the speed 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}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] \ .As a special example, if the particle moves in a circle of constant radius r , then d / d t = 0, v = v , 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] {\displaystyle =\mathbf {u} _{\rho }\left[-{\frac {v^{2}}{r}}\right]+\mathbf {u} _{\theta }\left[{\frac {\mathrm {d} v}{\mathrm {d} t}}\right]\ }Where from
v = v_{\theta}.These results agree with the above results for non-uniform circular motion. See also article on non-uniform circular motion. If this acceleration is multiplied by the particle mass, the leading term is the centripetal force and the negative of the second term related to the angular acceleration is sometimes called the Euler force. [21]
For trajectories other than circular motion, for example, above the more general trajectory, the instantaneous center of rotation and the radius of curvature of the trajectory to the coordinate system defined in Fig . and length to | r ( t )| = . _ Consequently, in the general case, it is not straightforward to separate the centripetal and Euler terms from the above general acceleration equation. [22] [23] To deal with this issue directly, local coordinates are preferable, as discussed further.
Local coordinates
Local coordinates mean a set of coordinates that the particle travels along, [24] and has an orientation determined by the particle’s path. [25] Unit vectors are formed as shown in the image on the right, both tangent and normal to the path. This coordinate system is sometimes referred to as intrinsic or path coordinates [26] [27] or nt-coordinates , for common-tangents, referring to these unit vectors. These coordinates are a very special example of the more general concept of local coordinates from the theory of differential forms. [28]
The distance along the path of the particle is the length of the arc s , which is assumed to be a known function of time.
s = s(t) \ .
A center of curvature is defined at each position s located a distance (on a line with radius of curvature normal) from the position u n ( s ). The required distance ( arc length at s) s is defined as 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 initial position is ( s ) , then ( s ) is defined by the derivative d / ds :
\frac{1} {\rho (s)} = \kappa (s) = \frac {\mathrm{d}\theta}{\mathrm{d}s}\ .The radius of curvature is usually taken (that is, as an absolute value), while the curvature is a signed quantity , taken as positive.
A geometric approach to find the center of curvature and radius of curvature uses a finite process that leads to the oscillating circle. [29] [30] See image above.
Using these coordinates, motion along the path is viewed as a succession of circular paths of ever-changing center, and each position constitutes non-uniform circular motion with radius at that position . Then the place value of the angular rate of rotation 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 local speed v 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 left-hand insert in the image above):
\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, velocity and acceleration are:
\mathbf{v}(t) = v \mathbf{u}_\mathrm{t}(s)\ ;and using the chain-law 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 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 for non-uniform circular motion with a local radius (s), and the centripetal acceleration is recognized as the second term . [34]
Extending this approach to three-dimensional space curves leads to the Frenet–Serat formulas. [35] [36]
Alternative approach
, given the image above, one might wonder whether sufficient account has been taken of the difference in curvature between ( s ) and ( s + ds as computing d in arc length) s = ( s ) d . Assurance at this point can be found using the more formal approach outlined below. This approach also ties in with the article on curvature.
To introduce the unit vectors of a local coordinate system, one approach is to start in Cartesian coordinates and describe the local coordinates in terms of these Cartesian coordinates. In terms of arc length s , the path may be described as:
\mathbf{r}(s) = \left[ x(s),\ y(s) \right] \ .Then an incremental displacement along the path d s 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 represent the derivatives with respect to s . The magnitude of this displacement is d s , indicating that:
\left[ x'(s)^2 + y'(s)^2 \right] = 1 \ .(Eq. 1)
This displacement is essentially the tangent to the curve at s , indicating that the unit vector tangent to the curve is:
\mathbf{u}_\mathrm{t}(s) = \left[ x'(s), \ y'(s) \right] \ ,Whereas the external unit vector common to the curve is
\mathbf{u}_\mathrm{n}(s) = \left[ y'(s),\ -x'(s) \right] \ ,The orthogonality can be verified by showing that the vector dot product is zero. The unit magnitude of these vectors is the result of Eq. 1. Using the tangent vector, the angle 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 purely formally (without the need for geometric interpretation):
\frac{1}{\rho} = \frac{\mathrm{d}\theta}{\mathrm{d}s}\ .The derivative of for sin sin can be found from:
\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 forms arises from the 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}^2s}{\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}^2s}{\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 for circular motion based on radius . Using this coordinate system in the inertial frame, it is easy to identify the force normal to the trajectory as the centripetal force and the tangential force parallel to the trajectory. From a qualitative point of view, the path can be approximated by the arc of a circle for a finite time, and a particular radius of curvature applied for a finite time, analysis of centrifugal and Euler forces based on circular motion along that radius. can be done. ,
This result for acceleration agrees with a previously found result. However, in this method, the question of the change in radius of curvature along s is completely formally handled in conformity with a geometric interpretation, but it is not relied upon, thereby neglecting the variation in the image above. Is to avoid any question .
Example: circular motion
To illustrate the above formulas, let x , 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 radius α . The position corresponds to s = 0 [ α , 0], or 3 o’clock. To use the above formalism, the derivative needs to be:
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} \ .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 show that s = 0 is located at the position [ , 0] and s = / 2 at [0, ] , for which the motif agrees with x and y . In other words, s is measured counterclockwise around the circle at 3 o’clock. In addition, 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, there is a time-dependence necessary for s . For counterclockwise motion at variable speed v ( t ):
s(t) = \int_0^t \ dt^{\prime} \ v(t^{\prime}) \ ,where v ( t ) is 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}{\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 is already established that α = . This acceleration is the standard result of non-uniform circular motion.
