Vector Calculus & General Coordinate Systems

Vector Calculus & General Coordinate Systems

Homework #4, Prob 2: 1
2
x1 = θ 1θ 2 cosθ 3, x2 = θ 1θ 2 sinθ 3, x3 = ⎣⎡(θ 1)2 − (θ 2 )2 ⎦⎤

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Curvilinear Systems: Spherical Coordinates

The curvilinear spherical coordinate system is probably
familiar to all of you. In engineering and physics this
coordinate system is used to take advantage of spherical
symmetry. Let’s examine this coordinate system in detail.

The curvilinear transformations and inverse transformations
that define the spherical system are given by,

q1 = r = x2 + y2 + z2

⎛ x2 + z + z2 ⎞ x = r sinθ cosφ
y2 ⎟⎠⎟ y = r sinθ sinφ
q2 = θ = cos−1 ⎝⎜⎜ z = r cosθ

q3 = φ = tan −1 ⎛ y ⎞ 128
⎝⎜ x ⎟⎠

Vector Calculus & General Coordinate Systems

and scale factors and fundamental metric components are

h1 = hr = ⎡⎛ ∂x ⎞2 + ⎛ ∂y ⎞2 + ⎛ ∂z ⎞2 ⎤1/ 2
⎢⎢⎣⎜⎝ ∂r ⎠⎟ ⎜⎝ ∂r ⎟⎠ ⎝⎜ ∂r ⎠⎟ ⎥
⎥⎦

= ⎡⎣(sinθ cosφ )2 + (sinθ cosφ )2 + cos2 θ ⎤⎦1/2 = 1

h2 = hθ = r

h3 = hφ = r sinθ

g11 = h12 = 1 g11 = 1 =1 Since this is an
g22 = h22 = r 2 h12 orthogonal
curvilinear
g 22 = 1 = 1 system,
h22 r2

g33 = h32 = r 2 sin2 θ g 33 = 1 = 1 gij = gij = 0, i ≠ j
h32
r2 sin2 θ 129

Vector Calculus & General Coordinate Systems

θ = const θ z eˆ r
eˆφ
r = const r θ
eˆθ y
φ = const r
φ
φ

x

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Vector Calculus & General Coordinate Systems

Spherical coordinates basis and dual basis:
ei = hieˆi (no summation)
e j = gijei

We can now write the basis and dual basis,

er = eˆr er = eˆr
eθ = reˆθ eθ = eˆθ

eφ = r sinθ eˆφ r
eφ = eˆφ

r sinθ

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Basis vectors in terms of the Cartesian basis:

eˆ i = 1 ∂x j ˆi j (no summation in i)
hi ∂qi

eˆ r = 1 ∂x j ˆi j , eˆθ = 1 ∂x j ˆi j , eˆφ = 1 ∂x j ˆi j
hr ∂qr hθ ∂qθ hφ ∂qφ

In matrix format, these three equations are,

⎛ eˆ r ⎞ = ⎡ sin θ cosφ sinθ sinφ cosθ ⎤ ⎛ ˆi x ⎞
⎜ eˆθ ⎟ ⎢⎢cosθ cosφ cosθ sinφ − sinθ ⎥ ⎜ ˆi y ⎟
⎜ ⎟ ⎥ ⎜ ˆi z ⎟
⎜⎝ eˆφ ⎟⎠ ⎢⎣ −sinφ cosφ 0 ⎥⎦ ⎜⎜⎝ ⎟⎟⎠

So this matrix equation gives spherical basis in terms of the

Cartesian basis.

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Now for the inverse transformation that gives the Cartesian
basis in terms of the spherical basis. We can start with the now
familiar relation,

a = (a ⋅ ei )ei → ˆii = (ˆii ⋅ eˆ j )eˆ j
Then, for example,

ˆix = (ˆix ⋅ eˆr )eˆr + (ˆix ⋅ eˆθ )eˆθ + (ˆix ⋅ eˆφ )eˆφ

In matrix format,

⎛ ˆi x ⎞ ⎡sinθ cosφ cosθ cosφ − sin φ ⎤ ⎛ eˆ r ⎞
⎜ ˆi y ⎟ cosθ sinφ cosφ ⎥ ⎜ eˆθ ⎟
⎜ ˆi z ⎟ = ⎢ sinθ sin φ ⎥ ⎜ ⎟
⎜⎝⎜ ⎟⎟⎠ ⎢ − sinθ
⎣⎢ cosθ 0 ⎦⎥ ⎝⎜ eˆφ ⎟⎠

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Note that if we designate the coefficient matrix of the
transformation as R, then the inverse transformation coefficient
matrix is R−1 = RT Thus, R is orthogonal (AEM Sec. 3.5, 3.8).
It can be shown that any orthogonal transformation represents
a rotation (possibly combined with a reflection).
Physical components of an arbitrary vector a:

aˆ i = aˆi = hi a i = 1 ai (no summation),
hi
A convenient

aˆr = aˆr = ar = ar , consequence of using
spherical coordinates
aˆθ = aˆθ = raθ = aθ , is that the position
r arrow has a single
component, i.e.,
aˆφ = aˆφ = r sinθ aφ = aφ . r = reˆr .

r sinθ

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Curvilinear Systems: Cylindrical Coordinates

Now we examine this eˆ Z eˆφ
familiar curvilinear z eˆ R
coordinate system in
detail. R

r
Z

φ y

x 135

Vector Calculus & General Coordinate Systems

Transformation and inverse transformation:

q1 = R = x2 + y2 x = R cosφ
y = R sinφ
q2 = φ = tan −1 ⎛ y ⎞
⎜⎝ x ⎠⎟ z=Z

q3 = Z = z

Scale factors:

h1 = hR = ⎡⎣⎢⎢⎛⎝⎜ ∂x ⎞2 + ⎛ ∂y ⎞2 + ⎛ ∂z ⎞2 ⎤1/ 2 =1
∂R ⎟⎠ ⎜⎝ ∂R ⎠⎟ ⎜⎝ ∂R ⎠⎟ ⎥
⎦⎥

h2 = hφ = R h3 = hZ = 1

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Fundamental metric components:

g11 = h12 = 1 g11 = 1 =1 Again, since this is an
h12 orthogonal curvilinear
system,
g22 = h22 = R2 g 22 = 1 = 1
h22 R2 gij = gij = 0, i ≠ j

g33 = h32 = 1 g 33 = 1 =1
h32

Basis and dual:

ei = hi eˆ i (no summation)⎫ → eR = eˆ R eR = eˆ R
ej = g ijei ⎬ eφ = Reˆφ eφ = eˆφ
⎭ eZ = eˆ z
R
eZ = eˆ z

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In terms of the Cartesian basis:

eˆ i = 1 ∂x j ˆi j (no summation in i)
hi ∂qi

In matrix format, these three equations are:

⎛ eˆ R ⎞ ⎡ cosφ sin φ 0⎤ ⎛ ˆi x ⎞
⎜ ⎟ ⎢⎢− cosφ 0⎥⎥ ⎜ ˆi y ⎟
⎜⎜⎝ eˆφ ⎟⎠⎟ = ⎣⎢ sin φ 1⎦⎥ ⎜ ˆi z ⎟
eˆ Z 0 0 ⎝⎜⎜ ⎠⎟⎟

and the inverse transformation is:

⎛ ˆi x ⎞ ⎡cosφ − sin φ 0⎤ ⎛ eˆ R ⎞
⎜ ˆi y ⎟ cosφ ⎜ ⎟
⎜ ˆi z ⎟ = ⎢ sin φ 0⎥⎥ ⎝⎜⎜ eˆφ ⎟⎟⎠
⎜⎜⎝ ⎟⎟⎠ ⎢ 0 1⎥⎦ eˆ Z

⎣⎢ 0

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Vector Calculus & General Coordinate Systems

You can again verify that the coefficient matrix R is
orthogonal, i.e., R−1 = RT.

The physical components of an arbitrary vector a:

aˆ i = aˆi = hi a i = 1 ai (no summation),
hi

aˆ R = aˆR = aR = aR ,

aˆφ = aˆφ = Raφ = aφ ,
R

aˆ Z = aˆZ = aZ = aZ .

Finally, the position arrow is r = ReˆR + ZeˆZ .

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Vector Calculus & General Coordinate Systems

General Transformation between Two Curvilinear Systems
Up to this point we have explored how to transform to and
from the Cartesian system to a curvilinear system and, in
particular, the spherical and cylindrical systems. But what of
the situation where we need a transformation between
curvilinear systems, say, cylindrical to spherical or vice versa?

We now present the rules for doing these types of
transformations. Consider two curvilinear systems, and as
before denote them as the “unbarred” and “barred” systems.
The transformations and inverse transformations are written
(for i = 1, 2, 3) as

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Vector Calculus & General Coordinate Systems

qi = qi (q j ), q i = q i (q j ),

{e1,e2 ,e3}, {e1, e2 , e3},

dr = dqiei dr = dq i ei

Since the vector dr must be the same, regardless of the

coordinate system,

es ⋅ (dqiei = dq j ej )

dqiδ s = dq j (es ⋅ ej)
i

dqs = (es ⋅ ej )dq j .

Similarly, if we dot with the dual of the barred system,
e s ⋅ (dqiei = dq j ej )
(e s ⋅ ei )dqi = dq s.

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Vector Calculus & General Coordinate Systems

Recall from multivariable differential calculus, the chain rule
for a differential,

dq s = ∂q s dq j and dq s = ∂q s dqi .
∂q j ∂q j

By comparison with the covariant and contravariant

transformation laws in the Vector Algebra section, we see that,

es ⋅ ej = ∂q s ≡ α s and e s ⋅ei = ∂q s ≡ βis .
∂q j j ∂qi

We can now write the

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Vector Calculus & General Coordinate Systems

Covariant Transformation Law:

es = ∂q j ej as = ∂q j aj.
∂q s ∂q s

Contravariant Transformation Law:

e s = ∂q s ei as = ∂q s ai.
∂qi ∂qi

An example of the summations:

e1 = ∂q1 e1 + ∂q2 e2 + ∂q3 e3 ,
∂q 1 ∂q 1 ∂q 1

e1 = ∂q 1 e1 + ∂q 1 e2 + ∂q 1 e3.
∂q1 ∂q2 ∂q 3
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Vector Calculus & General Coordinate Systems

Note these transformation laws allow one to determine the
basis and dual basis of one system in terms of the other by
using the given transformation relations,

qi = qi (q1, q 2 , q 3 ) and q j = q j (q1, q2 , q3 )

An additional relation can also be shown,

⎛ es = ∂q j ej ⎞ ⋅ ei Look familiar? In fact, everything
⎜ ∂q s ⎟ we’ve done here is similar to the
⎝ ⎠ transformations we derived between
the curvilinear system and Cartesian.
δ i = ∂q j (e j ⋅ ei ) In fact, we could write the present
s ∂q s relations in matrix format just as we
did in the previous section.
δ i = ∂q j ∂q i
s ∂q s ∂q j 144

Vector Calculus & General Coordinate Systems

Actually, none of this should surprise you since the Cartesian
system is just a particular (albeit special for we humans)
curvilinear system.

Analytical Definition of a Vector
If the ordered triples,

(a1, a2 , a3 ) (a1, a2 , a3 )
⇓ ⇓

(q1, q2 , q3 ) (q1,q 2,q3)

satisfy

aj = ∂qi ai ,
∂q j

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Vector Calculus & General Coordinate Systems

then a j and a iare the covariant components of vector a.
Note: If qi and q j are rectangular Cartesian coordinates, the
covariant and contravariant are identical.
A vector quantity is independent of any coordinate system, thus
is invariant to a coordinate transformation

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Vector Calculus & General Coordinate Systems

x3 x2
x1 a1

x3 a3 a a 2 2

a

Example: Vector a a1
in two Cartesian a3
systems.
x2

x1

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Vector Calculus & General Coordinate Systems

Example: Non- P x3 x2
invariance of the r
position “vector” r
x1
x3 r0
x2
x1

The barred coordinate system is translated from the unbarred,
r = r0 + r

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Vector Calculus & General Coordinate Systems

Both are rectangular Cartesian thus,

ˆii = ˆii and xi = x i + x0i .
The contravariant transformation law gives, for some a,

ai = ∂x i aj = ∂x i aj = δ i a j = ai.
∂x j ∂x j j

Similarly, for these Cartesian systems, the covariant
components are,

ai = ai .
Thus, the components of vector a are unchanged by the
coordinate transformation. But!

x j ≠ x j since x j = x j + x0j .

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Vector Calculus & General Coordinate Systems

Because the tail of the position “vector” is, by definition,
located at the origin of the coordinate system, it is tied to that
origin. Therefore, the position “vector” is not invariant to a
coordinate translation. Consequently, the position “vector” r
is not a vector under a translation transformation.

Some ordered triples are vectors P
for certain types of
transformations, but not others. r
For instance, r transforms as a r
vector for a rotation
transformation.

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Vector Calculus & General Coordinate Systems

Derivatives of an Orthonormal Basis

An orthonormal basis vector triad can be viewed as a rotating
rigid body, i.e., the orientation of the triad may change, but the
vectors remain fixed with respect to each other.

eˆ 3 The eˆi have constant unit magnitude
ω δΦ but variable orientation.

Note we use δΦ instead of δΦ since

Φ is a finite rotation, thus is not a
vector.

eˆ1 dφ

dφ eˆ 2

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Vector Calculus & General Coordinate Systems

Recall, for rigid-body rotation,
v =ω×r

then,
dr = δΦ × r → dr = δΦ × r
dt dt

Since |r| = const for rigid-body rotation, we can separately look
at each of the eˆi using,

r = eˆi → deˆi = δΦ × eˆi , i = 1, 2,3 (8)

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Vector Calculus & General Coordinate Systems

Example: Cylindrical coordinates

In the cylindrical system, a change in position causes a rigid-

body rotation of the basis-vector triad if it involves the angle φ

(a rotation about the z-axis). Based on this, we develop
expressions for the differential changes.

eˆ Z eˆ Z eˆφ
eˆ R
eˆ R eˆφ
r1 r2

φ

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Vector Calculus & General Coordinate Systems

eˆ Z δΦ = dφeˆZ arc-length formula:

s = rφ for constant r,

ds = r dφ

dφ

eˆ R dφ eˆφ deˆφ = dφ (−eˆ R )
ds = Rdφ = dφ = −dφeˆ R

deˆ R = dφeˆφ

For this simple rotation about the z-axis, the differential
change due to the rotation is δΦ. In the cylindrical coordinate
system, we have the following for the differentials of the basis
vectors:

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Vector Calculus & General Coordinate Systems

deˆi = δΦ × eˆi ,

deˆ R = δΦ × eˆ R = dφ (eˆφ × eˆ R ) = dφeˆφ ,
deˆφ = δΦ × eˆφ = dφ (eˆZ × eˆφ ) = −dφeˆ R ,
deˆZ = δΦ × eˆZ = dφ (eˆZ × eˆZ ) = 0.

Example: Spherical coordinates eˆ r
eˆφ
This case is a bit more θ r1 r2 eˆθ
complex, since two angles eˆφ
eˆ r 155
(θ,φ) are involved. A eˆθ φ

change in position results in
a superposition of two
angular rotations.

Vector Calculus & General Coordinate Systems

eˆ z dθ eˆφ
eˆ r
dθ
dφ

dφ + eˆφ dφeˆ z

eˆφ = δΦ

dθ

dφ

eˆθ

δΦ = dφeˆ z + dθ eˆφ (9)

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Vector Calculus & General Coordinate Systems

The vector eˆ z is not a member of the spherical triad, so we
write it in terms of the spherical basis,

eˆ z = (eˆ z ⋅ eˆr )eˆr + (eˆ z ⋅ eˆθ )eˆθ + (eˆ z ⋅ eˆ
φ )eˆφ = cosθ eˆr − sinθ eˆθ .

⊥ → =0

The superposition of differential rotations, Eq. (9), becomes,

δΦ = dφ cosθ eˆr − dφ sinθ eˆθ + dθ eˆφ .

Employing Eq. (8), we write the differentials of the
orthonormal spherical basis vectors,

deˆi = δΦ × eˆi ,

deˆr = δΦ × eˆr = dφ sinθ eˆφ + dθ eˆθ ,
deˆθ = δΦ × eˆθ = dφ cosθ eˆφ − dθ eˆr ,
deˆφ = δΦ × eˆφ = −dφ cosθ eˆθ − dφ sinθ eˆr .

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Vector Calculus & General Coordinate Systems

We will define the curl vector-differential operator soon,
however we will introduce it now, a bit prematurely perhaps,
to state that for a general orthogonal system,
δΦ = 1 curl(dr) = 1 (∇ × dr).

22
This expression is obtained by taking the curl of the equation
v = ω × r (how this is done will become clear later). Here we
have introduced the gradient or del operator “∇” and the curl
operation.

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Vector Calculus & General Coordinate Systems

Curl of a Vector
The curl of an arbitrary vector a is written as
curl a ≡ ∇ × a

where ∇ (del, nabla) is a vector differential operator, the
general form of which we will define. For now, in an
orthogonal curvilinear system,

h1eˆ1 h2eˆ 2 h3eˆ 3
∂ ∂
curl(dr) ≡ ∇ × dr = 1 ∂ ∂q2 ∂q3
h1h2h3 ∂q1
h2 (h2dq2 ) h3 (h3dq3 )
h1 (h1dq1 )

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Vector Calculus & General Coordinate Systems

curl(dr) = eˆ1 ⎡ dq3 ∂ (h32 ) − dq2 ∂ (h22 ⎤
h2h3 ⎢ ∂q2 ∂q3 )⎥⎦
⎣

+ eˆ 2 ⎢⎡dq1 ∂ (h12 ) − dq3 ∂ (h32 ) ⎤
h1h3 ⎣ ∂q3 ∂q1 ⎥
⎦

+ eˆ 3 ⎡ 2 ∂ (h22 ) − dq1 ∂ (h12 ⎤
h1h2 ⎢dq ∂q1 ∂q2 )⎥
⎣ ⎦

Now, for example: deˆ1 = δΦ × eˆ1 = 1 (∇ × dr) × eˆ1
2

= eˆ 2 ⎡ dq 2 ∂ (h22 ) − dq1 ∂ (h12 ) ⎤
2h1h2 ⎢ ∂q1 ∂q2 ⎥
⎣ ⎦

+ eˆ 3 ⎢⎡dq3 ∂ (h32 ) − dq1 ∂ (h12 ⎤
h1h3 ⎣ ∂q1 ∂q3 )⎥⎦

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Vector Calculus & General Coordinate Systems

The differentials deˆ2 and deˆ3 are computed similarly.
Comparing these results to the total differential

deˆ i = ∂eˆ i dq j ,
∂q j

We can identify the partial derivatives of the orthonormal base
vectors

∂eˆ1 = − eˆ1 ∂h1 − eˆ 3 ∂h1 ,
∂q1 h2 ∂q2 h3 ∂q3

∂eˆ1 = − eˆ2 ∂h2 , ∂eˆ1 = eˆ3 ∂h3 .
∂q2 h1 ∂q1 ∂q3 h1 ∂q1

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Vector Calculus & General Coordinate Systems

∂eˆ 2 = eˆ1 ∂h1 , ∂eˆ 2 = − eˆ1 ∂h2 − eˆ 3 ∂h2 ,
∂q1 h2 ∂q2 ∂q2 h1 ∂q1 h3 ∂q3

∂eˆ2 = eˆ3 ∂h3 .
∂q3 h2 ∂q2

∂eˆ 3 = eˆ1 ∂h1 , ∂eˆ3 = eˆ 2 ∂h2 ,
∂q1 h3 ∂q3 ∂q2 h3 ∂q3

∂eˆ3 = − eˆ1 ∂h3 − eˆ2 ∂h3 .
∂q3 h1 ∂q1 h2 ∂q2

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Vector Calculus & General Coordinate Systems

Rotating Reference Frames

A primary application for the following analysis is in classical

mechanics (dynamics). For this problem of relative motion,

the unbarred frame represents a fixed (inertial) reference

frame. The rotating “barred” frame may also be translating

with respect to the fixed frame.

x3

x3 b P e3 ω

e3 r e1 x2
r0 x1 e2
e1
x1 e2 163
x2

Vector Calculus & General Coordinate Systems

We will show how the rotation of an observer in the barred

frame affects the measurement of the time rate of change of an
arbitrary vector quantity b and how this relates to an observer

in the fixed frame.

For simplicity, assume the{e1,e2 ,e3} are a constant (magnitude
and direction) basis in the nonrotating frame. Since the basis is
constant, the time derivative of a vector b is

db = dbi ei .
dt dt

In the rotating frame, the basis vectors {e1, e2 , e3} have constant
magnitude also, but variable orientation. So, for an observer in

the rotating frame,

db = db i ei + bi d ei .
dt dt dt

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Vector Calculus & General Coordinate Systems

Since the ei have constant magnitude, the rate of change is due

only from the rigid-body rotation of the frame, i.e.,

d ei = ω × ei.
dt

Note that because the rotating observer is rotating with the

barred coordinates, the observer does not detect a change in

orientation (to this observer, the inertial frame appears to be
rotating). We then define the time rate of change of b, as

observed by the observer in the rotating frame,

db i ei ≡ db .
dt dt
rot

Thus we have,

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Vector Calculus & General Coordinate Systems

db = db + b i (ω × ei )
dt dt
rot

= db + (ω × b).
dt rot

What does this mean? An observer in the inertial frame is not
rotating so sees the absolute derivative db/dt. The rotating
observer, however, sees the derivative (db/dt)rot and an
additional part due to the fact that the observer is rotating.

Now define a vector differential operator:

d = d +ω× . (10)
dt dt rot

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Vector Calculus & General Coordinate Systems

Velocity and Acceleration in Rotating Frames

Velocity

We now inspect the determination of velocity and acceleration

at point P with respect to a rotating coordinate frame. We
apply the vector differential operator (10) to r = r0 + r ,

dr = ⎛ d ω × ⎞ (r0 + r)
dt ⎜ dt ⎟
⎝ ⎠
rot

= dr0 + dr + ω × r.
dt dt rot

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Vector Calculus & General Coordinate Systems

where,
dr = absolute velocity,
dt
dr0 = absolute velocity of the origin of the rotating frame
dt
dr = velocity of object at P with respect to rotating frame
dt rot
ω × r = velocity of point P in the rotating frame.

Note that when dr0/dt = 0 and (dr / dt)rot = 0 , the point P is
fixed with respect to the rotating frame and we have a rigid-
body rotation with respect to the inertial frame.

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Vector Calculus & General Coordinate Systems

Acceleration
Applying the operator (10) to the velocity, we determine the
acceleration,

d 2r = ⎛ d ω × ⎞ ⎛ dr0 + dr + ω×r ⎞
dt 2 ⎜ dt ⎟ ⎜ dt dt ⎟
⎝ ⎠⎝ ⎠
rot rot

= d 2r0 + d 2r + d ω×r +ω× dr + ω × (ω × r)
dt 2 dt 2 dt dt
rot rot rot

= d 2r0 + d 2r + dω × r + 2ω × dr + ω × (ω × r)
dt 2 dt 2 dt dt
rot rot

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Vector Calculus & General Coordinate Systems

where,

d 2r = absolute acceleration
dt 2

d 2r0 = absolute acceleration of the origin of the rotating frame
dt 2

dω × r = ⎧tangential component of acceleration in the plane
dt ⎨⎩of r and (dr / dt)rot and perpendicular to r

2ω × dr = Coriolis acceleration
dt rot

ω × (ω × r) = centripetal acceleration

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Vector Calculus & General Coordinate Systems

Example:

Y Given: At the instant shown,
B
ωOAB = −20 rad/s ˆj
A
αOAB = dωOAB = −200 rad/s2 ˆj
8″ 30° dt
D
The velocity and acceleration of D,

relative to the rod, are 50 in/s and 600

in/s2 upward, respectively.

X
O
Z Find: The velocity and acceleration of

the collar D.

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Vector Calculus & General Coordinate Systems

Solution:
Note that for this problem, r = r . First find position and
velocity at D,
r = (8 in)(sin30° ˆi + cos30° ˆj) = (4 in)ˆi + (6.93 in)ˆj
vD = (vD )OAB + ω × r
The two parts of the velocity are

(vD )OAB = (50 in/s)(sin30° ˆi + cos30° ˆj)
= (25 in/s) ˆi + (43.3in/s) ˆj

ω × r = (−20 rad/s) ˆj×[(4 in) ˆi + (6.93 in) ˆj]
= 80 in/s kˆ

vD = (25 in/s) ˆi + (43.3 in/s)ˆj + 80 in/s kˆ ⇐

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Vector Calculus & General Coordinate Systems

Now the acceleration,

aD = d 2r0 + d2r + dω ×r + 2ω × (vD )OAB + ω × (ω ×r)
dt 2 dt 2 dt
OAB

The individual terms are

d 2r0 = (600 in/s2 )(sin30° ˆi + cos30° ˆj)
dt 2
OAB

= (300 in/s2 ) ˆi + (520 in/s2 ) ˆj

dω × r = (−200 rad/s2 ) ˆj×[(4 in) ˆi + (6.93 in) ˆj]
dt

= (800 in/s2 ) kˆ

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Vector Calculus & General Coordinate Systems

2ω × (vD )OAB = 2(−20 rad/s) ˆj×[(25 in/s ˆi + (43.3 in/s) ˆj]
= (1000 in/s2 ) kˆ

ω × (ω × r) = (−20 rad/s) ˆj× (80 in/s) kˆ
= (−1600 in/s2 ) ˆi

aD = (−1300 in/s2 ) ˆi + (520 in/s2 ) ˆj× (1800 in/s2 ) kˆ ⇐

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