Tangential Normal Particle Kinematics

Tangential-Normal Particle Kinematics

Tangential-normal coordinates, also called path coordinates, describe particle motion using directions tied directly to the trajectory rather than to a fixed coordinate grid. At each point on a smooth path, one unit vector points tangent to the motion and another points toward the local center of curvature.

This viewpoint is especially useful when the shape of the path is known and the speed along that path is known or can be found. The velocity then lies entirely in the tangent direction, while the acceleration naturally separates into a tangential part that changes speed and a normal part that changes direction.

The central result is

|-----------------| | dv v2 | |a = ---et + ---en|, -----dt------ρc----

(1)

where ρ_c is the local radius of curvature. The subscript c is used here to distinguish radius of curvature from the cylindrical coordinate ρ used in M01-11.

1 Arc length as the path coordinate

Let the trajectory be described by a position vector

r = r(s),

(2)

where s is arc length measured along the path. By definition of arc length,

| | |dr | ||---|| = 1. ds

(3)

This motivates the unit tangent vector

|--------| | dr-| |et = ds . ---------

(4)

The tangent vector points in the direction of increasing s.

Figure 1. Local tangential-normal basis attached to a smooth trajectory. The unit tangent e_t points along increasing arc length, while e_n points toward the local center of curvature.

2 Velocity in path coordinates

Because

r = r(s(t)),

(5)

the chain rule gives

v = dr-= dr-ds. dt ds dt

(6)

Using the tangent vector definition,

v = et ⋅s.

(7)

Since ⋅
s is the scalar speed v,

|v-=-vet-. ---------|

(8)

Thus path coordinates encode an important geometric fact automatically: the velocity is always tangent to the trajectory.

3 Curvature

Although e_t has unit magnitude, its direction generally changes as the particle moves along a curved path. The rate at which the tangent direction changes with arc length defines the curvature

|----||---||-| |κ = |det| . -----|ds-|-|

(9)

The radius of curvature is

|-------| | 1 | |ρc = --. ------κ--

(10)

The principal unit normal is defined by

|-----------| |e = 1det-. --n----κ-ds-|

(11)

Therefore

|------------------| |de 1 | |--t-= κen = --en . -ds-----------ρc---

(12)

The normal direction points toward the local center of curvature.

Figure 2. Curvature measures how rapidly the unit tangent rotates with arc length. The osculating circle has radius ρ_c = 1∕κ and shares the trajectory’s local tangent and curvature.

4 Derivation of the acceleration formula

Differentiate the velocity

v = vet.

(13)

The product rule gives

dv det a = --et + v---. dt dt

(14)

Apply the chain rule to the tangent-vector derivative:

det-= detds-. dt ds dt

(15)

Since

det-= 1-e ds ρc n

(16)

and

ds dt-= v,

(17)

we obtain

de v --t-= -- en. dt ρc

(18)

Substitution gives

|-------------2---| |a = dvet + v-en . ------dt-----ρc---|

(19)

It is customary to define

|--------| | dv | |at = ---| ------dt-

(20)

and

|--------| | v2 | an = ---. ------ρc--

(21)

Thus

|---------------| -a-=-atet +-anen-.

(22)

Figure 3. Acceleration in path coordinates separates naturally into a tangential component that changes speed and a normal component that changes the velocity direction.

5 Physical interpretation

The tangential component

dv at = dt-

(23)

changes the magnitude of the velocity. If a_t > 0, the particle speeds up. If a_t < 0, it slows down.

The normal component

v2- an = ρc

(24)

changes the direction of the velocity. It always points toward the local center of curvature. Even if the speed is constant, a_n can be nonzero.

This makes uniform circular motion an immediate special case. For a circle of fixed radius R,

ρc = R,

(25)

and if the speed is constant,

at = 0.

(26)

Therefore

v2- a = R en,

(27)

which is the centripetal-acceleration result of M01-08.

6 Acceleration magnitude and direction

Because e_t and e_n are perpendicular,

∘ -------- |a | = a2+ a2 . t n

(28)

If α is the angle from the positive tangent direction toward the normal direction,

an- tan α = at ,

(29)

with the quadrant determined by the signs of the components.

7 Using speed as a function of arc length

Sometimes speed is supplied as a function of path position rather than time:

v = v (s).

(30)

Then

dv dvds ---= -----. dt ds dt

(31)

Since ds∕dt = v,

|---------| | dv | |at = vds-. -----------

(32)

Therefore if v(s) and the path curvature are known,

|----(-----)-------2---| | dv- v-- | |a = vds et + ρc en . -----------------------

(33)

This form is especially useful in particle dynamics problems where speed is found from energy as a function of position along a track.

8 Curvature of a planar Cartesian path

If the path is written as

y = y(x ),

(34)

its curvature magnitude is

|-----------′′-----| |κ = -----|y--|--- . | [1 + (y′)2]3∕2| -------------------

(35)

Hence

|------------------| | [1 + (y′)2]3∕2 | |ρc = ------------ . ----------|y′′|-----|

(36)

This allows path-coordinate acceleration to be computed directly from the geometry of a curve expressed in Cartesian form.

For a planar parametric path

r(u) = x(u)ex + y(u )ey,

(37)

the curvature can be written

|--------------------| | -|x′y′′ −-y′x′′|-| κ = ′2 ′2 3∕2 , -----[(x-)-+--(y-)-]----

(38)

where primes denote differentiation with respect to the parameter u.

9 Three-dimensional extension

For a smooth three-dimensional trajectory, define the binormal unit vector

eb = et × en.

(39)

The three vectors

(et,en, eb)

(40)

form the Frenet triad. The trajectory can twist out of the local osculating plane, a behavior measured by torsion. However, the particle acceleration itself still has only tangential and principal-normal components:

|---------------| |a = atet + anen|. -----------------

(41)

There is no independent binormal acceleration component for ordinary particle kinematics because de_t∕ds lies in the principal-normal direction.

For an arbitrary three-dimensional parameter u,

|------′----′′-| |κ = |r-×--r-|. | |r′|3 | --------------

(42)

10 Choosing path coordinates versus polar or Cartesian coordinates

Path coordinates are especially useful when the geometry of the trajectory is already known. Cartesian coordinates are often preferable when the acceleration components are naturally fixed in space. Polar or cylindrical coordinates are useful when the geometry is organized around a fixed origin or axis.

Figure 4. Coordinate-choice guide for particle kinematics. Tangential-normal coordinates are most natural when the trajectory shape and its local curvature are known.

No coordinate system changes the physics. The goal is to choose a basis that makes the known information and desired unknowns as simple as possible.

11 Worked example 1: vehicle on a curved road

A vehicle travels along a road whose local radius of curvature is

ρc = 50.0 m.

(43)

At one instant its speed is

v = 20.0 m ∕s,

(44)

and its speed is increasing at

dv- 2 dt = 2.0 m∕s .

(45)

The tangential component is

a = 2.0 m ∕s2. t

(46)

The normal component is

v2 (20.0 )2 2 an = ---= -------= 8.0 m ∕s . ρc 50.0

(47)

Therefore

|-----------------------| -a-=-2.0et-+-8.0en-m-∕s2 .

(48)

Its magnitude is

√ --2------2- 2 |a| = 2.0 + 8.0 = 8.25 m ∕s.

(49)

The acceleration points mostly inward because the normal component dominates.

12 Worked example 2: speed prescribed along a path

A particle moves along a known curve whose radius of curvature at s = 20 m is

ρ = 14.0 m. c

(50)

Its speed is prescribed, with s measured in meters, by

v (s ) = 5.0 + 0.10s m ∕s.

(51)

At s = 20 m,

v = 7.0 m ∕s.

(52)

Also,

dv ---= 0.10 s−1. ds

(53)

Therefore

at = vdv-= (7.0 )(0.10) = 0.70 m ∕s2. ds

(54)

The normal component is

7.02 2 an = 14.0 = 3.50 m∕s .

(55)

Thus

|------------------------2| -a-=-0.70et-+-3.50en-m-∕s- .

(56)

and

|a | = 3.57 m ∕s2.

(57)

13 Worked example 3: curvature from a Cartesian path

A particle moves along the path

2 y = x--, 2L

(58)

where

L = 1.0 m.

(59)

At the point x = L, its speed is 4.0 m/s and its speed is increasing at 1.0 m/s². Find the local radius of curvature and the tangential-normal acceleration components.

Differentiate the path:

y′ = x-, L

(60)

and

y′′ = 1-. L

(61)

At x = L,

′ ′′ − 1 y = 1, y = 1 m .

(62)

Therefore

[1 + (1)2]3∕2 √ -- ρc = ------------= 2 2L = 2.83 m. 1∕L

(63)

The tangential acceleration is

at = 1.0 m ∕s2.

(64)

The normal acceleration is

2 a = (4.0)-= 5.66 m ∕s2. n 2.83

(65)

Hence

|----------------------2-| -a =-1.0et +-5.66en-m-∕s-.

(66)

The principal normal points toward the concave side of the parabola.

14 Practice problems

A particle travels at 12 m/s along a path with local radius of curvature 30 m. Its speed is instantaneously constant. Find a_t and a_n.
A vehicle moves at 10 m/s on a path of radius of curvature 25 m while slowing at 1.5 m/s². Find a_t, a_n, and |a|.
At fixed speed, the radius of curvature doubles. By what factor does the normal acceleration change?
At fixed radius of curvature, the speed doubles. By what factor does the normal acceleration change?
A particle has v(s) = 3.0 + 0.20s m/s, where s is in meters. At s = 10 m the radius of curvature is 20 m. Find a_t, a_n, and the acceleration magnitude.
For the path y = x²∕(4L) with L = 1 m, find the radius of curvature at x = 2 m.
Show that uniform circular motion follows from the tangential-normal formula when v and ρ_c = R are constant.
A helix is parameterized by
$r(u ) = 3 cosu e + 3sinu e + 4u e x y z$ (67)

with lengths in meters. Show that its curvature radius is 25∕3 m. If a particle moves along the helix at constant speed 10 m/s, find its acceleration magnitude.

15 Answer check

a_t = 0, a_n = 4.8 m/s².
a_t = −1.5 m/s², a_n = 4.0 m/s², |a| = 4.27 m/s².
It is halved.
It increases by a factor of 4.
v = 5.0 m/s, a_t = 1.0 m/s², a_n = 1.25 m/s², |a| = 1.60 m/s².
ρ_c = 4 m ≃ 5.66 m.
a_t = 0 and a_n = v²∕R, directed toward the circle center.
ρ_c = 25∕3 m and |a| = v²∕ρ_c = 12.0 m/s².

16 Summary

Tangential-normal coordinates attach the basis directly to the trajectory. The central definitions are

e = dr, t ds

(68)

| | |det| 1- κ = ||ds ||, ρc = κ,

(69)

and

|--------------------------2---| |v = ve , a = dv-e + v--e . | t dt t ρc n | -------------------------------

(70)

The tangential component changes speed, while the normal component changes direction. This framework generalizes the centripetal-acceleration idea from circular motion to arbitrary smooth trajectories and provides a natural final bridge from kinematics into Newtonian particle dynamics.

References

[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[2] J. B. Marion and S. T. Thornton, Classical Dynamics of Particles and Systems, 5th ed., Brooks/Cole, 2004.

[3] J. Moore et al., Mechanics Map, Engineering LibreTexts, CC BY-SA 4.0.

[4] PhysicsLibrary, M01-08, Uniform Circular Motion.

[5] PhysicsLibrary, M01-10, Polar-Coordinate Particle Kinematics.

"Tangential Normal Particle Kinematics" is owned by bloftin.
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