26
Chapter 2. Kinematics
Question 2–11
A rod of length L with a wheel of radius R attached to one of its ends is rotating
about the vertical axis OA with a constant angular velocity Ω relative to a fixed
reference frame as shown in Fig. P2-11. The wheel is vertical and rolls without
slip along a fixed horizontal surface. Determine the angular velocity and angular
acceleration of the wheel as viewed by an observer in a fixed reference frame.
Ω
B
R
L
O
A
Figure P 2-11
Solution to Question 2–11
Let F be a reference frame fixed to the ground. Then choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along OB when t = 0
Along Ω
Ez × Ex
Next, let A be a reference frame fixed to the direction of OB. Then choose the
following coordinate system fixed in reference frame A:
ex
ez
ey
Origin at O
=
=
=
Along OB
Along Ω
ez × ex
Finally, let D be the reference frame of the wheel. Then choose the following
coordinate system fixed in reference frame D:
ux
uy
uz
Origin at B
=
=
=
Along OB
In the plane of the wheel
In the plane of the wheel = ux × uy
27
Now the angular velocity of the arm OB as viewed by an observer fixed to the
ground, denoted F ωA , is given as
F
ωA = Ω = Ωez
(2.136)
Next, the position of point B is given as
rB = Lex
(2.137)
Computing the rate of change of rB in reference frame F , we obtain the velocity
of point B in reference frame F as
F
vB =
A
drB F A
drB
=
+ ω × rB
dt
dt
F
(2.138)
Now we have
A
drB
dt
F A
ω × rB
= 0
(2.139)
= Ωez × Lex = LΩey
(2.140)
Consequently,
F
vB = LΩey
(2.141)
Next, suppose we let Q be the point of contact of the wheel with the ground.
Then, because the wheel rolls without slip along the ground, we know that
F D
vQ
=0
(2.142)
Then, using Eq. (2–517) on page 107, we can obtain a second expression for F vB
as
F
F D
vB = F vD
(2.143)
Q + ω × (rB − rQ )
Now we know that F ωD is given from the angular velocity addition theorem as
F
ωD = F ωA + AωD
(2.144)
We already have F ωA from earlier. Then, because the wheel rotates about the
ex -direction (≡ ux -direction) and ex = ux is fixed in reference frame A, we have
A
ωD = ωux
(2.145)
where ω is to be determined. Adding Eqs. (2.136) and (2.145), we obtain
F
Also, rB − rQ is given as
ωD = Ωez + ωux
rB − rQ = REz = Rez
(2.146)
(2.147)
28
Chapter 2. Kinematics
Therefore,
F
vB = (Ωez + ωex ) × Rez = −Rωey
(2.148)
Setting the expressions for F vB from Eqs. (2.148) and (2.148) equal, we obtain
LΩ = −Rω
(2.149)
from which we obtain ω as
L
(2.150)
ω=− Ω
R
The angular velocity of the wheel as viewed by an observer fixed to the ground
is then given as
L
D F
(2.151)
ω = Ωez − Ωex
R
The angular acceleration of the wheel in reference frame F is then given as
F
αD =
d F D Fd F A Fd A D ω
=
ω
+
ω
dt
dt
dt
F
(2.152)
Now, because F ωA = Ωez = ΩEz and Ez is fixed in reference frame F , we have
F
d F A ω
= Ω̇ez = 0
dt
(2.153)
because Ω is constant. Next, because AωD = −(L/R)Ωex and ex is fixed in reference frame A, we can apply the transport theorem to AωD between reference
frames A and F as
d A D Ad A D F A A D
ω
=
ω
+ ω × ω
dt
dt
F
(2.154)
Now we have
A
L
d A D ω
= − Ω̇ex = 0
dt
R
L
L
F A
A D
ω × ω
= Ωez × − Ωex = − Ω2 ey
R
R
(2.155)
(2.156)
where we have again used the fact that Ω is constant. Therefore,
F
d A D L
ω
= − Ω2 ey
dt
R
(2.157)
Consequently, the angular acceleration of the disk as viewed by an observer
fixed to the ground is given as
F
L
αD = − Ω2 ey
R
(2.158)
29
Question 2–13
A collar is constrained to slide along a track in the form of a logarithmic spiral
as shown in Fig. P2-13. The equation for the spiral is given as
r = r0 e−aθ
where r0 and a are constants and θ is the angle measured from the horizontal
direction. Determine (a) expressions for the intrinsic basis vectors et , en , and
eb in terms any other basis of your choosing, (b) the curvature of the trajectory
as a function of the angle θ, and (c) the velocity and acceleration of the collar as
viewed by an observer fixed to the track.
r
=
r0
e −a
θ
P
θ
O
Figure P 2-13
Solution to Question 2–13
(a) Intrinsic Basis
Let F be a reference frame fixed to the track. Then, choose the following coordinate system fixed in reference frame F :
Origin at O
Ex To the Right
Ez Out of Page
Ey = Ez × Ex
Next, let A be a reference frame that rotates with the direction along Om. Then,
choose the following coordinate system fixed in reference frame A:
Origin at O
er Along Om
Ez Out of Page
eθ = Ez × er
30
Chapter 2. Kinematics
The position of the particle is given in terms of the basis {er , eθ , ez } as
r = r er = r0 e−aθ er
(2.159)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇Ez
(2.160)
Applying the rate of change transport theorem between reference frame A and
reference frame F , the velocity of the particle in reference frame F is given as
F
v=
F
dr Adr F A
=
+ ω ×r
dt
dt
(2.161)
where
A
dr
= ṙ er = −ar0 θ̇e−aθ er
dt
F A
ω × r = θ̇ez × r er = θ̇ez × r0 e−aθ er = r0 θ̇e−aθ eθ
(2.162)
(2.163)
Adding Eq. (2.162) and Eq. (2.163), we obtain the velocity of the particle in reference frame F as
F
v = −ar0 θ̇e−aθ er + r0 θ̇e−aθ eθ
(2.164)
Simplifying Eq. (2.161), we obtain F v as
F
v = r0 θ̇e−aθ [−aer + eθ ]
(2.165)
The tangent vector in reference frame F is then given as
et =
Fv
kF vk
=
Fv
kF vk
(2.166)
where F v is the speed of the particle in reference frame F . Now the speed of
the particle in reference frame F is given as
p
F
v = kF vk = r0 θ̇e−aθ 1 + a2
(2.167)
Dividing F v in Eq. (2.165) by
reference frame F as
Fv
in Eq. (2.167), we obtain the tangent vector in
−aer + eθ
et = √
1 + a2
(2.168)
Next, the principle unit normal vector is obtained as
en =
F
F
det /dt
k det /dtk
(2.169)
31
Now we have from the rate of change transport theorem that
F
A
det F A
det
=
+ ω × et
dt
dt
(2.170)
where
F
F
det
dt
ωA × et
= 0
(2.171)
−aer + eθ
er + aeθ
= θ̇ez × √
= −θ̇ √
1 + a2
1 + a2
(2.172)
Adding Eq. (2.171) and Eq. (2.172), we obtain
F
det
er + aeθ
= −θ̇ √
dt
1 + a2
(2.173)
Consequently,
F
Dividing
F
det
dt
= θ̇
(2.174)
det /dt in Eq. (2.173) by kF det /dtk in Eq. (2.174), we obtain en as
er + aeθ
en = − √
1 + a2
(2.175)
Finally, the bi-normal vector is obtained as
er + aeθ
−aer + eθ
×−√
= Ez
eb = et × en = √
1 + a2
1 + a2
(2.176)
(b) Curvature
The curvature of the trajectory in reference frame F is then obtained as
κ=
kF det /dtk
Fv
(2.177)
Substituting kF det /dtk from Eq. (2.174) and F v from Eq. (2.167), we obtain κ
as
1
√
κ=
(2.178)
−aθ
r0 e
1 + a2
(c) Velocity and Acceleration
The velocity of the particle in reference frame F can be expressed in the intrinsic
basis as
F
v = F vet
(2.179)
32
Chapter 2. Kinematics
Using the expression for F v from Eq. (2.167), we obtain
p
F
v = r0 θ̇e−aθ 1 + a2 et
(2.180)
Next, the acceleration of the particle in reference frame F can be expressed in
terms of the intrinsic basis as
F
Now we have that
Furthermore,
κ
F
v
2
=
a=
2
d F v et + κ F v en
dt
(2.181)
p
d F v = r0 (θ̈ − aθ̇ 2 )e−aθ 1 + a2
dt
(2.182)
1
√
(2.183)
r0 e−aθ 1 + a2
p
r02 θ̇ 2 e−2aθ (1 + a2 ) = r0 θ̇ 2 e−aθ 1 + a2
The acceleration in reference frame F is then given as
p
p
F
a = r0 e−aθ 1 + a2 (θ̈ − aθ̇ 2 )et + r0 θ̇ 2 e−aθ 1 + a2 en
(2.184)
33
Question 2–15
A circular disk of radius R is attached to a rotating shaft of length L as shown in
Fig. P2-15. The shaft rotates about the vertical direction with a constant angular
velocity Ω relative to the ground. The disk, in turn, rotates about its center
about an axis orthogonal to the shaft. Knowing that the angle θ describes the
position of a point P located on the edge of the disk relative to the center of the
disk, determine the following quantities as viewed by an observer fixed to the
ground: (a) the angular velocity of the disk and (b) the velocity and acceleration
of point P .
Ω
θ
A
P
O
R
L
Figure P 2-15
Solution to Question 2–15
First, let F be a reference frame fixed to the ground. Then, we choose the
following coordinate system fixed in reference frame F :
Ex
Ey
Ez
Origin at Point A
=
Along Ω
=
Along AO at t = 0
=
Ex × Ey
Next, let A be a reference frame fixed to the horizontal shaft. Then, we choose
the following coordinate system fixed in reference frame F :
ex
ey
ez
Origin at Point A
=
Along Ω
=
Along AO
=
ex × ey
34
Chapter 2. Kinematics
Lastly, let B be a reference frame fixed to the disk. Then, choose the following
coordinate system fixed in reference frame B:
er
ez
eθ
Origin at Point O
=
=
=
Along OP
Same as Reference Frame A
ez × er
The geometry of the bases {ex , ey , ez } and {er , eθ , ez } is shown in Fig. (2.185).
In particular, using Fig. (2.185), we have that
ex
ey
= cos θ er − sin θ eθ
= sin θ er + cos θ eθ
(2.185)
ex
er
θ
ez ⊗
ey
θ
eθ
Figure 2-3
tion 2–15
Relationship Between Basis {ex , ey , ez } and {er , eθ , ez } for Ques-
Now, since the shaft rotates with angular velocity Ω about the ey -direction
relative to the ground, the angular velocity of reference frame A in reference
frame F is given as
F A
ω = Ω = Ωex
(2.186)
Next, since the disk rotates with angular rate θ̇ relative to the shaft in the ez direction, the angular velocity of reference frame B in reference frame A is
given as
A B
ω = θ̇ez
(2.187)
The angular velocity of reference frame B in reference frame F is then obtained
using the theorem of angular velocity addition as
F
ωB = F ωA + AωB = Ωex + θ̇ez
(2.188)
35
Then, using the relationship between {ex , ey , ez } and {er , eθ , ez } from Eq. (2.185),
we obtain F ωB in terms of the basis {er , eθ , ez } as
F
ωB = Ω(cos θ er − sin θ eθ ) + θ̇ez = Ω cos θ er − Ω sin θ eθ + θ̇ez
(2.189)
Next, the position of point P is given as
rP = rO + rP /O
(2.190)
Now, in terms of the basis {ex , ey , ez }, the position of point O is given as
rO = Ley
(2.191)
Also, in terms of the basis {er , eθ , ez } we have that
rP /O = Rer
(2.192)
rP = Ley + Rer
(2.193)
Consequently,
The velocity of point P in reference frame F is then given as
F
vP =
F
F
d
d
d
rO/P = F vO + F vP /O
=
+
(rP )
(rO )
dt
dt
dt
F
(2.194)
First, since rO is expressed in terms of the basis {ex , ey , ez } and {ex , ey , ez } is
fixed in reference frame A, we can apply the rate of change transport theorem
to rO between reference frames A and F as
F
vO =
A
d
d
(rO ) =
(rO ) + F ωA × rO
dt
dt
F
(2.195)
Now we have that
A
d
(rO ) = 0
dt
F A
ω × rO = Ωex × Ley = LΩez
(2.196)
(2.197)
Adding Eq. (2.196) and Eq. (2.197), we obtain
F
vO = LΩez
(2.198)
Next, since rP /O is expressed in terms of the basis {er , eθ , ez } and {er , eθ , ez } is
fixed in reference frame B, we can apply the rate of change transport theorem
to rP /O between reference frames B and F as
F
vP /O =
Bd
d
rP /O =
rP /O + F ωB × rP /O
dt
dt
F
(2.199)
36
Chapter 2. Kinematics
Now we have that
B
d
= 0
rP /O
dt
F B
ω × rP /O = (Ω cos θ er − Ω sin θ eθ + θ̇ez ) × Rer
= R θ̇eθ + RΩ sin θ ez
(2.200)
(2.201)
Adding Eq. (2.200) and Eq. (2.201), we obtain
F
vP /O = R θ̇eθ + RΩ sin θ ez
(2.202)
The velocity of point P in reference frame F is then obtained by adding Eq. (2.198)
and Eq. (2.202) as
F
vP = F vO + F vP /O = LΩez + R θ̇eθ + RΩ sin θ ez
(2.203)
Simplifying Eq. (2.203), we obtain
F
vP = R θ̇eθ + (L + R sin θ )Ωez
(2.204)
The acceleration of point P in reference frame F is obtained in the same
manner as was used to obtain the velocity in reference frame F . First, we have
from Eq. (2.203) that
F
vP = F vO + F vP /O
(2.205)
Now, since F vO is expressed in the basis {ex , ey , ez } and {ex , ey , ez } is fixed
in reference frame A, the acceleration of point O in reference frame F can
be obtained by applying the rate of change transport theorem to F vO between
reference frames A and F as
F
aO =
F
d F Ad F F A F
vO =
vO + ω × vO
dt
dt
(2.206)
Now we have that
A
d F = 0
vO
dt
F A
ω × F vO = Ωex × LΩez = −LΩ2 ey
(2.207)
(2.208)
Then, adding Eq. (2.207) and Eq. (2.208), we obtain F aO as
F
aO = −LΩ2 ey
(2.209)
Next, since F vP /O is expressed in terms of the basis {er , eθ , ez }, the acceleration
of point P relative to point O in reference frame F is obtained by applying the
rate of change transport theorem to F vP /O between reference frames B and F
as
F
Bd d F
F
F
(2.210)
vP /O =
vP /O + F ωB × F vP /O
aP /O =
dt
dt
37
B
d F
= R θ̈eθ + RΩθ̇ cos θ ez
(2.211)
vP /O
dt
F B
ω × F vO = (Ω cos θ er − Ω sin θ eθ + θ̇ez ) × (R θ̇eθ + RΩ sin θ ez )
= RΩθ̇ cos θ ez − RΩ2 cos θ sin θ eθ
−RΩ2 sin2 θer − R θ̇ 2 er
(2.212)
Simplifying Eq. (2.212), we obtain
F
ωB × F vO = −(R θ̇ 2 + RΩ2 sin2 θ)er − RΩ2 cos θ sin θ eθ + RΩθ̇ cos θ ez (2.213)
Then, adding Eq. (2.211) and Eq. (2.213), we obtain F aP /O as
aP /O = −(R θ̇ 2 + RΩ2 sin2 θ)er + (R θ̈ − RΩ2 cos θ sin θ )eθ + 2RΩθ̇ cos θ ez
(2.214)
F
Finally, adding Eq. (2.209) and Eq. (2.214), we obtain aP as
F
aP = −LΩ2 ey −(R θ̇ 2 +RΩ2 sin2 θ)er +(R θ̈ −RΩ2 cos θ sin θ )eθ +2RΩθ̇ cos θ ez
(2.215)
F
Now it is seen from Eq. (2.215) that some of the terms in aP are expressed
in the basis {ex , ey , ez } while other terms are expressed in the basis {er , eθ , ez }.
However, using Eq. (2.185), we can obtain an expression for F aP in terms of
a single basis. Now, while it is possible to write F aP in terms of the basis
{ex , ey , ez }, it is preferable (and simpler) to write both quantities in terms of
the basis {er , eθ , ez }. First, substituting the expression for ey from Eq. (2.185)
into Eq. (2.215), we obtain F aP in terms of {er , eθ , ez } as
F
F
aP = −LΩ2 (sin θ er + cos θ eθ ) − (R θ̇ 2 + RΩ2 sin2 θ)er
+ (R θ̈ − RΩ2 cos θ sin θ )eθ + 2RΩθ̇ cos θ ez
(2.216)
Simplifying Eq. (2.216) gives
F
aP = −(LΩ2 sin θ + R θ̇ 2 + RΩ2 sin2 θ)er
+ (R θ̈ − LΩ2 cos θ − RΩ2 cos θ sin θ )eθ
+ 2RΩθ̇ cos θ ez
(2.217)
38
Chapter 2. Kinematics
Question 2–16
A disk of radius R rotates freely about its center at a point located on the end of
an arm of length L as shown in Fig. P2-16. The arm itself pivots freely at its other
end at point O to a vertical shaft. Finally, the shaft rotates with constant angular
velocity Ω relative to the ground. Knowing that φ describes the location of a
point P on the edge of the disk relative to the direction OQ and that θ is formed
by the arm with the downward direction, determine the following quantities as
viewed by an observer fixed to the ground: (a) the angular velocity of the disk
and (b) the velocity and acceleration of point P .
Ω
O
L
θ
P
φ
Q
Figure P 2-16
Solution to Question 2–16
Let F be a reference frame fixed to the ground. Then choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along OQ when θ = 0
Orthogonal to plane of shaft and arm and out of page at t = 0
Ez × Ex
39
Next, let A be a reference frame fixed to the vertical shaft. Then choose the
following coordinate system fixed in reference frame A:
ex
ez
ey
Origin at O
=
=
=
Ex
Orthogonal to plane of shaft and arm
ez × ex
We note that ez and Ez are equal when t = 0. Next, let B be a reference frame
fixed to the rod OQ. Then choose the following coordinate system fixed in
reference frame B:
ux
uz
uy
Origin at O
=
=
=
Along OQ
Orthogonal to plane of shaft, arm, and disk
uz × ux
Finally, let D be a reference frame fixed to the disk. Then choose the following
coordinate system fixed in reference frame D:
ir
iz
iφ
Origin at O
=
=
=
Along OQ
−ez
uz × ur
The geometry of the bases {ex , ey , ez }, {ux , uy , uz }, and {ir , iφ , iz } is shown in
Fig. 2-4.
ir
φ
uy
−ux
ez (out of page)
θ
⊗
φ
ey
iφ
iz (into page)
θ
ex
ux
Figure 2-4
Geometry of bases {ex , ey , ez }, {ux , uy , uz }, and {ir , iφ , iz } for
Question 2–16.
The angular velocity of reference frame A in reference frame F is then given as
F
ωA = −Ω = −Ωex
(2.218)
40
Chapter 2. Kinematics
where the negative sign arises from the fact that the positive sense of Ω is
vertically upward while the direction Ex is downward. Next, the angular velocity
of reference frame B in reference frame A is given as
A
ωB = θ̇uz
(2.219)
Next, the angular velocity of reference frame D in reference frame B is given as
B
ωD = φ̇iz = −φuz
(2.220)
The angular velocity of the disk as viewed by an observer fixed to the ground is
then obtained from the angular velocity addition theorem as
F
ωD = F ωA + AωB + BωD = −Ωex + θ̇uz − φ̇uz
(2.221)
Now from the geometry of the bases, it is seen that
ex = cos θ ux − sin θ uy
(2.222)
which implies that
F
ωD = −Ω(cos θ ux − sin θ uy ) + (θ̇ − φ̇)uz
= −Ω cos θ ux + Ω sin θ uy + (θ̇ − φ̇)uz
(2.223)
Now because point P (i.e., the point for which we want the velocity) is fixed to
the disk, it is helpful to obtain an expression for F ωD in terms of the basis
{ir , iφ , iz }. In order to obtain such an expression, it is first important to see
from Fig. 2-4 that
−ux
uy
= cos φir − sin φiφ =⇒ ux = − cos φir + sin φiφ
= sin φir + cos φiφ
(2.224)
(2.225)
where it is observed that diagramatically it is first easier to determine −ux in
terms of ir and iφ and then take the negative sign of the result. Consequently,
F
ωD = −Ω cos θ (− cos φir + sin φiφ ) + Ω sin θ (sin φir + cos φiφ ) + (θ̇ − φ̇)uz
= Ω(cos θ cos φ + sin θ sin φ)ir + Ω(cos θ sin φ − sin θ cos φ)iφ + (θ̇ − φ̇)iz
= Ω cos(θ − φ)ir + Ω sin(θ − φ)iφ + (θ̇ − φ̇)iz
(2.226)
where we have used the two trigonometric identities
cos θ cos φ + sin θ sin φ = cos(θ − φ)
cos θ sin φ − sin θ cos φ = sin(θ − φ)
(2.227)
(2.228)
41
Now we know that the position of point P is given as
rP = rQ + rP /Q
(2.229)
= Lux
(2.230)
where
rQ
rP /Q
= Rir
(2.231)
Because the basis {ux , uy , uz } is fixed in reference frame B, we can apply the
transport theorem to rQ between reference frames B and F as
F
vQ =
B
drQ F B
drQ
=
+ ω × rQ
dt
dt
F
(2.232)
Now we have
F
ωB = F ωA + AωB = −Ωex + θ̇uz = −Ω cos θ ux + Ω sin θ uy + θ̇uz
(2.233)
Furthermore,
B
drQ
dt
F B
ω × rQ
= 0
(2.234)
= (−Ω cos θ ux + Ω sin θ uy + θ̇uz ) × Lux
(2.235)
which gives
F
ωB × rQ = (−Ω cos θ ux + Ω sin θ uy + θ̇uz ) × Lux = Lθ̇uy − LΩ sin θ uz (2.236)
Therefore,
F
vQ = Lθ̇uy − LΩ sin θ uz
(2.237)
Next, because rP /Q is fixed in reference frame D, we can apply the transport
theorem to rP /Q between reference frames D and F as
F
vP /Q =
Now we have
F
Dd
d
rP /Q =
rP /Q + F ωD × rP /Q
dt
dt
(2.238)
D
d
= 0
(2.239)
rP /Q
dt
F B
ω × rP /Q = [Ω cos(θ − φ)ir + Ω sin(θ − φ)iφ + (θ̇ − φ̇)iz ]
×Rir
(2.240)
which implies that
F
ωB × rP /Q = [Ω cos(θ − φ)ir + Ω sin(θ − φ)iφ + (θ̇ − φ̇)iz ] × Rir
= R(θ̇ − φ̇)iφ − RΩ sin(θ − φ)iz
(2.241)
42
Chapter 2. Kinematics
Therefore,
F
vP /Q = R(θ̇ − φ̇)iφ − RΩ sin(θ − φ)iz
(2.242)
The velocity of point P in reference frame F is then obtained by adding Eqs. (2.237)
and (2.242) as
F
vP = Lθ̇uy − LΩ sin θ uz + +R(θ̇ − φ̇)iφ − RΩ sin(θ − φ)iz
(2.243)
It is noted that this last expression can be converted to an expression in terms
of a single basis using the relationships between the bases as given in Fig. 2-4.
43
Question 2–17
A particle slides along a track in the form of a spiral as shown in Fig. P2-17. The
equation for the spiral is
r = aθ
where a is a constant and θ is the angle measured from the horizontal. Determine (a) expressions for the intrinsic basis vectors et , en , and eb in terms any
other basis of your choosing, (b) determine the curvature of the trajectory as a
function of the angle θ, and (c) determine the velocity and acceleration of the
collar as viewed by an observer fixed to the track.
P
r
θ
O
Figure P 2-17
Solution to Question 2–17
First, let F be a reference frame fixed to the spiral. Then, choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
To the Right
Out of Page
Ez × Ex
Next, letA be a reference frame fixed to direction of OP . Then, choose the
following coordinate system fixed in reference frame A:
er
Ez
eθ
Origin at Point O
=
=
=
Along OP
Out of The Page
Ez × e r
44
Chapter 2. Kinematics
Determination of Intrinsic Basis
The position of the particle in terms of the basis {er , eθ , Ez } is given as
r = r er = aθer
(2.244)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇Ez
(2.245)
The velocity of the particle in reference frame F is then obtained using the rate
of change transport theorem as
F
v=
F
dr A dr F A
=
+ ω ×r
dt
dt
(2.246)
Now we have that
A
dr
dt
F ωA
= aθ̇er
(2.247)
× r = θ̇Ez × aθer = aθ̇θeθ
Adding the two expressions in Eq. (2.247), the velocity of the particle in reference frame F is obtained as
F
v = aθ̇er + aθ̇θeθ
(2.248)
Eq. (2.248) can be re-written as
F
v = aθ̇ (er + θeθ )
(2.249)
The speed of the particle in reference frame F is then obtained from Eq. (2.249)
as
1/2
p
F
v = kF vk = aθ̇ 1 + θ 2 = aθ̇ 1 + θ 2
(2.250)
Then, the tangent vector in reference frame F is obtained as
et =
Fv
Fv
(2.251)
Then, using F v from Eq. (2.249) and F v from Eq. (2.250), we obtain the tangent
vector in reference frame F as
et =
−1/2
er + θeθ
aθ̇ (er + θeθ )
√
= √
= 1 + θ2
(er + θeθ )
1 + θ2
aθ̇ 1 + θ 2
(2.252)
Next, the principle unit normal vector in reference frame F is obtained as
en =
F
F
det /dt
k det /dtk
(2.253)
45
Now, using the rate of change transport theorem, we can compute
reference frame F as
F
A
det F A
det
=
+ ω × et
dt
dt
F
det /dt in
(2.254)
Using the expression for et from Eq. (2.252), we have that
A
−3/2
−1/2
det
1
1 + θ2
(2θ θ̇) (er + θeθ ) + 1 + θ 2
θ̇eθ
=−
dt
2
(2.255)
Eq. (2.255) simplifies to
A
−3/2
det
= θ̇ 1 + θ 2
(−θer + eθ )
dt
(2.256)
Next, the second term in Eq. (2.254) is obtained as
F
−1/2
ωA × et = θ̇Ez × 1 + θ 2
(er + θeθ )
(2.257)
Eq. (2.257) simplifies to
F
ωA × et = θ̇(1 + θ 2 )−1/2 (−θer + eθ )
(2.258)
Eq. (2.258) can be re-written as
F
ωA × et = θ̇(1 + θ 2 )(1 + θ 2 )−3/2 (−θer + eθ )
(2.259)
Then, adding Eq. (2.256) and Eq. (2.259), we obtain
F
−3/2 det
2 + θ 2 (−θer + eθ )
= θ̇ 1 + θ 2
dt
(2.260)
Then the magnitude of F det /dt is obtained as
F
det
dt
p
−3/2 2 + θ2 1 + θ2
= θ̇ 1 + θ 2
(2.261)
Then, dividing Eq. (2.260) by Eq. (2.261), we obtain the principle unit normal in
reference frame F as
−θer + eθ
en = √
(2.262)
1 + θ2
Finally, the principle unit bi-normal vector in reference frame F is obtained as
−θer + eθ
er + θeθ
× √
= ez
eb = et × en = √
2
1+θ
1 + θ2
(2.263)
46
Chapter 2. Kinematics
Curvature of Trajectory in Reference Frame F
First, we know that
F
det
= κ F ven
dt
(2.264)
Taking the magnitude of both sides, we have that
F
det
dt
= κ Fv
(2.265)
kF det /dtk
Fv
(2.266)
Solving for κ, we have that
κ=
Substituting the expression for kF det /dtk from Eq. (2.261) and the expression
for F v from Eq. (2.250) into Eq. (2.266), we obtain κ as
κ=
θ̇ 1 + θ 2
√
−3/2
2 + θ2
2 + θ2
1 + θ2
√
=
a(1 + θ 2 )3/2
aθ̇ 1 + θ 2
(2.267)
Velocity and Acceleration of Particle
The velocity of the particle in reference frame F is given in intrinsic coordinates
as
F
v = F vet
(2.268)
Using the expression for F v from Eq. (2.250), we obtain F v as
p
F
v = aθ̇ 1 + θ 2 et
(2.269)
Furthermore, the acceleration in reference frame F is obtained in intrinsic coordinates as
2
d F F
v et + κ F v en
(2.270)
a=
dt
Differentiating F v from Eq. (2.250), we have that
p
d F v = aθ̈ 1 + θ 2 + aθ̇(1 + θ 2 )−1/2 2θ θ̇
dt
(2.271)
Simplifying Eq. (2.271), we obtain
−1/2 h i
d F v = a 1 + θ2
θ̈ 1 + θ 2 + θ̇ 2 θ
dt
(2.272)
Next, using the expression for κ from Eq. (2.267), we have that
κ
F
v
2
=
p
a(2 + θ 2 )θ̇ 2
2 + θ2
2 =
√
a
θ̇
1
+
θ
a(1 + θ 2 )3/2
1 + θ2
(2.273)
47
Substituting the results of Eq. (2.272) and Eq. (2.273) into Eq. (2.270), we obtain
the acceleration of the particle in reference frame F as
F
−1/2 h i
a(2 + θ 2 )θ̇ 2
en
a = a 1 + θ2
θ̈ 1 + θ 2 + θ̇ 2 θ et + √
1 + θ2
(2.274)
Simplifying Eq. (2.274) gives
F
a= √
a
1 + θ2
h
i
θ̈(1 + θ 2 ) + θ̇ 2 θ et + (2 + θ 2 )θ̇ 2 en
(2.275)
48
Chapter 2. Kinematics
Question 2–19
A particle P slides without friction along the inside of a fixed hemispherical bowl
of radius R as shown in Fig. P2-19. The basis {Ex , Ey , Ez } is fixed to the bowl.
Furthermore, the angle θ is measured from the Ex -direction to the direction OQ,
where point Q lies on the rim of the bowl while the angle φ is measured from
the OQ-direction to the position of the particle. Determine the velocity and
acceleration of the particle as viewed by an observer fixed to the bowl. Hint:
Express the position in terms of a spherical basis that is fixed to the direction
OP ; then determine the velocity and acceleration as viewed by an observer fixed
to the bowl in terms of this spherical basis.
O
φ
Ey
θ
Ex
Q
R
P
Ez
Figure P 2-18
Solution to Question 2–19
Let F be a reference frame fixed to the bowl. Then choose the following coordinate system fixed in reference frame F :
Ex
Ey
Ez
Origin at O
=
=
=
Given
Given
Ex × Ey = Given
Next, let A be a reference frame fixed to the plane defined by the points O and
Q and the direction Ez . Then choose the following coordinate system fixed in
reference frame A:
Origin at O
er
=
Along OQ
ez
=
Ez
eθ
=
ez × er
49
Finally, let B be a reference frame fixed to the direction OP . Then choose the
following coordinate system fixed in reference frame B:
ur
uθ
uφ
Origin at O
=
=
=
Along OP
eθ
ur × uθ
The relationship between the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in
Fig. 2-5 while the relationship between the bases {er , eθ , ez } and {ur , uθ , uφ }
is shown in Fig. 2-6.
e z , Ez
⊗
Ex
θ
eθ
θ
er
Ey
Figure 2-5
tion 2–19.
Relationship between bases {Ex , Ey , Ez } and {er , eθ , ez } for Quesuθ , eθ
φ
φ
uφ
er
ur
ez
Figure 2-6
tion 2–19.
Relationship between bases {er , eθ , ez } and {ur , uθ , uφ } for Ques-
The position of the particle is then given as
r = Rur
(2.276)
Next, the angular velocity of reference frame A in reference frame F is given as
F
ωA = θ̇ez
(2.277)
Furthermore, the angular velocity of reference frame B in reference frame A is
given as
A B
ω = −φ̇uθ
(2.278)
where the negative sign on AωB is due to the fact that the angle φ is measured positively about the negative uθ -direction (see Fig. 2-6). Then, applying
the angular velocity addition theorem, we have
F
ωB = F ωA + AωB = θ̇ez − φ̇uθ
(2.279)
50
Chapter 2. Kinematics
Now we can obtain an expression for F ωB in terms of the basis {ur , uθ , uφ } by
expressing ez in terms of ur and uφ as
ez = sin φur + cos φuφ
(2.280)
Consequently,
F
ωB = θ̇(sin φur + cos φuφ ) − φ̇uθ = θ̇ sin φur − φ̇uθ + θ̇ cos φuφ
(2.281)
Then, the velocity of point P in reference frame F is obtained by applying the
transport theorem between reference frames B and F as
F
v=
F
dr Bdr F B
=
+ ω ×r
dt
dt
(2.282)
Now we have
B
dr
= 0
dt
F B
ω × r = (θ̇ sin φur − φ̇uθ + θ̇ cos φuφ ) × Rur
= R θ̇ cos φuθ + R φ̇uφ
(2.283)
(2.284)
Therefore,
F
v = R θ̇ cos φuθ + R φ̇uφ
(2.285)
The acceleration of point P in reference frame F is obtained by applying the
transport theorem to F v between reference frames B and F as
F
a=
F
d F B d F F B F
v =
v + ω × v
dt
dt
(2.286)
Now we have
B
d F v
= R(θ̈ cos φ − θ̇ φ̇ sin φ)uθ + R φ̈uφ
(2.287)
dt
F B
ω × F v = (θ̇ sin φur − φ̇uθ + θ̇ cos φuφ ) × (R θ̇ cos φuθ + R φ̇uφ )
= R θ̇ 2 cos φ sin φuφ − R θ̇ φ̇ sin φuθ
−R φ̇2 ur − R θ̇ 2 cos2 φur
(2.288)
Adding these last two equations and simplifying gives
F
a = −(R φ̇2 + R θ̇ 2 cos2 φ)ur
+ (R θ̈ cos φ − 2R θ̇ φ̇ sin φ)uθ
+ (R φ̈ + R θ̇ 2 cos φ sin φ)uφ
(2.289)
51
Question 2–20
A particle P slides along a circular table as shown in Fig. P2-20. The table is
rigidly attached to two shafts such that the shafts and table rotate with angular
velocity Ω about an axis along the direction of the shafts. Knowing that the
position of the particle is given in terms of a polar coordinate system relative to
the table, determine (a) the angular velocity of the table as viewed by an observer
fixed to the ground, (b) the velocity and acceleration of the particle as viewed
by an observer fixed to the table, and (c) the velocity and acceleration of the
particle as viewed by an observer fixed to the ground.
A
r
O
P
θ
B
Ω
Figure P 2-19
Solution to Question 2–20
Let F be a reference frame fixed to the ground. Then choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at 0
=
Along OB
=
Vertically Upward
=
Ez × Ex
Next, let A be a reference frame fixed to the table. Then choose the following
coordinate system fixed in reference frame A:
ex
ez
ey
Origin at 0
=
Along OB
=
Orthogonal to Table and = Ez When t = 0
=
ez × ex
Finally, let B be a reference frame fixed to the direction of OP . Then choose the
following coordinate system fixed in reference frame B:
er
ez
eθ
Origin at 0
=
Along OB
=
Same as in Reference Frame B
=
ez × er
52
Chapter 2. Kinematics
The position of the particle is then given as
r = r er
(2.290)
Now because the position is expressed in terms of the basis {er , eθ , ez } and
{er , eθ , ez } is fixed in reference frame B, the velocity of the particle as viewed by
an observer fixed to the ground is obtained by applying the transport theorem
between reference frames B and F as
F
dr Bdr F B
=
+ ω ×r
dt
dt
F
v=
(2.291)
First, the angular velocity of B in F is obtained from the angular velocity addition theorem as
F B
ω = F ωA + AωB
(2.292)
Now we have
F
ωA
A
B
(2.293)
= Ω = Ωex
= θ̇ez
(2.294)
ωB = Ωex + θ̇ez
(2.295)
ω
which implies that
F
Next, because the position is expressed in terms of the basis {er , eθ , ez }, the
unit vector ex must also be expressed in terms of the basis {er , eθ , ez }. The
relationship between the bases {ex , ey , ez } and {er , eθ , ez } is shown in Fig. 2-7.
Using Fig. 2-7, it is seen that
ey
θ
eθ
er
θ
ez
Figure 2-7
ex
Geometry of bases {ex , ey , ez } and {er , eθ , ez } for Question P2–20.
ex
ey
= cos θ er − sin θ eθ
= sin θ er + cos θ eθ
(2.296)
(2.297)
Therefore,
F
ωB = Ω(cos θ er − sin θ eθ ) + θ̇ez = Ω cos θ er − Ω sin θ eθ + θ̇ez
(2.298)
53
Now the two terms required to obtain F v are given as
F
dr
= ṙ er
(2.299)
dt
F B
ω × r = (Ω cos θ er − Ω sin θ eθ + θ̇ez ) × r er = r θ̇eθ + r Ω sin θ (2.300)
ez
Therefore, the velocity of the particle in reference frame F is
F
v = ṙ er + r θ̇eθ + r Ω sin θ ez
(2.301)
Next, the acceleration of the particle as viewed by an observer fixed to the
ground is given from the transport theorem as
F
a=
F
d F Bd F F B F
v =
v + ω × v
dt
dt
(2.302)
Now we have
B
h
i
d F v
= r̈ er + (ṙ θ̇ + r θ̈)eθ + ṙ Ω sin θ + r (Ω̇ sin θ + Ωθ̇ cos θ )(2.303)
ez
dt
F B
ω × F v = (Ω cos θ er − Ω sin θ eθ + θ̇ez ) × (ṙ er + r θ̇eθ + r Ω sin θ ez )
= r θ̇Ω cos θ ez − r Ω2 cos θ sin θ eθ + ṙ Ω sin θ ez
−r Ω2 sin2 θer + ṙ θ̇eθ − r θ̇ 2 er
= −(r θ̇ 2 + r Ω2 sin2 θ)er + (ṙ θ̇ − r Ω2 cos θ sin θ )eθ
+ (ṙ Ω sin θ + r θ̇Ω cos θ )ez
(2.304)
Adding these last two equations, we obtain the acceleration as viewed by an
observer fixed to the ground as
F
a = (r̈ − r θ̇ 2 − r Ω2 sin2 θ)er
+ (2ṙ θ̇ + r θ̈ − r Ω2 cos θ sin θ )eθ
h
i
+ r Ω̇ sin θ + 2(ṙ Ω sin θ + r Ωθ̇ cos θ ) ez
(2.305)
54
Chapter 2. Kinematics
Question 2–21
A slender rod of length l is hinged to a collar as shown in Fig. P2-21. The collar
slides freely along a fixed horizontal track. Knowing that x is the horizontal
displacement of the collar and that θ describes the orientation of the rod relative
to the vertical direction, determine the velocity and acceleration of the free end
of the rod as viewed by an observer fixed to the track.
P
θ
l
O
x
Figure P 2-20
Solution to Question 2–21
Let F be a reference frame fixed to the horizontal track. Then, choose the
following coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O at t = 0
=
To the Right
=
Into Page
=
Ez × Ex
Next, let A be a reference frame fixed to the rod. Then, choose the following
coordinate system fixed in reference frame A:
er
ez
eθ
Origin at O
=
=
=
Along OP
Into Page
Ez × er
We note that the relationship between the basis {Ex , Ey , Ez } and {er , eθ , ez } is
given as
Ex =
sin θ er + cos θ eθ
(2.306)
Ey = − cos θ er + sin θ eθ
as
Using the bases {Ex , Ey , Ez } and {er , eθ , ez }, the position of point P is given
rP = rO + rP /O = xEx + ler
(2.307)
55
Next, the angular velocity of reference frame A in reference frame F is given as
F
ωA = θ̇ez
(2.308)
The velocity of point P in reference frame F is then given as
F
vP =
F
d
d
rP /O = F vO + F vP /O
(rO ) +
dt
dt
F
(2.309)
Now since rO is expressed in the basis {Ex , Ey , Ez } and {Ex , Ey , Ez } is fixed, we
have that
F
d
F
vO =
(2.310)
(rO ) = ẋEx
dt
Next, since rP /O is expressed in the basis {er , eθ , ez } and {er , eθ , ez } rotates
with angular velocity F ωA , we can apply the rate of change transport theorem
to rP /O between reference frame A and reference frame F as
F
vP /O =
F
Ad
d
rP /O =
rP /O + F ωA × rP /O
dt
dt
Now we have that
A
d
rP /O
= 0
dt
F A
ω × rP /O = θ̇ez × ler = lθ̇eθ
(2.311)
(2.312)
(2.313)
Adding Eq. (2.312) and Eq. (2.313) gives
F
vP /O = lθ̇eθ
(2.314)
Therefore, the velocity of point P in reference frame F is given as
F
vP = ẋEx + lθ̇eθ
(2.315)
Next, the acceleration of point P in reference frame F is obtained as
F
aP =
F
d F vP
dt
(2.316)
Now we have that
F
vP = F vO + F vP /O
(2.317)
where
F
= ẋEx
(2.318)
= lθ̇eθ
(2.319)
d F Fd F
vO +
vP /O
dt
dt
(2.320)
F
vO
vP /O
Consequently,
F
aP =
F
56
Chapter 2. Kinematics
Now, since F vO is expressed in the basis {Ex , Ey , Ez }, we have that
F
aO =
F
d F vO = ẍEx
dt
(2.321)
Furthermore, since F vP /O is expressed in the basis {er , eθ , ez } and {er , eθ , ez }
rotates with angular velocity F ωA , we can obtain F aP /O by applying the rate of
change transport theorem between reference frame A and reference frame F
as
F
d F Ad F F A F
F
(2.322)
vO =
vO + ω × vO
aP /O =
dt
dt
Now we have that
A
d F = lθ̈eθ
vO
dt
F A
ω × F vO = θ̇ez × lθ̇eθ = −lθ̇ 2 er
(2.323)
(2.324)
Adding Eq. (2.323) and Eq. (2.324) gives
F
aP /O = −lθ̇ 2 er + lθ̈eθ
(2.325)
Then, adding Eq. (2.321) and Eq. (2.325), we obtain the velocity of point P in
reference frame F as
F
aP = ẍEx − lθ̇ 2 er + lθ̈eθ
(2.326)
Finally, substituting the expression for Ex from Eq. (2.306), we obtain
terms of the basis {er , eθ , ez } as
F
Fa
P
in
aP = ẍ(sin θ er + cos θ eθ ) − lθ̇ 2 er + lθ̈eθ = (ẍ sin θ − lθ̇ 2 )er + (ẍ cos θ + lθ̈)eθ
(2.327)
57
Question 2–23
A particle slides along a fixed track y = − ln cos x as shown in Fig. P2-23 (where
−π /2 < x < π /2). Using the horizontal component of position, x, as the variable to describe the motion and the initial condition x(t = 0) = x0 , determine
the following quantities as viewed by an observer fixed to the track: (a) the arclength parameter s as a function of x, (b) the intrinsic basis {et , en , eb } and the
curvature κ, and (c) the velocity and acceleration of the particle.
Ey
P
y = − ln cos x
O
Ex
Figure P 2-21
Solution to Question 2–23
For this problem, it is convenient to use a reference frame F that is fixed to the
track. Then, we choose the following coordinate system fixed in reference frame
F:
Origin at O
=
=
=
Ex
Ey
Ez
Along Ox
Along Oy
Ex × Ey
The position of the particle is then given as
r = xEx − ln cos xEy
(2.328)
Now, since the basis {Ex , Ey , Ez } does not rotate, the velocity in reference frame
F is given as
F
v = ẋEx + ẋ tan xEy
(2.329)
Using the velocity from Eq. (2.329), the speed of the particle in reference frame
F is given as
q
F
F
(2.330)
v = k vk = ẋ 1 + tan2 x = ẋ sec x
58
Chapter 2. Kinematics
Arc-length Parameter as a Function of x
Now we recall the arc-length equation as
q
d F F
s = v = ẋ 1 + tan2 x = ẋ sec x
dt
(2.331)
Separating variables in Eq. (2.331), we obtain
F
ds = sec xdx
(2.332)
Integrating both sides of Eq. (2.332) gives
Zx
F
F
sec xdx
s − s0 =
(2.333)
x0
Using the integral given for sec x, we obtain
F
F
s − s0 = ln [sec x
+ tan x]x
x0
sec x + tan x
= ln
sec x0 + tan x0
(2.334)
Noting that F s(0) = F s0 = 0, the arc-length is given as
F
s = ln [sec x + tan x]x
x0
(2.335)
Simplifying Eq. (2.335), we obtain
F
s = ln
sec x + tan x
sec x0 + tan x0
(2.336)
Intrinsic Basis
Next, we need to compute the intrinsic basis. First, we have the tangent vector
as
Fv
ẋ(Ex + tan xEy )
tan x
1
Ex +
Ey
(2.337)
=
et = F =
sec x
sec x
ẋ sec x
v
Now we note that sec x = 1/ cos x. Therefore,
tan x
= sin x
sec x
(2.338)
et = cos xEx + sin xEy
(2.339)
Eq. (2.337) then simplifies to
Next, the principle unit normal is given as
F
det
= κ F ven
dt
(2.340)
59
Differentiating et in Eq. (2.339), we obtain
F
det
= −ẋ sin xEx + ẋ cos xEy
dt
(2.341)
Consequently,
F
det
dt
= ẋ = κ F v
(2.342)
which implies that
en =
F
det /dt
F
det /dt
=
−ẋ sin xEx + ẋ cos xEy
= − sin xEx + cos xEy
ẋ
(2.343)
Then, using F v from Eq. (2.330), we obtain the curvature as
κ=
1
ẋ
=
= cos x
ẋ sec x
sec x
(2.344)
Finally, the principle unit bi-normal is given as
eb = et × en = (cos xEx + sin xEy ) × (− sin xEx + cos xEy ) = Ez
(2.345)
Velocity and Acceleration in Terms of Intrinsic Basis
Using the speed from Eq. (2.330), the velocity of the particle in reference frame
F is given in terms of the instrinsic basis as
F
v = ẋ sec xet
(2.346)
Next, the acceleration is given in terms of the intrinsic basis as
F
a=
d F v et + κ F v en
dt
(2.347)
Now, using F v from Eq. (2.330), we obtain d(F v)/dt as
h
i
d F v = ẍ sec x + ẋ 2 sec x tan x = sec x ẍ + ẋ 2 tan x
dt
Also, using κ from Eq. (2.344) we obtain
κ F v = cos x(ẋ sec x)2 = ẋ 2 sec x
The acceleration of the particle in reference frame F is then given as
h
i
F
a = sec x ẍ + ẋ 2 tan x et + ẋ 2 sec xen
(2.348)
(2.349)
(2.350)
60
Chapter 2. Kinematics