126 Vector Calculus & General Coordinate Systems 1 2 3 1 2 3 1 2 2 2 1 2 3 1 cos , sin , ( ) ( ) 2 x x x θθ θ θθ θ θ θ ⎡ ⎤ = = = − ⎣ ⎦ Homework #4, Prob 2:
126
Vector Calculus & General Coordinate Systems
1 2 3 1 2 3 1 2 2 21 2 3
1cos , sin , ( ) ( )2
x x xθ θ θ θ θ θ θ θ⎡ ⎤= = = −⎣ ⎦
Homework #4, Prob 2:
127
Vector Calculus & General Coordinate Systems
128
Curvilinear Systems: Spherical CoordinatesThe 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,
Vector Calculus & General Coordinate Systems
1 2 2 2
2 1
2 2 2
3 1
cos
tan
q r x y z
zqx y z
yqx
θ
φ
−
−
= = + +
⎛ ⎞= = ⎜ ⎟
⎜ ⎟+ +⎝ ⎠⎛ ⎞= = ⎜ ⎟⎝ ⎠
sin cossin sincos
x ry rz r
θ φθ φθ
===
129
and scale factors and fundamental metric components are
Vector Calculus & General Coordinate Systems
1/ 22 2 2
1
1/ 22 2 2
2
3
(sin cos ) (sin cos ) cos 1
sin
rx y zh hr r r
h h rh h r
θ
φ
θ φ θ φ θ
θ
⎡ ⎤∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
⎡ ⎤= + + =⎣ ⎦= == =
2 1111 1 2
1
2 2 2222 2 2 2
2
2 2 2 3333 3 2 2 2
3
11 1
1 1
1 1sinsin
g h gh
g h r gh r
g h r gh r
θθ
= = = =
= = = =
= = = =
Since this is an orthogonal curvilinear system,
gij = gij = 0, i ≠ j
130
Vector Calculus & General Coordinate Systems
φ
θ
rr = const
= const
= constφ
θ θ
r
φx
y
z ˆ re
ˆφe
ˆθe
131
Spherical coordinates basis and dual basis:
We can now write the basis and dual basis,
Vector Calculus & General Coordinate Systems
ˆ (no summation)i i i
j iji
h
g
=
=
e e
e e
ˆ ˆˆˆ
ˆˆsin
sin
rr r r
rr
rr
θ θθ θ
φφφ φθ
θ
= =
= =
= =
e e e eee e e
ee e e
132
Basis vectors in terms of the Cartesian basis:
In matrix format, these three equations are,
So this matrix equation gives spherical basis in terms of the Cartesian basis.
Vector Calculus & General Coordinate Systems
1 ˆˆ (no summation in )
1 1 1ˆ ˆ ˆˆ ˆ ˆ, ,
j
i jii
j j j
r j j jrr
x ih q
x x xh q h q h qθ φθ φ
θ φ
∂=
∂
∂ ∂ ∂= = =
∂ ∂ ∂
e i
e i e i e i
ˆˆ sin cos sin sin cosˆˆ cos cos cos sin sin
ˆ ˆsin cos 0
xr
y
z
θ
φ
θ φ θ φ θθ φ θ φ θ
φ φ
⎛ ⎞⎛ ⎞ ⎡ ⎤ ⎜ ⎟⎜ ⎟ ⎢ ⎥= − ⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎜ ⎟ −⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠
iee ie i
133
Now for the inverse transformation that gives the Cartesian basis in terms of the spherical basis. We can start with the now familiar relation,
Then, for example,
In matrix format,
Vector Calculus & General Coordinate Systems
ˆ ˆ ˆ ˆ( ) ( )ii i i j j= ⋅ → = ⋅a a e e i i e e
ˆ ˆsin cos cos cos sinˆ ˆsin sin cos sin cosˆ ˆcos sin 0
x r
y
z
θ
φ
θ φ θ φ φθ φ θ φ φθ θ
⎛ ⎞ ⎛ ⎞−⎡ ⎤⎜ ⎟ ⎜ ⎟⎢ ⎥=⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟−⎢ ⎥⎜ ⎟ ⎣ ⎦ ⎝ ⎠⎝ ⎠
i ei e
ei
ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( )x x r r x xθ θ φ φ= ⋅ + ⋅ + ⋅i i e e i e e i e e
134
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:
Vector Calculus & General Coordinate Systems
1ˆ ˆ (no summation),
ˆ ˆ ,
ˆ ˆ ,
ˆ ˆ sin .sin
i ii i i
i
r rr r
a a h a ah
a a a aaa a rar
aa a r a
r
θ θ θθ
φφ φφ θ
θ
= = =
= = =
= = =
= = =
A convenient consequence of using spherical coordinates is that the position arrow has a single component, i.e.,
ˆ .rr=r e
135
Curvilinear Systems: Cylindrical Coordinates
Vector Calculus & General Coordinate Systems
r
φ
x
y
z
R
Z
Now we examine this familiar curvilinear coordinate system in detail.
ˆZe
ˆφe
ˆRe
136
Transformation and inverse transformation:
Scale factors:
Vector Calculus & General Coordinate Systems
1 2 2
2 1
3
tan
q R x y
yqx
q Z z
φ −
= = +
⎛ ⎞= = ⎜ ⎟⎝ ⎠
= =
cossin
x Ry Rz Z
φφ
===
1/ 22 2 2
1
2 3
1
1
R
Z
x y zh hR R R
h h R h hφ
⎡ ⎤∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦= = = =
137
Fundamental metric components:
Basis and dual:
Vector Calculus & General Coordinate Systems
2 1111 1 2
1
2 2 2222 2 2 2
2
2 3333 3 2
3
11 1
1 1
11 1
g h gh
g h R gh R
g h gh
= = = =
= = = =
= = = =
Again, since this is an orthogonal curvilinear system,
gij = gij = 0, i ≠ j
ˆ (no summation)i i ij ij
i
h
g
= ⎫→⎬
= ⎭
e e
e e
ˆ ˆˆ
ˆ
ˆ ˆ
RR R R
ZZ z z
RRφφ
φ φ
= =
= =
= =
e e e ee
e e e
e e e e
138
In terms of the Cartesian basis:
In matrix format, these three equations are:
and the inverse transformation is:
Vector Calculus & General Coordinate Systems
1 ˆˆ (no summation in )j
i jii
x ih q∂
=∂
e i
ˆˆ cos sin 0ˆˆ sin cos 0
ˆ ˆ0 0 1
xR
y
Z z
φ
φ φφ φ
⎛ ⎞⎛ ⎞ ⎡ ⎤ ⎜ ⎟⎜ ⎟ ⎢ ⎥= − ⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎝ ⎠ ⎣ ⎦ ⎝ ⎠
iee ie i
ˆ ˆcos sin 0ˆ ˆsin cos 0ˆ ˆ0 0 1
x R
y
Zz
φ
φ φφ φ
⎛ ⎞ −⎡ ⎤⎛ ⎞⎜ ⎟ ⎜ ⎟⎢ ⎥=⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎣ ⎦ ⎝ ⎠⎝ ⎠
i ei e
ei
139
You can again verify that the coefficient matrix R is orthogonal, i.e., R−1 = RT.
The physical components of an arbitrary vector a:
Finally, the position arrow is
Vector Calculus & General Coordinate Systems
1ˆ ˆ (no summation),
ˆ ˆ ,
ˆ ˆ ,
ˆ ˆ .
i ii i i
iR R
R R
Z ZZ Z
a a h a ah
a a a aa
a a RaR
a a a a
φφ φφ
= = =
= = =
= = =
= = =
ˆ ˆ .R ZR Z= +r e e
140
General Transformation between Two Curvilinear SystemsUp 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
Vector Calculus & General Coordinate Systems
141
Since the vector dr must be the same, regardless of the coordinate system,
Similarly, if we dot with the dual of the barred system,
Vector Calculus & General Coordinate Systems
1 2 3
( ),{ , , },
i i j
ii
q q q
d dq
=
=
e e e
r e1 2 3
( ),{ , , },
i i j
ii
q q q
d dq
=
=
e e e
r e
( )
( )
( ) .
s i ji j
i s j si j
s s jj
dq dq
dq dq
dq dq
δ
⋅ =
= ⋅
= ⋅
e e e
e e
e e
( )
( ) .
s i ji j
s i si
dq dq
dq dq
⋅ =
⋅ =
e e e
e e
142
Recall from multivariable differential calculus, the chain rule for a differential,
By comparison with the covariant and contravarianttransformation laws in the Vector Algebra section, we see that,
We can now write the
Vector Calculus & General Coordinate Systems
and .s s
s j s ij j
q qdq dq dq dqq q∂ ∂
= =∂ ∂
and .s s
s s s sj j i ij i
q qq q
α β∂ ∂⋅ = ≡ ⋅ = ≡
∂ ∂e e e e
143
Covariant Transformation Law:
Contravariant Transformation Law:
An example of the summations:
Vector Calculus & General Coordinate Systems
.j j
s j s js s
q qa aq q∂ ∂
= =∂ ∂
e e
.s s
s i s ii i
q qa aq q
∂ ∂= =∂ ∂
e e
1 2 3
1 1 2 31 1 1
1 1 11 1 2 3
1 2 3
,
.
q q qq q qq q qq q q
∂ ∂ ∂= + +∂ ∂ ∂
∂ ∂ ∂= + +∂ ∂ ∂
e e e e
e e e e
144
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,
An additional relation can also be shown,
Vector Calculus & General Coordinate Systems
1 2 3 1 2 3( , , ) and ( , , )i i j jq q q q q q q q q q= =
( )
ji
s js
ji is js
j iis s j
qqq qq q
δ
δ
⎛ ⎞∂= ⋅⎜ ⎟∂⎝ ⎠∂
= ⋅∂
∂ ∂=∂ ∂
e e e
e e
Look familiar? In fact, everything we’ve done here is similar to the transformations we derived between the curvilinear system and Cartesian. In fact, we could write the present relations in matrix format just as we did in the previous section.
145
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,
satisfy
Vector Calculus & General Coordinate Systems
1 2 3 1 2 3
1 2 3 1 2 3
( , , ) ( , , )
( , , ) ( , , )
a a a a a a
q q q q q q⇓ ⇓
,i
j ij
qa aq∂
=∂
146
then and are the covariant components of vector a.
Note: If and 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
Vector Calculus & General Coordinate Systemsja ia
iq jq
147
Vector Calculus & General Coordinate Systems
1a
2x
1x
3x 2a
3a
1x2x
3x
1a
3a 2a
Example: Vector ain two Cartesian systems.
a
148
The barred coordinate system is translated from the unbarred,
Vector Calculus & General Coordinate Systems
2x
3x
1x
2x
3xExample: Non-invariance of the position “vector” P
1x
rr
0r
0= +r r r
149
Both are rectangular Cartesian thus,
The contravariant transformation law gives, for some a,
Similarly, for these Cartesian systems, the covariant components are,
Thus, the components of vector a are unchanged by the coordinate transformation. But!
Vector Calculus & General Coordinate Systems
0ˆ ˆ and .i i ii i x x x= = +i i
.i i
i j j i j ijj j
x xa a a a ax x
δ∂ ∂= = = =∂ ∂
.i ia a=
0 since .j j j j jx x x x x≠ = +
150
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.
Vector Calculus & General Coordinate Systems
PSome ordered triples are vectors for certain types of transformations, but not others. For instance, r transforms as a vector for a rotation transformation.
rr
151
Derivatives of an Orthonormal BasisAn 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.
Vector Calculus & General Coordinate Systems
The have constant unit magnitude but variable orientation.
Note we use δΦ instead of δΦ since Φ is a finite rotation, thus is not a vector.
1e
2e
3e
dφ dφ
ω δΦ
ˆ ie
152
Recall, for rigid-body rotation,
then,
Since |r| = const for rigid-body rotation, we can separately look at each of the using,
(8)
Vector Calculus & General Coordinate Systems
= ×v ω r
d ddt dt
= × → = ×r δΦ r r δΦ r
ˆ ie
ˆ ˆ ˆ , 1,2,3i i id i= → = × =r e e δΦ e
153
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.
Vector Calculus & General Coordinate Systems
ˆZe
φ
ˆZe
ˆRe ˆφe
ˆφe
ˆRe
1r 2r
154
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 basisvectors:
Vector Calculus & General Coordinate SystemsˆZe
ˆRe ˆφedφ
dφ
arc-length formula:s = rφ for constant r, ds = r dφ
ˆZdφ=δΦ e
ˆ ˆ( )ˆ
R
R
d d
dφ φ
φ
= −
= −
e e
e
ˆ ˆR
ds Rd dd d φ
φ φφ
= ==e e
155
Example: Spherical coordinates
Vector Calculus & General Coordinate Systems
ˆ ˆ ,ˆ ˆ ˆ ˆ ˆ( ) ,ˆ ˆ ˆ ˆ ˆ( ) ,ˆ ˆ ˆ ˆ( ) .
i i
R R R
Z R
Z Z Z Z
dd d d
d d d
d d
φ φ
φ φ φ
φ φ
φ φ
φ
= ×= × = × =
= × = × = −
= × = × =
e δΦ ee δΦ e e e e
e δΦ e e e e
e δΦ e e e 0
This case is a bit more complex, since two angles (θ,φ) are involved. A change in position results in a superposition of two angular rotations.
θ
φˆθe
ˆ re
ˆφe
ˆ reˆφe
ˆθe1r
2r
156
(9)
Vector Calculus & General Coordinate Systems
ˆ ze
ˆ re
ˆφe ˆ zdφe
ˆφe
ˆθe
ˆd φθe
dθ
dθ
dφ
dφ
dφ + =δΦ
ˆ ˆzd d φφ θ= +δΦ e e
157
The vector is not a member of the spherical triad, so we write it in terms of the spherical basis,
The superposition of differential rotations, Eq. (9), becomes,
Employing Eq. (8), we write the differentials of the orthonormal spherical basis vectors,
Vector Calculus & General Coordinate Systemsˆ ze
0
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) cos sin .z z r r z z rθ θ φ φ θθ θ⊥→=
= ⋅ + ⋅ + ⋅ = −e e e e e e e e e e e e
ˆ ˆ ˆcos sin .rd d dθ φφ θ φ θ θ= − +δΦ e e e
ˆ ˆ ,ˆ ˆ ˆ ˆsin ,ˆ ˆ ˆ ˆcos ,ˆ ˆ ˆ ˆcos sin .
i i
r r
r
r
dd d dd d d
d d d
φ θ
θ θ φ
φ φ θ
φ θ θ
φ θ θ
φ θ φ θ
= ×= × = +
= × = −
= × = − −
e δΦ ee δΦ e e ee δΦ e e e
e δΦ e e e
158
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,
This expression is obtained by taking the curl of the equation(how this is done will become clear later). Here we
have introduced the gradient or del operator “∇” and the curl operation.
Vector Calculus & General Coordinate Systems
1 1curl( ) ( ).2 2
d d= = ∇×δΦ r r
= ×v ω r
159
Curl of a VectorThe curl of an arbitrary vector a is written as
where ∇ (del, nabla) is a vector differential operator, the general form of which we will define. For now, in an orthogonal curvilinear system,
Vector Calculus & General Coordinate Systems
1 1 2 2 3 3
1 2 31 2 3
1 2 31 1 2 2 3 3
ˆ ˆ ˆ1curl( )
( ) ( ) ( )
h h h
d dh h h q q q
h h dq h h dq h h dq
∂ ∂ ∂≡ ∇× =
∂ ∂ ∂
e e e
r r
curl ≡ ∇×a a
160
Now, for example:
Vector Calculus & General Coordinate Systems3 2 2 21
3 22 32 3
1 2 3 221 33 1
1 3
2 2 1 232 11 2
1 2
ˆcurl( ) ( ) ( )
ˆ( ) ( )
ˆ( ) ( )
d dq h dq hh h q q
dq h dq hh h q q
dq h dq hh h q q
⎡ ⎤∂ ∂= −⎢ ⎥∂ ∂⎣ ⎦
⎡ ⎤∂ ∂+ −⎢ ⎥∂ ∂⎣ ⎦
⎡ ⎤∂ ∂+ −⎢ ⎥∂ ∂⎣ ⎦
er
e
e
1 1 1
2 2 1 222 11 2
1 2
3 2 1 233 11 3
1 3
1ˆ ˆ ˆ( )2
ˆ( ) ( )
2
ˆ( ) ( )
d d
dq h dq hh h q q
dq h dq hh h q q
= × = ∇× ×
⎡ ⎤∂ ∂= −⎢ ⎥∂ ∂⎣ ⎦
⎡ ⎤∂ ∂+ −⎢ ⎥∂ ∂⎣ ⎦
e δΦ e r e
e
e
161
The differentials and are computed similarly. Comparing these results to the total differential
We can identify the partial derivatives of the orthonormal base vectors
Vector Calculus & General Coordinate Systems2ˆde 3ˆde
ˆˆ ,jii jd dq
q∂
=∂
ee
31 1 1 11 2 3
2 3
3 31 2 2 12 1 3 1
1 1
ˆˆ ˆ,
ˆˆ ˆ ˆ, .
h hq h q h q
hhq h q q h q
∂ ∂ ∂= − −
∂ ∂ ∂∂∂ ∂ ∂
= − =∂ ∂ ∂ ∂
ee e
ee e e
162
Vector Calculus & General Coordinate Systems
3 31 1 2 21 3 2 3
3 3
3 3 31 23 1 2
1 2
ˆ ˆˆ ˆ, ,
ˆ ˆ ˆ.
h hq h q q h q
h hq h q h q
∂ ∂∂ ∂= =
∂ ∂ ∂ ∂∂ ∂ ∂
= − −∂ ∂ ∂
e ee e
e e e
32 1 1 2 1 2 21 2 2 1 3
2 1 3
3 323 2
2
ˆˆ ˆ ˆ ˆ, ,
ˆˆ.
h h hq h q q h q h q
hq h q
∂ ∂ ∂ ∂ ∂= = − −
∂ ∂ ∂ ∂ ∂∂∂
=∂ ∂
ee e e e
ee
163
Rotating Reference Frames A primary application for the following analysis is in classicalmechanics (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.
Vector Calculus & General Coordinate Systems
Px3
x2x1
1x
2x
3xb
r
0r
1e
3e
2e
2e1e
3e ω
164
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 are a constant (magnitude and direction) basis in the nonrotating frame. Since the basis is constant, the time derivative of a vector b is
In the rotating frame, the basis vectors have constant magnitude also, but variable orientation. So, for an observer in the rotating frame,
Vector Calculus & General Coordinate Systems
1 2 3{ , , }e e e
1 2 3{ , , }e e e
.i
id dbdt dt
=b e
.i
i ii
dd db bdt dt dt
= +eb e
165
Since the have constant magnitude, the rate of change is dueonly from the rigid-body rotation of the frame, i.e.,
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,
Thus we have,
Vector Calculus & General Coordinate Systems
rot
.i
idb ddt dt
≡be
.ii
ddt
= ×e ω e
ie
166
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:
(10)
Vector Calculus & General Coordinate Systems
rot
.d ddt dt
= + ×ω
rot
rot
( )
( ).
ii
d d bdt dt
ddt
= + ×
= + ×
b b ω e
b ω b
167
Velocity and Acceleration in Rotating FramesVelocityWe 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 ,
Vector Calculus & General Coordinate Systems
0rot
0
rot
( )
.
d ddt dt
d ddt dt
⎛ ⎞= × +⎜ ⎟⎝ ⎠
= + + ×
r ω r r
r r ω r
0= +r r r
168
where,
Note that when dr0/dt = 0 and , the point P is fixed with respect to the rotating frame and we have a rigid-body rotation with respect to the inertial frame.
Vector Calculus & General Coordinate Systems
0
rot
absolute velocity,
absolute velocity of the origin of the rotating frame
velocity of object at with respect to rotating frame
velocity of point in the rotating frame.
ddtddtd Pdt
P
=
=
=
× =
r
r
r
ω r
rot( / )d dt =r 0
169
AccelerationApplying the operator (10) to the velocity, we determine the acceleration,
Vector Calculus & General Coordinate Systems
20
2rot rot
2 20
2 2rot rotrot
2 20
2 2rotrot
( )
2 ( )
dd d ddt dt dt dt
d d d ddt dt dt dt
d d d ddt dt dt dt
⎛ ⎞⎛ ⎞= × + + ×⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + + × + × + × ×
= + + × + × + × ×
rr rω ω r
r r rω r ω ω ω r
r r ω rr ω ω ω r
170
where,
Vector Calculus & General Coordinate Systems
2
2
20
2
rot
absolute acceleration
absolute acceleration of the origin of the rotating frame
tangential component of acceleration in the plane
of and ( / ) and perpendicular to
2
ddtddtd
d dtdt
=
=
⎧× = ⎨
⎩
×
r
r
ω rr r r
ωrot
Coriolis acceleration
( ) centripetal acceleration
ddt
=
× × =
r
ω ω r
171
Example:
Vector Calculus & General Coordinate Systems
D8″ 30°
AB
Y
X
ZO
Given: At the instant shown,
The velocity and acceleration of D, relative to the rod, are 50 in/s and 600 in/s2 upward, respectively.
Find: The velocity and acceleration of the collar D.
2
ˆ20 rad/s
ˆ200 rad/s
OAB
OABOAB
ddt
= −
= = −
ω jωα j
172
Solution: Note that for this problem, . First find position and velocity at D,
The two parts of the velocity are
Vector Calculus & General Coordinate Systems
ˆ ˆ ˆ ˆ (8 in)(sin30 + cos30 ) (4 in) + (6.93 in) ( )D D OAB
= ° ° == + ×
r i j i jv v ω r
=r r
ˆ ˆ( ) (50 in/s)(sin30 + cos30 )ˆ ˆ(25 in/s) (43.3in/s)
ˆ ˆ ˆ( 20 rad/s) [(4 in) (6.93 in) ]ˆ80 in/s
ˆ ˆ ˆ(25 in/s) (43.3 in/s) 80 in/s
D OAB
D
= ° °
= +
× = − × +
=
= + + ⇐
v i j
i j
ω r j i j
k
v i j k
173
Now the acceleration,
The individual terms are
Vector Calculus & General Coordinate Systems
220
2
2 2
2
2
ˆ ˆ(600 in/s )(sin30 + cos30 )
ˆ ˆ(300 in/s ) (520 in/s )
ˆ ˆ ˆ( 200 rad/s ) [(4 in) (6.93 in) ]
ˆ(800 in/s )
OAB
ddt
ddt
= ° °
= +
× = − × +
=
r i j
i jω r j i j
k
20
2Dddt
=ra
2
2 2 ( ) ( )D OABOAB
d ddt dt
+ + × + × + × ×r ω r ω v ω ω r
174
Vector Calculus & General Coordinate Systems
2
2
2 2 2
ˆ ˆ ˆ2 ( ) 2( 20 rad/s) [(25 in/s (43.3 in/s) ]ˆ(1000 in/s )
ˆ ˆ( ) ( 20 rad/s) (80 in/s) ˆ( 1600 in/s )
ˆ ˆ ˆ( 1300 in/s ) (520 in/s ) (1800 in/s )
D OAB
D
× = − × +
=
× × = − ×
= −
= − + × ⇐
ω v j i j
k
ω ω r j k
i
a i j k