DYNAMICS OF PARTICLES AND RIGID BODIES:
A SYSTEMATIC APPROACH
SOLUTION MANUAL TO TEXTBOOK PROBLEMS
ANIL V. RAO
Department of Aerospace & Mechanical Engineering
Boston University
ii
Anil V. Rao
Department of Aerospace & Mechanical Engineering
Boston University
110 Cummington Street
Boston, MA 02215
c
Copyright 2004
Anil Vithala Rao. All Rights Reserved.
This solution manual may not be reproduced in part or in whole without the
expressed written consent of Anil Vithala Rao.
Image on Cover Page Used with Permission via Unlimited Royalty-Free License
from Foto Search Stock Footage and Stock Photography, 21155 Watertown Road,
Waukesha, WI 53186-1898, USA.
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Contents
2 Kinematics
Question 2–1 .
Question 2–2 .
Question 2–3 .
Question 2–4 .
Question 2–5 .
Question 2–8 .
Question 2–9 .
Question 2–10
Question 2–11
Question 2–13
Question 2–15
Question 2–16
Question 2–17
Question 2–19
Question 2–20
Question 2–21
Question 2–23
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1
1
3
6
9
11
13
18
23
27
30
34
39
44
49
52
55
58
3 Kinetics of Particles
Question 3–1 . . . . .
Question 3–2 . . . . .
Question 3–3 . . . . .
Question 3–5 . . . . .
Question 3–7 . . . . .
Question 3–9 . . . . .
Question 3–10 . . . .
Question 3–11 . . . .
Question 3–12 . . . .
Question 3–13 . . . .
Question 3–17 . . . .
Question 3–19 . . . .
Question 3–20 . . . .
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61
61
68
71
80
87
92
98
102
105
111
116
121
126
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iv
Contents
Question 3–22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
Question 3–23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
Question 3–25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
4 Kinetics of a System of Particles
Question 4–1 . . . . . . . . . . . . .
Question 4–2 . . . . . . . . . . . . .
Question 4–3 . . . . . . . . . . . . .
Question 4–8 . . . . . . . . . . . . .
Question 4–17 . . . . . . . . . . . .
5 Kinetics of Rigid
Question 5–1 . .
Question 5–2 . .
Question 5–3 . .
Question 5–4 . .
Question 5–5 . .
Question 5–6 . .
Question 5–7 . .
Question 5–8 . .
Question 5–10 .
Question 5–11 .
Question 5–12 .
Question 5–17 .
Question 5–20 .
Bodies
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147
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. 154
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. 165
. 169
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179
. 179
. 185
. 193
. 199
. 204
. 214
. 220
. 225
. 232
. 237
. 246
. 254
. 262
Chapter 2
Kinematics
Question 2–1
A bug B crawls radially outward at constant speed v0 from the center of a rotating disk as shown in Fig. P2-1. Knowing that the disk rotates about its center O with constant absolute angular velocity Ω relative to the ground (where
Ω = Ω), determine the velocity and acceleration of the bug as viewed by an
observer fixed to the ground.
r
O
v0
B
Ω
Figure P 2-1
Solution to Question 2–1
For this problem it is convenient to choose a fixed reference frame F and a noninertial reference frame A that is fixed in the disk. Corresponding to reference
frame F we choose the following coordinate system:
Ex
Ez
Ey
Origin at Point O
=
=
=
Along OB at Time t = 0
Out of Page
Ez × Ex
2
Chapter 2. Kinematics
Corresponding to the reference frame A that is fixed in the disk, we choose the
following coordinate system
Origin at Point O
=
=
=
ex
ez
ey
Along OB
Out of Page (= Ez )
ez × ex
The position of the bug is then resolved in the basis {ex , ey , ez } as
r = r ex
(2.1)
Now, since the platform rotates about the ez -direction relative to the ground,
the angular velocity of reference frame A in reference frame F is given as
F
ωA = Ωez
(2.2)
The velocity is found by applying the basic kinematic equation. This gives
F
v=
F
dr Adr F A
=
+ ω ×r
dt
dt
(2.3)
Now we have
A
dr
= ṙ ex = v0 ex
dt
F A
ω × r = Ωez × r ex
= Ωr ey
(2.4)
(2.5)
Adding Eqs. (2.4) and (2.5), we obtain the velocity of the bug in reference frame
F as
F
v = v0 ex + Ωr ey
(2.6)
The acceleration is found by applying the basic kinematic equation to
gives
F
d F A d F F A F
F
a=
v =
v + ω × v
dt
dt
Using
Fv
F v.
This
(2.7)
from Eq. (2.6) and noting that v0 and Ω are constant, we have that
d F v
dt
A
F ωA
= Ωṙ ey = Ωv0 ey
× F v = Ωez × [v0 ex + r Ωey ]
= −Ω2 r ex + Ωv0 ey
(2.8)
Therefore, the acceleration in reference frame F is given as
F
a = −Ω2 r ex + 2Ωv0 ey
(2.9)
3
Question 2–2
A particle, denoted by P , slides on a circular table as shown in Fig. P2-2. The
position of the particle is known in terms of the radius r measured from the
center of the table at point O and the angle θ where θ is measured relative to the
direction of OQ where Q is a point on the circumference of the table. Knowing
that the table rotates with constant angular rate Ω, determine the velocity and
acceleration of the particle as viewed by an observer in a fixed reference frame.
r
P
Q
θ
O
Ω
Figure P 2-2
Solution to Question 2–2
For this problem it is convenient to define a fixed inertial reference frame F
and two non-inertial reference frames A and B. The first non-inertial reference
frame A is fixed to the disk while the second non-inertial reference frame B is
fixed to the direction of OP . Corresponding to the fixed inertial reference frame
F , we choose the following coordinate system:
Ex
Ez
Ey
Origin at point O
=
=
=
Along Ox at t = 0
Out of Page
Ez × Ex
Corresponding to non-inertial reference frame A, we choose the following coordinate system:
ex
ez
ey
Origin at point O
=
=
=
Along OQ
Out of Page (= Ez )
ez × ex
4
Chapter 2. Kinematics
Finally, corresponding to reference frame B, we choose the following coordinate
system:
Origin at point O
=
Along OP
er
=
Out of Page
ez
=
ez × er
eθ
Then, the position of the particle can be desribed in terms of the basis {er , eθ , ez }
as
(2.10)
r = r er .
Now, in order to compute the velocity of the particle, it is necessary to apply
the basic kinematic equation. In this case since we are interested in motion
as viewed by an observer in the fixed inertial reference frame F , we need to
determine the angular velocity of B in F . First, since A rotates relative to F
with angular velocity Ω, we have that
F
ωA = Ω = Ωez
(2.11)
Next, since B rotates relative to A with angular rate θ̇ about the ez -direction,
we have that
A B
ω = θ̇ez
(2.12)
Then, applying the theorem of addition of angular velocities, we have that
F
ωB = F ωA + A ωB = Ωez + θ̇ez = (Ω + θ̇)ez
(2.13)
The velocity in reference frame is then found by applying the rate of change
transport theorem as
F
dr Bdr F B
F
=
+ ω ×r
v=
(2.14)
dt
dt
Now we have
B
dr
= ṙ er
dt
F B
ω × r = (Ω + θ̇)ez × r er = r (Ω + θ̇)eθ
(2.15)
(2.16)
Adding Eqs. (2.15) and (2.16), we obtain the velocity of the particle in reference
frame F as
F
v = ṙ er + r (Ω + θ̇)eθ
(2.17)
The acceleration is found by applying the rate of change transport theorem toF v
. This gives
F
d F Bd F F B F
F
a=
v =
v + ω × v
(2.18)
dt
dt
5
Using F v from Eq. (2.17) and noting again that Ω is constant, we have
d F v
= r̈ er + ṙ (Ω + θ̇) + r θ̈ eθ
dt
B
F
ωB × F v = (Ω + θ̇)ez × ṙ er + r (Ω + θ̇)eθ
= −r (Ω + θ̇)2 er + ṙ (Ω + θ̇)eθ
(2.19)
(2.20)
Adding Eqs. (2.19) and (2.20), we obtain the acceleration of the particle in reference frame F as
F
a = r̈ − r (Ω + θ̇)2 er + r θ̈ + 2ṙ (Ω + θ̇) eθ
(2.21)
6
Chapter 2. Kinematics
Question 2–3
A collar slides along a rod as shown in Fig. P2-3. The rod is free to rotate about
a hinge at the fixed point O. Simultaneously, the rod rotates about the vertical
direction with constant angular velocity Ω relative to the ground. Knowing that
r describes the location of the collar along the rod, that θ is the angle measured
from the vertical, and that Ω = Ω, determine the velocity and acceleration of
the collar as viewed by an observer fixed to the ground.
P
Ω
r
θ
O
Figure P 2-3
Solution to Question 2–3
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Ex
Ez
Origin at point O
=
=
Ey
=
Along Ω
Orthogonal to Plane of
Shaft and Arm at t = 0
Ez × Ex
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
Origin at point O
=
=
ey
=
Along Ω
Orthogonal to Plane of
Shaft and Arm
ez × ex
7
Finally, let B be a reference frame fixed to the rod. Then, choose the following
coordinate system fixed in reference frame B:
er
ez
eθ
Origin at point O
=
=
=
Along OP
uz
ez × er
The geometry of the bases {ex , ey , ez } and {er , eθ , ez } is shown in Fig. 2-1. Using
Fig. 2-1, the relationship between the basis {ex , ey , ez } and {er , eθ , ez } is given
as
ex = cos θ er − sin θ eθ
(2.22)
ey = sin θ er + cos θ eθ
ex
er
θ
ez ⊗
ey
θ
eθ
Figure 2-1
Geometry of Bases {ex , ey , ez } and {er , eθ , ez } for Question 2–3.
The position of the particle can then be expressed in the basis {er , eθ , ez } as
r = r er
(2.23)
Now, since {er , eθ , ez } is fixed in reference frame B, and we are interested in
obtaining the velocity and acceleration as viewed by an observer fixed in the
ground (i.e., reference frame F ), we need to obtain an expression for the angular
velocity of reference frame B in reference frame F . First, since reference frame
A rotates relative to reference frame F with angular velocity Ω and Ω lies along
the ex -direction, we have that
F
ωA = Ω = Ωex
(2.24)
Next, since reference frame B rotates relative to reference frame A with angular
rate θ̇ about the ez -direction. Therefore,
A
ωB = θ̇ez
(2.25)
8
Chapter 2. Kinematics
Then, using the angular velocity addition theorem, we have the angular velocity
of reference frame B in reference frame F as
F
ωB = F ωA + AωB = Ωex + θ̇ez
(2.26)
Now, since we have determined that the position of the collar is expressed most
conveniently in terms of the basis {er , eθ , ez }, it is also most convenient to express F ωB in terms of the basis {er , eθ , ez }. In particular, substituting the expression for ex from Eq. (2.22) into Eq. (2.26), we obtain F ωB as
F
ωB = Ω(cos θ er − sin θ eθ ) + θ̇ez = Ω cos θ er − Ω sin θ eθ + θ̇ez
(2.27)
The velocity in reference frame F is then found by applying the rate of change
transport theorem between reference frames B and F as
F
v=
F
dr Bdr F B
=
+ ω ×r
dt
dt
(2.28)
Now we have that
B
dr
= ṙ er
dt
F B
ω × r = (Ω cos θ er − Ω sin θ eθ + θ̇ez ) × r er
= Ωr sin θ Ez + r θ̇eθ
(2.29)
(2.30)
Adding Eq. (2.29) and Eq. (2.30), we obtain the velocity of the collar in reference
frame F as
F
v = ṙ er + r θ̇eθ + r Ω sin θ ez
(2.31)
The acceleration of the collar is then obtained by applying the rate of change
transport theorem to F v between reference frames B and F as
F
d F Bd F F B F
F
a=
v =
v + ω × v
(2.32)
dt
dt
Now we have
B
d F v
= r̈ er + (ṙ θ̇ + r θ̈)eθ + Ω(ṙ sin θ + r θ̇ cos θ ) ez
(2.33)
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 Ωθ̇ 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
+(r Ωθ̇ cos θ + ṙ Ω sin θ )ez
(2.34)
Adding Eqs. (2.33) and (2.34), we obtain the acceleration of the collar in reference frame F as
F
a = (r̈ − r θ̇ 2 − r Ω2 sin2 θ)er + (2ṙ θ̇ + r θ̈ − r Ω2 cos θ sin θ )eθ
+2Ω(ṙ sin θ + r θ̇ cos θ )ez
(2.35)
9
Question 2–4
A particle slides along a track in the form of a parabola y = x 2 /a as shown
in Fig. P2-4. The parabola rotates about the vertical with a constant angular
velocity Ω relative to a fixed reference frame (where Ω = Ω). Determine the
velocity and acceleration of the particle as viewed by an observer in a fixed
reference frame.
y
Ω
P
y = x 2 /a
O
Q
x
Figure P 2-4
Solution to Question 2–4
For this problem it is convenient to define a fixed inertial reference frame F
and a non-inertial reference frame A. Corresponding to reference frame F , we
choose the following coordinate system:
Ex
Ey
Ez
Origin at Point O
=
=
=
Along OQ When t = 0
Along Oy When t = 0
E x × Ey
Furthermore, corresponding to reference frame A, we choose the following coordinate system:
Origin at Point O
=
Along OQ
ex
=
Along Oy
ey
=
ex × ey
ez
The position of the particle is then given in terms of the basis {ex , ey , ez } as
r = xex + yey = xex + (x 2 /a)ey
(2.36)
Furthermore, since the parabola spins about the ey-direction, the angular velocity of reference frame A in reference frame F is given as
F
ωA = Ω = Ωey
(2.37)
10
Chapter 2. Kinematics
The velocity in reference frame F is then found using the rate of change transport theorem as
F
dr Adr F A
F
=
+ ω ×r
v=
(2.38)
dt
dt
Using r from Eq. (2.36) and
F ωA
from Eq. (2.37), we have
A
dr
= ẋex + (2x ẋ/a)ey
dt
F A
ω × r = Ωey × (xex + (x 2 /a)ey ) = −Ωxez
(2.39)
(2.40)
Adding Eqs. (2.39) and (2.40), we obtain F v as
F
v = ẋex + (2x ẋ/a)ey − Ωxez
(2.41)
The acceleration in reference frame F is found by applying the rate of change
transport theorem to F v as
F
d F Ad F F A F
a=
v =
v + ω × v
dt
dt
F
Using F v from Eq. (2.41) and
F ωA
(2.42)
from Eq. (2.37), we have
d F v
= ẍex + 2(ẋ 2 + x ẍ)/a ey − Ωẋez
(2.43)
dt
F A
ω × F v = Ωey × (ẋex + (2x ẋ/a)ey − Ωxez ) = −Ωẋez − Ω2 xex(2.44)
A
Adding Eq. (2.43) and (2.44), we obtain F a as
F
a = (ẍ − Ω2 x)ex + 2(ẋ 2 + x ẍ)/a ey − 2Ωẋez
(2.45)
11
Question 2–5
A satellite is in motion over the Earth as shown in Fig. P2-5. The Earth is modeled as a sphere of radius R that rotates with constant angular velocity Ω in
a direction ez where ez lies along a radial line that lies in the direction from
the center of the Earth at point O to the North Pole of the Earth at point N.
Furthermore, the center of the Earth is assumed to be an absolutely fixed point.
The position of the satellite is known in terms of an Earth-centered Earth-fixed
Cartesian coordinate system whose right-handed basis {ex , ey , ez } is defined as
follows:
• The direction ex lies orthogonal to ez in the equatorial plane of the Earth
along the line from O to P where P lies at the intersection of the equator
with the great circle called the Prime Meridian
• The direction ey lies orthogonal to both ex and ez in the equatorial plane
of the Earth such that ey = ez × ex
Using the basis {ex , ey , ez } to express all quantities, determine the velocity and
acceleration of the spacecraft (a) as viewed by an observer fixed to the Earth and
(b) as viewed by an observer in a fixed inertial reference frame.
Spacecraft
ez
Ω
Prime Meridian
r
N
×
O
Q
P
ey
ex
Equator
Figure P 2-5
Solution to Question 2–5
First, let F be a fixed inertial reference frame. Next, let A be a reference frame
that is fixed in the planet. Corresponding to reference frame A, we choose the
12
Chapter 2. Kinematics
following coordinate system:
ex
ez
ey
Origin at point O
=
=
=
Along OP
Along ON
ez × ex (= Along OQ)
The position of the spacecraft is then given in terms of the basis {ex , ey , ez } as
r = xex + yey + zez
(2.46)
Now, since the planet rotates with constant angular velocity Ω about the ONdirection relative to reference frame F , we have that
F
ωA = Ωez
(2.47)
The velocity of the spacecraft is then found by applying the rate of change
transport theorem as
F
dr Adr F A
F
=
+ ω ×r
v=
(2.48)
dt
dt
Now we have
A
dr
= ẋex + ẏey + żez
dt
F A
ω × r = Ωez × (xex + yey + zez )
= Ωxey − Ωyex
(2.49)
(2.50)
Adding Eqs. (2.49) and (2.50), we obtain F v as
F
v = (ẋ − Ωy)ex + (ẏ + Ωx)ey + żez
(2.51)
Next, the acceleration of the spacecraft in reference frame F is found by applying the rate of change transport theorem to F v as
F
a=
d F Ad F F A F
v =
v + ω × v
dt
dt
F
(2.52)
Now we have
d F v
= (ẍ − Ωẏ)ex + (ÿ + Ωẋ)ey + z̈ez
dt
F A
ω × F v = Ωez × [(ẋ − Ωy)ex + (ẏ + Ωx)ey + żez ]
A
= Ω(ẋ − Ωy)ey − Ω(ẏ + Ωx)ex
(2.53)
(2.54)
Adding Eqs. (2.53) and (2.54), we obtain F a as
F
a = (ẍ − 2Ωẏ − Ω2 x)ex + (ÿ + 2Ωẋ − Ω2 y)ey + z̈ez
(2.55)
13
Question 2–8
A bead slides along a fixed circular helix of radius R and helical inclination
angle φ as shown in Fig. P2-8. Knowing that the angle θ measures the position
of the bead and is equal to zero when the bead is at the base of the helix,
determine the following quantities relative to an observer fixed to the helix: (a)
the arclength parameter s as a function of the angle θ, (b) the intrinsic basis
{et , en , eb } and the curvature of the trajectory as a function of the angle θ, and
(c) the position, velocity, and acceleration of the particle in terms of the intrinsic
basis {et , en , eb }.
A
R
O P
φ
θ
Figure P 2-8
Solution to Question 2–8
Let F be a reference frame fixed to the helix. Then, choose the following coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along OA
Out of page
Ez × Ex
Next, let A be a reference
frame that rotates with the projection of the position of particle into the Ex , Ey -plane. Corresponding to A, we choose the
14
Chapter 2. Kinematics
following coordinate system to describe the motion of the particle:
Origin at O
er
ez
eθ
=
=
=
Along O to projection of P into
Ez
ez × er
Ex , Ey
plane
Now, since φ is the angle formed by the helix with the horizontal, we have from
the geometry that
z = Rθ tan φ
(2.56)
Suppose now that we make the following substitution:
α tan φ
(2.57)
Then the position of the bead can be written as
r = Rer + tan φRθez = Rer + αRθez
(2.58)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇ez
(2.59)
Then, applying the rate of change transport theorem to r between reference
frames A and F , we have
F
v=
F
dr Adr F A
=
+ ω ×r
dt
dt
(2.60)
where
A
dr
= αR θ̇ez
dt
F A
ω × r = θ̇ez × (Rer + αRθez ) = R θ̇eθ
(2.61)
(2.62)
Adding Eqs. (2.61) and (2.62), we obtain
F
v = R θ̇eθ + αR θ̇ez
(2.63)
The speed in reference frame F is then given as
F
Consequently,
d F v = F v = R θ̇ 1 + α2 ≡
s
dt
F
ds = R 1 + α2 dθ
(2.64)
(2.65)
Integrating both sides of Eq. (2.65), we obtain
Fs
Fs
0
ds =
θ
θ0
R 1 + α2 dθ
(2.66)
15
We then obtain
s − F s0 = R 1 + α2 (θ − θ0 )
(2.67)
Solving Eq. (2.67) for s, the arclength is given as
F
s = F s0 + R 1 + α2 (θ − θ0 )
(2.68)
F
Intrinsic Basis and Curvature of Trajectory
The intrinsic basis is obtained as follows. First, the tangent vector et is given as
et =
Fv
Substituting the expressions for
obtain
et =
Fv
(2.69)
Fv
and
Fv
from part (a) into Eq. (2.69), we
R θ̇eθ + αR θ̇ez
√
R θ̇ 1 + α2
(2.70)
Simplifying this last expression, we obtain
eθ + αez
et = √
1 + α2
Next, we have that
F
where
where
F
A
det
dt
F ωA
Therefore,
det
= κ F ven
dt
(2.72)
A
det
det F A
=
+ ω × et
dt
dt
(2.73)
= 0
× et
(2.71)
eθ + αez
θ̇
= θ̇ez × √
= −√
er
1 + α2
1 + α2
F
θ̇
det
= −√
er
dt
1 + α2
(2.74)
(2.75)
The principle unit normal is then given as
en =
F
det /dt
F
det /dt
= −er
(2.76)
Furthermore, the curvature is given as
κ=
F
1
det /dt
=
Fv
R(1 + α2 )
(2.77)
16
Chapter 2. Kinematics
Finally, the principle unit bi-normal vector is given as
eθ + αez
√
1 + α2
eb = et × en =
ez − αeθ
× (−er ) = √
1 + α2
(2.78)
Rearranging this last equation, we obtain
αeθ − ez
eb = − √
1 + α2
(2.79)
Position, Velocity, and Acceleration of Bead
First, we can solve for the basis {er , eθ , ez } in terms of {et , en , eb } by using
Eqs. (2.71), (2.76), and (2.79). First, from Eq. (2.76), we have
er = −en
(2.80)
Next, restating Eqs. (2.71) and (2.79), we have
et
eb
eθ + αez
√
1 + α2
αeθ − ez
= −√
1 + α2
=
(2.81)
(2.82)
Solving Eqs. (2.81) and (2.82) simultaneously for eθ and ez , we obtain
eθ
=
ez
=
αet + eb
√
1 + α2
et − αeb
√
1 + α2
(2.83)
(2.84)
Then, substituting the expressions for er and ez from Eqs. (2.80) and (2.84) into
Eq. (2.58), we have the position in terms of {et , en , eb } as
r = −Ren + Rαθ
et − αeb
√
1 + α2
(2.85)
Next, the velocity in reference frame F is given in terms of {et , en , eb } as
F
v = F vet
(2.86)
Susstituting the expression for F v from Eq. (2.64) into Eq. (2.86), we have
F
v = R θ̇ 1 + α2 et
(2.87)
Finally, the acceleration in reference frame F is given as
F
a=
2
d F v et + κ F v en
dt
(2.88)
17
Computing the rate of change of F v using the expression for F v from Eq. (2.64),
we have
d F v = R θ̈ 1 + α2
(2.89)
dt
Then, subsituting the expresion for κ from Eq. (2.77) into Eq. (2.88), we obtain
F
a = R θ̈ 1 + α2 et +
2
1
2
1
+
α
en
R
θ̇
R(1 + α2 )
(2.90)
Simplifying Eq. (2.90) gives
F
a = R θ̈ 1 + α2 et + R θ̇ 2 en
(2.91)
18
Chapter 2. Kinematics
Question 2–9
Arm AB is hinged at points A and B to collars that slide along vertical and horizontal shafts, respectively, as shown in Fig. P2-9. The vertical shaft rotates with
angular velocity Ω relative to a fixed reference frame (where Ω = Ω) and point
B moves with constant velocity v0 relative to the horizontal shaft. Knowing that
point P is located at the center of the arm and the angle θ describes the orientation of the arm with respect to the vertical shaft, determine the velocity and
acceleration of point P as viewed by an observer fixed to the ground. In simplifying your answers, find an expression for θ̇ in terms of v0 and l and express
your answers in terms of only l, Ω, Ω̇, θ, and v0 .
Ω
A
l/2
θ
P
l/2
v0
O
B
Figure P 2-9
Solution to Question 2–9
Let F be the ground. Then choose the following coordinate system fixed in
reference frame F :
Ex
Ey
Ez
Origin at O
=
=
=
Along OB when t = 0
Along OA
Ex × Ey
19
Next, let A be the L-shaped assembly. Then choose the following coordinate
system fixed in A:
Origin at O
=
Along OB
ex
ey
=
Along OA
=
ex × ey
ez
Finally, let B be the rod. Then choose the
B:
Origin at A
=
ur
=
uz
uθ
=
following coordinate system fixed in
Along AB
ez
uz × ur
From the geometry of the coordinate systems, we have
ex
ey
= sin θ ur + cos θ uθ
= − cos θ ur + sin θ uθ
(2.92)
Next, because we must measure all distances from point O (because point O is
fixed to the ground and we want all rates of change as viewed by an observer
fixed to the ground), the position of the center of the rod is given as
rP /O = rA/O + rP /A ≡ r
(2.93)
Using the coordinates systems defined for this problem, we have
rA/O
rP /A
= l cos θ ey
l
= 2 ur
(2.94)
Consequently,
l
ur
(2.95)
2
Because rA/O is expressed in the basis {ex , ey , ez } while rP /A is expressed in
the basis {ur , uθ , uz }, it is convenient to differentiate each piece of the vector
rP /O separately. First, the velocity of point A relative to point O as viewed by
an observer fixed to the ground is obtained by applying the transport theorem
from A to F as
rP /O = l cos θ ey +
F
First, we have
Next,
vA/O =
F
A
d
d
rA/O =
rA/O + F ωA × rA/O
dt
dt
F
ωA = Ω = Ωey
A d
dt rA/O
F ωA × r
A/O
= −lθ̇ sin θ ey
= Ωey × l cos θ ey = 0
(2.96)
(2.97)
(2.98)
20
Chapter 2. Kinematics
Consequently,
F
vA/O = −lθ̇ sin θ ey
(2.99)
The acceleration of point A relative to point O as viewed by an observer fixed to
the ground is then given as
F
aA/O =
F
A d d F
F
vA/O =
vA/O + F ωA × F vA/O
dt
dt
(2.100)
Now we have
A d
Fv
A/O
dt
F ωA × F v
A/O
Therefore,
F
= −l(θ̈ sin θ + θ̇ 2 cos θ )ey
= Ωey × (−lθ̇ sin θ )ey = 0
aA/O = −l(θ̈ sin θ + θ̇ 2 cos θ )ey
(2.101)
(2.102)
The velocity of point P relative to point A as viewed by an observer fixed to the
ground is obtained by applying the transport theorem from reference frame B
to reference frame F as
F
vP /A =
F
B
d
d
rP /A =
rP /A + F ωB × rP /A
dt
dt
Now
F
where
ωB = F ωA + A ωB
A
(2.103)
(2.104)
ωB = θ̇uz
(2.105)
Therefore,
F
ωB = Ωey + θ̇uz = Ω(− cos θ ur + sin θ uθ ) + θ̇uz
= −Ω cos θ ur + Ω sin θ uθ + θ̇uz
(2.106)
Now we have
B
d
rP /A
dt
F ωB × r
P /A
= 0
l
= (−Ω cos θ ur + Ω sin θ uθ + θ̇uz ) × 2 ur
lθ̇
lΩ sin θ
uθ −
uz
=
2
2
(2.107)
Therefore,
lθ̇
lΩ sin θ
uθ −
uz
(2.108)
2
2
The acceleration of point P relative to point A as viewed by an observer fixed to
the ground is then given as
F
F
aP /A =
F
vP /A =
B d d F
F
vP /A =
vP /A + F ωB × F vP /A
dt
dt
(2.109)
21
Now we have
B
d F
vP /A
dt
F ωB
× F vP /A
=
lΩθ̇ cos θ
lθ̈
uθ −
uz
2
2
= (−Ω cos θ ur + Ω sin θ uθ + θ̇uz ) × (
lθ̇
lΩ sin θ
uθ −
uz )
2
2
(2.110)
The second term in Eq. (2.110) can be simplified to
F
B
F
ω × vP /A
lΩθ̇ cos θ
lΩ2 cos θ sin θ
uz −
uθ −
=−
2
2
lΩ2 sin2 θ + lθ̇ 2
2
ur
(2.111)
Adding the first term in Eq. (2.110) to the result of Eq. (2.111), we obtain the
acceleration of point P relative to point A as viewed by an observer fixed to the
ground as
lΩ2 cos θ sin θ
lΩ2 sin2 θ + lθ̇ 2
lθ̈
F
−
aP /A = −
ur +
uθ − lΩθ̇ cos θ uz
2
2
2
(2.112)
Using the aforementioned results, we obtain the velocity and acceleration of
point P relative to point O as viewed by an observer fixed to the ground as
follows. First, adding the results of Eqs. (2.99) and (2.108), we obtain
F
vP /O = −lθ̇ sin θ ey +
lθ̇
lΩ sin θ
uθ −
uz
2
2
(2.113)
Finally, adding the results of Eqs. (2.102) and (2.112), we obtain
F
aP /O = −l(θ̈ sin θ + θ̇ 2 cos θ )ey
lΩ2 cos θ sin θ
lΩ2 sin2 θ + lθ̇ 2
lθ̈
ur +
−
uθ
−
2
2
2
(2.114)
− lΩθ̇ cos θ uz
A last point pertains to the velocity of point B. It was stated in the problem
that, “point B moves with constant velocity v0 relative to the horizontal shaft.”
Now, because the horizontal shaft is fixed in reference frame A, we have
A
Another expression for the
vB = v0 ex = constant
Av
B
(2.115)
is obtained as follows. First,
rB = l sin θ ex
(2.116)
vB = lθ̇ cos θ ex
(2.117)
Therefore,
A
22
Chapter 2. Kinematics
Equating the expressions in Eq. (2.115) and (2.117), we obtain
which implies that
θ̇ =
v0 = lθ̇ cos θ
(2.118)
v0
v0
=
sec θ
l cos θ
l
(2.119)
Differentiating Eq. (2.119) with respect to time, we have
θ̈ =
v2
v0
θ̇ sec θ tan θ = 20 sec2 θ tan θ
l
l
(2.120)
The expressions for θ̇ and θ̈ given in Eqs. (2.119) and (2.120), respectively, can
be substituted into the expressions for F vP /O and F aP /O to obtain expressions
that do not involve either θ̇ or θ̈.
23
Question 2–10
A circular disk of radius R is attached to a rotating shaft of length L as shown
in Fig. P2-10. The shaft rotates about the horizontal 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 velocity and acceleration of point P relative to the
ground.
θ
A
P
O
R
Ω
L
Figure P 2-10
Solution to Question 2–10
First, let F be a reference frame fixed to the ground. Then, we choose the
following coordinate system fixed in reference frame F :
E2
E3
Origin at Point O
=
=
E1
=
Along AO
Orthogonal to Disk
and Into Page at t = 0
E2 × E3
Next, let A be a reference frame fixed to the horizontal shaft. Then, we choose
the following coordinate system fixed in reference frame F :
e2
e3
Origin at Point O
=
=
e1
=
Along AO
Orthogonal to Disk
and Into Page
e2 × e3
24
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:
u1
u3
Origin at Point O
=
=
u2
=
Along OP
Orthogonal to Disk
and Into Page
u3 × u1
Now, since the shaft rotates with angular velocity Ω relative to the ground, the
angular velocity of reference frame A in reference frame F is given as
F
ωA = Ω = Ωe2
(2.121)
Furthermore, since the disk rotates with angular rate θ̇ relative to the shaft, the
angular velocity of reference frame B in reference frame A is given as
A
ωB = θ̇u3
(2.122)
Finally, the geometry of the bases {e1 , e2 , e3 } and {u1 , u2 , u3 } is shown in Fig. (2.123).
Using Fig. (2.123), we have that
e1
e2
= cos θ u1 − sin θ u2
= sin θ u1 + cos θ u2
(2.123)
e1
u1
θ
u3 , e3 ⊗
e2
θ
u2
Figure 2-2
tion 2–10
Relationship Between Basis {e1 , e2 , e3 } and {u1 , u2 , u3 } for Ques-
Given the definitions of the reference frames and coordinate systems, the
position of point P is given as
r = Ru1
(2.124)
25
The velocity of point P in reference frame F is then given as
F
v=
F
dr Fd
=
(Ru1 )
dt
dt
(2.125)
Now, since the basis {u1 , u2 , u3 } is fixed in reference frame F , it is convenient
to apply the rate of change transport theorem of Eq. (2–128) between reference
frame B and reference frame F as
F
B
d
d
(Ru1 ) =
(Ru1 ) + F ωB × Ru1
dt
dt
(2.126)
First, since R is constant and the basis {u1 , u2 , u3 } is fixed in reference frame B,
we have that
B
d
(2.127)
(Ru1 ) = 0
dt
Next, applying the angular velocity addition rule of Eq. (2–136), we obtain F ωB
as
F B
ω = F ωA + A ωB = Ωe2 + θ̇u3
(2.128)
Using
F ωB
F
F ωB
from Eq. (2.128), we obtain
× Ru1 as
ωB × Ru1 = (Ωe2 + θ̇u3 ) × Ru1 = RΩe2 × u1 + R θ̇u2
(2.129)
Then, from Eq. (2.123), we have that
e2 × u1 = (sin θ u1 + cos θ u2 ) × u1 = − cos θ u3
(2.130)
Substituting the result of Eq. (2.130) into Eq. (2.129), we obtain
F
ωB × Ru1 = −RΩ cos θ u3 + R θ̇u2
(2.131)
Adding Eq. (2.127) and Eq. (2.131), we obtain the velocity of point P in reference
frame F as
F
v = R θ̇u2 − RΩ cos θ u3
(2.132)
Next, the acceleration of point P in reference frame F is given as
F
a=
d F v
dt
F
(2.133)
It is seen that the expression for F v is given in terms of the basis {u1 , u2 , u3 }
where {u1 , u2 , u3 } is fixed in reference frame B. Thus, applying the rate of
change transport theorem of Eq. (2–128) between reference frame B and F to
F v, we obtain
F
d F Bd F F B F
F
a=
v =
v + ω × v
(2.134)
dt
dt
26
Chapter 2. Kinematics
Now, observing that R and Ω are constant, the first term in Eq. (2.134) is given
as
B
d F v = R θ̈u2 + RΩθ̇ sin θ u3
(2.135)
dt
Next, using
F ωB
F
from Eq. (2.128), we obtain the second term in Eq. (2.134) as
ωB × F v = (Ωe2 + θ̇u3 ) × (−RΩ cos θ u3 + R θ̇u2 )
(2.136)
Expanding Eq. (2.136), we obtain
F
ωB × F v = RΩθ̇e2 × u2 − RΩ2 cos θ e2 × u3 − R θ̇ 2 u1
(2.137)
Then, using the expression for e2 from Eq. (2.123), we obtain
e2 × u2
e2 × u3
= (sin θ u1 + cos θ u2 ) × u2 = sin θ u3
= (sin θ u1 + cos θ u2 ) × u3 = cos θ u1 − sin θ u2
(2.138)
Substituting the results of Eq. (2.138) into Eq. (2.137), we obtain
F
ωB × F v = RΩθ̇ sin θ u3 − RΩ2 cos θ (cos θ u1 − sin θ u2 ) − R θ̇ 2 u1
(2.139)
Adding the expressions in Eq. (2.135) and Eq. (2.139), we obtain the acceleration
of point P in reference frame F as
F
a = R θ̈u2 + RΩθ̇ sin θ u3 + RΩθ̇ sin θ u3 − RΩ2 cos θ (cos θ u1 − sin θ u2 ) − R θ̇ 2 u1
(2.140)
Simplifying Eq. (2.140), we obtain
F
a = −(RΩ2 cos2 θ + R θ̇ 2 )u1 + (R θ̈ + RΩ2 cos θ sin θ )u2 + 2RΩθ̇ sin θ u3 (2.141)
27
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
28
Chapter 2. Kinematics
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
(2.142)
Next, the position of point B is given as
rB = Lex
(2.143)
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 =
F
A
drB
drB F A
=
+ ω × rB
dt
dt
(2.144)
Now we have
A
drB
dt
F A
ω × rB
Consequently,
= 0
(2.145)
= Ωez × Lex = LΩey
(2.146)
F
vB = LΩey
(2.147)
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
(2.148)
Then, using Eq. (2–517) on page 106, we can obtain a second expression for F vB
as
F
F D
vB = F vD
(2.149)
Q + ω × (rB − rQ )
Now we know that F ωD is given from the angular velocity addition theorem as
F
ωD = F ωA + AωD
(2.150)
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
(2.151)
where ω is to be determined. Adding Eqs. (2.142) and (2.151), we obtain
F
ωD = Ωez + ωux
(2.152)
Also, rB − rQ is given as
rB − rQ = REz = Rez
(2.153)
29
Therefore,
F
vB = (Ωez + ωex ) × Rez = −Rωey
(2.154)
Setting the expressions for F vB from Eqs. (2.154) and (2.154) equal, we obtain
LΩ = −Rω
(2.155)
from which we obtain ω as
L
(2.156)
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
ω = Ωez − Ωex
(2.157)
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
(2.158)
Now, because F ωA = Ωez = ΩEz and Ez is fixed in reference frame F , we have
d F A ω
= Ω̇ez = 0
dt
F
(2.159)
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
(2.160)
Now we have
L
d A D ω
= − Ω̇ex = 0
dt
R
L
F A
ω × AωD = Ωez × − Ωex
R
A
(2.161)
L
= − Ω2 ey
R
(2.162)
where we have again used the fact that Ω is constant. Therefore,
L
d A D ω
= − Ω2 ey
dt
R
F
(2.163)
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
(2.164)
30
Chapter 2. Kinematics
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 = E z × E x
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
31
The position of the particle is given in terms of the basis {er , eθ , ez } as
r = r er = r0 e−aθ er
(2.165)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇Ez
(2.166)
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
(2.167)
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
(2.168)
(2.169)
Adding Eq. (2.168) and Eq. (2.169), we obtain the velocity of the particle in reference frame F as
F
v = −ar0 θ̇e−aθ er + r0 θ̇e−aθ eθ
(2.170)
Simplifying Eq. (2.167), we obtain F v as
F
v = r0 θ̇e−aθ [−aer + eθ ]
(2.171)
The tangent vector in reference frame F is then given as
et =
Fv
F v
=
Fv
F v
(2.172)
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
F
v = F v = r0 θ̇e−aθ 1 + a2
(2.173)
Dividing F v in Eq. (2.171) by
reference frame F as
Fv
in Eq. (2.173), we obtain the tangent vector in
−aer + eθ
et = √
1 + a2
(2.174)
Next, the principle unit normal vector is obtained as
en =
F
F
det /dt
det /dt
(2.175)
32
Chapter 2. Kinematics
Now we have from the rate of change transport theorem that
F
A
det
det F A
=
+ ω × et
dt
dt
(2.176)
where
F
F
det
dt
ωA × et
= 0
(2.177)
−aer + eθ
er + aeθ
= θ̇ez × √
= −θ̇ √
1 + a2
1 + a2
(2.178)
Adding Eq. (2.177) and Eq. (2.178), we obtain
F
Consequently,
Dividing
F
er + aeθ
det
= −θ̇ √
dt
1 + a2
F de t
= θ̇
dt (2.179)
(2.180)
det /dt in Eq. (2.179) by F det /dt in Eq. (2.180), we obtain en as
er + aeθ
en = − √
1 + a2
(2.181)
Finally, the bi-normal vector is obtained as
−aer + eθ
er + aeθ
eb = et × en = √
×−√
= Ez
1 + a2
1 + a2
(2.182)
(b) Curvature
The curvature of the trajectory in reference frame F is then obtained as
κ=
F det /dt
Fv
(2.183)
Substituting F det /dt from Eq. (2.180) and F v from Eq. (2.173), we obtain κ
as
1
√
(2.184)
κ=
−aθ
1 + a2
r0 e
(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
(2.185)
33
Using the expression for F v from Eq. (2.173), we obtain
F
v = r0 θ̇e−aθ 1 + a2 et
(2.186)
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
a=
2
d F v et + κ F v en
dt
d F v = r0 (θ̈ − aθ̇ 2 )e−aθ 1 + a2
dt
(2.187)
(2.188)
Furthermore,
κ
F
v
2
=
1
√
r0 e−aθ 1 + a2
r02 θ̇ 2 e−2aθ (1 + a2 ) = r0 θ̇ 2 e−aθ 1 + a2
The acceleration in reference frame F is then given as
F
a = r0 e−aθ 1 + a2 (θ̈ − aθ̇ 2 )et + r0 θ̇ 2 e−aθ 1 + a2 en
(2.189)
(2.190)
34
Chapter 2. Kinematics
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
35
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. (2.191).
In particular, using Fig. (2.191), we have that
ex
ey
= cos θ er − sin θ eθ
= sin θ er + cos θ eθ
(2.191)
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
(2.192)
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
(2.193)
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
(2.194)
36
Chapter 2. Kinematics
Then, using the relationship between {ex , ey , ez } and {er , eθ , ez } from Eq. (2.191),
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
(2.195)
Next, the position of point P is given as
rP = rO + rP /O
(2.196)
Now, in terms of the basis {ex , ey , ez }, the position of point O is given as
rO = Ley
(2.197)
Also, in terms of the basis {er , eθ , ez } we have that
rP /O = Rer
(2.198)
rP = Ley + Rer
(2.199)
Consequently,
The velocity of point P in reference frame F is then given as
F
vP =
F
F
F
d
d
d
=
+
rO/P = F vO + F vP /O
(rP )
(rO )
dt
dt
dt
(2.200)
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 =
F
A
d
d
(rO ) =
(rO ) + F ωA × rO
dt
dt
(2.201)
Now we have that
A
d
(rO ) = 0
dt
F A
ω × rO = Ωex × Ley = LΩez
(2.202)
(2.203)
Adding Eq. (2.202) and Eq. (2.203), we obtain
F
vO = LΩez
(2.204)
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 =
F
B
d
d
rP /O =
rP /O + F ωB × rP /O
dt
dt
(2.205)
37
Now we have that
B
d
rP /O
dt
F B
ω × rP /O
= 0
(2.206)
= (Ω cos θ er − Ω sin θ eθ + θ̇ez ) × Rer
= R θ̇eθ + RΩ sin θ ez
(2.207)
Adding Eq. (2.206) and Eq. (2.207), we obtain
F
vP /O = R θ̇eθ + RΩ sin θ ez
(2.208)
The velocity of point P in reference frame F is then obtained by adding Eq. (2.204)
and Eq. (2.208) as
F
vP = F vO + F vP /O = LΩez + R θ̇eθ + RΩ sin θ ez
(2.209)
Simplifying Eq. (2.209), we obtain
F
vP = R θ̇eθ + (L + R sin θ )Ωez
(2.210)
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. (2.209) that
F
vP = F vO + F vP /O
(2.211)
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 =
d F Ad F F A F
vO =
vO + ω × vO
dt
dt
F
(2.212)
Now we have that
d F vO
= 0
dt
F A
ω × F vO = Ωex × LΩez = −LΩ2 ey
A
(2.213)
(2.214)
Then, adding Eq. (2.213) and Eq. (2.214), we obtain F aO as
F
aO = −LΩ2 ey
(2.215)
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
aP /O =
vP /O =
vP /O + F ωB × F vP /O
(2.216)
dt
dt
38
Chapter 2. Kinematics
d F
vP /O
(2.217)
= R θ̈eθ + RΩθ̇ cos θ ez
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
(2.218)
B
Simplifying Eq. (2.218), we obtain
F
ωB × F vO = −(R θ̇ 2 + RΩ2 sin2 θ)er − RΩ2 cos θ sin θ eθ + RΩθ̇ cos θ ez (2.219)
Then, adding Eq. (2.217) and Eq. (2.219), we obtain F aP /O as
F
aP /O = −(R θ̇ 2 + RΩ2 sin2 θ)er + (R θ̈ − RΩ2 cos θ sin θ )eθ + 2RΩθ̇ cos θ ez
(2.220)
F
Finally, adding Eq. (2.215) and Eq. (2.220), 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
(2.221)
F
Now it is seen from Eq. (2.221) 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. (2.191), 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. (2.191)
into Eq. (2.221), we obtain F aP in terms of {er , eθ , ez } as
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
(2.222)
Simplifying Eq. (2.222) 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
(2.223)
39
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
40
Chapter 2. Kinematics
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
θ
⊗
ez (out of page)
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
(2.224)
41
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
(2.225)
Next, the angular velocity of reference frame D in reference frame B is given as
B
ωD = φ̇iz = −φuz
(2.226)
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
(2.227)
Now from the geometry of the bases, it is seen that
ex = cos θ ux − sin θ uy
(2.228)
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
(2.229)
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
(2.230)
= sin φir + cos φiφ
(2.231)
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
(2.232)
where we have used the two trigonometric identities
cos θ cos φ + sin θ sin φ = cos(θ − φ)
(2.233)
cos θ sin φ − sin θ cos φ = sin(θ − φ)
(2.234)
42
Chapter 2. Kinematics
Now we know that the position of point P is given as
rP = rQ + rP /Q
(2.235)
= Lux
(2.236)
= Rir
(2.237)
where
rQ
rP /Q
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
F
vQ =
B
drQ
drQ F B
=
+ ω × rQ
dt
dt
(2.238)
Now we have
F
ωB = F ωA + AωB = −Ωex + θ̇uz = −Ω cos θ ux + Ω sin θ uy + θ̇uz
(2.239)
Furthermore,
B
drQ
dt
F B
ω × rQ
= 0
(2.240)
= (−Ω cos θ ux + Ω sin θ uy + θ̇uz ) × Lux
(2.241)
which gives
F
ωB × rQ = (−Ω cos θ ux + Ω sin θ uy + θ̇uz ) × Lux = Lθ̇uy − LΩ sin θ uz (2.242)
Therefore,
F
vQ = Lθ̇uy − LΩ sin θ uz
(2.243)
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 =
F
D
d
d
rP /Q =
rP /Q + F ωD × rP /Q
dt
dt
(2.244)
Now we have
D
d
rP /Q
dt
F B
ω × rP /Q
= 0
(2.245)
= [Ω cos(θ − φ)ir + Ω sin(θ − φ)iφ + (θ̇ − φ̇)iz ]
×Rir
(2.246)
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
(2.247)
43
Therefore,
F
vP /Q = R(θ̇ − φ̇)iφ − RΩ sin(θ − φ)iz
(2.248)
The velocity of point P in reference frame F is then obtained by adding Eqs. (2.243)
and (2.248) as
F
vP = Lθ̇uy − LΩ sin θ uz + +R(θ̇ − φ̇)iφ − RΩ sin(θ − φ)iz
(2.249)
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.
44
Chapter 2. Kinematics
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
45
Determination of Intrinsic Basis
The position of the particle in terms of the basis {er , eθ , Ez } is given as
r = r er = aθer
(2.250)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇Ez
(2.251)
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
(2.252)
Now we have that
A
dr
dt
F ωA
= aθ̇er
(2.253)
× r = θ̇Ez × aθer = aθ̇θeθ
Adding the two expressions in Eq. (2.253), the velocity of the particle in reference frame F is obtained as
F
v = aθ̇er + aθ̇θeθ
(2.254)
Eq. (2.254) can be re-written as
F
v = aθ̇ (er + θeθ )
(2.255)
The speed of the particle in reference frame F is then obtained from Eq. (2.255)
as
1/2
F
v = F v = aθ̇ 1 + θ 2 = aθ̇ 1 + θ 2
(2.256)
Then, the tangent vector in reference frame F is obtained as
et =
Fv
Fv
(2.257)
Then, using F v from Eq. (2.255) and F v from Eq. (2.256), we obtain the tangent
vector in reference frame F as
et =
−1/2
aθ̇ (er + θeθ )
er + θeθ
√
= √
= 1 + θ2
(er + θeθ )
1 + θ2
aθ̇ 1 + θ 2
(2.258)
Next, the principle unit normal vector in reference frame F is obtained as
en =
F
F
det /dt
det /dt
(2.259)
46
Chapter 2. Kinematics
Now, using the rate of change transport theorem, we can compute
reference frame F as
F
A
det
det F A
=
+ ω × et
dt
dt
F
det /dt in
(2.260)
Using the expression for et from Eq. (2.258), we have that
−3/2
−1/2
1
det
=−
θ̇eθ
(2θ θ̇) (er + θeθ ) + 1 + θ 2
1 + θ2
dt
2
A
(2.261)
Eq. (2.261) simplifies to
−3/2
det
= θ̇ 1 + θ 2
(−θer + eθ )
dt
A
(2.262)
Next, the second term in Eq. (2.260) is obtained as
F
−1/2
ωA × et = θ̇Ez × 1 + θ 2
(er + θeθ )
(2.263)
Eq. (2.263) simplifies to
F
ωA × et = θ̇(1 + θ 2 )−1/2 (−θer + eθ )
(2.264)
Eq. (2.264) can be re-written as
F
ωA × et = θ̇(1 + θ 2 )(1 + θ 2 )−3/2 (−θer + eθ )
(2.265)
Then, adding Eq. (2.262) and Eq. (2.265), we obtain
F
−3/2 det
= θ̇ 1 + θ 2
2 + θ 2 (−θer + eθ )
dt
(2.266)
Then the magnitude of F det /dt is obtained as
F de −3/2 t
= θ̇ 1 + θ 2
2 + θ2 1 + θ2
dt (2.267)
Then, dividing Eq. (2.266) by Eq. (2.267), we obtain the principle unit normal in
reference frame F as
−θer + eθ
en = √
(2.268)
1 + θ2
Finally, the principle unit bi-normal vector in reference frame F is obtained as
er + θeθ
−θer + eθ
eb = et × en = √
× √
= ez
2
1+θ
1 + θ2
(2.269)
47
Curvature of Trajectory in Reference Frame F
First, we know that
F
det
= κ F ven
dt
(2.270)
Taking the magnitude of both sides, we have that
F de t
= κ Fv
dt Solving for κ, we have that
κ=
(2.271)
F det /dt
Fv
(2.272)
Substituting the expression for F det /dt from Eq. (2.267) and the expression
for F v from Eq. (2.256) into Eq. (2.272), we obtain κ as
κ=
θ̇ 1 + θ 2
−3/2
2 + θ2
√
aθ̇ 1 + θ 2
√
1 + θ2
=
2 + θ2
a(1 + θ 2 )3/2
(2.273)
Velocity and Acceleration of Particle
The velocity of the particle in reference frame F is given in intrinsic coordinates
as
F
v = F vet
(2.274)
Using the expression for F v from Eq. (2.256), we obtain F v as
F
v = aθ̇ 1 + θ 2 et
(2.275)
Furthermore, the acceleration in reference frame F is obtained in intrinsic coordinates as
2
d F F
a=
v et + κ F v en
(2.276)
dt
Differentiating F v from Eq. (2.256), we have that
d F v = aθ̈ 1 + θ 2 + aθ̇(1 + θ 2 )−1/2 2θ θ̇
dt
(2.277)
Simplifying Eq. (2.277), we obtain
−1/2 d F v = a 1 + θ2
θ̈ 1 + θ 2 + θ̇ 2 θ
dt
(2.278)
Next, using the expression for κ from Eq. (2.273), we have that
κ
F
v
2
=
a(2 + θ 2 )θ̇ 2
2 + θ2
2 =
√
1
+
θ
a
θ̇
a(1 + θ 2 )3/2
1 + θ2
(2.279)
48
Chapter 2. Kinematics
Substituting the results of Eq. (2.278) and Eq. (2.279) into Eq. (2.276), we obtain
the acceleration of the particle in reference frame F as
F
−1/2 a(2 + θ 2 )θ̇ 2
a = a 1 + θ2
en
θ̈ 1 + θ 2 + θ̇ 2 θ et + √
1 + θ2
(2.280)
Simplifying Eq. (2.280) gives
F
a= √
a
1 + θ2
θ̈(1 + θ 2 ) + θ̇ 2 θ et + (2 + θ 2 )θ̇ 2 en
(2.281)
49
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
θ
φ
Ey
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
=
Along OQ
er
=
Ez
ez
eθ
=
ez × er
50
Chapter 2. Kinematics
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
eθ
⊗
Ex
θ
θ
er
Ey
Figure 2-5
tion 2–19.
Relationship between bases {Ex , Ey , Ez } and {er , eθ , ez } for Ques-
uθ , 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
(2.282)
Next, the angular velocity of reference frame A in reference frame F is given as
F
ωA = θ̇ez
(2.283)
Furthermore, the angular velocity of reference frame B in reference frame A is
given as
A B
ω = −φ̇uθ
(2.284)
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
(2.285)
51
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
(2.286)
Consequently,
F
ωB = θ̇(sin φur + cos φuφ ) − φ̇uθ = θ̇ sin φur − φ̇uθ + θ̇ cos φuφ
(2.287)
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
(2.288)
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
Therefore,
F
v = R θ̇ cos φuθ + R φ̇uφ
(2.289)
(2.290)
(2.291)
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
(2.292)
Now we have
d F v
= R(θ̈ cos φ − θ̇ φ̇ sin φ)uθ + R φ̈uφ
(2.293)
dt
F B
ω × F v = (θ̇ sin φur − φ̇uθ + θ̇ cos φuφ ) × (R θ̇ cos φuθ + R φ̇uφ )
B
= R θ̇ 2 cos φ sin φuφ − R θ̇ φ̇ sin φuθ
−R φ̇2 ur − R θ̇ 2 cos2 φur
(2.294)
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
(2.295)
52
Chapter 2. Kinematics
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
53
The position of the particle is then given as
r = r er
(2.296)
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
F
dr Bdr F B
=
+ ω ×r
dt
dt
v=
(2.297)
First, the angular velocity of B in F is obtained from the angular velocity addition theorem as
F B
ω = F ωA + AωB
(2.298)
Now we have
F
ωA
= Ω = Ωex
(2.299)
= θ̇ez
(2.300)
ωB = Ωex + θ̇ez
(2.301)
A
B
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
= cos θ er − sin θ eθ
(2.302)
ey
= sin θ er + cos θ eθ
(2.303)
Therefore,
F
ωB = Ω(cos θ er − sin θ eθ ) + θ̇ez = Ω cos θ er − Ω sin θ eθ + θ̇ez
(2.304)
54
Chapter 2. Kinematics
Now the two terms required to obtain F v are given as
F
dr
= ṙ er
(2.305)
dt
F B
ω × r = (Ω cos θ er − Ω sin θ eθ + θ̇ez ) × r er = r θ̇eθ + r Ω sin θ (2.306)
ez
Therefore, the velocity of the particle in reference frame F is
F
v = ṙ er + r θ̇eθ + r Ω sin θ ez
(2.307)
Next, the acceleration of the particle as viewed by an observer fixed to the
ground is given from the transport theorem as
F
a=
d F Bd F F B F
v =
v + ω × v
dt
dt
F
(2.308)
Now we have
d F v
= r̈ er + (ṙ θ̇ + r θ̈)eθ + ṙ Ω sin θ + r (Ω̇ sin θ + Ωθ̇ cos θ )(2.309)
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 )
B
= 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
(2.310)
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
+ r Ω̇ sin θ + 2(ṙ Ω sin θ + r Ωθ̇ cos θ ) ez
(2.311)
55
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
sin θ er + cos θ eθ
Ex =
(2.312)
Ey = − cos θ er + sin θ eθ
Using the bases {Ex , Ey , Ez } and {er , eθ , ez }, the position of point P is given
as
rP = rO + rP /O = xEx + ler
(2.313)
56
Chapter 2. Kinematics
Next, the angular velocity of reference frame A in reference frame F is given as
F
ωA = θ̇ez
(2.314)
The velocity of point P in reference frame F is then given as
F
vP =
F
F
d
d
rP /O = F vO + F vP /O
(rO ) +
dt
dt
(2.315)
Now since rO is expressed in the basis {Ex , Ey , Ez } and {Ex , Ey , Ez } is fixed, we
have that
F
d
F
vO =
(2.316)
(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
A
d
d
rP /O =
rP /O + F ωA × rP /O
dt
dt
(2.317)
Now we have that
A
d
rP /O
dt
F A
ω × rP /O
= 0
(2.318)
= θ̇ez × ler = lθ̇eθ
(2.319)
Adding Eq. (2.318) and Eq. (2.319) gives
F
vP /O = lθ̇eθ
(2.320)
Therefore, the velocity of point P in reference frame F is given as
F
vP = ẋEx + lθ̇eθ
(2.321)
Next, the acceleration of point P in reference frame F is obtained as
F
Now we have that
F
aP =
d F vP
dt
F
vP = F vO + F vP /O
(2.322)
(2.323)
where
F
vO
= ẋEx
(2.324)
vP /O
= lθ̇eθ
(2.325)
d F Fd F
vO +
vP /O
dt
dt
(2.326)
F
Consequently,
F
aP =
F
57
Now, since F vO is expressed in the basis {Ex , Ey , Ez }, we have that
F
aO =
d F vO = ẍEx
dt
F
(2.327)
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
aP /O =
vO =
vO + ω × vO
(2.328)
dt
dt
Now we have that
d F vO
= lθ̈eθ
dt
F A
ω × F vO = θ̇ez × lθ̇eθ = −lθ̇ 2 er
A
(2.329)
(2.330)
Adding Eq. (2.329) and Eq. (2.330) gives
F
aP /O = −lθ̇ 2 er + lθ̈eθ
(2.331)
Then, adding Eq. (2.327) and Eq. (2.331), we obtain the velocity of point P in
reference frame F as
F
aP = ẍEx − lθ̇ 2 er + lθ̈eθ
(2.332)
Finally, substituting the expression for Ex from Eq. (2.312), 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
(2.333)
58
Chapter 2. Kinematics
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
(2.334)
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
(2.335)
Using the velocity from Eq. (2.335), the speed of the particle in reference frame
F is given as
F
v = F v = ẋ 1 + tan2 x = ẋ sec x
(2.336)
59
Arc-length Parameter as a Function of x
Now we recall the arc-length equation as
d F F
s = v = ẋ 1 + tan2 x = ẋ sec x
dt
(2.337)
Separating variables in Eq. (2.337), we obtain
F
ds = sec xdx
(2.338)
Integrating both sides of Eq. (2.338) gives
F
s − F s0 =
x
sec xdx
(2.339)
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
(2.340)
Noting that F s(0) = F s0 = 0, the arc-length is given as
F
s = ln [sec x + tan x]x
x0
(2.341)
Simplifying Eq. (2.341), we obtain
F
s = ln
sec x + tan x
sec x0 + tan x0
(2.342)
Intrinsic Basis
Next, we need to compute the intrinsic basis. First, we have the tangent vector
as
Fv
ẋ(Ex + tan xEy )
1
tan x
=
Ex +
Ey
(2.343)
et = F =
ẋ sec x
v
sec x
sec x
Now we note that sec x = 1/ cos x. Therefore,
tan x
= sin x
sec x
(2.344)
et = cos xEx + sin xEy
(2.345)
Eq. (2.343) then simplifies to
Next, the principle unit normal is given as
F
det
= κ F ven
dt
(2.346)
60
Chapter 2. Kinematics
Differentiating et in Eq. (2.345), we obtain
F
det
= −ẋ sin xEx + ẋ cos xEy
dt
Consequently,
F de t
= ẋ = κ F v
dt (2.347)
(2.348)
which implies that
F
−ẋ sin xEx + ẋ cos xEy
det /dt
=
= − sin xEx + cos xEy
en = F
ẋ
det /dt (2.349)
Then, using F v from Eq. (2.336), we obtain the curvature as
κ=
1
ẋ
=
= cos x
ẋ sec x
sec x
(2.350)
Finally, the principle unit bi-normal is given as
eb = et × en = (cos xEx + sin xEy ) × (− sin xEx + cos xEy ) = Ez
(2.351)
Velocity and Acceleration in Terms of Intrinsic Basis
Using the speed from Eq. (2.336), the velocity of the particle in reference frame
F is given in terms of the instrinsic basis as
F
v = ẋ sec xet
(2.352)
Next, the acceleration is given in terms of the intrinsic basis as
F
a=
d F v et + κ F v en
dt
(2.353)
Now, using F v from Eq. (2.336), we obtain d(F v)/dt as
d F v = ẍ sec x + ẋ 2 sec x tan x = sec x ẍ + ẋ 2 tan x
dt
(2.354)
Also, using κ from Eq. (2.350) we obtain
κ F v = cos x(ẋ sec x)2 = ẋ 2 sec x
(2.355)
The acceleration of the particle in reference frame F is then given as
F
a = sec x ẍ + ẋ 2 tan x et + ẋ 2 sec xen
(2.356)
Chapter 3
Kinetics of Particles
Question 3–1
A particle of mass m moves in the vertical plane along a track in the form of a
circle as shown in Fig. P3-1. The equation for the track is
r = r0 cos θ
Knowing that gravity acts downward and assuming the initial conditions θ(t =
0) = 0 and θ̇(t = 0) = θ̇0 , determine (a) the differential equation of motion for
the particle and (b) the force exerted by the track on the particle as a function
of θ.
r=
r0
c
θ
os
θ
O
Figure P 3-1
m
g
62
Chapter 3. Kinetics of Particles
Solution to Question 3–1
Kinematics
Let F be a reference frame fixed to the track. Then, choose the following coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at point O
=
=
=
Along OP when θ = 0
Out of page
Ez × Ex
Next, let A be a reference frame fixed to the direction OP . Then, choose the
following coordinate system fixed in reference frame A:
er
ez
eθ
Origin at point O
=
=
=
Along OP
Ez
ez × er
The geometry of the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in Fig. 3-1. Using
Fig. 3-1, we have that
Ex
= cos θ er − sin θ eθ
(3.1)
Ey
= sin θ er + cos θ eθ
(3.2)
Ey
er
eθ
θ
θ
Ex
Figure 3-1
Geometry of Coordinate System for Question 3.1
Next, the position of the particle is given in terms of the basis {er , eθ , ez } as
r = r er = r0 cos θ er
(3.3)
Furthermore, since the angle θ is measured from the fixed horizontal direction,
the angular velocity of A in F is given as
F
ωA = θ̇ez
(3.4)
63
Applying the transport theorem to r from reference frame A to F , the velocity
of the particle in reference frame F as
F
v=
F
dr Adr F A
=
+ ω ×r
dt
dt
(3.5)
Now we have
A
dr
= −r0 θ̇ sin θ er
dt
F A
ω × r = θ̇Ez × r0 cos θ er = r0 θ̇ cos θ eθ
(3.6)
(3.7)
Adding the expressions in Eq. (3.6) and Eq. (3.7), we obtain the velocity in reference frame F as
F
v = −r0 θ̇ sin θ er + r0 θ̇ cos θ eθ
(3.8)
Re-writing Eq. (3.8), we obtain
F
v = r0 θ̇(− sin θ er + cos θ eθ )
(3.9)
The speed in reference frame F is then given as
F
v = F v = r0 θ̇
(3.10)
Dividing F v by F v, we obtain the tangent vector as
et = − sin θ er + cos θ eθ
(3.11)
Next, the principal unit normal vector is computed as
en =
F
det /dt
F
det /dt
(3.12)
Applying the transport theorem to et , we have
F
A
det
det F A
=
+ ω × et
dt
dt
(3.13)
Now
A
det
dt
F A
ω × et
Consequently,
F
= −θ̇ cos θ er − θ̇ sin θ eθ
(3.14)
= θ̇ez × (− sin θ er + cos θ eθ )
= −θ̇ cos θ er − θ̇ sin θ eθ
(3.15)
det
= −2θ̇ cos θ er − 2θ̇ sin θ eθ
dt
(3.16)
64
Chapter 3. Kinetics of Particles
which implies that
−2θ̇ cos θ er − 2θ̇ sin θ eθ
en =
− 2θ̇ cos θ er − 2θ̇ sin θ eθ = − cos θ er − sin θ eθ
(3.17)
The principal unit bi-normal vector to the track is then obtained as
eb = et × en = (− sin θ er + cos θ eθ ) × (− cos θ er − sin θ eθ ) = ez
(3.18)
The acceleration as viewed by an observer fixed to the track is then obtained as
F
a=
F
d F A d F F A F
v =
v + ω × v
dt
dt
(3.19)
Now we have
A
d F det
v
= r0 θ̈et + r0 θ̇
dt
dt
= r0 θ̈et + r0 θ̇(−θ̇ cos θ er − θ̇ sin θ eθ )
A
= r0 θ̈et + r0 θ̇ 2 (− cos θ er − sin θ eθ )
= r0 θ̈et + r0 θ̇ 2 en
F
A
ω
(3.20)
F
× v = θ̇ez × r0 θ̇et
= θ̇eb × r0 θ̇et = r0 θ̇ 2 en
(3.21)
where we note that the results of Eqs. (3.14) and (3.17) have been used to obtain
the result given in Eq. (3.20). Therefore,
F
a = r0 θ̈et + 2r0 θ̇ 2 en
(3.22)
Kinetics
Next, in order to obtain the differential equation of motion, we need to apply
Newton’s 2nd Law to the particle. The free body diagram of the particle is given
in Fig. 3-2 as where
m
N
mg
Figure 3-2
Free Body Diagram for Question 3–1.
N
= Reaction Force of Track on Particle
mg = Force of Gravity
65
Now we know that the reaction force is orthogonal to the track while gravity
acts vertically downward. Consequently, we have that
N = Nn en + Nb
mg = −mgEy
(3.23)
(3.24)
Then, using the expression for Ey from Eq. (3.2), we obtain the force of gravity
as
(3.25)
mg = −mg(sin θ er + cos θ eθ ) = −mg sin θ er − mg cos θ eθ
The total force on the particle is then given as
F = N + mg = Nn en + Nb eb − mg sin θ er − mg cos θ eθ
(3.26)
Applying Newton’s 2nd Law using the acceleration from Eq. (3.22), we obtain
Nn en + Nb eb − mg sin θ er − mg cos θ eθ = mr0 θ̈et + 2mr0 θ̇ 2 en
(3.27)
Now it is seen that the unknown reaction forces exerted by the track lie in the
directions of en and eb . Therefore, the reaction force exerted by the track can
be eliminated if the scalar product with et is taken with both sides of Eq. (3.27)
as
(Nn en +Nb eb −mg sin θ er −mg cos θ eθ )·et = (mr0 θ̈et +2mr0 θ̇ 2 en )·et (3.28)
Then, observing that en · et = eb · et = 0, Eq. (3.28) simplifies to
−mg sin θ er · et − mg cos θ eθ · et = mr0 θ̈
(3.29)
Now, using the expression for et from Eq. (3.11), we have that
er · et
= er · (− sin θ er + cos θ eθ ) = − sin θ
(3.30)
eθ · et
= eθ · (− sin θ er + cos θ eθ ) = cos θ
(3.31)
Substituting the results of Eq. (3.30) and Eq. (3.31) into Eq. (3.29), we obtain
mg sin2 θ − mg cos2 θ = mr0 θ̈
(3.32)
cos2 θ − sin2 θ = cos 2θ
(3.33)
Now we also note that
Therefore, Eq. (3.32) can be written as
−mg cos 2θ = mr0 θ̈
(3.34)
Next, taking the scalar product of Eq. (3.28) in the en direction, we obtain
Nn − mg sin θ er · en − mg cos θ eθ · en = 2mr0 θ̇ 2
(3.35)
66
Chapter 3. Kinetics of Particles
Then, using the expression for en from Eq. (3.17), we have that
er · en
= er · (− cos θ er − sin θ eθ ) = − cos θ
(3.36)
eθ · en
= eθ · (− cos θ er − sin θ eθ ) = − sin θ
(3.37)
Substituting the results of Eq. (3.36) and Eq. (3.37) into Eq. (3.35) gives
Nn + mg sin θ cos θ + mg cos θ sin θ = 2mr0 θ̇ 2
(3.38)
Now we have that
sin θ cos θ + cos θ sin θ = 2 sin θ cos θ = sin 2θ
(3.39)
Consequently, Eq. (3.38 can be written as
N + mg sin 2θ = 2mr0 θ̇ 2
(3.40)
Finally, taking the scalar product of Eq. (3.28) in the eb direction, we obtain
Nb = 0
(3.41)
The following three scalar equations then result from Eqs. (3.34), Eq. (3.40), and
Eq. (3.41):
mr0 θ̈
2mr0 θ̇
2
= −mg cos 2θ
(3.42)
= Nn + mg sin 2θ
(3.43)
0 = Nb
(3.44)
Since Eq. (3.43) contains no reaction forces, it is the differential equation of
motion, i.e. the differential equation of motion is given as
mr0 θ̈ = −mg cos 2θ
(3.45)
Rearranging this last equation, we obtain the differential equation of motion for
the particle as
g
θ̈ +
cos 2θ = 0
(3.46)
r0
Force Exerted by Track on Particle As a Function of θ
First we note the following:
θ̈ =
dθ̇
dθ̇ dθ
dθ̇
=
= θ̇
dt
dθ dt
dθ
(3.47)
Substituting Eq. (3.47) into Eq. (3.46), we obtain
θ̇
dθ̇
g
cos 2θ = 0
+
dθ
r0
(3.48)
67
Rearranging Eq. (3.48) and separating variables, we obtain
θ̇dθ̇ = −
g
cos 2θdθ
r0
(3.49)
Integrating this last equation, we obtain
1 2
g
θ̇ − θ̇02 = −
[sin 2θ]θθ0
2
2r0
(3.50)
Noting that θ(t = 0) = 0, this last equation simplifies to
θ̇ 2 = θ̇02 −
g
sin 2θ
r0
(3.51)
Solving for the reaction force using Eq. (3.42), we obtain
We then obtain
Nn = 2mr0 θ̇ 2 − mg sin 2θ
(3.52)
g
sin 2θ − mg sin 2θ
Nn = 2mr0 θ̇02 −
r0
(3.53)
Simplifying this last equation, we obtain
Nn = 2mr0 θ̇02 − 3mg sin 2θ
The force exerted by the track on the particle is then given as
N = 2mr0 θ̇02 − 3mg sin 2θ en
(3.54)
(3.55)
68
Chapter 3. Kinetics of Particles
Question 3–2
A collar of mass m slides without friction along a rigid massless rod as shown
in Fig. P3-2. The collar is attached to a linear spring with spring constant K and
unstretched length L. Assuming no gravity, determine the differential equation
of motion for the collar.
O
K
L
m
x
Figure P 3-2
Solution to Question 3–2
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Ey
Ez
Ex
Origin at Attachment
Point of Spring
=
Up
=
Out of Page
=
Ey × E z
Then, in terms of the basis {Ex , Ey , Ez }, the position of the collar is given as
r = xEx − LEy
(3.56)
Since reference frame F is fixed and L is constant, the velocity of the collar in
reference frame F is given as
F
v=
F
dr
= ẋEx
dt
(3.57)
Furthermore, the acceleration of the collar in reference frame F is given as
F
a=
d F v = ẍEx
dt
F
(3.58)
Next, using the free body diagram of the collar as shown in Fig. 3-3, we have
that
Fs = Spring Force
N = Reaction Force of Rod on Collar
69
N
Fs
Figure 3-3
Free Body Diagram for Question 3.2
Since the reaction force acts in the Ey direction, we have that
N = NEy
(3.59)
Next, the force in a linear spring is given as
Fs = −K( −
0 )us
(3.60)
First, the stretched length of the spring is
= r − rA (3.61)
where the position of the attachment point is zero, i.e., rA = 0. Therefore, the
stretched length of the spring is given as
= r = xEx − LEy = x 2 + L2
(3.62)
Furthermore, the unstretched length of the spring is given as
0
=L
(3.63)
Finally, the direction from the attachment point to the particle, us , is given as
us =
xEx − LEy
r − rA
= √
r − rA x 2 + L2
(3.64)
Consequently, the force of the spring is given as
Fs = −K
xEx − LEy
x 2 + L2 − L √
x 2 + L2
(3.65)
Grouping this last expression into components, we obtain
Fs = −K
x
L
x 2 + L2 − L √
Ex + K x 2 + L2 − L √
Ey
2
2
2
x +L
x + L2
(3.66)
The resultant force acting on the particle is then given as
Fs = −K
x
L
x 2 + L2 − L √
Ex + N + K x 2 + L 2 − L √
Ey
x 2 + L2
x 2 + L2
(3.67)
70
Chapter 3. Kinetics of Particles
Applying Newton’s 2nd Law, we obtain
x
L
x 2 + L2 − L √
Ex + N + K x 2 + L2 − L √
Ey = mẍEx
2
2
2
x +L
x + L2
(3.68)
Using the Ex -component of the last equation, we obtain
−K
mẍ = −K
x 2 + L2 − L √
x
x 2 + L2
(3.69)
Rearranging this last equation, we obtain the differential equation of motion as
mẍ + K
x 2 + L2 − L √
x
x2
+ L2
=0
(3.70)
71
Solution to Question 3–3
A bead of mass m slides along a fixed circular helix of radius R and constant
helical inclination angle φ as shown in Fig. P3-3. The equation for the helix is
given in cylindrical coordinates as
z = Rθ tan φ
(3.71)
Knowing that gravity acts vertically downward, determine the differential equation of motion for the bead in terms of the angle θ using (a) Newton’s 2nd law
and (b) the work-energy theorem for a particle. In addition assuming the initial
conditions θ(t = 0) = θ0 and θ̇(t = 0) = θ̇0 , determine (c) the displacement
attained by the bead when it reaches its maximum height on the helix.
g
R
A
Om P
θ
φ
z
Figure P 3-3
Solution to Question 3–3
Kinematics
Let F be a reference frame fixed to the helix. Then, choose the following coordinate system fixed in reference frame F :
Ex
Ey
Ez
Origin at O
=
=
=
Along er at t = 0
Along eθ at t = 0
er × eθ
Next, let A be a reference
frame that rotates with the projection of the position of particle into the Ex , Ey -plane. Corresponding to A, we choose the
72
Chapter 3. Kinetics of Particles
following coordinate system to describe the motion of the particle:
er
ez
eθ
Origin at O
=
=
=
Along Radial Direction of Circle
Ez from Reference Frame F
ez × er
Now, since φ is the angle formed by the helix with the horizontal, we have from
the geometry that
z = Rθ tan φ
(3.72)
Suppose now that we make the following substitution:
α tan φ
(3.73)
Then the position of the bead can be written as
r = Rer + tan φRθez = Rer + αRθez
(3.74)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇ez
(3.75)
Then, differentiating Eq. (3.74) in reference frame F , the velocity of the bead is
given as
F
dr Adr F A
F
=
+ ω ×r
v=
(3.76)
dt
dt
where
A
dr
= αR θ̇ez
dt
(3.77)
F ωA × r = θ̇e × (Re + αRθe ) = R θ̇e
z
r
z
θ
Adding the two expressions in Eq. (3.77), we obtain
F
v = R θ̇eθ + αR θ̇ez
(3.78)
The speed in reference frame F is then given as
F
Consequently,
d F v = F v = R θ̇ 1 + α2 ≡
s
dt
F
ds = R 1 + α2 dθ
(3.79)
(3.80)
Integrating both sides of Eq. (3.80), we obtain
Fs
Fs
0
ds =
θ
θ0
R 1 + α2 dθ
(3.81)
73
We then obtain
s − F s0 = R 1 + α2 (θ − θ0 )
(3.82)
Solving Eq. (3.82) for s, the arclength is given as
F
s = F s0 + R 1 + α2 (θ − θ0 )
(3.83)
F
Now the tangent vector in reference frame F is given as
et =
Fv
(3.84)
Fv
Using the speed from Eq. (3.79) and the velocity from Eq. (3.78), the tangent
vector in reference frame F is obtained as Substituting the expressions for F v
and F v from part (a) into Eq. (3.84), we obtain
et =
R θ̇eθ + αR θ̇ez
√
R θ̇ 1 + α2
(3.85)
Simplifying Eq. (3.85), we have
eθ + αez
et = √
1 + α2
Next, we have
F
det
= κ F ven
dt
(3.86)
(3.87)
Applying the rate of change transport theorem between reference frames A and
F , we have
F
A
det
det F A
=
+ ω × et
(3.88)
dt
dt
where
A
det
dt
F
ωA × et
= 0
(3.89)
eθ + αez
θ̇
= θ̇ez × √
= −√
er
2
1+α
1 + α2
(3.90)
Adding Eqs. (3.89) and (3.90) gives
F
θ̇
det
= −√
er
dt
1 + α2
(3.91)
The principal unit normal is then given as
en =
F
F
det /dt
det /dt
= −er
(3.92)
74
Chapter 3. Kinetics of Particles
We then obtain the principal unit bi-normal vector as
eθ + αez
αeθ − ez
eb = et × en = √
× (−er ) = − √
1 + α2
1 + α2
(3.93)
Furthermore, the curvature is given as
κ=
F
1
det /dt
=
Fv
R(1 + α2 )
(3.94)
The acceleration in reference frame F is then given as
F
a=
2
d F v et + κ F v en
dt
(3.95)
Using the expression for F v from Eq. (3.79) we have that
d F v = R θ̈ 1 + α2
dt
(3.96)
Also, using the curvature from Eq. (3.94), we have that
κ
F
v
2
=
2
1
2
1
+
α
= R θ̇ 2
R
θ̇
R(1 + α2 )
(3.97)
Thus, the acceleration is given as
F
a = R θ̈ 1 + α2 et + R θ̇ 2 en
(3.98)
Kinetics
Using the free body diagram in Fig. 3-4, we have that
Nn = Reaction Force of Track on Bead in en Direction
Nb = Reaction Force of Track on Bead in eb Direction
mg = Force of Gravity
Therefore,
Nb
⊗
Nn
mg
Figure 3-4
Free Body Diagram for Question 3.4
F = Nn + Nb + mg
(3.99)
75
From the geometry we have that
Nn
= Nn en
(3.100)
Nb
= Nb en
(3.101)
mg = −mgez
(3.102)
F = Nn en + Nb eb − mgez
(3.103)
Consequently,
Now from Eqs. (3.86) and (3.93) we have
et
eb
eθ + αez
√
1 + α2
αeθ − ez
= −√
1 + α2
=
(3.104)
Using Eq. (3.104) we√can obtain an expression for ez √
in terms of et and eb . First,
multiplying et by α 1 + α2 and multiplying eb by − 1 + α2 , we obtain
√
α 1 + α2 et
√
− 1 + α2 eb
= αeθ + α2 ez
= αeθ − ez
Subtracting these last two equations gives
α 1 + α2 et + 1 + α2 eb = (1 + α2 )ez
(3.105)
(3.106)
Solving this last equation for ez , we obtain
αet + eb
ez = √
1 + α2
The force F given in Eq. (3.103) can then be written as
αet + eb
√
F = Nn en + Nb eb − mg
1 + α2
(3.107)
(3.108)
Separating this last equation into components, we obtain
mgα
mg
F = −√
et + Nn en + Nb − √
eb
1 + α2
1 + α2
(3.109)
(a) Differential Equation Using Newton’s 2nd Law
Setting F = mF a using the expression for F a from Eq. (3.98), we obtain
mgα
mg
−√
et + Nn en + Nb − √
eb = mR θ̈ 1 + α2 et + mR θ̇ 2 en (3.110)
1 + α2
1 + α2
76
Chapter 3. Kinetics of Particles
Equating components in Eq. (3.110) yields the following three scalar equations:
mgα
−√
1 + α2
Nn
Nb
= mR θ̈ 1 + α2
= mR θ̇ 2
mg
= √
1 + α2
(3.111)
(3.112)
(3.113)
It is noted that, because it contains no reaction forces, Eq. (3.111) is the differential equation of motion for the particle, i.e., the differential equation of motion
is given as
mgα
=0
(3.114)
mR θ̈ 1 + α2 + √
1 + α2
Eq. (3.114) can be rewritten as
mR(1 + α2 )θ̈ + mgα = 0
(3.115)
Simplifying Eq. (3.115), we obtain
θ̈ +
g
α=0
R(1 + α2 )
(3.116)
Then, using the the fact that α = tan φ from Eq. (3.73), we obtain
θ̈ +
g
tan φ = 0
R(1 + tan2 φ)
(3.117)
Now from trigonometry we have that
1 + tan2 φ = sec2 φ
(3.118)
Using the result of Eq. (3.118) in Eq. (3.117), we obtain the differential equation
of motion as
g tan φ
=0
(3.119)
θ̈ +
R sec2 φ
(b) Differential Equation Using Work-Energy Theorem
Applying the work-energy theorem to the bead, we have
d F T = F · Fv
dt
(3.120)
Using the expression for F v from Eq. (3.78), the kinetic energy in reference frame
F is given as
F
T =
1 F
1
1
m v · F v = m(R 2 θ̇ 2 α2 R 2 θ̇ 2 ) = mR 2 (1 + α2 )θ̇ 2
2
2
2
(3.121)
77
Computing the rate of change of kinetic energy, we obtain
d F T = mR 2 (1 + α2 )θ̇ θ̈
dt
(3.122)
Next, using the resultant force acting on the bead as given in Eq. (3.109), the
power produced by all forces is given as
mgα
mg
et + Nn en + Nb − √
F · Fv = − √
eb · F v
2
2
1+α
1+α
(3.123)
Recalling by definition that F v = F vet , Eq. (3.123) simplifies to
mgα F
v
F · Fv = − √
1 + α2
Then, substituting the expression for F v from Eq. (3.79), we have
mgα
R θ̇ 1 + α2
F · Fv = − √
1 + α2
(3.124)
(3.125)
Setting Eq. (3.122) equal to Eq. (3.125), we obtain
mgα
R θ̇ 1 + α2
mR 2 (1 + α2 )θ̇ θ̈ = − √
1 + α2
Rearranging Eq. (3.126) yields
mgα
R 1 + α2 = 0
θ̇ mR 2 (1 + α2 )θ̈ + √
1 + α2
(3.126)
(3.127)
Observing that θ̇ ≠ 0 as a function of time, the differential equation of motion
is obtained as
mgα
mR 2 (1 + α2 )θ̈ + √
R 1 + α2 = 0
(3.128)
1 + α2
(c) Maximum Displacement of Bead
For this particular problem, we can obtain the maximum distance traveled using
the alternate form of the work-energy theorm for a particle. In particular, we
know that
d F E = Fnc · F v
(3.129)
dt
Now since the force of gravity is conservative, we know that the only possible
non-conservative forces are due to the reaction of the track on the bead, i.e.,
Fnc = Nn + Nb
(3.130)
Using the expressions for Nn and Nb from Eq. (3.100) and Eq. (3.101), we have
that
Fnc = Nn en + Nb eb
(3.131)
78
Chapter 3. Kinetics of Particles
Furthermore, since en and eb both lie in the direction orthogonal to et , we have
that
en · et = 0
(3.132)
eb · et = 0
Furthermore, since F v = F vet , we know that
Fnc = (Nn en + Nb eb ) · F vet = 0
(3.133)
d F E =0
dt
(3.134)
Consequently,
Integrating Eq. (3.134), we obtain
F
E = constant
(3.135)
Now since F E = F T + F U, we have
F
T + F U = constant
(3.136)
Next, we know that the bead will attain its maximum distance when its velocity is
zero, i.e., the maximum distance will be attained when θ̇ = 0. Using Eq. (3.136),
we have that
F
T0 + F U 0 = F T1 + F U 1
(3.137)
where the subscript “0” is at time t = t0 = 0, and the subscript “1” is at time t =
t1 when θ̇ = 0. We already have the kinetic energy of the bead from Eq. (3.121).
Next, since the only conservative force acting on the bead is due to gravity, the
potential energy of the bead in reference frame F is given as
F
U = F Ug = −mg · r
(3.138)
Substituting the expression for r from Eq. (3.74) and the expression for mg from
Eq. (3.102) into Eq. (3.138), we obtain
F
U = F Ug = mgez · (Rer + αRθez ) = mgRθα
Then, substituting the expression for
into Eq. (3.136), we obtain
F
T and
F
(3.139)
U from Eqs. (3.121) and 3.139
1
mR 2 θ̇ 2 (1 + α2 ) + mgRθα = constant
2
(3.140)
Furthermore, applying Eq. (3.137), we obtain
1
1
mR 2 θ̇02 (1 + α2 ) + mgRθ0 α = mR 2 θ̇12 (1 + α2 ) + mgRθ1 α
2
2
(3.141)
79
Now we know that since the maximum distance is obtained when the velocity of
the bead is zero, we must have that θ̇1 = 0. Furthermore, since the initial value
of θ is zero, we have that θ0 = 0. Consequently, Eq. (3.141) reduces to
1
mR 2 θ̇02 (1 + α2 ) = mgRθ1 α
2
(3.142)
Solving Eq. (3.142) for θ1 , we obtain
θ1 =
R θ̇02 (1 + α2 )
2gα
(3.143)
Finally, since the distance traveled along the helix is equivalent to the arclength,
the distance traveled along the helix is given from Eq. (3.83) as
F
R θ̇02 (1 + α2 )
s = R 1 + α2
2gα
(3.144)
Simplifying Eq. (3.144), we obtain the maximum distance traveled along the incline as
3/2
R 2 θ̇02 F
s=
(3.145)
1 + α2
2gα
80
Chapter 3. Kinetics of Particles
Question 3–5
A collar of mass m is constrained to move along a frictionless track in the form
of a logarithmic spiral as shown in Fig. P3-5. The equation for the spiral is given
as
r = r0 e−aθ
where r0 and a are constants and θ is the angle as shown in the figure. Assuming that gravity acts downward, determine the differential equation of motion in
terms of the angle θ using (a) Newton’s 2nd law and (b) the work-energy theorem
for a particle.
θ
m
r
=
r0
e −a
g
θ
O
Figure P 3-5
Solution to Question 3–5
Kinematics
Let F be a reference frame fixed to the track. 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, let A be a reference frame fixed to the direction of Om. Then, we choose
the following coordinate system fixed in reference frame F :
er
Ez
eθ
Origin at O
=
=
=
Along Om
Out of Page
Ez × e r
81
The relationship between the bases {Ex , Ey , Ez } and {er , eθ , ez } is given as
er
= cos θ Ex + sin θ Ey
(3.146)
eθ
= − sin θ Ex + cos θ Ey
(3.147)
The position of the particle is then given as
r = r er = r0 e−aθ er
(3.148)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇Ez
(3.149)
Applying the rate of change transport theorem between reference frames A and
F , we obtain the velocity of the particle in reference frame F as
F
v=
F
dr Adr F A
=
+ ω ×r
dt
dt
(3.150)
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
(3.151)
(3.152)
Adding the expressions in Eq. (3.151) and Eq. (3.152), we obtain the velocity of
the particle in reference frame F as
F
v = −ar0 θ̇e−aθ er + r0 θ̇e−aθ eθ
(3.153)
Simplifying Eq. (3.150), we obtain F v as
F
v = r0 θ̇e−aθ [−aer + eθ ]
(3.154)
Now we need the acceleration of the collar in reference frame F . For this problem it is most convenient to obtain F a in terms of an intrinsic basis as viewed
by an observer fixed to the track. First, the tangent vector is given as
et =
Fv
F v
=
Fv
F v
(3.155)
where F v is the speed of the particle in reference frame F . Now we know that
the speed of the particle in reference frame F is given as
F
v = F v = r0 θ̇e−aθ a2 + 1
(3.156)
Dividing F v in Eq. (3.154) by F v in Eq. (3.156), we obtain et as
−aer + eθ
et = √
a2 + 1
(3.157)
82
Chapter 3. Kinetics of Particles
Then, the principle unit normal vector is obtained as
en =
F
det /dt
F
det /dt
(3.158)
Now we have from the basic kinematic equation that
F
where
F
det
dt
A
det
det F A
=
+ ω × et
dt
dt
(3.159)
= 0
−aer + eθ
er + aeθ
× et = θ̇Ez × √
= −θ̇ √
2
a +1
a2 + 1
Adding the expressions in Eq. (3.160), we obtain
(3.160)
F ωA
F
Consequently,
Dividing
F
er + aeθ
det
= −θ̇ √
dt
a2 + 1
F de t
= θ̇
dt (3.161)
(3.162)
det /dt in Eq. (3.161) by F det /dt in Eq. (3.162), we obtain en as
er + aeθ
en = − √
a2 + 1
(3.163)
Furthermore, the curvature, κ, is obtained as
κ=
F det /dt
Fv
(3.164)
Substituting F det /dt from Eq. (3.162) and F v from Eq. (3.156), we obtain κ
as
1
√
(3.165)
κ=
−aθ
r0 e
a2 + 1
The acceleration is then given in terms of the intrinsic basis as
2
d F F
a=
v et + κ F v en
(3.166)
dt
Now we have that
Furthermore,
2
κ Fv =
d F v = r0 (θ̈ − aθ̇ 2 )e−aθ a2 + 1
dt
1
√
r0 e−aθ a2 + 1
r02 θ̇ 2 e−2aθ (a2 + 1) = r0 θ̇ 2 e−aθ a2 + 1
The acceleration in reference frame F is then given as
F
a = r0 e−aθ a2 + 1(θ̈ − aθ̇ 2 )et + r0 θ̇ 2 e−aθ a2 + 1en
(3.167)
(3.168)
(3.169)
83
Kinetics
The free body diagram of the particle is shown in Fig. 3-5. It can be seen that
N
Figure 3-5
mg
Free Body Diagram fof Question 3–5.
the following two forces act on the collar: (1) the reaction force of the track, N,
and (2) gravity, mg. Since N acts in the direction normal to the track, we have
N = Nn en
(3.170)
Now, since gravity acts vertically downward, we have that
mg = −mgEy
(3.171)
where Ey is the vertically upward direction. Resolving Ey in the basis {er , eθ , Ez },
we have that
(3.172)
Ey = sin θ er + cos θ eθ
Consequently, the force of gravity can be written as
mg = −mg sin θ er − mg cos θ eθ
(3.173)
Then the total force acting on the collar is given as
F = N + mg = Nn en − mg sin θ er − mg cos θ eθ
(3.174)
(a) Differential Equation Using Newton’s 2nd Law
Setting F equal to mF a using F a from Eq. (3.169), we obtain
Nn en − mg sin θ er − mg cos θ eθ = mr0 e−aθ a2 + 1(θ̈ − aθ̇ 2 )et
+ mr0 θ̇ 2 e−aθ a2 + 1en
(3.175)
Now we know that, in order to obtain the differential equation of motion, we
need to eliminate the reaction force exerted by the track. An easy way to
eliminate N is to take the scalar product in the et -direction on both sides of
Eq. (3.175). Noting that en · et = 0, we then obtain
(3.176)
−mg sin θ er · et − mg cos θ eθ · et = mr0 e−aθ a2 + 1(θ̈ − aθ̇ 2 )
84
Chapter 3. Kinetics of Particles
Now, using the expression for et from Eq. (3.157) and the expression for en from
Eq. (3.163), we have that
er · et
eθ · et
a
−aer + eθ
= −√
= er · √
2
2
a +1
a +1
1
−aer + eθ
=√
= eθ · √
2
2
a +1
a +1
(3.177)
Substituting the expressions in Eq. (3.177) into Eq. (3.176), we obtain
mg sin θ √
a
a2 + 1
− mg cos θ √
1
a2 + 1
= mr0 e−aθ a2 + 1(θ̈ − aθ̇ 2 )
(3.178)
Rearranging and simplifying Eq. (3.178), we obtain the differential equation of
motion as
g
(3.179)
eaθ (cos θ − a sin θ ) = 0
θ̈ − aθ̇ 2 +
r0 (a2 + 1)
(b) Differential Equation Using Work-Energy Theorem for a Particle
To obtain the differential equation of motion using the work-energy, we choose
to apply the alternate form of the work-energy theorem for a particle. The
alternate form of the work-energy theorem is given in reference frame F as
d F E = Fnc · F v
dt
(3.180)
Now for this problem we know that the only two forces acting on the particle are
the force of gravity and the reaction force of the track. Moreover, we know that
the force of gravity is conservative while the reaction force is non-conservative.
Therefore, we have Fnc as
Fnc = N
(3.181)
Now, since N acts in the direction of en and F v acts in the direction of et , we
have that
(3.182)
Fnc · F v = N · F v = Nen · F vet = 0
Consequently,
d F E =0
dt
(3.183)
Now the total energy in reference frame F is given as
F
E = FT + FU
(3.184)
First, the kinetic in reference frame F is given as
F
T =
1 F
m v · Fv
2
(3.185)
85
Using the expression for
reference frame F as
F
Fv
from Eq. (3.154), we obtain the kinetic energy in
1 m r0 θ̇e−aθ [−aer + eθ ] · r0 θ̇e−aθ [−aer + eθ ]
2
1
= mr02 (a2 + 1)θ̇ 2 e−2aθ
2
T =
(3.186)
Next, since gravity is the only conservative force acting on the particle, the potential energy in reference frame F is given as
F
U = F Ug
(3.187)
Now since gravity is a constant force, we have that
F
Ug = −mg · r
(3.188)
Using the expression for r from Eq. (3.148) and the expression for mg from
Eq. (3.171), we obtain F Ug as
F
Ug = −mg · r = −(−mgEy ) · r0 e−aθ er = mgr0 e−aθ Ey · er
(3.189)
Using the expression for er from Eq. (3.146), we have that
Ey · er = Ey · (cos θ Ex + sin θ Ey ) = sin θ
(3.190)
Consequently, F Ug can be written as
F
Ug = mgr0 e−aθ sin θ
(3.191)
Then, adding Eq. (3.186) and Eq. (3.191), the total energy in reference frame F
is given as
F
E = FT + FU =
1
mr02 (a2 + 1)θ̇ 2 e−2aθ + mgr0 e−aθ sin θ
2
(3.192)
Then, computing the rate of change of F E, we obtain
d F E = mr02 (a2 + 1) θ̇ θ̈e−2aθ = aθ̇ 2 θ̇e−2aθ
dt
+ mgr0 −aθ̇e
−aθ
sin θ + θ̇e
−aθ
(3.193)
cos θ
Eq. (3.193) can be re-written as
d F E = mr02 (a2 + 1)θ̇e−2aθ θ̈ − aθ̇ 2 + mgr0 θ̇e−aθ (−a sin θ + cos θ )
dt
(3.194)
86
Chapter 3. Kinetics of Particles
Simplifying Eq. (3.194) and setting the result equal to zero, we obtain
d F E = θ̇ mr02 (a2 + 1)e−2aθ (θ̈ − aθ̇ 2 ) + mgr0 e−aθ (cos θ − a sin θ ) = 0
dt
(3.195)
Now since θ̇ ≠ 0 as a function of time (otherwise the particle would not be
moving), the term in the square brackets must be zero, i.e.,
mr02 (a2 + 1)e−2aθ (θ̈ − aθ̇ 2 ) + mgr0 e−aθ (cos θ − a sin θ ) = 0
(3.196)
Then, dividing Eq. (3.196) by mr02 (a2 + 1)e−2aθ , we obtain the differential equation of motion as
θ̈ − aθ̇ 2 +
g
eaθ (cos θ − a sin θ ) = 0
r0 (a2 + 1)
(3.197)
It is seen that the result of Eq. (3.197) is identical to that obtained in part (a).
87
Question 3–7
A particle of mass m slides without friction along the inner surface of a fixed
cone of semi-vertex angle β as shown in the Fig. P3-7. The equation for the cone
is given in cylindrical coordinates as
z = r cot β
Knowing that the basis {Ex , Ey , Ez } is fixed to the cone, that θ is the angle between the Ex -direction and the direction OQ where Q is the projection of the
particle into the {Ex , Ey }-plane, and that gravity acts vertically downward, determine a system of two differential equations in terms of r and θ that describe
the motion of the particle.
Ez
g
m
β
P
O
Ey
θ
r
Q
Ex
Figure P 3-7
Solution to Question 3–7
Kinematics
First, let F be a reference frame fixed to the cone. Then, choose the following
coordinate system fixed in reference frame F :
Ex
Ey
Ez
Origin at O
=
=
=
As Given
As Given
Ex × Ey = As Given
Next, let A be a reference frame fixed to the plane formed by the points O, Q,
and P . Then, choose the following coordinate system fixed in reference frame
88
Chapter 3. Kinetics of Particles
A:
er
ez
eθ
Origin at O
=
=
=
Along OQ
Ez
Ez × er
The position of the particle is then given as
r = r er + zez = r er + r cot βez
(3.198)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇Ez
(3.199)
The velocity of the particle in reference frame F is then obtained from the rate
of change transport theorem as
F
F
dr Adr F A
=
+ ω ×r
v=
dt
dt
(3.200)
Now we have that
A
dr
= ṙ er + ṙ cot βez
dt
F A
ω × r = θ̇ez × (r er + r cot βez ) = r θ̇eθ
(3.201)
(3.202)
Adding the expressions in Eq. (3.201) and Eq. (3.202), we obtain the velocity of
the particle in reference frame F as
F
v = ṙ er + r θ̇eθ + ṙ cot βez
(3.203)
Next, applying the rate of change transport theorem to F v, we obtain the acceleration of the particle in reference frame F as
F
a=
d F Ad F F A F
v =
v + ω × v
dt
dt
F
(3.204)
Now we have that
d F v
= r̈ er + (ṙ θ̇ + r θ̈)eθ + r̈ cot βez
(3.205)
dt
F A
ω × F v = θ̇Ez × ṙ er + r θ̇eθ + ṙ cot βEz = ṙ θ̇eθ − r θ̇ 2 er (3.206)
A
Adding the expressions in Eq. (3.205) and Eq. (3.206), we obtain the acceleration
of the particle in reference frame F as
F
a = (r̈ − r θ̇ 2 )er + (r θ̈ + 2ṙ θ̇)eθ + r̈ cot βez
(3.207)
89
N
m
mg
Figure 3-6
Free Body Diagram of Particle for Question 3–7.
Kinetics
In order to determine the differential equations of motion, we need to apply
Newton’s 2nd Law, i.e., we need to apply F = mF a. The free body diagram of
the particle is shown in Fig. 3-6. Using Fig. 3-6, we see that the only two forces
acting on the particle are
N
= Reaction Force of Cone on Particle
mg = Force of Gravity
Since we now that N must lie in the direction orthogonal to the surface of the
cone while the force of gravity acts vertically downward, we can write N and
mg, respectively, as
N = Nn
mg = −mgez
(3.208)
(3.209)
where n is the direction orthogonal to the surface of the cone at the location of
the particle. Now we know that the direction orthogonal to the surface of the
cone is the same as the direction of the gradient of the function that describes
the cone. In particular, the function that describes the surface of the cone is
given as
z = r cot β
(3.210)
Rearranging Eq. (3.210), the function that describes the surface of the cone is
given in cylindrical coordinates as
f (r , θ, z) = z − r cot β = 0
(3.211)
The gradient of f in cylindrical coordinates is then given as
∇f =
∂f
1 ∂f
∂f
er +
eθ +
ez
∂r
r ∂θ
∂z
(3.212)
90
Chapter 3. Kinetics of Particles
where
∂f
∂r
∂f
∂θ
∂f
∂z
= − cot β
(3.213)
= 0
(3.214)
= 1
(3.215)
∇f = − cot βer + ez
(3.216)
We then obtain ∇f as
The unit normal to the surface of the cone is then given as
n=
− cot βer + ez
∇f
= ∇f 1 + cot2 β
(3.217)
Now from trigonometry we have that
1 + cot2 β = csc2 β = 1/ sin2 β
(3.218)
Substituting the result of Eq. (3.218) into Eq. (3.217), we obtain the unit normal
to the surface of the cone as
n = sin β(− cot βer + ez ) = − cos βer + sin βez
(3.219)
Then, substituting the expression for n from Eq. (3.219) into Eq. (3.208), we
obtain the reaction force of the cone on the particle as
N = N(− cos βer + sin βez ) = −N cos βer + N sin βez
(3.220)
The resultant force on the particle is then given as
F = N + mg = −N cos βer + N sin βez − mgez = −N cos βer + (N sin β − mg)ez
(3.221)
Setting F in Eq. (3.221) equal to mF a using the expression for F a from Eq. (3.207),
we have that
−N cos βer + (N sin β − mg)ez = m(r̈ − r θ̇ 2 )er + m(r θ̈ + 2ṙ θ̇)eθ + mr̈ cot βez
(3.222)
Equating components in Eq. (3.222), we obtain the following three scalar equations:
−N cos β = m(r̈ − r θ̇ 2 )
0 = m(r θ̈ + 2ṙ θ̇)
N sin β − mg
= mr̈ cot β
(3.223)
(3.224)
(3.225)
91
Now, it is seen that Eq. (3.224) has no unknown reaction forces. Consequently,
Eq. (3.224) is one of the differential equations of motion. Dropping m from
Eq. (3.224), the first differential equation of motion can be written as
r θ̈ + 2ṙ θ̇ = 0
(3.226)
The second differential equation of motion can be obtained using Eq. (3.223)
and Eq. (3.223). In particular, we can rearrange Eq. (3.225) as
N sin β = mr̈ cot β + mg
(3.227)
Then, dividing Eq. (3.227) by Eq. (3.223), we obtain
mr̈ cot β + mg
N sin β
=
−N cos β
m(r̈ − r θ̇ 2 )
Eq. (3.228) simplifies to
tan β = −
r̈ cot β + g
(3.228)
(3.229)
r̈ − r θ̇ 2
Rearranging Eq. (3.229) gives
(r̈ − r θ̇ 2 ) tan β = −r̈ cot β − g
(3.230)
Then, dividing Eq. (3.230) by tan β, we obtain
(r̈ − r θ̇ 2 ) = r̈ cot2 β + g cot β = 0
(3.231)
Rearranging Eq. (3.231) gives
(1 + cot2 β)r̈ − r θ̇ 2 + g cot β = 0
(3.232)
Once again, using the fact that 1 + cot2 β = csc2 β, Eq. (3.232) simplifies to
csc2 βr̈ − r θ̇ 2 + g cot β = 0
(3.233)
Dividing Eq. (3.233) by csc2 β, we obtain the second differential equation of motion as
r̈ − r θ̇ 2 sin2 β + g cos β sin β = 0
(3.234)
The two differential equations that govern the motion of the particle are then
given from Eq. (3.226) and Eq. (3.234) as
r θ̈ + 2ṙ θ̇
= 0
(3.235)
r̈ − r θ̇ sin β + g cos β sin β = 0
(3.236)
2
2
92
Chapter 3. Kinetics of Particles
Question 3–9
A particle of mass m is attached to one end of a flexible but inextensible massless rope as shown in Fig. P3-9. The rope is wrapped around a cylinder of radius
R where the cylinder rotates with constant angular velocity Ω relative to the
ground. The rope unravels from the cylinder in such a manner that it never
becomes slack. Furthermore, point A is fixed to the cylinder and corresponds
to a configuration where no portion of the rope is exposed while point B is the
instantaneous point of contact of the exposed portion of the rope with the cylinder. Knowing that the exposed portion of the rope is tangent to the cylinder
at every instant of the motion, that θ is the angle between points A and B, and
assuming the initial conditions θ(t = 0) = 0, θ̇(t = 0) = Ω (where Ω = Ω),
determine (a) the angular velocity of the exposed portion of the rope as viewed
by an observer fixed to the ground, (b) the acceleration of the particle as viewed
by an observer fixed to the ground, (c) the differential equation for the particle
in terms of the variable θ, and (d) the tension in the rope as a function of time.
m
P
B
θ
A
O
R
Ω
Figure P 3-9
Solution to Question 3–9
Kinematics
First, let F be a reference frame that is fixed in the ground. Then, choose the
following coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along OA at t = 0
Out of Page
Ez × Ex
93
Next, let A be a reference frame fixed to the cylinder. Then, choose the following
coordinate system fixed in reference frame A:
ux
uz
uy
Origin at O
=
=
=
Along OA
Out of Page
uz × ux
Finally, let B be a reference frame fixed to the rope. Then, choose the following
coordinate system fixed in reference frame B:
ex
ez
ey
Origin at B
=
=
=
Along OB
Out of Page
ez × ex
Now, we note that the cylinder rotates with constant angular velocity Ω
about the uz -direction. Consequently, the angular velocity of the cylinder in
reference frame F is given as
F
ωA = Ω = Ωuz
(3.237)
Next, since A is fixed in the cylinder and B is fixed in the rope, the angular
velocity of the rope relative to the cylinder is equivalent to the angular velocity
of reference frame B relative to reference frame A. Observing from Fig. 3-7 that
θ defines the rotation of the rope relative to the cylinder, we have that
A
ωB = θ̇ez
(3.238)
Then, applying the angular velocity addition theorem, the angular velocity of
the rope relative to the ground is obtained by adding the results of Eq. (3.237)
and Eq. (3.238) to obtain
F
ωB = F ωA + A ωB = Ωuz + θ̇ez = (Ω + θ̇)ez
(3.239)
where we note that uz = ez . Next, we know that, when no portion of the rope is
exposed (i.e., s = 0), the particle is in contact with point A on the cylinder. Using
Fig. 3-7 along with the fact that the cylinder is circular, the arclength along the
cylinder from point A to point B is given as
s = Rθ
(3.240)
Differentiating Eq. (3.240), we obtain
ṡ = R θ̇
(3.241)
Next, the position of the particle is given in terms of the basis {ex , ey , ez } as
r = Rex − sey = Rex − Rθey
(3.242)
94
Chapter 3. Kinetics of Particles
s=
Rθ
m
θ
B
A
R
O
Figure 3-7
Geometry of Rope and Cylinder for Question 3–9.
The velocity of the particle in reference frame F is then given as
F
v=
F
dr Bdr F B
=
+ ω ×r
dt
dt
(3.243)
where
B
dr
dt
= −R θ̇ey
F B
ω ×r =
Ω + θ̇ ez × Rex − Rθey
= Rθ Ω + θ̇ ex + R Ω + θ̇ ey
(3.244)
(3.245)
Adding the expressions in Eq. (3.244) and Eq. (3.245), we obtain the velocity of
the particle in reference frame F as
F
v = −R θ̇ey + Rθ Ω + θ̇ ex + R Ω + θ̇ ey
(3.246)
Simplifying Eq. (3.246), we obtain
F
v = Rθ Ω + θ̇ ex + RΩey
(3.247)
Then, the acceleration in reference frame F is given as
F
v=
d F B d F F B F
v =
v + ω × v
dt
dt
F
(3.248)
where
d F v
=
R θ̇ Ω + θ̇ + Rθ θ̈ ex
dt
F B
ω × Fv =
Ω + θ̇ ez × Rθ Ω + θ̇ ex + RΩey
B
(3.249)
95
2
= −RΩ Ω + θ̇ ex + Rθ Ω + θ̇ ey
(3.250)
Adding Eq. (3.249) and Eq. (3.250), we obtain the acceleration of the particle in
reference frame F
F
2
a = R θ̇ Ω + θ̇ + Rθ θ̈ ex − RΩ Ω + θ̇ ex + Rθ Ω + θ̇ ey
(3.251)
Simplifying Eq. (3.251
F
2
a = R θ̇ 2 + Rθ θ̈ − RΩ2 ex + Rθ Ω + θ̇ ey
(3.252)
Kinetics and Differential Equation of Motion
We need to apply Newton’s 2nd law to the particle. Using the free body diagram
as shown in Fig. 3-8, it can be seen that the only force acting on the particle is
due to the tension in the rope. Since the tension must act along the direction of
the rope, we have that
(3.253)
T = T ey
Therefore, the resultant force acting on the particle is given as
m
T
Figure 3-8
Free Body Diagram for Question 3.5
F = T = T ey
(3.254)
Setting F = mF a using F a from Eq. (3.252), we obtain
2
T ey = m R θ̇ 2 + Rθ θ̈ − RΩ2 ex + mRθ Ω + θ̇ ey
We then obtain the following two scalar equations:
m R θ̇ 2 + Rθ θ̈ − RΩ2
= 0
2
= T
mRθ Ω + θ̇
(3.255)
(3.256)
(3.257)
From Eq (3.256) we have
R θ̇ 2 + Rθ θ̈ − RΩ2 = 0
(3.258)
Simplifying this last expression, we obtain the differential equation of motion
as
(3.259)
θ̇ 2 + θ θ̈ − Ω2 = 0
96
Chapter 3. Kinetics of Particles
Tension in Rope As a Function of Time
Eq. (3.259) can be solved for θ. This is done as follows. First, we note that
θ̇ 2 + θ θ̈ =
d θ θ̇
dt
(3.260)
Substituting this last result into Eq. (3.259), we obtain
d θ θ̇ − Ω2 = 0
dt
(3.261)
Integrating Eq. (3.261) once with respect to time, we obtain
θ θ̇ = Ω2 t + c1
(3.262)
where c1 is an arbitrary constant of integration. Then, applying the initial condition θ(t = 0) = 0, we have that
c1 = 0
(3.263)
θ θ̇ = Ω2 t
(3.264)
Therefore, we have
Then, separating variables in Eq. (3.264), we obtain
θdθ = Ω2 tdt
(3.265)
Integrating both sides of Eq. (3.265) gives
θ2
Ω2 t 2
=
+ c2
2
2
(3.266)
where c2 is an arbitrary constant of integration. Then, again applying the initial
condition θ(t = 0) = 0, we obtain
c2 = 0
(3.267)
Ω2 t 2
θ2
=
2
2
(3.268)
θ 2 = Ω2 t 2
(3.269)
Consequently,
which gives
Since θ has to be positive, we can take the principal square root of Eq. (3.269)
to obtain
θ = Ωt
(3.270)
Differentiating with respect to time, we have
θ̇ = Ω
(3.271)
97
Then, substituting θ from Eq. (3.270) and θ̇ from Eq. (3.271) into Eq. 3.257 gives
T = mRΩt (Ω + Ω)2 = 4mRΩ3 t
The tension in the rope is then given as
T = 4mRΩ3 t ey
(3.272)
(3.273)
98
Chapter 3. Kinetics of Particles
Question 3–10
A particle of mass m moves under the influence of gravity in the vertical plane
along a track as shown in Fig. P3-10. The equation for the track is given in
Cartesian coordinates as
y = − ln cos x
where −π /2 < x < π /2. Using the horizontal component of position, x, as the
variable to describe the motion determine the differential equation of motion
for the particle using (a) Newton’s 2nd law and (b) one of the forms of the workenergy theorem for a particle.
Ey
g
m
y = − ln cos x
O
Ex
Figure P 3-10
Solution to Question 3–10
Kinematics
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
=
Along Ox
Ex
Ey
=
Along Oy
=
Ex × Ey
Ez
The position of the particle is then given as
r = xEx − ln cos xEy
(3.274)
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
(3.275)
Using the velocity from Eq. (3.275), the speed of the particle in reference frame
F is given as
F
v = F v = ẋ 1 + tan2 x = ẋ sec x
(3.276)
99
Arclength Parameter as a Function of x
Now we recall the arclength equation as
d F F
s = v = ẋ 1 + tan2 x = ẋ sec x
dt
(3.277)
Separating variables in Eq. (3.277), we obtain
F
ds = sec xdx
(3.278)
Integrating both sides of Eq. (3.278) gives
F
s − F s0 =
x
sec xdx
(3.279)
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
(3.280)
Noting that F s(0) = F s0 = 0, the arclength is given as
F
s = ln [sec x + tan x]x
x0
(3.281)
Simplifying Eq. (3.281), we obtain
F
s = ln
sec x + tan x
sec x0 + tan x0
(3.282)
Intrinsic Basis
Next, we need to compute the intrinsic basis. First, we have the tangent vector
as
Fv
ẋ(Ex + tan xEy )
1
tan x
=
Ex +
Ey
(3.283)
et = F =
ẋ sec x
v
sec x
sec x
Now we note that sec x = 1/ cos x. Therefore,
tan x
= sin x
sec x
(3.284)
et = cos xEx + sin xEy
(3.285)
Eq. (3.283) then simplifies to
Next, the principle unit normal is given as
F
det
= κ F ven
dt
(3.286)
100
Chapter 3. Kinetics of Particles
Differentiating et in Eq. (3.285), we obtain
F
det
= −ẋ sin xEx + ẋ cos xEy
dt
Consequently,
F de t
= ẋ = κ F v
dt (3.287)
(3.288)
which implies that
F
−ẋ sin xEx + ẋ cos xEy
det /dt
=
= − sin xEx + cos xEy
en = F
ẋ
det /dt (3.289)
Then, using F v from Eq. (3.276), we obtain the curvature as
κ=
1
ẋ
=
= cos x
ẋ sec x
sec x
(3.290)
Finally, the principle unit bi-normal is given as
eb = et × en = (cos xEx + sin xEy ) × (− sin xEx + cos xEy ) = Ez
(3.291)
Differential Equation of Motion in Terms of x
The differential equation of motion is obtained using Newton’s 2nd Law for a
particle. First we obtain F using the free body diagram shown below: Using the
N
mg
free body diagram, we can see that
N
= Nen
mg = −mgEy
Therefore, the resultant force acting on the particle is
F = N + mg = Nen − mgEy
(3.292)
Next, the acceleration is given in the intrinsic basis as
F
a=
d F v et + κ F v en
dt
(3.293)
101
Now, using F v from Eq. (3.276), we obtain d(F v)/dt as
d F v = ẍ sec x + ẋ 2 sec x tan x = sec x ẍ + ẋ 2 tan x
dt
(3.294)
Then, using κ from Eq. (3.290) we obtain
κ F v = cos x(ẋ sec x)2 = ẋ 2 sec x
(3.295)
The acceleration of the particle in reference frame F is then given as
F
a = sec x ẍ + ẋ 2 tan x et + ẋ 2 sec xen
(3.296)
Setting F from Eq. (3.292) equal to mF a using F a from Eq. (3.296), we obtain
(3.297)
Nen − mgEy = m sec x ẍ + ẋ 2 tan x et + mẋ 2 sec xen
Now we can take the scalar products on both sides of Eq. (3.297) in the et and
en directions. Taking the scalar product on both sides of Eq. (3.297) in the
et -direction, we obtain
(3.298)
−mgEy · en = m sec x ẍ + ẋ 2 tan x
Taking the scalar product on both sides in the en -direction, we obtain
N − mgEy · en = mẋ 2 sec x
(3.299)
= Ey · (cos xEx + sin xEy ) = sin x
= Ey · (− sin xEx + cos xEy ) = cos x
(3.300)
Now we note that
Ey · et
Ey · en
Substituting Ey · et and Ey · en into Eq. (3.298) and Eq. (3.299), respectively, we
obtain the following two scalar equations:
m sec x ẍ + ẋ 2 tan x = −mg sin x
(3.301)
= N − mg cos x
mẋ 2 sec x
Seeing that the first equation in Eq. (3.301) has no reaction forces, the differential equation of motion of the particle is given as
(3.302)
m sec x ẍ + ẋ 2 tan x = −mg sin x
Eq. (3.302) can be rearranged to give
ẍ sec x + ẋ 2 sec x tan x + g sin x = 0
(3.303)
102
Chapter 3. Kinetics of Particles
Question 3–11
A particle of mass m moves in the horizontal plane as shown in Fig. P3-11. The
particle is attached to a linear spring with spring constant K and unstretched
length while the spring is attached at its other end to the fixed point O. Assuming no gravity, (a) determine a system of two differential equations of motion
for the particle in terms of the variables r and θ, (b) show that the total energy
of the system is conserved, and (c) show that the angular momentum relative to
point O is conserved.
O
r
K
m
θ
Figure P 3-11
Solution to Question 3–11
Acceleration of Particle
First, let F be a reference frame fixed to the ground. Next, let A be a reference
frame that is fixed to the direction Om. Corresponding to reference frame A, we
choose the following coordinate system to describe the motion of the particle:
er
Ez
eθ
Origin at O (Corner)
=
Along Om
=
Out of Page
=
Ez × er
The position of the particle is then given as
r = r er
(3.304)
Now the angular velocity of reference frame A in reference frame F is given as
F
ωA = θ̇Ez
(3.305)
103
The velocity in reference frame F is computed from the rate of change transport
theorem as
F
dr Adr F A
F
=
+ ω ×r
v=
(3.306)
dt
dt
where
Adr
dt
= ṙ er
F ωA
(3.307)
× r = θ̇Ez × r er = r θ̇eθ
Adding the two expressions in Eq. (3.307), we obtain F v as
F
v = ṙ er + r θ̇eθ
Applying the rate of change transport theorem to
particle in reference frame F is given as
F
a=
(3.308)
F v,
the acceleration of the
d F Ad F F A F
v =
v + ω × v
dt
dt
F
(3.309)
where
Ad
dt
Fv
F ωA
= r̈ er + (ṙ θ̇ + r θ̈)eθ
× F v = θ̇Ez × (ṙ er + r θ̇eθ ) = −r θ̇ 2 er + ṙ θ̇eθ
(3.310)
Adding the two expressions in Eq. (3.310), we obtain F a as
F
a = (r̈ − r θ̇ 2 )er + (r θ̈ + 2ṙ θ̇)eθ
(3.311)
System of Two Differential Equations
To obtain a system of two differential equations, we need to apply Newton’s
2nd Law to the particle. We already have F a from Eq. (3.311). Next, in order
to obtain an expression for the resultant force, F, we need to examine the free
body diagram as shown in Fig. 3-9 where Fs is the force due to the spring. Now
Fs
Figure 3-9
Free Body Diagram for Question 2.10.
we note that the general form for the force of a linear spring is
Fs = −K( −
0 )us
(3.312)
104
Chapter 3. Kinetics of Particles
Now since the attachment point of the spring for this problem is rA = 0, we
have that
(3.313)
= r − rA = r = r
Furthermore, the direction us is given as
us =
r er
r − rA
=
= er
r − rA r
(3.314)
Finally, the unstretched length of the spring is given as
0
=L
(3.315)
Therefore, we obtain the spring force as
Fs = −K (r − L) er
(3.316)
Next, since the only force acting on the particle is that of the spring, we can set
Fs from Eq. (3.316) equal to mF a from Eq. (3.311) to give
−K [r − L] er = m(r̈ − r θ̇ 2 )er + m(r θ̈ + 2ṙ θ̇)eθ
(3.317)
Equating components, we obtain the following two scalar equations:
= −K [r − L]
m(r̈ − r θ̇ 2 )
m(r θ̈ + 2ṙ θ̇) =
0
(3.318)
Since there are no reaction forces in either of the equations in Eq. (3.318), these
two equations are the differential equations of motion for the particle.
Conservation of Energy
From the work-energy theorem for a particle, we have that
d F E = Fnc · F v
dt
(3.319)
For this problem, the only force acting on the particle is that of the linear spring.
Since the spring force is conservative, we have that Fnc = 0. Therefore,
d F E =0
dt
which implies that
F
E = constant
which implies that energy is conserved.
(3.320)
(3.321)
105
Question 3–12
A particle of mass m is attached to a linear spring with spring constant K and
unstretched length r0 as shown in Fig. P3-12. The spring is attached at its other
end at point P to the free end of a rigid massless arm of length l. The arm
is hinged at its other end and rotates in a circular path at a constant angular
rate ω. Knowing that the angle θ is measured from the downward direction
and assuming no friction, determine a system of two differential equations of
motion for the particle in terms of r and θ.
O
r
l
P
K
ωt
m
θ
Figure P 3-12
Solution to Question 3–12
Kinematics
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along OP When t = 0
Out of Page
Ez × Ex
Next, let A be a reference frame fixed to the arm. Then, choose the following
coordinate system fixed in reference frame A:
ex
ez
ey
Origin at O
=
=
=
Along OP
Out of Page (= Ez )
ez × ex
Finally, let B be a reference frame fixed to the direction along which the spring
lies (i.e., the direction P m). Then, choose the following coordinate system fixed
106
Chapter 3. Kinetics of Particles
in reference frame B:
ur
uz
uθ
Origin at O
=
=
=
Along P m
Out of Page (= Ez = ez )
uz × ur
The geometry of the bases {Ex , Ey , Ez }, {ex , ey , ez }, and {ur , uθ , uz } is shown in
Fig. 3-10. Using Fig. 3-10, we have the following relationship between the basis
uθ
θ
ey
ωt
uz , ez , Ez
Ey
ωt
θ
Ex
ur
ex
Figure 3-10
Geometry of Bases {Ex , Ey , Ez }, {ex , ey , ez }, and {ur , uθ , uz } for
Question 3–12 .
{ex , ey , ez } and the basis {ur , uθ , uz }:
ex
ey
= cos(θ − ωt)ur − sin(θ − ωt)uθ
= sin(θ − ωt)ur + cos(θ − ωt)uθ
(3.322)
Next, observing that the basis {ex , ey , ez } rotates with angular rate ω relative
to the basis {Ex , Ey , Ez }, the angular velocity of reference frame A in reference
frame F is given as
F A
ω = ωEz = ωez
(3.323)
Next, using Eq. (3.322), we observe that the angle formed between the basis
vectors ur and ex (and similarly between uθ and ey ) is θ − ωt. Consequently,
the angular velocity of reference frame B in reference frame A is given as
A
ωB = (θ̇ − ω)ez = (θ̇ − ω)uz
(3.324)
Finally, observing that the basis {ur , uθ , uz } rotates with angular rate θ̇ relative
to the basis {Ex , Ey , Ez }, we obtain the angular velocity of reference frame B in
107
reference frame F as
F
ωB = θ̇uz
(3.325)
The position of the particle can be written as
r = rP + rm/P
(3.326)
where rP is the position of point P and rm/P is the postion of the particle relative
to point P . In terms of the bases defined above, we have that
rP
rm/P
= Rex
= r ur
(3.327)
Substituting the expressions from Eq. (3.327) into Eq. (3.326), we obtain
r = Rex + r ur
(3.328)
Differentiating the expression for the position as given in Eq. (3.328) in reference
frame F , we have that
F
v=
F
F
dr Fd
d
=
rm/P = F vP + F vm/P
(rP ) +
dt
dt
dt
(3.329)
Now since rP is expressed in 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 rP
between reference frames A and F to give
F
vP =
F
A
d
d
(rP ) =
(rP ) + F ωA × rP
dt
dt
(3.330)
Now we have that
A
d
(rP )
dt
F
ω
A
× rP
=
0
(3.331)
= ωez × Rex = Rωey
Adding the two expressions in Eq. (3.331), we obtain
F
vP = Rωey
(3.332)
Next, since rm/P is expressed in the basis {ur , uθ , uz } and {ur , uθ , uz } is fixed in
reference frame B, we can apply the rate of change transport theorem to rm/P
between reference frames B and F to give
F
vm/P =
F
B
d
d
rm/P =
rm/P + F ωB × rm/P
dt
dt
(3.333)
108
Chapter 3. Kinetics of Particles
Now we have that
B
d
rm/P
dt
F
=
ωB × rm/P
ṙ ur
(3.334)
= θ̇uz × r ur = r θ̇uθ
Adding the two expressions in Eq. (3.334), we have that
F
vm/P = ṙ ur + r θ̇uθ
(3.335)
Then, adding Eq. (3.332) and Eq. (3.335), we obtain the velocity of the particle in
reference frame F as
F
v = F vP + F vm/P = Rωey + ṙ ur + r θ̇uθ
(3.336)
Now the acceleration of the particle in reference frame F is given as
F
a=
d F F d F F d F
v =
vP +
vm/P = F aP + F am/P
dt
dt
dt
F
(3.337)
Observing that the expression for F vP as given in Eq. (3.332) is expressed in 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 F vP between reference frames A and
F to give
F
d F Ad F A F F
F
aP =
vP =
vP + ω × vP
(3.338)
dt
dt
Now since R and ω are constant, we have that
d F vP
dt
A
F
A
ω
F
× vP
=
0
(3.339)
2
= ωez × Rωex = −Rω ey
Adding the two expressions in Eq. (3.339), we obtain the acceleration of point P
in reference frame F as
F
aP = −Rω2 ex
(3.340)
Next, since F vm/P is expressed in the basis {ur , uθ , uz } and {ur , uθ , uz } is fixed
in reference frame B, the acceleration of the particle relative to point P in reference frame F can be obtained by applying the rate of change transport theorem
to F vm/P between reference frames B and F as
F
am/P =
Bd d F
F
vm/P =
vm/P + F ωB × F vm/P
dt
dt
F
(3.341)
Now we have that
d F
vm/P
dt
B
F ωB
× Fv
m/P
=
r̈ ur + (ṙ θ̇ + r θ̈)uθ
= θ̇uz × (ṙ ur + r θ̇uθ ) =
−r θ̇ 2 ur
+ ṙ θ̇uθ
(3.342)
109
Adding the two expressions in Eq. (3.342), we obtain the acceleration of the
particle relative to point P in reference frame F as
F
am/P = (r̈ − r θ̇ 2 )ur + (2ṙ θ̇ + r θ̈)uθ
(3.343)
Then, adding Eq. (3.340) and Eq. (3.343), we obtain the acceleration of the particle in reference frame F as
F
a = −Rω2 ex + (r̈ − r θ̇ 2 )ur + (2ṙ θ̇ + r θ̈)uθ
(3.344)
Finally, using the expression for ex in terms of {ur , uθ } from Eq. (3.322), the
acceleration of the particle in reference frame F can be written in terms of the
basis {ur , uθ , uz } as
F
a = −Rω2 [cos(θ−ωt)ur −sin(θ−ωt)uθ ]+(r̈ −r θ̇ 2 )ur +(2ṙ θ̇+r θ̈)uθ (3.345)
Simplifying Eq. (3.345), we obtain
F
a = [r̈ − r θ̇ 2 − Rω2 cos(θ − ωt)]ur + [2ṙ θ̇ + r θ̈ + Rω2 sin(θ − ωt)]uθ (3.346)
Kinetics and Differential Equations of Motion
In order to obtain the two differential equation of motion for the particle, we
need to apply Newton’s 2nd law, i.e., F = mF a. The free body diagram of the
particle is shown in Fig. 3-11. It can be seen that the only force acting on the
Fs
Figure 3-11
Free Body Diagram of Particle for Question 3–12 .
particle is due to the linear spring, Fs . Consequently, we have that
Fs = −K
−
0
us
(3.347)
Now we are given that the unstretched length of the spring is rO which implies
that 0 = r0 . Furthermore, the attachment point of the spring is rA = rP . Consequently, the stretched length of the spring is given as
= r − rA = r − rP (3.348)
Using the expression for r from Eq. (3.328) and the expression for rP from
Eq. (3.327), we obtain
= r ur + Rex − Rex = r ur = r
(3.349)
110
Chapter 3. Kinetics of Particles
Finally, we have that
us =
r − rP
r ur
r − rA
=
=
= ur
r − rA r − rP r
(3.350)
The spring force is then given as
Fs = −K(r − r0 )ur
(3.351)
The resultant force acting on the particle is then given as
F = Fs = −K(r − r0 )ur
(3.352)
Then, setting F in Eq. (3.352) equal to mF a where F a is obtained from Eq. (3.346),
we obtain
−K(r −r0 )ur = m[r̈ −r θ̇ 2 −Rω2 cos(θ−ωt)]ur +m[2ṙ θ̇+r θ̈+Rω2 sin(θ−ωt)]uθ
(3.353)
Equating components in Eq. (3.353), we obtain the following two scalar equations:
m[r̈ − r θ̇ 2 − Rω2 cos(θ − ωt)] = −K(r − r0 )
2
m[2ṙ θ̇ + r θ̈ + Rω sin(θ − ωt)] = 0
(3.354)
(3.355)
It can be seen that neither Eq. (3.354) nor Eq. (3.355) contains any unknown reactions forces. Consequently, Eq. (3.354) and Eq. (3.355) are the two differential
equations of motion for the particle. We can re-write Eq. (3.354) and Eq. (3.355)
in a slightly different form to give
m[r̈ − r θ̇ 2 − Rω2 cos(θ − ωt)] + K(r − r0 ) = 0
2
m[2ṙ θ̇ + r θ̈ + Rω sin(θ − ωt)] = 0
(3.356)
(3.357)
111
Question 3–13
A particle of mass m slides without friction along a surface in form of a paraboloid
as shown in Fig. P3-13. The equation for the paraboloid is
z=
r2
2R
where z is the height of the particle above the horizontal plane, r is the distance
from O to Q where Q is the projection of P onto the horizontal plane, and R is
a constant. Knowing that θ is the angle formed by the direction OQ with the
x-axis and that gravity acts downward, determine a system of two differential
equations.
z
g
m P
z=
O
r
θ
Q
r2
2R
y
x
Figure P 3-13
Solution to Question 3–13
Kinematics
First, let F be a reference frame fixed to the paraboloid. Then, choose the
following coordinate system fixed in reference frame F :
Ex
Ey
Ez
Origin at O
=
=
=
Along Ox
Along Oy
Ex × Ey
112
Chapter 3. Kinetics of Particles
Next, let A be a reference frame fixed to the plane formed by the vectors Ez and
OQ. Then, choose the following coordinate system fixed in reference frame A:
er
Ez
eθ
Origin at O
=
=
=
Along OQ
Up
Ez × e r
The position of the particle is then given as
r2
Ez
(3.358)
2R
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇Ez
(3.359)
r = r er +
The velocity of the particle in reference frame F is then obtained from the rate
of change transport theorem as
F
v=
F
dr Adr F A
=
+ ω ×r
dt
dt
(3.360)
Now we note that
A
dr
dt
F ωA
=
× r = θ̇Ez × r er +
r ṙ
ṙ er +
Ez
R
r2
Ez
2R
(3.361)
= r θ̇eθ
Adding the two expressions in Eq. (3.361), we obtain the velocity of the particle
in reference frame F as
F
v = ṙ er + r θ̇eθ +
r ṙ
Ez
R
(3.362)
Next, applying the rate of change transport theorem to F v, we obtain the acceleration of the particle in reference frame F as
F
Now we have that
A
d F v
=
dt
F ωA
a=
d F Ad F F A F
v =
v + ω × v
dt
dt
F
r̈ er + (ṙ θ̇ + r θ̈)eθ +
ṙ 2 + r r̈
Ez
R
r ṙ
Ez = ṙ θ̇eθ − r θ̇ 2 er
= θ̇Ez × ṙ er + r θ̇eθ +
R
× Fv
(3.363)
(3.364)
Adding the expressions in Eq. (3.364), we obtain the acceleration of the particle
in reference frame F as
F
a = (r̈ − r θ̇ 2 )er + (r θ̈ + 2ṙ θ̇)eθ +
ṙ 2 + r r̈
Ez
R
(3.365)
113
Kinetics
We need to apply Newton’s 2nd Law, i.e. F = mF a. The free body diagram of
the particle is shown in Fig. 3-12. We note from Fig. 3-12 that N is the reaction
N
mg
Figure 3-12
Free Body Diagram of Particle for Question 3–13.
force of the paraboloid on the particle and that mg is the force due to gravity.
We further note that N must lie normal to the surface at the point of contact.
Now we note that the vector that is normal to a surface is in the direction of
the gradient of the function that defines the surface. In order to compute the
gradient, we re-write the equation for the paraboloid in the following form:
f (r , θ, z) = z −
r2
=0
2R
(3.366)
Then, from calculus, the gradient in obtained in cylindrical coordinates as
∇f =
∂f
1 ∂f
∂f
er +
eθ +
Ez
∂r
r ∂θ
∂z
(3.367)
Computing the gradient, we obtain
r
∇f = − er + Ez
R
(3.368)
The unit vector in the direction of the gradient is then given as
r
− er + Ez
∇f
= R
en =
∇f r 2
1+
R
(3.369)
which implies that the normal force is
⎡
⎤
r
⎢ − er + Ez ⎥
⎢ R
⎥
⎥
N = Nen = N ⎢
⎢
⎥
2⎦
⎣
r
1+
R
(3.370)
Next, the force of gravity is
mg = −mgEz
(3.371)
114
Chapter 3. Kinetics of Particles
The resultant force on the particle is then given as
⎡
⎤
r
⎢ − er + Ez ⎥
⎢ R
⎥
⎥ − mgEz
F = N + mg = N ⎢
⎢
⎥
2⎦
⎣
r
1+
R
which can be re-written as
⎛
(3.372)
⎞
r
R
⎜
⎜
1
F = −N er + ⎜
N
⎜
⎝
r 2
r
1+
1+
R
R
2
⎟
⎟
− mg ⎟
⎟ Ez
⎠
(3.373)
Setting F = mF a using F a from part (a), we obtain
⎛
⎞
r
⎜
⎟
⎜
⎟
1
ṙ 2 + r r̈
R
2
⎜
Ez
e r +⎜N − mg ⎟
=
m(r̈
−r
θ̇
)e
+m(r
θ̈+2ṙ
θ̇)e
+m
−N E
r
θ
⎟ z
R
⎝
⎠
r 2
r 2
1+
1+
R
R
(3.374)
We then obtain the following three scalar equations:
−N r
R
r
1+
R
2
m(r̈ − r θ̇ 2 )
⎞ = m(r θ̈ + 2ṙ θ̇)
0
⎛
⎜
⎜
1
⎜N ⎜
⎝
r
1+
R
=
2
⎟
⎟
− mg ⎟
⎟ =
⎠
m
(3.375)
ṙ 2 + r r̈
R
A system of two differential equations can then be obtained as follows. The first
differential equation is simply the second equation in Eq. (3.375), i.e.
m(r θ̈ + 2ṙ θ̇) = 0
(3.376)
r θ̈ + 2ṙ θ̇ = 0
(3.377)
Dropping m, we obtain
Next, rearranging the third equation in Eq. (3.375) by adding mg to both sides,
we obtain
ṙ 2 + r r̈
1
(3.378)
= mg + m
N
R
r 2
1+
R
115
Then, dividing the first equation in Eq. (3.375) by this last result, we obtain
−
r
r̈ − r θ̇ 2
= 2
R
ṙ + r r̈
+g
R
(3.379)
Rearranging and simplifying this last equation, we obtain the second differential
equation as
$
#
r
gr
r 2
r̈ + 2 ṙ 2 − r θ̇ 2 +
=0
(3.380)
1+
R
R
R
The system of two differential equations is then given as
r
#θ̈ + 2ṙ θ̇ $
r
gr
r 2
r̈ + 2 ṙ 2 − r θ̇ 2 +
1+
R
R
R
= 0
= 0
(3.381)
Conservation of Energy
Two forces act on the particle: N and mg. We know that the force of gravity
is conservative, but we do not know anything about N. However, we note the
following about N:
⎡
⎤
r
⎢ − er + Ez ⎥ ⎢ R
⎥
r ṙ
F
⎢
⎥
N· v=N=N⎢
(3.382)
⎥ · ṙ er + r θ̇eθ + R Ez
⎣
r 2⎦
1+
R
Simplifying Eq. (3.382), we obtain
⎧
⎪
⎪
r
⎪
⎪
−
⎨
R
N · Fv = N ⎪
⎪
⎪
r
⎪
⎩ 1+
R
2
ṙ + ⎫
⎪
⎪
⎪
⎪
⎬
r
R
r
1+
R
2
ṙ
⎪
⎪
⎪
⎪
⎭
=0
(3.383)
Therefore, N does no work. Consequently, from the work-energy theorem we
have that
d F E =N·v=0
(3.384)
dt
which implies that F E = constant, i.e. energy is conserved.
116
Chapter 3. Kinetics of Particles
Question 3–17
A particle of mass m is attached to an inextensible massless rope of length l as
shown in Fig. P3-17. The rope is attached at its other end to point A located at
the top of a fixed cylinder of radius R. As the particle moves, the rope wraps
itself around the cylinder and never becomes slack. Knowing that θ is the angle
measured from the vertical to the point of tangency of the exposed portion
of the rope with the cylinder and that gravity acts downward, determine the
differential equation of motion for the particle in terms of the angle θ. You may
assume in your solution that the angle θ is always positive.
A
B
g
θ
R
O
m
Figure P 3-17
Solution to Question 3–17
Kinematics
First, let F be a reference frame fixed to the circular track. Then, choose the
following coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along OA
Out of Page
Ez × Ex
Next, let A be a reference frame fixed to the exposed portion of the rope. Then,
choose the following coordinate system fixed in reference frame A:
er
ez
eθ
Origin at O
=
=
=
Along OB
Out of Page
ez × er
The geometry of the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in Fig. 3-13.
Using Fig. 3-13, we have that
Ex
= cos θ er − sin θ eθ
(3.385)
Ey
= sin θ er + cos θ eθ
(3.386)
117
Ex
er
Ey
θ
e z , Ez
θ
eθ
Figure 3-13
Geometry of Question 3–17.
Now we note that the rope has a fixed length l. Since the length of the portion
of the rope wrapped around the cylinder is Rθ, the exposed portion of the rope
must have length l − Rθ. Furthermore, since the exposed portion of the rope
lies along the direction from B to m, the position of the particle is given as
r = Rer + (l − Rθ)eθ
(3.387)
Furthermore, since the direction along OB is fixed to reference frame A, the
angular velocity of reference frame A in reference frame F is given as
F
ωA = θ̇ez
(3.388)
The velocity of the particle is then computed using the rate of change transport
theorem as
F
dr Adr F A
F
=
+ ω ×r
v=
(3.389)
dt
dt
Now we have that
A
dr
= −R θ̇eθ
dt
F A
ω × r = θ̇ez × [Rer + (l − Rθ)eθ ] = R θ̇eθ − (l − Rθ)θ̇er
(3.390)
(3.391)
Adding Eq. (3.390) and Eq. (3.391), we obtain the velocity of the particle in reference frame F as
F
v = −(l − Rθ)θ̇er
(3.392)
The acceleration of the particle in reference frame F is then obtained by applying the rate of change transport theorem to F v as
F
a=
d F A d F F A F
v =
v + ω × v
dt
dt
F
(3.393)
118
Chapter 3. Kinetics of Particles
Now we have that
d F v = − (−R θ̇)θ̇ + (l − Rθ)θ̈ er
dt F A
F
ω × v = θ̇ez × −(l − Rθ)θ̇er = −(l − Rθ)θ̇ 2 eθ
A
Adding Eq. (3.394) and Eq. (3.395), we obtain
F
a = − −R θ̇ 2 + (l − Rθ)θ̈ er − (l − Rθ)θ̇ 2 eθ
(3.394)
(3.395)
(3.396)
Eq. (3.396) simplifies to
F
a = R θ̇ 2 − (l − Rθ)θ̈ er − (l − Rθ)θ̇ 2 eθ
(3.397)
Kinetics
The free body diagram of the particle is shown in Fig. 3-14. From Fig. 3-14 it can
T
mg
Figure 3-14
Free Body Diagram for Question 3–17.
be seen that the two forces acting on the particle are
T
= Tension in Rope
mg = Force of Gravity
Since the tension must act along the direction of the exposed portion of the
rope and gravity acts vertically downward, we have that
T = T eθ
mg = −mgEx
(3.398)
(3.399)
Then, using the expresion for Ex from Eq. (3.385), the force of gravity can be
expressed in the basis {er , eθ , ez } as
mg − mg(cos θ er − sin θ eθ ) = −mg cos θ er + mg sin θ eθ
(3.400)
The resultant force on the particle is then given as
F = T + mg = T eθ − mg cos θ er + mg sin θ eθ = −mg cos θ er + (T + mg sin θ )eθ
(3.401)
119
Determination of Differential Equation Using Newton’s 2nd Law
Setting F from Eq. (3.401) equal to mF a using F a from Eq. (3.397), we have that
−mg cos θ er + (T + mg sin θ )eθ = m R θ̇ 2 − (l − Rθ)θ̈ er − m(l − Rθ)θ̇ 2 eθ
(3.402)
Equating components in Eq. (3.402) results in the following two scalar equations:
= −mg cos θ
(3.403)
m R θ̇ 2 − (l − Rθ)θ̈
m(l − Rθ)θ̇ 2
= T + mg sin θ
(3.404)
Then, since Eq. (3.403) has no unknown reaction forces, it is the differential
equation, i.e., the differential equation of motion is given as
(3.405)
m R θ̇ 2 − (l − Rθ)θ̈ = −mg cos θ
Simplifying Eq. (3.405) by dropping m and rearranging, we obtain the differential equation as
(3.406)
(l − Rθ)θ̈ − R θ̇ 2 − g cos θ = 0
Determination of Differential Equation Using Alternate Form of Work-Energy
Theorem
Since the motion of the particle can be described using a single variable (namely,
θ), we can apply the work-energy theorem for a particle to obtain the differential
equation of motion. In particular, we will use the alternate form of the workenergy theorem for a particle in reference frame F as
d F E = Fnc · F v
dt
(3.407)
Now the total energy in reference frame F is given as
F
E = FT + FU
(3.408)
First, we have the kinetic energy in reference frame F as
F
T =
1 F
m v · Fv
2
(3.409)
Substituting F v from Eq. (3.392), we have that
F
T =
1
m(l − Rθ)θ̇eθ · (l − Rθ)θ̇eθ
2
(3.410)
1
m(l − Rθ)2 θ̇ 2
2
(3.411)
Eq. (3.410) simplifies to
F
T =
120
Chapter 3. Kinetics of Particles
Furthermore, since the only conservative force acting on the particle is due to
gravity and gravity is a constant force, the potential energy is given as
F
U = F Ug = −mg · r
(3.412)
Substituting mg from Eq. (3.400) and r from Eq. (3.387), we obtain
F
U = −(−mg cos θ er + mg sin θ eθ ) · (Rer + (l − Rθ)eθ )
= mgR cos θ − mg(l − Rθ) sin θ
(3.413)
The total energy of the system is then obtained by adding T from Eq. (3.411)
and U from Eq. (3.413) as
F
E = FT + FU =
1
m(l − Rθ)2 θ̇ 2 + mgR cos θ − mg(l − Rθ) sin θ
2
(3.414)
Next, the only force other than gravity acting on the particle is that due to the
tension in the rope. Using the expression for the tension in the rope from
Eq. (3.398) and the velocity of the particle from Eq. (3.392), the power of the
tension force is given as
T · F v = T eθ · (l − Rθ)θ̇er = 0
(3.415)
Consequently, the power produced by all non-conservative forces is zero which
implies that
d F E =0
(3.416)
dt
Differentiating F E from Eq. (3.414) and setting the result equal to zero, we obtain
d F E = m(l − Rθ)(−R θ̇)θ̇ 2 + m(l − Rθ)2 θ̇ θ̈ − mgR θ̇ sin θ
dt
+ mgR θ̇ sin θ − mg(l − Rθ)θ̇ cos θ = 0
Factoring out m and θ̇ from Eq. (3.417), we obtain
mθ̇ −(l − Rθ)2 θ̈ − R(l − Rθ)θ̇ 2 − g(l − Rθ) cos θ = 0
(3.417)
(3.418)
Noting that θ̇ ≠ 0, we can drop m and θ̇ from Eq. (3.418) to give
(l − Rθ)θ̈ − R θ̇ 2 − g cos θ = 0
(3.419)
It can be seen that Eq. (3.419) is identical the result obtained using Newton’s 2nd
law as shown in Eq. (3.406).
121
Question 3–19
A collar of mass m slides without friction along a circular track of radius R
as shown in Fig. P3-19. Attached to the collar is a linear spring with spring
constant K and unstretched length zero. The spring is attached at the fixed
point A located a distance 2R from the center of the circle. Assuming no gravity
and the initial conditions θ(t = 0) = θ0 and θ̇(t = 0) = θ̇0 , determine (a) the
differential equation of motion for the collar in terms the angle θ and (b) the
reaction force exerted by the track on the collar as a function of the angle θ.
m
K
R
θ
A
O
2R
Figure P 3-19
Solution to Question 3–19
Kinematics
First, let F be a reference frame fixed to the track. Then, choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along Om at t = 0
Out of Page
Ez × Ex
Next, let A be a reference frame fixed to the direction of Om. Then, choose the
following coordinate system fixed in reference frame A:
er
ez
eθ
Origin at O
=
=
=
Along Om
Out of Page
Ez × e r
122
Chapter 3. Kinetics of Particles
The geometry of the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in Fig. 3-15 from
which we obtain
Ex
= cos θ er − sin θ eθ
(3.420)
Ey
= sin θ er + cos θ eθ
(3.421)
Ey
eθ
θ
er
θ
Ez
Figure 3-15
Ex
Geometry of Question .
The position of the particle is then given as
r = Rer
(3.422)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F
ωA = θ̇ez
(3.423)
The velocity of the particle in reference frame F is then obtained from the rate
of change transport theorem as
F
v=
F
dr Adr F
=
+ ωA × r
dt
dt
(3.424)
Now we have that
A
dr
= 0
dt
F
ωA × r = θ̇ez × Rer = R θ̇eθ
(3.425)
(3.426)
Adding the expressions in Eq. (3.425) and Eq. (3.426), we obtain the velocity of
the collar in reference frame F as
F
v = R θ̇eθ
(3.427)
The acceleration of the collar in reference frame F is obtained by applying the
rate of change transport theorem to F v as
F
a=
d F F
v + ωA × F v
dt
F
(3.428)
123
Now we have that
d F v
= R θ̈eθ
dt
F
ωA × F v = θ̇ez × (R θ̇eθ ) = −R θ̇ 2 er
F
(3.429)
(3.430)
Adding the expressions in Eq. (3.429) and Eq. (3.430), we obtain the acceleration
of the collar in reference frame F as
F
v = −R θ̇ 2 er + R θ̈eθ
(3.431)
Kinetics
The differential equation of motion will be determined by applying Newton’s
2nd law. First, the free body diagram of the collar is shown in Fig. 3-16. Using
N
Fs
Figure 3-16
Free Body Diagram for Question 3–19.
Fig. 3-16, we see that the following two forces act on the collar:
N
Fs
= Reaction force of Track on Particle
= Spring Force
Now we know that the reaction force N is orthogonal to the track. Furthermore,
since the collar is undergoing circular motion, the tangent vector to the track
is in the direction eθ . Consequently, the direction normal to the track is er .
Therefore, the reaction force can be written as
N = Ner
(3.432)
Next, the general expression for a spring force is
Fs = −K( −
0 )us
(3.433)
Now we recall for a spring that
us
= r − rA r − rA
=
r − rA (3.434)
(3.435)
124
Chapter 3. Kinetics of Particles
where A is the attachment point of the spring. It is seen from the geometry that
the spring is attached a distance 2a from the center of the circle. In terms of
the direction Ex , we have that
rA = −2REx
(3.436)
Then, using the expression for r from Eq. (3.422), we obtain
r − rA = Rer + 2REx
(3.437)
Therefore, the stretched length of the spring is obtained as
= r − rA = Rer + 2REx (3.438)
Then, substituting the result of Eq. (3.434) into Eq. (3.435), we obtain
us =
Rer + 2REx
r − rA
=
r − rA Rer + 2REx (3.439)
Substituting the results of Eq. (3.438) and Eq. (3.439) into Eq. (3.433) and using
the fact that 0 = 0, we have that
Rer + 2REx
= −K(Rer + 2REx ) = −KR(er + 2Ex )
Rer + 2REx (3.440)
Then, substituting the expression for Ex from Eq. (3.420) into Eq. (3.440), we
obtain
Fs = −KRer + 2REx Fs = −KRer − 2KR(cos θ er − sin θ eθ ) = −KR(1 + 2 cos θ )er + 2KR sin θ eθ
(3.441)
The resultant force acting on the particle is then given as
F = N + Fs = [N − KR(1 + 2 cos θ )] er + 2KR sin θ eθ
(3.442)
Setting F in Eq. (3.442) equal to mF a using F a from Eq. (3.431), we obtain
[N − KR(1 + 2 cos θ )] er + 2KR sin θ eθ = −mR θ̇ 2 + mR θ̈eθ
(3.443)
We then obtain the following two scalar equations:
−mR θ̇ 2
mR θ̈
= N − KR(1 + 2 cos θ )
(3.444)
= 2KR sin θ
(3.445)
(a) Differential Equation of Motion
Since Eq. (3.445) has no reaction forces and the motion of the particle is described using a single variable (i.e., θ), the differential equation of motion is
given as
mR θ̈ = 2KR sin θ
(3.446)
Simplifying Eq. (3.446), we obtain
θ̈ −
2K
sin θ = 0
m
(3.447)
125
(b) Reaction Force of Track on Particle As a Function of θ
Solving for the reaction force using Eq. (3.444), we obtain
N = −mR θ̇ 2 + KR(1 + 2 cos θ )
(3.448)
It is seen from Eq. (3.448) that, in order to obtain N as a function of θ, it is
necessary to find θ̇ 2 as a function of θ. We can obtain θ̇ 2 in terms of θ using
the differential equation in Eq. (3.447). First, we have from the chain rule that
θ̈ = θ̇
2K
dθ̇
=
sin θ
dθ
m
(3.449)
Separating variables in Eq. (3.449), we obtain
θ̇dθ̇ =
2K
sin θ dθ
m
(3.450)
Integrating Eq. (3.450) gives
θ̇
θ̇0
νdν =
θ
θ0
2K
sin ηdη
m
(3.451)
where ν and η are dummy variables of integration. We then obtain
θ̇ 2 − θ̇02
2K
=−
(cos θ − cos θ 0 )
2
m
(3.452)
Rearranging and simplifying Eq. (3.452) gives
θ̇ 2 = θ̇02 +
4K
(cos θ 0 − cos θ )
m
Substituting θ̇ 2 from Eq. (3.453) into Eq. (3.448), we obtain
4K
2
N = −mR θ̇0 +
(cos θ 0 − cos θ ) + KR(1 + 2 cos θ )
m
(3.453)
(3.454)
Simplifying Eq. (3.454) gives
N = −mR θ̇02 − 4KR(cos θ 0 − cos θ ) + KR(1 + 2 cos θ )
(3.455)
Eq. (3.455) simplifies further to
N = −mR θ̇02 − 4KR cos θ 0 + 6KR cos θ + KR
The reaction force exerted by the track on the particle is then given as
N = −mR θ̇02 − 4KR cos θ 0 + 6KR cos θ + KR er
(3.456)
(3.457)
126
Chapter 3. Kinetics of Particles
Question 3–20
A particle of mass m slides without friction along a fixed horizontal table as
shown in Fig. P3-20. The particle is attached to an inextensible rope. The rope
itself is threaded through a tiny hole in the table at point O such that the portion
of the rope that hangs below the table remains vertical. Knowing that a constant
vertical force F is applied to the rope, that the rope remains taut, and that gravity
acts vertically downward, (a) determine a system of two differential equations in
terms of r and θ describing the motion of the particle, (b) show that the angular
momentum of the particle relative to point O is conserved, and (c) show that
the total energy of the system is conserved.
r
O
m
θ
F
g
Figure P 3-20
Solution to Question 3–20
Kinematics
First, let F be a reference frame fixed to the table. Then, choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along Om When θ = 0
Orthogonal to Table
Ez × Ex
Next, let A be a reference frame fixed to the portion of the rope that lies on the
table. Then, choose the following coordinate system fixed in reference frame A:
er
ez
eθ
Origin at O
=
=
=
Along Om
Orthogonal to Table
ez × er
127
Then the position of the particle is given as
r = r er
(3.458)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F
ωA = θ̇ez
(3.459)
The velocity of the particle in reference frame F is then obtained from the rate
of change transport theorem as
F
v=
F
dr Adr F A
=
+ ω ×r
dt
dt
(3.460)
Now we have that
A
dr
= ṙ er
dt
F A
ω × r = θ̇ez × r er = r θ̇eθ
(3.461)
(3.462)
Adding Eq. (3.461) and Eq. (3.462), we obtain the velocity of the particle in reference frame F as
F
v = ṙ er + r θ̇eθ
(3.463)
The acceleration of the particle in reference frame F is then obtained by applying the rate of change transport theorem to F v as
F
a=
d F A d F F A F
v =
v + ω × v
dt
dt
F
(3.464)
Now we have that
d F v = r̈ er + (ṙ θ̇ + r θ̈)eθ
dt
F A
ω × F v = θ̇ez × (ṙ er + r θ̇eθ ) = ṙ θ̇eθ − r θ̇ 2 er
A
(3.465)
(3.466)
Adding Eq. (3.465) and Eq. (3.466), we obtain
F
a = (r̈ − r θ̇ 2 )er + (2ṙ θ̇ + r θ̈)eθ
(3.467)
Kinetics
The free body diagram of the particle is shown in Fig. 3-17. From Fig. 3-17 it can
be seen that the three forces acting on the particle are given as
N
= Force of Table on Particle
mg = Force of Gravity
F
= Force Exerted by Rope on Particle
128
Chapter 3. Kinetics of Particles
N
F
mg
Figure 3-17
Free Body Diagram for Question 3–20.
Now we have that
N = NEz
(3.468)
mg = −mgEz
(3.469)
F = −F er
(3.470)
It is noted that, because the rope exerts a known force on the particle, it is
necessary that the force F in Eq. (3.470) be in the negative er -direction. Then,
the resultant force acting on the particle (which we denote as R in order to avoid
confusion with the given force F) is given as
R = N + mg + F = NEz − mgEz + F er = F er + (N − mg)Ez
Then, setting R in Eq. (3.471) equal to mF a using the expression for
Eq. (3.467), we obtain
−F er + (N − mg)Ez = m(r̈ − r θ̇ 2 )er + m(2ṙ θ̇ + r θ̈)eθ
(3.471)
Fa
from
(3.472)
Equating components in Eq. (3.472), we obtain the following three scalar equations:
−F
= m(r̈ − r θ̇ 2 )
(3.473)
0 = m(2ṙ θ̇ + r θ̈)
(3.474)
0 = N − mg
(3.475)
From Eq. (3.475) we have that
N = mg
(3.476)
Furthermore, since neither Eq. (3.473) nor Eq. (3.474) has any unknown reaction
forces, these two equations are the differential equations of motion for the particle, i.e., a system of two differential equations of motion for the particle are
given as
−F
= m(r̈ − r θ̇ 2 )
0 = m(2ṙ θ̇ + r θ̈)
(3.477)
(3.478)
129
Conservation of Angular Momentum Relative to Point O
First, it is important to observe that O is fixed in the inertial reference frame F .
Then, using the definition of the angular momentum of a particle relative to an
inertially fixed point O, we have that
F
HO = (r − r0 ) × mF v
Then, noting that rO = 0 and substituting the expressions for r and
Eq. (3.458) and Eq. (3.463), respectively, into Eq. (3.479), we have that
F
HO = r er × mr θ̇eθ = mr 2 θ̇Ez
(3.479)
Fv
from
(3.480)
Now, in order to show that the angular momentum of the particle relative to
point O will be conserved, we need to show that
F
d F
HO = 0
(3.481)
dt
Differentiating F HO in Eq. (3.479) in reference frame F , we have that
F
d F
HO = m(2r ṙ θ̇ + r 2 θ̈)Ez
(3.482)
dt
where we note that Ez is a non-rotating direction. Then, using the second differential equation as given in Eq. (3.478) in Eq. (3.482), we obtain
F
d F
HO = 0
(3.483)
dt
which implies that F HO is conserved.
Conservation of Energy
Applying the alternate form of the work-energy theorem for a particle in reference frame F , we have that
d F E = Fnc · F v
(3.484)
dt
Now, examining the free body diagram of the particle as given in Fig. 3-17, we
see that the forces mg and F are conservative (since both of these forces are constant). Therefore, the only possible non-conservative force is N. Consequently,
we have that
(3.485)
Fnc · F v = N · F v
Substituting the expression for N from Eq. (3.468) and Eq. (3.463), we have that
Fnc · F v = NEz · (ṙ er + r θ̇eθ ) = 0
Substituting the result of Eq. (3.486) into Eq. (3.484), we have that
d F E =0
dt
which implies that energy is conserved.
(3.486)
(3.487)
130
Chapter 3. Kinetics of Particles
Question 3–22
A particle of mass m is attached to a linear spring with spring constant K and
unstretched length r0 as shown in Fig. P3-22. The spring is attached at its other
end to a massless collar where the collar slides along a frictionless horizontal
track with a known displacement x(t). Knowing that gravity acts downward,
determine a system of two differential equations in terms of the variables r and
θ that describe the motion of the particle.
x(t)
O
A
B
K
g
r
m
P
θ
Figure P 3-22
Solution to Question 3–22
Kinematics
First, let F be a reference frame fixed to the track. Then, choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at Q
When x = 0
=
To The Right
=
Out of Page
=
Ez × Ex
Next, let A be a reference frame fixed to the direction of QP such that Q is a
point fixed in reference frame A. Then, choose the following coordinate system
fixed in reference frame A:
er
ez
eθ
Origin at O
=
=
=
Along QP
Out of Page
Ez × e r
131
The geometry of the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in Fig. 3-18.
Using Fig. 3-18, we have that
Ex
= sin θ er + cos θ eθ
(3.488)
Ey
= − cos θ er + sin θ eθ
(3.489)
Ey
eθ
θ
ez , Ez
Ex
θ
er
Figure 3-18
Geometry of Bases {Ex , Ey , Ez } and {er , eθ , ez } for Question 3–22.
The position of the particle is then given as
r = rQ + rP /Q
(3.490)
rQ
= xEx
(3.491)
rP /Q
= r er
(3.492)
r = xEx + r er
(3.493)
Now we have that
Consequently, we obtain
Next, the angular velocity of reference frame A in reference frame F is given as
F
ωA = θ̇ez
(3.494)
Then, the velocity of the particle in reference frame is obtained as
F
F
F
dr Fd
d
=
v=
rQ +
rP /Q = F vQ + F vP /Q
dt
dt
dt
(3.495)
Now since rQ is expressed in terms of the basis {Ex , Ey , Ez } and {Ex , Ey , Ez } is
fixed in reference frame F , we have that
F
vQ = ẋEx
(3.496)
132
Chapter 3. Kinetics of Particles
Furthermore, since rP /Q is expressed in terms of the basis {er , eθ , ez } and {er , eθ , ez }
is fixed in reference frame A, we can obtain F vP /Q using the rate of change
transport theorem between reference frame A and reference frame F as
F
vP /Q =
F
A
d
d
rP /Q =
rP /Q + F ωA × rP /Q
dt
dt
(3.497)
Now we have that
A
d
rP /Q
dt
F A
ω × rP /Q
= ṙ er
(3.498)
= θ̇ez × r er = r θ̇eθ
(3.499)
Adding Eq. (3.498) and Eq. (3.499), we obtain F vP /Q as
F
vP /Q = ṙ er + r θ̇eθ
(3.500)
Then, adding Eq. (3.496) and Eq. (3.500), the velocity of the particle in reference
frame F is obtained as
F
v = ẋEx + ṙ er + r θ̇eθ
(3.501)
The acceleration of the particle in reference frame F is then obtained as
F
a=
d F F d F Fd F
v =
vQ +
vP /Q = F aQ + F aP /Q
dt
dt
dt
F
Now we have that
F
aQ = ẍEx
(3.502)
(3.503)
Furthermore, since F vP /Q is expressed in the basis {er , eθ , ez } and {er , eθ , ez } is
fixed in reference frame A, we obtain F aP /Q using the rate of change transport
theorem between reference frame A and reference frame F as
F
aP /Q =
Ad d F
F
vP /Q =
vP /Q + F ωA × F vP /Q
dt
dt
F
(3.504)
Now we have that
d F
vP /Q
= r̈ er + (ṙ θ̇ + r θ̈)eθ
dt
F A
ω × F vP /Q = θ̇ez × (ṙ er + r θ̇eθ ) = −r θ̇ 2 er + ṙ θ̇eθ
A
(3.505)
(3.506)
Adding Eq. (3.505) and Eq. (3.506), we obtain
F
aP /Q = (r̈ − r θ̇ 2 )er + (2ṙ θ̇ + r θ̈)eθ
(3.507)
Then, adding Eq. (3.503) and Eq. (3.507), we obtain the acceleration of the particle in reference frame F as
F
a = ẍEx + (r̈ − r θ̇ 2 )er + (2ṙ θ̇ + r θ̈)eθ
(3.508)
133
Fs
mg
Figure 3-19
Free Body Diagram of Particle for Question 3–22.
Kinetics
The free body diagram of the particle is shown in Fig. 3-19.
Using Fig. 3-19. we see that the following forces act on the particle:
= Force of Spring
Fs
mg = Force of Gravity
First, we know that the model for a linear spring is
Fs = −K( −
where
that
0 )us
(3.509)
= r − rQ and us = (r − rQ )/r − rQ . Now for this problem we have
r − rQ = r er
(3.510)
r − rQ = r
(3.511)
which implies that
Therefore, we obtain
us =
r er
= er
r
Furthermore, the unstretched length of the spring is
spring force is obtained as
(3.512)
0
= r0 . Consequently, the
Fs = −K(r − r0 )er
(3.513)
Next, the force of gravity is given as
mg = −mgEy
(3.514)
Using the expression for Ey in terms of the basis {er , eθ , ez } from Eq. (3.489),
we obtain the force of gravity as
mg = −mg(− cos θ er + sin θ eθ ) = mg cos θ er − mg sin θ eθ
(3.515)
134
Chapter 3. Kinetics of Particles
Then, adding Eq. (3.513) and Eq. (3.515), the resultant force acting on the particle
is given as
F = Fs + mg = −K(r − r0 )er + mg cos θ er − mg sin θ eθ
(3.516)
Simplifying Eq. (3.516), we obtain
F = mg sin θ − K(r − r0 ) er − mg sin θ eθ
(3.517)
Then, setting F in Eq. (3.517) equal to mF a using the expression for F a from
Eq. (3.508), we obtain
mg sin θ − K(r − r0 ) er −mg sin θ eθ = m ẍEx + (r̈ − r θ̇ 2 )er + (2ṙ θ̇ + r θ̈)eθ
(3.518)
Then, substituting the expression for Ex from Eq. (3.488) into Eq. (3.518), we
obtain
mg sin θ − K(r − r0 ) er − mg sin θ eθ = m [ẍ(sin θ er + cos θ eθ )
+ (r̈ − r θ̇ 2 )er + (2ṙ θ̇ + r θ̈)eθ
(3.519)
Rearranging Eq. (3.519) gives
mg sin θ − K(r − r0 ) er − mg sin θ eθ = m(ẍ sin θ + r̈ − r θ̇ 2 )er
+ m(ẍ cos θ + 2ṙ θ̇ + r θ̈)eθ
(3.520)
Equating components in Eq. (3.520), we obtain
mẍ sin θ + mr̈ − mr θ̇ 2
mẍ cos θ + 2mṙ θ̇ + mr θ̈
= mg sin θ − K(r − r0 )
(3.521)
= −mg sin θ
(3.522)
Simplifying and rearranging Eq. (3.521) and Eq. (3.522), we obtain a system of
two differential equations describing the motion of the particle as
K
(r − r0 ) = −ẍ sin θ
m
r θ̈ + 2ṙ θ̇ + g sin θ = −ẍ cos θ
r̈ − r θ̇ 2 − g sin θ +
(3.523)
(3.524)
135
Question 3–23
A collar of mass m slides with friction along a rod that is welded rigidly at a
constant angle β with the vertical to a shaft as shown in Fig. P3-23. The shaft
rotates about the vertical with constant angular velocity Ω (where Ω = Ω).
Knowing that r is the radial distance from point of the weld to the collar, that
the friction is viscous with viscous friction coefficient c, and that gravity acts
vertically downward, determine the differential equation of motion for the collar
in terms of r .
Viscous Friction, c
P
Ω
r
β
m
O
g
Figure P 3-23
Solution to Question 3–23
Kinematics
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
To the Right att = 0
Into Page at t = 0
Ez × Ex = Up
Next, let A be a reference frame fixed to the shaft and tube. Then, choose the
following coordinate system fixed in reference frame A:
er
ez
Origin at O
=
=
eθ
=
along OP
Orthogonal to Plane of
Shaft and Arm and Into Page
ez × er
136
Chapter 3. Kinetics of Particles
The geometry of the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in Fig. 3-20.
Using Fig. 3-20, the relationship between the basis {Ex , Ey , Ez } and {er , eθ , ez }
er
β
ez , Ez ⊗
β
Ey
Figure 3-20
Ex
eθ
Geometry for Question 3–23.
is
Ex
= sin βer + cos βeθ
(3.525)
Ey
= − cos βer + sin βeθ
(3.526)
The position of the particle is then given in terms of the basis {er , eθ , ez } as
r = r er
(3.527)
Furthermore, the angular velocity of reference frame A in reference frame F is
then given as
F A
ω = Ω = −ΩEy
(3.528)
Then, substituting the expression for Ey from Eq. (3.526) into Eq. (3.528), we
obtain
F
ωA = −Ω(− cos βer + sin βeθ ) = Ω cos βer − Ω sin βeθ
(3.529)
The velocity of the particle in reference frame F is then obtained from the
transport theorem as
F
dr Adr F A
F
=
+ ω ×r
v=
(3.530)
dt
dt
Now we have that
A
dr
= ṙ er
dt
F A
ω × r = (Ω cos βer − Ω sin βeθ ) × r er = r Ω sin βez
(3.531)
(3.532)
137
Adding the expressions in Eq. (3.531) and Eq. (3.532), we obtain the velocity of
the collar in reference frame F as
F
v = ṙ er + r Ω sin βez
(3.533)
The acceleration of the collar in reference frame F is obtained by applying the
rate of change transport theorem as
F
a=
d F A d F F A F
v =
v + ω × v
dt
dt
F
(3.534)
Now we have that
d F v
= r̈ er + ṙ Ω sin βez
dt
F A
ω × F v = (Ω cos βer − Ω sin βeθ ) × (ṙ er + r Ω sin βez )
= −r Ω2 cos β sin βeθ + ṙ Ω sin βez − r Ω2 sin2 βer
A
(3.535)
(3.536)
Adding the expressions in Eq. (3.535) and Eq. (3.536), we obtain the acceleration
of the collar in reference frame F as
F
a = r̈ er + ṙ Ω sin βez − r Ω2 cos β sin βeθ + ṙ Ω sin βez − r Ω2 sin2 βer
(3.537)
Simplifying Eq. (3.537) gives
F
a = (r̈ − r Ω2 sin2 β)er − r Ω2 cos β sin βeθ + 2ṙ Ω sin βez
(3.538)
Kinetics
The free body diagram of the particle is shown in Fig. 3-21 Using Fig. 3-21, we
⊗ Nz
Ff
Nθ
mg
Figure 3-21
Free Body Diagram for Question 3–23.
have that
Nθ
Nz
mg
Ff
=
=
=
=
Reaction Force of Rod on Particle in eθ − Direction
Reaction Force of Rod on Particle in ez − Direction
Force of Gravity
Force of Viscous Friction
138
Chapter 3. Kinetics of Particles
Given the geometry of the problem, we have that
Nθ
= Nθ eθ
(3.539)
Nz
= Nz ez
(3.540)
mg = −mgEy
Ff
where vrel is given as
(3.541)
= −cvrel
(3.542)
vrel = F v − F vA
P
(3.543)
F
where vA
P is the velocity of the point P on the rod that is instantaneously in
contact with the collar. Now we have that
F A
vP
However, since
have that
F A
vP
F A
= AvA
P + ω ×r
(3.544)
corresponds to the velocity of a point fixed to the rod, we
A A
vP
=0
(3.545)
Therefore,
F A
vP
= F ωA × r = (Ω cos βer − Ω sin βeθ ) × r er = r Ω sin βez
Then, substituting the result of Eq. (3.546) and the expression for
Eq. (3.533) into Eq. (3.543), we obtain
vrel = ṙ er + r Ω sin βez − r Ω sin βez = ṙ er
(3.546)
Fv
from
(3.547)
The force of viscous friction acting on the collar is then given as
Ff = −cvrel = −c ṙ er
(3.548)
The resultant force acting on the particle is then given as
F = Nθ + Nz + mg + Ff = Nθ eθ + Nz ez + mgEy − c ṙ er
(3.549)
Then, substituting the expression for Ey from Eq. (3.526) into Eq. (3.549), we
obtain
F = Nθ eθ + Nz ez + mg(− cos βer + sin βeθ ) − c ṙ er
= −(mg cos β + c ṙ )er + (Nθ + mg sin β)eθ + Nz ez
(3.550)
Then, applying Newton’s 2nd Law F from Eq. (3.550) and F a from Eq. (3.538), we
obtain
−(mg cos β + c ṙ )er + (Nθ + mg sin β)eθ + Nz ez = m(r̈ − r Ω2 sin2 β)er
− mr Ω2 cos β sin βeθ
+ 2mṙ Ω sin βez
(3.551)
139
Equating components in Eq. (3.551), we obtain the following three scalar equations:
m(r̈ − r Ω2 sin2 β) = −(mg cos β + c ṙ )
(3.552)
2
(3.553)
2mṙ Ω sin β = Nz
(3.554)
−mr Ω cos β sin β = Nθ − mg sin θ
It can be seen that Eq. (3.552) has no reaction forces. Furthermore, in this problem only one variable is required to describe the motion. Consequently, the
differential equation of motion is given as
m(r̈ − r Ω2 sin2 β) = −(mg cos β + c ṙ )
(3.555)
Rearranging and simplifying Eq. (3.555), we obtain the differential equation of
motion as
(3.556)
r̈ − r Ω2 sin2 β + c ṙ + g cos β = 0
140
Chapter 3. Kinetics of Particles
Question 3–25
A particle of mass m slides without friction along a track in the form of a
parabola as shown in Fig. P3-25. The equation for the parabola is
y=
r2
2a
where a is a constant, r is the distance from point O to point Q, point Q is
the projection of point P onto the horizontal direction, and y is the vertical
distance. Furthermore, the particle is attached to a linear spring with spring
constant K and unstretched length x0 . The spring is always aligned horizontally
such that its attachment point is free to slide along a vertical shaft through
the center of the parabola. Knowing that the parabola rotates with constant
angular velocity Ω (where Ω = Ω) about the vertical direction and that gravity
acts vertically downward, determine the differential equation of motion for the
particle in terms of the variable r .
Ω
g
m
Q
y
O
r
P
Figure P 3-25
Solution to Question 3–25
Kinematics
For this problem it is convenient to define a fixed inertial reference frame F
and a non-inertial reference frame A. Corresponding to reference frame F , we
141
choose the following coordinate system:
Origin at Point O
=
=
=
Ex
Ey
Ez
Along OP When t = 0
Along Oy When t = 0
Ex × Ey
Furthermore, corresponding to reference frame A, we choose the following coordinate system:
Origin at Point O
=
Along OP
ex
=
Along Oy
ey
=
ex × ey
ez
The position of the particle is then given in terms of the basis {ex , ey , ez } as
r = xex + yey = xex + (x 2 /a)ey
(3.557)
Furthermore, since the parabola spins about the ey-direction, the angular velocity of reference frame A in reference frame F is given as
F
ωA = Ω = Ωey
(3.558)
The velocity in reference frame F is then found using the rate of change transport theorem as
F
dr Adr F A
F
=
+ ω ×r
v=
(3.559)
dt
dt
Using r from Eq. (3.557) and F ωA from Eq. (3.558), we have that
A
dr
= ẋex + (2x ẋ/a)ey
dt
F A
ω × r = Ωey × (xex + (x 2 /a)ey ) = −Ωxez
(3.560)
(3.561)
Adding the expressions in Eq. (3.560) and Eq. (3.561), we obtain F v as
F
v = ẋex + (2x ẋ/a)ey − Ωxez
(3.562)
The acceleration in reference frame F is found by applying the rate of change
transport theorem to F v as
F
a=
d F Ad F F A F
v =
v + ω × v
dt
dt
F
Using F v from Eq. (3.562) and
F ωA
(3.563)
from Eq. (3.558), we have that
d F v
= ẍex + 2(ẋ 2 + x ẍ)/a ey − Ωẋez
(3.564)
dt
F A
ω × F v = Ωey × (ẋex + (2x ẋ/a)ey − Ωxez ) = −Ωẋez − Ω2 xe(3.565)
x
A
Adding the expressions in Eq. (3.564) and Eq. (3.565) and, we obtain F a as
F
a = (ẍ − Ω2 x)ex + 2(ẋ 2 + x ẍ)/a ey − 2Ωẋez
(3.566)
142
Chapter 3. Kinetics of Particles
Kinetics
The free body diagram of the particle is shown in Fig. 3-22. Using Fig. 3-22, it is
Nn
⊗
Fs
Nb
mg
Figure 3-22
Free Body Diagram of Particle for Question 3–25.
seen that the following forces act on the particle:
= Reaction Force of Track on Particle
Normal to Track and In Plane of Parabola
Nb = Reaction Force of Track on Particle
Normal to Track and Orthogonal to Plane of Parabola
= Spring Force
Fs
mg = Force of Gravity
Nn
Given the description of the two reaction forces Nn and Nb , we have that
Nn
= Nn en
(3.567)
Nb
= Nb eb
(3.568)
where en and eb are the principle unit normal and principle unit bi-normal vector to the parabola. Now since reference frame A is the reference frame of the
parabola, the tangent vector to the parabola is given as
et =
Now we have
Av
Av
Av
(3.569)
from Eq. (3.560) as
A
v = ẋex + (2x ẋ/a)ey
Consequently,
et =
ẋex + (2x ẋ/a)ey
ex + (2x/a)ey
= 2x 2
2x 2
ẋ 1 +
1+
a
a
(3.570)
(3.571)
Next, we know that eb must lie orthogonal to the plane of the parabola. Consequently, we have that
eb = ez
(3.572)
143
Therefore,
ex + (2x/a)ey
−(2x/a)ex + ey
en = eb × et = ez × = 2
2x
2x 2
1+
1+
a
a
(3.573)
Suppose now that we define
γ=
Then we can write
en =
1+
2x
a
2
−(2x/a)ex + ey
γ
(3.574)
(3.575)
The reaction force exerted by the parabola on the particle is then given as
N = Nn + Nb = Nn
−(2x/a)ex + ey
+ Nb ez
γ
(3.576)
Next, the spring force is given as
Fs = −K( −
0 )us
(3.577)
Now for this problem we know that the unstretched length of the spring is
0 = x0 . Furthermore, the stretched length of the spring is given as
= r − rQ (3.578)
where P is the attachment point of the spring. Now since the attachment point
lies on the ey -axis at the same value of y as the particle, we have that
rQ = yey =
x2
ey
a
(3.579)
Therefore,
x2
x2
ey −
ey = xex
a
a
The stretched length of the spring is then given as
r − rQ = xex +
= xex = x
(3.580)
(3.581)
Finally, the direction of the spring force is given as
us =
r − rQ
xex
=
= ex
r − rQ x
(3.582)
The force of the linear spring is then given as
Fs = −K(x − x0 )ex
(3.583)
144
Chapter 3. Kinetics of Particles
Finally, the force of gravity is given as
mg = −mgey
(3.584)
Then, adding Eq. (3.576), Eq. (3.583), and Eq. (3.584), the resultant force acting
on the particle is then obtained as
−(2x/a)ex + ey
+ Nb ez − K(x − x0 )ex − mgey
γ
(3.585)
Then, combining terms with common components in Eq. (3.585), we have that
$
$
#
#
Nn
2x/a
+ K(x − x0 ) ex +
− mg ey + Nb ez
(3.586)
F = − Nn
γ
γ
F = Nn + Nb + Fs + mg = Nn
Then, setting F from Eq. (3.586) equal to mF a using the expression for F a from
Eq. (3.566), we obtain
$
$
#
#
Nn
2x/a
+ K(x − x0 ) ex +
− mg ey + Nb ez = m(ẍ − Ω2 x)ex
− Nn
γ
γ
+ m 2(ẋ 2 + x ẍ)/a ey
− 2mΩẋez
(3.587)
Equating components in Eq. (3.587), we obtain the following three scalar equations:
m(ẍ − Ω2 x) = −Nn
=
m 2(ẋ 2 + x ẍ)/a
−2mΩẋ
2x/a
− K(x − x0 )
γ
Nn
− mg
γ
= Nb
(3.588)
(3.589)
(3.590)
Now the differential equation of motion for the particle is obtained as follows. Rearranging Eq. (3.588) and Eq. (3.589), we have that
Nn
2x/a
γ
Nn
γ
= −m(ẍ − Ω2 x) − K(x − x0 )
(3.591)
= m 2(ẋ 2 + x ẍ)/a + mg
(3.592)
Then, dividing Eq. (3.591) by Eq. (3.592), we obtain
Nn
2x/a
−m(ẍ − Ω2 x) − K(x − x0 )
γ
=
Nn
m 2(ẋ 2 + x ẍ)/a + mg
γ
(3.593)
145
Simplifying Eq. (3.593) gives
−m(ẍ − Ω2 x) − K(x − x0 )
2x
=
a
m 2(ẋ 2 + x ẍ)/a + mg
(3.594)
Multiplying Eq. (3.594) by m 2(ẋ 2 + x ẍ)/a + mg, we obtain
2x
= −m(ẍ − Ω2 x) − K(x − x0 )
m 2(ẋ 2 + x ẍ)/a + mg
a
(3.595)
Rearranging Eq. (3.595), we have that
2x
=0
m(ẍ − Ω2 x) + K(x − x0 ) + m 2(ẋ 2 + x ẍ)/a + mg
a
Simplifying further, we obtain the differential equation of motion as
#
$
ẋ 2
2x 2
2g
2
mẍ 1 +
− Ω x + K(x − x0 ) + 4mx
=0
+m
a
a
a
(3.596)
(3.597)
146
Chapter 3. Kinetics of Particles
Chapter 4
Kinetics of a System of Particles
Question 4–1
A particle of mass m is connected to a block of mass M via a rigid massless
rod of length l as shown in Fig. P4-1. The rod is free to pivot about a hinge
attached to the block at point O. Furthermore, the block rolls without friction
along a horizontal surface. Knowing that a horizontal force F is applied to the
block and that gravity acts downward, determine a system of two differential
equations describing the motion of the block and the particle.
x
m
g
θ
l
O
F
M
Figure P 4-1
P
148
Chapter 4. Kinetics of a System of Particles
Solution to Question 4–1
Kinematics
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 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
(4.1)
Ey
= − cos θ er + sin θ eθ
(4.2)
er
= sin θ Ex − cos θ Ey
(4.3)
eθ
= cos θ Ex + sin θ Ey
(4.4)
Also, we have that
Using the bases {Ex , Ey , Ez } and {er , eθ , ez }, the position of the block is given
as
rO = xEx
(4.5)
Then the velocity and acceleration of the block in reference frame F are given,
respectively, as
F
vO
F
aO
= ẋEx
(4.6)
= ẍEx
(4.7)
Next, the position of the particle is given as
r = rP = rO + rP /O = xEx + ler
(4.8)
Next, the angular velocity of reference frame A in reference frame F is given as
F
ωA = θ̇ez
(4.9)
149
The velocity of point P in reference frame F is then given as
F
vP =
F
F
d
d
rP /O = F vO + F vP /O
(rO ) +
dt
dt
(4.10)
Now we already have F vO from Eq. (4.6). 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
A
d
d
rP /O =
rP /O + F ωA × rP /O
dt
dt
(4.11)
Now we have that
A
d
rP /O
dt
F A
ω × rP /O
= 0
(4.12)
= θ̇ez × ler = lθ̇eθ
(4.13)
Adding Eq. (4.12) and Eq. (4.13) gives
F
vP /O = lθ̇eθ
(4.14)
Therefore, the velocity of the particle in reference frame F is given as
F
vP = ẋEx + lθ̇eθ
(4.15)
Next, the acceleration of point P in reference frame F is obtained as
F
Now we have that
F
aP =
d F vP
dt
F
vP = F vO + F vP /O
(4.16)
(4.17)
where we have from Eq. (4.6) and Eq. (4.14) that
F
vO
= ẋEx
(4.18)
vP /O
= lθ̇eθ
(4.19)
d F Fd F
vO +
vP /O
dt
dt
(4.20)
F
Consequently,
F
aP =
F
Now we already have F aO from Eq. (4.7. 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
aP /O =
Ad d F
F
vP /O =
vP /O + F ωA × F vP /O
dt
dt
F
(4.21)
150
Chapter 4. Kinetics of a System of Particles
Now we have that
d F
vP /O
= lθ̈eθ
dt
F A
ω × F vP /O = θ̇ez × lθ̇eθ = −lθ̇ 2 er
A
(4.22)
(4.23)
Adding Eq. (4.22) and Eq. (4.23) gives
F
aP /O = −lθ̇ 2 er + lθ̈eθ
(4.24)
Then, adding Eq. (4.7) and Eq. (4.24), we obtain the acceleration of point P in
reference frame F as
F
aP = ẍEx − lθ̇ 2 er + lθ̈eθ
(4.25)
The acceleration of the center of mass of the system is then computed using
the expressions for F aO and F aP from Eqs. (4.7) and (4.25), respectively. In
particular, we have that
M F aO + m F aP
F
(4.26)
ā =
M +m
Substituting the results of Eqs. (4.7) and (4.25) into Eq. (4.26), we obtain
F
ā =
M ẍEx + m(ẍEx − lθ̇ 2 er + lθ̈eθ )
m = ẍEx +
−lθ̇ 2 er + lθ̈eθ (4.27)
M +m
M +m
Kinetics
The free body diagram of the block is shown in Fig. 4-1.
R
F
N
Mg
Figure 4-1
Free Body Diagram of Block for Question. 4–1
Using Fig. 4-1, the forces acting on the block are given as
F
R
N
Mg
=
=
=
=
External Force
Reaction Force of Particle on Block
Reaction Force of Ground on Block
Force of Gravity
151
Now we have that
F = F Ex
(4.28)
R = Rer
(4.29)
N = NEy
(4.30)
Mg = MgEy
(4.31)
Then the resultant force acting on the block is given as
FO = F + R + N + Mg = F Ex + NEy + Rer + MgEy
(4.32)
Using the expression for er from Eq. (4.3), we have that
FO = F Ex +NEy +R(sin θ Ex −cos θ Ey )+MgEy = (F +R sin θ )Ex +(N−R cos θ +Mg)Ey
(4.33)
Setting FO equal to M F aO , we obtain the following two scalar equations:
F + R sin θ
N − R cos θ + Mg
= M ẍ
(4.34)
= 0
(4.35)
The free body diagram of the block-particle system is shown in Fig. 4-2.
(M + m)g
F
N
Figure 4-2
Free Body Diagram of Block-Particle System for Question. 4–1
Using Fig. 4-2, the forces acting on the block-particle system are given as
F
= External Force
N
= Reaction Force of Ground on Block
(M + m)g = Force of Gravity
Now we have that
F = F Ex
(4.36)
N = NEy
(4.37)
(M + m)g = (M + m)gEy
(4.38)
152
Chapter 4. Kinetics of a System of Particles
Therefore, the resultant force acting on the block-particle system is given as
FT = F + N + (M + m)g = F Ex + NEy + (M + m)gEy
(4.39)
Eq. (4.39) simplifies to
FT = F Ex + N + (M + m)g Ey
Setting FT equal to (M + m)F ā, we obtain
F Ex + N + (M + m)g Ey = (M + m) ẍ +
m −lθ̇ 2 er + lθ̈eθ
M +m
(4.40)
(4.41)
Eq. (4.41) can be rewritten as
F Ex + N + (M + m)g Ey = (M + m)ẍ + m −lθ̇ 2 er + lθ̈eθ
(4.42)
Substituting the expressions for er and eθ from Eqs. (4.3) and (4.4) into Eq. (4.42),
we obtain
F Ex + N + (M + m)g Ey = (M + m)ẍ + m −lθ̇ 2 (sin θ Ex − cos θ Ey )
(4.43)
+lθ̈(cos θ Ex + sin θ Ey )
Eq. (4.43) simplifies to
F Ex + N + (M + m)g Ey = (M + m)ẍ − mlθ̇ 2 sin θ + mlθ̈ cos θ Ex
+ (mlθ̇ 2 cos θ + mlθ̈ sin θ )Ey
(4.44)
Equating components in Eq. (4.44), we obtain the following two scalar equations:
F
N + (M + m)g
= (M + m)ẍ − mlθ̇ 2 sin θ + mlθ̈ cos θ
2
= mlθ̇ cos θ + mlθ̈ sin θ
(4.45)
(4.46)
System of Two Differential Equations
The first differential equation is Eq. (4.45), i.e.,
F = (M + m)ẍ − mlθ̇ 2 sin θ + mlθ̈ cos θ
(4.47)
The second differential equation is obtained by using Eqs. (4.34), (4.35) and
(4.46). Solving Eq. (4.46) for N, we obtain
N = mlθ̇ 2 cos θ + mlθ̈ sin θ − (M + m)g
(4.48)
Substituting N from Eq. (4.48) into Eq. (4.35) gives
mlθ̇ 2 cos θ + mlθ̈ sin θ − (M + m)g − R cos θ + Mg = 0
(4.49)
153
Eq. (4.49) simplifies to
mlθ̇ 2 cos θ + mlθ̈ sin θ − mg − R cos θ = 0
(4.50)
Then, multiplying Eq. (4.50) by sin θ , we obtain
mlθ̇ 2 cos θ sin θ + mlθ̈ sin2 θ − mg sin θ − R sin θ = 0
(4.51)
Next, multiplying Eq. (4.34) by cos θ , we obtain
F cos θ + R sin θ cos θ = M ẍ cos θ
(4.52)
Rearranging Eq. (4.52) gives
F cos θ + R sin θ cos θ − M ẍ cos θ = 0
(4.53)
Adding Eq. (4.51) and Eq. (4.53), we have that
mlθ̇ 2 cos θ sin θ + mlθ̈ sin2 θ − M ẍ cos θ − mg sin θ + F cos θ = 0
(4.54)
Rearranging Eq. (4.54), we obtain the second differential equation of motion as
M ẍ cos θ − mlθ̈ sin2 θ − mlθ̇ 2 cos θ sin θ + mg sin θ = −F cos θ
(4.55)
The system of two differential equations is then given from Eqs. (4.47) and (4.55)
as
(M + m)ẍ − mlθ̇ 2 sin θ + mlθ̈ cos θ
2
2
M ẍ cos θ − mlθ̈ sin θ − mlθ̇ cos θ sin θ + mg sin θ = F cos θ
= F(4.56)
(4.57)
The system in Eqs. (4.56) and (4.57) can be written in a slightly more elegant
form as follows. Multiplying Eq. (4.56) by cos θ and subtracting the result from
Eq. (4.57), we obtain
mẍ cos θ + mlθ̈ + mg sin θ = 0
(4.58)
An alternate system of differential equations is then obtained from Eqs. 4.56)
and (4.58) as
(M + m)ẍ − mlθ̇ 2 sin θ + mlθ̈ cos θ
= F
(4.59)
mẍ cos θ + mlθ̈ + mg sin θ
= 0
(4.60)
154
Chapter 4. Kinetics of a System of Particles
Solution to Question 4–2
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Origin at Location of Block
Ex To The Left
Ez Into Page
Ey = E z × E x
(4.61)
Now in order to solve this problem we need to apply the principle of linear
impulse and momentum to the system and/or subsystems of the system. For
this problem, it is convenient to choose the following systems:
• System 1: Ball Bearing
• System 2: Block
• System 3: Bullet and Ball Bearing
Application of Linear Impulse and Linear Momentum to Ball Bearing
Applying linear impulse and momentum to the ball bearing, we have that
F
F̂m = m vm − mF vm
(4.62)
Now the free body Diagram of the ball bearing is shown in Fig. 4-3 where
F̂m
Figure 4-3
Free Body Diagram of Ball Bearing for Question 4–2.
F̂m = Impulse Applied on Block on Ball Bearing
Now we have that
Fv
m
F vm
= v0 = v0 cos θ Ex − v0 sin θ Ey
E + v E
=
vmx
x
my y
(4.63)
Also, let
F̂m ≡ F̂ = F̂x Ex + F̂y Ey
(4.64)
Then, substituting the results of Eq. (4.63) and the result of Eq. (4.64) into
Eq. (4.62), we obtain
Ex + vmy
Ey )
m(v0 cos θ Ex − v0 sin θ Ey ) + F̂x Ex + F̂y Ey = m(vmx
(4.65)
155
Rearranging, we have
(mv0 cos θ + F̂x )Ex + (F̂y − mv0 sin θ )Ey = mvmx
Ex + mvmy
Ey
(4.66)
We then obtain the following two scalar equations:
mv0 cos θ + F̂x
= mvmx
(4.67)
F̂y − mv0 sin θ
mvmy
(4.68)
=
Application of Linear Impulse and Linear Momentum to the Entire System
For the entire system we have that
F
F̂ = G − F G
(4.69)
F
where F G and G are the linear momenta of the system before and after impact.
Now we have that
F
F
G = m F vm + M F vM
G
=
F
m vm
F
+ M vM
(4.70)
(4.71)
F
We already have expressions for F vm and vm from Eq. (4.63). In addition, we
have that
Fv
M = 0
(4.72)
F vM = vMx
Ex + vMy
Ey
Next, the free body diagram of the entire system is shown in Fig. 4-4 It is seen
P̂
Figure 4-4
Free Body Diagram of Entire System for Question 4–2.
that the only impulse applied to the entire system is P̂ where P̂ is the impulse
applied by the ground on the system. Therefore,
F̂ = P̂ = P̂ Ey
(4.73)
Substituting the results of Eq. (4.63), Eq. (4.72), and Eq. (4.73) into Eq. (4.69), we
obtain
Ex + vmy
Ey ) + M(vMx
Ex + vMy
Ey ) − m(v0 cos θ Ex − v0 sin θ Ey )
P̂ Ey = m(vmx
(4.74)
156
Chapter 4. Kinetics of a System of Particles
Rearranging Eq. (4.74), we obtain
mv0 cos θ Ex + (P̂ − mv0 sin θ )Ey = (mvmx
+ MvMx
)Ex + (mvmy
+ MvMy
)Ey
(4.75)
We then obtain the following two scalar equations:
mv0 cos θ
P̂ − mv0 sin θ
= mvmx
+ MvMx
(4.76)
+ MvMy
(4.77)
=
mvmy
Kinematic Constraints
We know that, because the bullet embeds itself in the block immediately after
impact the velocity of the bullet and the block must be the same. Therefore,
F vm
For convenience, let
F
F
= vM = v
(4.78)
F v = vx Ex + vy Ey
(4.79)
Second, we know that the bullet and the block must move horizontally after
impact. Therefore,
(4.80)
vy = 0
Solving for the Impulse Exerted By Block on Ball Bearing During Impact
The impulse exerted by the block on the ball bearing can be solved for using the
results of Eq. (4.67), Eq. (4.68), Eq. (4.76), and Eq. (4.77). First, it is convenient
to solve for the post-impact velocity of the block and ball bearing. Because the
post-impact velocity of the block and the ball bearing are the same, we have
from Eq. (4.76) that
mv0 cos θ = mvmx
+ MvMx
= (m + M)vx
(4.81)
Solving this last equation for vx , we obtain
vx =
m
v0 cos θ
m+M
(4.82)
Then from Eq. (4.67) we have that
mv0 cos θ + F̂x = mvmx
= mvx = m
m
v0 cos θ
m+M
(4.83)
Solving this last equation for F̂x , we obtain
F̂x = mv0 cos θ
m
− mv0 cos θ
m+M
(4.84)
157
Simplifying this last expression, we have that
F̂x = −mv0 cos θ
M
m+M
(4.85)
Also, from Eq. (4.68) we have that
F̂y − mv0 sin θ = mvmy
= mvy = 0
(4.86)
Solving this last equation for F̂y , we obtain
F̂y = mv0 sin θ
(4.87)
Consequently, the impulse exerted by the block on the bullet is given as
F̂m = F̂ = mv0 cos θ
M
Ex + mv0 sin θ Ey
m+M
(4.88)
Post-Impact Velocity of Block and Ball Bearing
As stated above, the post-impact velocity of the block and the ball bearing are
the same. From Eq. (4.80) and Eq. (4.82) we have that
F v =
Consequently,
F vm
F vM
m
v0 cos θ Ex
m+M
=
m
v0 cos θ Ex
m+M
=
m
v0 cos θ Ex
m+M
(4.89)
(4.90)
158
Chapter 4. Kinetics of a System of Particles
Question 4–3
A block of mass m is dropped from a height h above a plate of mass M as shown
in Fig. P4-3. The plate is supported by three linear springs, each with spring
constant K, and is initially in static equilibrium. Assuming that the compression
of the springs due to the weight of the plate is negligible, that the impact is
perfectly inelastic, that the block strikes the vertical center of the plate, and
that gravity acts downward, determine (a) the velocity of the block and plate
immediately after impact and (b) the maximum compression, xmax , attained by
the springs after impact.
m
g
h
M
K
K
K
Figure P 4-2
Solution to Question 4–3
Kinematics
Let F be a fixed reference frame. Then, choose the following coordinate system
fixed in reference frame F :
Ex
Ez
Ey
Origin at m at t = 0
=
Down
=
Out of Page
=
Ez × Ex
Then the velocity of the block and plate in reference frame F are given, respectivley, as
F
v1
= v1 Ex
(4.91)
F
v2
= v2 Ex
(4.92)
159
The velocity of the center of mass of the block-plate system in reference frame
F is then given as
F
v̄ =
mv1 + Mv2
m F v1 + M F v2
==
Ex
m+M
m+M
(4.93)
Kinetics
Phase 1: During Descent of Block
The only force acting on the block during its descent is that of gravity. Since
gravity is conservative, we know that energy must be conserved. Consequently,
F
E(t0 ) = F E(t1 )
(4.94)
where t0 and t1 are the times of the start and end of the descent. Now we have
that
F
E = FT + FU
(4.95)
Since the block is dropped from rest, we have that
F
T (t0 ) = 0
(4.96)
Next, the kinetic energy of the block at the end of the descent is given as
F
T (t0 ) =
1 F
1
1
m v1 (t1 ) · mF v1 (t1 ) = mv12 (t1 )
2
2
2
(4.97)
Also, the initial potential energy is given as
F
U (t0 ) = −mg · r1 (t0 )
(4.98)
where r1 (t0 ) is the position of the block at time t = t0 . Now since the origin of
the coordinate system is located at the block at t = 0, we know that r(t0 ) = 0.
Consequently,
F
U (t0 ) = 0
(4.99)
Finally, the potential energy at the end of the descent is given as
F
U (t1 ) = −mg · r1 (t1 )
(4.100)
Now since the block is dropped from a height h above the plate, we have that
r1 (t1 ) = hEx
(4.101)
Noting that g = gEx , we have that
F
U (t1 ) = −mgEx · hEx = −mgh
(4.102)
160
Chapter 4. Kinetics of a System of Particles
Equating the energy of the block at times t0 and t1 we have that
1
mv12 (t1 ) − mgh
2
0=
(4.103)
Solving Eq. (4.103) for v1 (t1 ), we obtain
v1 (t1 ) = 2gh
(4.104)
From Eq. (4.104), the velocity of the block in reference frame F at the end of the
descent is given as
F
v1 (t1 ) = 2ghEx
(4.105)
Phase 2: During Impact
Applying the principle of linear impulse and linear momentum to the entire
system during impact, we have that
F
F
F̂ = G(t1+ ) − G(t1− )
(4.106)
Now, because the impact is assumed to occur instantaneously and neither gravity nor the spring can apply an instantaneous impulse, during impact there are
no external impulses applied to the system, i.e.,
F̂ = 0
Therefore,
F
F
G(t1+ ) = G(t1− )
(4.107)
(4.108)
Now the linear momentum of the system is given as
F
G = (m + M)F v̄
(4.109)
Using the expression for F v̄ from Eq. (4.93), we obtain
F
G = (m + M)
mv1 + Mv2
Ex = (mv1 + Mv2 )Ex
m+M
(4.110)
Then, noting that the plate is motionless before impact and using the expression
for F v1 (t1 ) from Eq. (4.105), we have that
F
G(t1− ) = m 2ghEx
(4.111)
Consequently, we have that
F
(m + M) v̄(t1+ ) = m 2ghEx
(4.112)
161
F
Solving for v̄(t1+ ), we obtain the velocity of the center of mass of the system
the instant after impact as
m F
v̄(t1+ ) =
2ghEx
(4.113)
m+M
Next, we know that the impact is perfectly inelastic. Consequently, the coefficient of restitution is zero, i.e., e = 0. Now we have the coefficient of restitution
condition as
F F
v · n − v1 · n
(4.114)
e= F 2
v1 · n − F v2 · n
Now since e = 0, Eq. (4.114) becomes
F
F
F
v2 (t1+ ) · n − v1 (t1+ ) · n
=0
(4.115)
v2 (t1+ ) · n − v1 (t1+ ) · n = 0
(4.116)
F
v1 (t1− ) · n − v2 (t1− ) · n
Eq. (4.115) implies that
F
F
Now for this problem is it seen that the direction of impact is along Ex , i.e., n =
Ex . Furthermore, since the velocities of the block and the plate are also along
Ex , we have that
F
v1 (t1+ ) · n = v1 (t1+ )
(4.117)
F
v2 (t1+ ) · n = v2 (t1+ )
(4.118)
v2 (t1+ ) − v1 (t1+ ) = 0
(4.119)
v2 (t1+ ) = v1 (t1+ )
(4.120)
Eq. (4.116) then reduces to
Eq. (4.119) implies that
It is seen from Eq. (4.120) that the post-impact velocities of the block and plate
are the same. Now we have that
(m + M)F v̄ = mF v1 + M F v2
F
(4.121)
Consequently, using the expression for v̄(t1+ ) from Eq. (4.113), we obtain
m 2gh
F
+
Ex = m 2ghEx = mv1 (t1+ )Ex + Mv2 (t1+ )
(m + M) v̄(t1 ) = (m + M)
m+M
(4.122)
+
+
+
But since v1 (t1 ) = v2 (t1 ) ≡ v(t1 ), we have that
m 2gh = (m + M)v(t1+ )
(4.123)
162
Chapter 4. Kinetics of a System of Particles
Solving for v(t1+ ) gives
m (4.124)
2ghEx
m+M
Therefore, the post-impact velocities of the block and plate in reference frame
F are given as
m F
v1 (t1+ ) =
2ghEx
(4.125)
m+M
m
F
v2 (t1+ ) =
2ghEx
(4.126)
m+M
v(t1+ ) =
Phase 3: Maximum Compression of Spring After Impact
First, it is important to recognize that, because the impact between the block
and the plate was perfectly inelastic, the block and plate will move together
after impact, i.e., the block and plate will move as if they are a single body after
impact. Next, for this part of the problem we see that the only forces acting on
the block and plate are those of the three springs. Since the attachment points
of the springs are inertially fixed (in this case they are fixed in reference frame
F ), we know that the spring forces will be conservative. Thus, the total energy
of the system after impact will be conserved. Now the total energy is given as
F
E = FT + FU
(4.127)
Now, the velocity of the block and plate is given as
F
v = vEx
(4.128)
Therefore, the kinetic energy of the system is given as
F
T =
1
1
(m + M)F v · F v = (m + M)v 2
2
2
Next, the potential energy is due to the three springs and is given as
1
3
F
K( − 0 )2 = K( − 0 )2
Us = 3
2
2
(4.129)
(4.130)
Now since the springs are initially uncompressed, we have that
(t1 ) =
0
(4.131)
Now since the springs are compressing in this phase, we know that − 0 will
be less than zero for t > t1 . In order to account for the fact that − 0 < 0,
let xmax = 0 − (t2 ). Then the potential energy in the springs at the point of
maximum compression is given as
1
3
F
2
2
K( (t2 ) − 0 ) = Kxmax
Us (t2 ) = 3
(4.132)
2
2
163
Then the total potential energy at the point of maximum compression is given
as
3
F
2
U (t2 ) = −(m + M)g(h + xmax ) + Kxmax
(4.133)
2
where we note that the potential energy due to gravity at t2 is different from
the potential energy due to gravity at t1 because the position of the particle and
block has changed from −hEx to −(h + xmax )Ex in moving from t1 to t2 . Now,
using the results from phase 2, we know at t1 that
m 1
(m + M)
2gh
2
m+M
F
U (t1 ) = −(m + M)gh
F
2
T (t1 ) =
=
m2 gh
m+M
(4.134)
(4.135)
Then the total energy of the system at t1 is given as
F
E(t1 ) = F T (t1 ) + F U (t1 ) = 0
(4.136)
Furthermore, at t2 we have that
F
T (t2 ) = 0
(4.137)
3
F
2
U (t2 ) = −(m + M)g(h + xmax ) + Kxmax
2
(4.138)
where we note that the kinetic energy at t2 is zero because t2 is the time at
which the springs have attained their maximum compression. Then the total
energy of the system at t2 is given as
F
E(t2 ) = F T (t2 ) + F U (t2 ) = −(m + M)g(h + xmax ) +
3
2
Kxmax
2
(4.139)
Setting F E(t2 ) equal to F E(t1 ), we have that
−(m + M)g(h + xmax ) +
3
2
Kxmax
=0
2
(4.140)
Eq. (4.140) can be simplified to
3
2
Kxmax
− (m + M)gxmax − (m + M)gh = 0
2
(4.141)
Simplifying Eq. (4.141) further, we obtain
2
−
xmax
2(m + M)g
2(m + M)gh
xmax −
=0
3K
3K
(4.142)
Eq. (4.142) is a quadratic in xmax . The roots of Eq. (4.142) are given from the
quadratic formula as
2(m + M)g 2 8(m + M)gh
2(m + M)g
±
+
3K
3K
3K
(4.143)
xmax =
2
164
Chapter 4. Kinetics of a System of Particles
For simplicity, let
a=
2(m + M)g
3K
(4.144)
Then Eq. (4.143) can be written as
xmax
a ± a2 + 4ah
=
2
Eq. (4.145) can be rewritten as
a ± a 1 + 4h/a
a
xmax =
=
1 ± 1 + 4h/a
2
2
(4.145)
(4.146)
Now since h and a are positive, we know that the quantity 1 + 4h/a must be
greater than unity. Consequently, one of the roots of Eq. (4.146) must be negative. Choosing the positive root, we obtain
a
(4.147)
1 + 1 + 4h/a
xmax =
2
Substituting the expression for a from Eq. (4.144), we obtain the maximum compression of the springs after impact as
#
$
(m + M)g
6Kh
xmax =
1+ 1+
(4.148)
3K
(m + M)g
165
Question 4–8
A particle of mass mA slides without friction along a fixed vertical rigid rod. The
particle is attached via a rigid massless arm to a particle of mass mB where mB
slides without friction along a fixed horizontal rigid rod. Assuming that θ is the
angle between the horizontal rod and the arm and that gravity acts downward,
determine the differential equation of motion for the system in terms of the
angle θ.
mB
O
B
θ
l
mA
A
g
Figure P 4-3
Solution to Question 4–8
Before beginning the kinematics for this problem, it is important to know which
balance laws will be used. Since this problem consists of two particles, it will
be solved using the following balance laws: (1) a force balance on the entire
system and (2) Newton’s second law applied to mA . Consequently, we will need
to compute the acceleration of each particle.
Kinematics
Let F be a fixed reference frame. Then, choose the following coordinate system
fixed in reference frame F :
Ex
Ey
Ez
Origin at Point O
=
=
=
Along OA
Along OB
E x × Ey
Then the position of each particle is given in terms of the basis {Ex , Ey , Ez } as
rA
= l sin θ Ex
(4.149)
rB
= l cos θ Ey
(4.150)
166
Chapter 4. Kinetics of a System of Particles
The position of the center of mass of the system is then given as
r̄ =
mA rA + mB rB
mA + m B
(4.151)
Substituting the expressions for rA and rB from Eq. (4.149) and Eq. (4.150), respectively, into Eq. (4.151), we obtain
r̄ =
mA l sin θ Ex + mB l cos θ Ey
mA + m B
(4.152)
Computing the rates of change of rA and rB in reference frame F , we obtain
F
vA
F
vB
F
d
(rA ) = lθ̇ cos θ Ex
dt
F
d
=
(rA ) = −lθ̇ sin θ Ey
dt
=
(4.153)
(4.154)
Computing the rates of change of F vA and F vB in reference frame F , we obtain
F
aA
F
aB
d F vA = l(θ̈ cos θ − θ̇ 2 sin θ )Ex
dt
F
d F =
vB = −l(θ̈ sin θ + θ̇ 2 cos θ )Ey
dt
=
F
(4.155)
(4.156)
Kinetics
Application of a Force Balance to the System
The free body diagram of the system is shown in Fig. 4-5. It can be seen from
NB
NA
Figure 4-5
(mA + mB )g
Free Body Diagram of System for Question 4–8.
the free body diagram that the following three forces act on the system:
= Reaction Force of Vertical Track on mA
NA
NB
= Reaction Force of Horizontal Track on mB
(mA + mB )g = Force of Gravity at Center of Mass
167
Now we know that NA and NB act orthogonal to the horizontal and vertical
tracks, respectively, i.e.,
NA
= NA Ey
(4.157)
NB
= NB Ex
(4.158)
Furthermore, the force of gravity acts vertically downward, i.e.,
(mA + mB )g = (mA + mB )gEx
(4.159)
Then the total force acting on the system is given as
F = NA + NB + (mA + mB )g = NA Ey + NB Ex + (mA + mB )gEx
(4.160)
Eq. (4.160) simplifies to
F = [NB + (mA + mB )g]Ex + NA Ey
(4.161)
Applying a force balance to the system, we obtain
[NB + (mA + mB )g]Ex + NA Ey = mA l(θ̈ cos θ − θ̇ 2 sin θ )Ex
− mB l(θ̈ sin θ + θ̇ 2 cos θ )Ey
(4.162)
Equating components, we obtain the following two scalar equations:
NB + (mA + mB )g
NA
= mA l(θ̈ cos θ − θ̇ 2 sin θ )
2
= −mB l(θ̈ sin θ + θ̇ cos θ )
(4.163)
(4.164)
Application of Newton’s 2nd Law to mA
The free body diagram of collar mA is shown in Fig. 4-6. We have already acR
NA
mA g
Figure 4-6
Free Body Diagram of System for Question 4–8.
counted for the force NA . The force of gravity and the force exerted by mB on
mA are given as
mA g = mA gEx
(4.165)
R
= Rer
168
Chapter 4. Kinetics of a System of Particles
where er is the direction from mA to mB . Now we have
er = − sin θ Ex + cos θ Ey
(4.166)
Also, defining eθ = Ez × er , we obtain
eθ = Ez × er = Ez × (− sin θ Ex + cos θ Ey ) = − cos θ Ex − sin θ Ey
(4.167)
Then the resultant force applied to collar mA is given as
FA = NA + R + mA g = NA Ey + Rer + mA gEx
(4.168)
Setting FA equal to mA F aA using the expression for F aA from Eq. (4.155), we
obtain
NA Ey + Rer + mA gEx = mA l(θ̈ cos θ − θ̇ 2 sin θ )Ex
(4.169)
Now it is seen that we have added another unknown reaction force R, but R does
not appear in either of the equations (4.163) or (4.163). Consequently, we need
to eliminate R directly using Eq. (4.169). The elimination of R is done by taking
the scalar product of Eq. (4.169) in the direction of eθ , resulting in
NA Ey · eθ + Rer · eθ + mA gEx · eθ = mA l(θ̈ cos θ − θ̇ 2 sin θ )Ex · eθ
(4.170)
Now we have
er · eθ
Ex · eθ
Ey · e θ
= 0
= Ex · (− cos θ Ex − sin θ Ey ) = − sin θ
= Ey · (− cos θ Ex − sin θ Ey ) = − cos θ
(4.171)
Eq. (4.170) then simplifies to
−NA cos θ − mA g sin θ = −mA l(θ̈ cos θ − θ̇ 2 sin θ ) sin θ
(4.172)
Determination of the Differential Equation of Motion
The differential equation of motion can be obtained using Eqs. (4.164) and
(4.172). First, multiplying Eqs. (4.164) by cos θ , we obtain
NA cos θ = −mB l(θ̈ sin θ + θ̇ 2 cos θ ) cos θ
(4.173)
Then, adding Eqs. (4.172) and (4.173), we obtain
−mA g sin θ = −mA l(θ̈ cos θ − θ̇ 2 sin θ ) sin θ − mB l(θ̈ sin θ + θ̇ 2 cos θ ) cos θ
(4.174)
Rearranging this last result, we obtain the differential equation of motion in
terms of θ as
mA l(θ̈ cos θ − θ̇ 2 sin θ ) sin θ + mB l(θ̈ sin θ + θ̇ 2 cos θ ) cos θ − mA g sin θ = 0
(4.175)
169
Question 4–17
A dumbbell consists of two particles A and B each of mass m connected by a
rigid massless rod of length 2l. Each end of the dumbbell slides without friction
along a fixed circular track of radius R as shown in Fig. P4-17. Knowing that
θ is the angle from the vertical to the center of the rod and that gravity acts
downward, determine the differential equation of motion for the dumbbell.
g
O
R
Bm
a
θ
A
m
C
2l
Figure P 4-12
Solution to Question 4–17
This problem can be solved using many different approaches. For the purposes
of this solution, we will show the following three two most prominent of these
approaches: (1) using point O as the reference point or (2) using the center of
mass as the reference point. In order to apply all of these approaches, we will
need to compute the following kinematic quantities: (a) the acceleration of each
particle, the angular momentum of the system relative to point O, and (c) the
angular momentum of the system relative to the center of mass.
170
Chapter 4. Kinetics of a System of Particles
Kinematics
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Ex
Ez
Ey
Origin at Point O
=
=
=
Along OC When θ = 0
Out of Page
Ez × Ex
Next, let A be a reference frame that rotates with the dumbbell. Then, choose
the following coordinate system fixed in reference frame A:
er
ez
eθ
Origin at Point O
=
=
=
Along OC
Ez (= Out of Page )
ez × er
The geometry of the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in Fig. 4-7 from
which we have that
eθ
θ
Ez , ez
Ey
θ
er
Ex
Figure 4-7
Geometry of Bases {Ex , Ey , Ez } and {er , eθ , ez } for 4–17.
Ex
= cos θ er − sin θ eθ
(4.176)
Ey
= − sin θ er + cos θ eθ
(4.177)
It is seen that the angular velocity of reference frame A in reference frame F is
given as
F A
ω = θ̇ez
(4.178)
Furthermore, the position of each particle is then given in terms of the basis
{er , eθ , ez } as
rA
= aer − leθ
(4.179)
rB
= aer + leθ
(4.180)
171
The velocity of each particle can then be obtained using the transport theorem
as
F
vA
F
vB
A
drA F A
+ ω × rA
dt
A
drB F A
+ ω × rB
=
dt
=
(4.181)
(4.182)
Now since a and l are constant, we have that
A
drA
dt
A
drB
dt
= 0
(4.183)
= 0
(4.184)
Furthermore,
F
F
ωA × rA
A
ω
× rB
= θ̇ez × (aer − leθ ) = lθ̇er + aθ̇eθ
(4.185)
= θ̇ez × (aer + leθ ) = −lθ̇er + aθ̇eθ
(4.186)
The velocity of each particle is then given as
F
vA
= lθ̇er + aθ̇eθ
(4.187)
F
vA
= −lθ̇er + aθ̇eθ
(4.188)
The acceleration of each particle is then obtained from the transport theorem
as
F
aA
F
aB
d F F A F
vA + ω × vA
dt
A
d F F A F
=
vB + ω × vB
dt
=
A
(4.189)
(4.190)
Now we have that
d F vA
= lθ̈er + aθ̈eθ
dt
A
d F vB
= −lθ̈er + aθ̈eθ
dt
A
(4.191)
(4.192)
Furthermore,
F
F
ωA × F vA
A
ω
F
× vB
= θ̇ez × (lθ̇er + aθ̇eθ ) = −aθ̇ 2 er + lθ̇ 2 eθ
2
2
= θ̇ez × (−lθ̇er + aθ̇eθ ) = −aθ̇ er − lθ̇ eθ
(4.193)
(4.194)
Therefore, the acceleration of each particle is given as
F
aA
F
aB
= (lθ̈ − aθ̇ 2 )er + (aθ̈ + lθ̇ 2 )eθ
2
2
= −(lθ̈ + aθ̇ )er + (aθ̈ − lθ̇ )eθ
(4.195)
(4.196)
172
Chapter 4. Kinetics of a System of Particles
Using the accelerations obtained in Eq. (4.195) and Eq. (4.196), the angular
momentum of the system relative to the (inertially fixed) point O is obtained as
F
HO = (rA − rO ) × mA F vA + (rB − rO ) × mB F vB
(4.197)
Noting that rO = 0 and substituting the results from Eq. (4.187) and Eq. (4.188)
into Eq. (4.197), we obtain
F
HO = (aer −leθ )×mA (lθ̇er +aθ̇eθ )+(aer +leθ )×mB (−lθ̇er +aθ̇eθ ) (4.198)
Simplifying Eq. (4.198), we obtain
F
HO = mA (a2 + l2 )θ̇ + mB (a2 + l2 )θ̇ ez
(4.199)
Finally, noting that mA = mB = m, we obtain F HO as
F
HO = 2m(a2 + l2 )θ̇ez
(4.200)
Similarly, the angular momentum of the system relative to the center of mass
of the system is given as
F
H̄ = (rA − r̄) × mA (F vA − F v̄) + (rB − r̄) × mB (F vB − F vA )
(4.201)
Now we have that
r̄ =
mA rA + mB rB
m(aer − leθ ) + m(aer + leθ )
= aer
=
mA + m B
2m
(4.202)
Therefore,
rA − r̄ = aer − leθ − aer = −leθ
(4.203)
rB − r̄ = aer + leθ − aer = leθ
(4.204)
(4.205)
Furthermore, from the transport theorem we have that
F
v̄ =
F
dr̄ Adr̄ F A
=
+ ω × r̄
dt
dt
(4.206)
Now since a is constant, we have that
A
Furthermore,
Consequently,
F
dr̄
=0
dt
(4.207)
ωA × r̄ = θ̇ez × aer = aθ̇eθ
(4.208)
F
v̄ = aθ̇eθ
(4.209)
173
We then have that
F
vA − F v̄ = lθ̇er + aθ̇eθ − aθ̇eθ = lθ̇er
F
F
vB − v̄ = −lθ̇er + aθ̇eθ − aθ̇eθ = −lθ̇er
(4.210)
(4.211)
(4.212)
Substituting the results of Eq. (4.203), Eq. (4.204), Eq. (4.210), Eq. (4.211), and
Eq. (4.202) into Eq. (4.201), we obtain
F
H̄ = −leθ ×mA (lθ̇er )+leθ ×mB (−lθ̇er ) = mA l2 θ̇ez +mB l2 θ̇ez = (mA +mB )l2 θ̇ez
(4.213)
Again, noting that mA = mB = m, Eq. (4.213) simplifies to
F
H̄ = 2ml2 θ̇ez
(4.214)
Kinetics
The free body diagram of the system is shown in Fig. 4-8. It can be seen that the
B
NB
A
2mg
NA
Figure 4-8
Free Body Diagram of System for Question 4–17.
following three forces act on the system:
NA
= Reaction Force of Track on Particle A
= Reaction Force of Track on Particle B
NB
2mg = Force of Gravity
Now we know that NA and NB must be orthogonal to the track at points A and
B, respectively. Consequently, we can express NA and NB as
NA
NB
= NA uA
= NB uB
(4.215)
where uA and uB are the directions orthogonal to the track at points A and
B, respectively. Now, since the track is circular, we know that the directions
174
Chapter 4. Kinetics of a System of Particles
orthogonal to the track at points A and B are along rA and rB , respectively, i.e.,
uA
=
uB
=
rA
rA rB
rB (4.216)
(4.217)
Consequently, we can write NA and NB as
NA
NB
rA
rA rB
= NB
rB = NA
(4.218)
(4.219)
Next, the force due to gravity can be written as
2mg = 2mgEx
(4.220)
where Ex is the vertically downward direction. Substituting the result for Ex in
terms of er and eθ from Eq. (4.177), we have that
2mg = 2mg(cos θ er − sin θ eθ ) = 2mg cos θ er − 2mg sin θ eθ
(4.221)
With the resolution of forces completed, we can now proceed to solve the
problem using the two approaches stated at the beginning of this solution,
namely (1) applying a moment balance using point O as the reference point
and (2) applying a force balance and a moment balance using the center of mass
as the reference point.
Method 1: Point O as Reference Point
Since O is a point fixed in the inertial reference frame F , we have that
d F
HO = MO
dt
F
(4.222)
First, differentiating the expression for F HO in reference frame F using F HO in
F reference frame F from Eq. (4.200), we obtain d F HO /dt as
d F
HO = 2m(a2 + l2 )θ̈ez
dt
F
(4.223)
Next, using the free body diagram of Fig. 4-8), we have have the moment relative
to point O as
MO = (rA − rO ) × NA + (rB − rO ) × NB + (rg − rO ) × 2mg
(4.224)
175
Now since rO = 0, Eq. (4.224) can be written as
MO = rA × NA + rB × NB + rg × 2mg
(4.225)
Now from Eq. (4.218) and Eq. (4.219) we have that the forces NA and NB lie in
the direction of rA and rB , respectively. Consequently, we have that
rA × NA
= 0
(4.226)
rB × NB
= 0
(4.227)
MO = rg × 2mg
(4.228)
Therefore, MO reduces to
Now we note that gravity acts at the center of mass, i.e., rg = r̄ = aer . Using this
last fact together with the expression for 2mg from Eq. (4.221), we have MO as
MO = aer × (2mg cos θ er − 2mg sin θ eθ ) = −2mga sin θ ez
(4.229)
F Equating MO from Eq. (4.229) with d F HO /dt from Eq. (4.223), we obtain
2m(a2 + l2 )θ̈ez = −2mga sin θ ez
(4.230)
Simplifying Eq. (4.230), we obtain the differential equation of motion as
θ̈ +
ag
sin θ = 0
+ l2
a2
(4.231)
Method 2: Center of Mass as Reference Point
Using the center of mass as the reference point, we need to apply a force balance
and a moment balance. Applying a force balance, we have that
F = 2mF ā
(4.232)
Now the resultant force acting on the particle is given as
F = NA + NB + 2mg
(4.233)
Using the expressions for NA and NB from Eq. (4.218) and Eq. (4.219), respectively, we have NA and NB as
NA
NB
rA
aer − leθ
= NA √
rA a2 + l2
rB
aer + leθ
= NB √
= NB
rB a2 + l2
= NA
(4.234)
(4.235)
Then, adding the expressions in Eq. (4.234) and Eq. (4.235) to the expression for
2mg from Eq. (4.221), we have that
aer − leθ
aer + leθ
+ NB √
+ 2mg cos θ er − 2mg sin θ eθ
F = NA √
2
2
a +l
a2 + l2
(4.236)
176
Chapter 4. Kinetics of a System of Particles
Simplifying Eq. (4.236), we obtain
a
l
F = (NA + NB ) √
+ 2mg cos θ er + (NB − NA ) √
− 2mg sin θ eθ
a2 + l2
a2 + l2
(4.237)
Now the acceleration of the center of mass is given as
F
ā =
d F Ad F F A F
v̄ =
v̄ + ω × v̄
dt
dt
F
(4.238)
Using the expression for F ā from Eq. (4.209), we have that
d F v̄
= aθ̈eθ
dt
F A
ω × F v̄ = θ̇ez × aθ̇eθ = −aθ̇ 2 er
A
(4.239)
(4.240)
The acceleration of the center of mass in reference frame F is then given as
F
ā = −aθ̇ 2 er + aθ̈eθ
(4.241)
Setting F in Eq. (4.237) equal to 2mF ā using F ā from Eq. (4.241), we obtain
a
l
+ 2mg cos θ er + (NB − NA ) √
− 2mg sin θ eθ
(NA + NB ) √
a2 + l2
a2 + l2
= 2m(−aθ̇ 2 er + aθ̈eθ )
(4.242)
Equating components, we obtain the following two scalar equations:
a
+ 2mg cos θ
(NA + NB ) √
2
a + l2
l
(NB − NA ) √
− 2mg sin θ
a2 + l2
= −2maθ̇ 2
(4.243)
= 2maθ̈
(4.244)
Next, we need to apply a balance of angular momentum relative to the center
of mass, i.e.,
F
d F H̄
(4.245)
M̄ =
dt
F F F
First, using the expression for H̄ from Eq. (4.214), we obtain d H̄ /dt
d F H̄ = 2ml2 θ̈ez
dt
F
(4.246)
Next, since gravity passes through the center of mass, the moment due to all
forces relative to the center of mass of the system is given as
M̄ = (rA − r̄) × NA + (rB − r̄) × NB
(4.247)
177
Then, using the expressions for rA −r̄ and rB −r̄ from Eq. (4.203) and Eq. (4.204),
respectively, and the expressions for NA and NB from Eq. (4.234) and Eq. (4.235),
respectively, we obtain M̄ as
aer − leθ
aer + leθ
+ lθ̇eθ × NB √
M̄ = −leθ × NA √
2
2
a +l
a2 + l2
(4.248)
Simplifying Eq. (4.248), we obtain
al
M̄ = (NA − NB ) √
ez
(4.249)
2
a + l2
F F Then, setting M̄ from Eq. (4.249) equal to d H̄ /dt from Eq. (4.246), we obtain
al
ez = 2ml2 θ̈ez
(4.250)
(NA − NB ) √
a2 + l2
We then obtain the following scalar equation:
al
= 2ml2 θ̈
(NA − NB ) √
2
2
a +l
(4.251)
Eq. (4.244), Eq. (4.244), and Eq. (4.251) can now be used together to obtain
the differential equation. First, multiplying Eq. (4.244) by a, we obtain
al
(NB − NA ) √
− 2mga sin θ = ma2 θ̈
a2 + l2
(4.252)
Next, adding Eq. (4.252) to Eq. (4.251), we obtain
−2mga sin θ = (2ml2 + 2ma2 )θ̈
(4.253)
Simplifying Eq. (4.253), we obtain the differential equation of motion as
θ̈ +
ag
sin θ = 0
a2 + l2
(4.254)
178
Chapter 4. Kinetics of a System of Particles
Chapter 5
Kinetics of Rigid Bodies
Question 5–1
A homogeneous circular cylinder of mass m and radius r is at rest atop a thin
sheet of paper as shown in Fig. P5-1. The paper lies flat on a horizontal surface.
Suddenly, the paper is pulled with a very large velocity to the right and is removed from under the cylinder. Assuming that the removal of the paper takes
place in a time t, that the coefficient of dynamic friction between all surfaces
is μ, and that gravity acts downward, determine (a) the velocity of the center of
mass of the cylinder and (b) the angular velocity of the cylinder the instant that
the paper is removed.
r
ω
O
g
x
Friction (μ)
Paper
Figure P 5-1
Solution to Question 5–1
First, let F be a fixed reference frame. Then, it is convenient to choose the
following coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
To The Right
Into Page
Ez × Ex
180
Chapter 5. Kinetics of Rigid Bodies
Next, in order to solve this problem, we need to apply linear impulse and momentum to the center of mass of the cylinder and angular impulse and momentum about the center of mass of the cylinder. In order to apply these two
principles, we use the free body diagram shown in Fig. 5-1 where
N
= Reaction Force of Paper (Surface) on Disk
mg = Force of Gravity
Ff
= Force of Friction
O
mg
P
Ff
N
Figure 5-1
Free Body Diagram of Disk
Now from the geometry we have that
= NEy
N
mg = mgEy
Ff
vrel
= −μN
vrel (5.1)
Now we need to determine vrel . Denoting the point of contact between the disk
and the paper by P (see Fig. 5-1), we note that
vrel = F vP − F vpaper
(5.2)
where F vpaper is the velocity of the paper in reference frame F . Since the paper
is pulled in the positive Ex -direction, we have that
F
vpaper = vpaper Ex
(5.3)
Next, we need to determine F vP . From kinematics of rigid bodies we have that
F
vP − F vO = F ωR × (rP − rO )
(5.4)
where R denotes the reference frame of the cylinder and F ωR is the angular
velocity of reference frame R in reference frame F . From the geometry we have
that
F R
ω = ωEz
(5.5)
181
and
rP − rO = r Ey
Consequently,
F
(5.6)
vP − F vO = ωEz × r Ey = −r ωEx
(5.7)
Furthermore,
rO = xEx
which implies that
F
We then have that
F
vO =
F
(5.8)
d
(rO ) = ẋEx
dt
vP = F vO − r ωEx = (ẋ − r ω)Ex
(5.9)
(5.10)
Now, since the paper is “suddenly” pulled to the right, it implies that the paper
is pulled such that its speed is extremely large. Therefore,
vpaper
ẋ − r ω
(5.11)
which implies that
ẋ − r ω − vpaper
0
(5.12)
Now we have that
vrel = vP − vpaper = (ẋ − r ω)Ex − vpaper Ex = (ẋ − r ω − vpaper )Ex
(5.13)
But from Eq. (5.12) we see that
which implies that
vrel = −|ẋ − r ω − vpaper |Ex
(5.14)
vrel
= −Ex
vrel (5.15)
The force of friction is then given as
Ff = μNEx
(5.16)
Now that we have expressions for all of the forces, we move on to the application
of linear impulse and momentum and angular impulse and momentum
(a) Velocity of Center of Mass of Cylinder at Instant When Paper is Removed
We have that
F = m F aO
(5.17)
Differentiating the velocity of the center of mass from Eq. (5.9) in reference
frame F , we have that
F
aO = ẍEx
(5.18)
182
Chapter 5. Kinetics of Rigid Bodies
Furthermore, using the result of the force resolution from above, we have that
F = N + mg + Ff = NEy + mgEy + μNEx = (N + mg)Ey + μNEx
(5.19)
Equating F and mF aO , we obtain
(N + mg)Ey + μNEx = mẍEx
(5.20)
which yields the following two scalar equations:
N + mg
μN
=
0
= mẍ
(5.21)
Therefore,
N = −mg
(5.22)
N = −mgEy
(5.23)
N = mg
(5.24)
μmg = mẍ
(5.25)
which implies that
which further implies that
We then have that
This last equation can be integrated from 0 to Δt to give
Δt
0
μmgdt =
Δt
mẍdt
(5.26)
0
We then have that
μmgΔt = mẋ(Δt) − mẋ(t = 0) = mvO (Δt) − mvO (t = 0)
(5.27)
Noting that the disk is initially stationary, we have that
vO (t = 0) = 0
(5.28)
μmgΔtmvO (Δt)
(5.29)
which implies that
Solving this last equation for vO (Δt), we obtain
vO (Δt) = μgΔt
(5.30)
Therefore, the velocity of the center of mass at the instant that the paper is
completely pulled out is
vO (Δt) = μgΔtEx
(5.31)
183
(b) Angular Velocity of Cylinder at Instant When Paper is Removed
Applying Euler’s law about the center of mass of the cylinder, we have that
MO =
d F
HO
dt
F
(5.32)
Now since N and mg pass through point O, the moment about O is due to only
Ff . Consequently,
(5.33)
MO = (rP − rO ) × Ff
Substituting the earlier expressions for rP − rO and Ff , we obtain
MO = r Ey × μmgEx = −r μmgEz
Furthermore,
F
F R
HO = IR
O · ω
(5.34)
(5.35)
Now since {Ex , Ey , Ez } is a principle-axis basis, we have that
O
O
O
IR
O = Ixx Ex ⊗ Ex + Iyy Ey ⊗ Ey + Iyy Ey ⊗ Ey
Using the angular velocity
F ωR
(5.36)
from Eq. (5.5), we obtain F HO as
F
HO =
mr 2
ωEz
2
(5.37)
Differentiating F HO with respect to time in reference frame F , we have that
mr 2
d F
ω̇Ez
HO =
dt
2
and F dF HO /dt , we obtain
F
Equating MO
−r μmg =
mr 2
ω̇
2
(5.38)
(5.39)
Integrating this last equation from 0 to Δt, we obtain
Δt
0
−r μmgdt =
Δt
0
mr 2
ω̇dt
2
(5.40)
We then obtain
−r μmgΔt =
mr 2
mr 2
ω(Δt) −
ω(t = 0)
2
2
(5.41)
Noting that the disk is initially stationary, we have that
ω(t = 0) = 0
(5.42)
184
Chapter 5. Kinetics of Rigid Bodies
Therefore,
−r μmgΔt =
mr 2
ω(Δt)
2
(5.43)
Solving for ω(Δt), we obtain
ω(Δt) = −
2μg
Δt
r
(5.44)
The angular velocity of the disk at the instant that the paper is pulled out is
then given as
2μg
F R
ΔtEz
ω (Δt) = −
(5.45)
r
185
Question 5–2
A collar of mass m1 is attached to a rod of mass m2 and length l as shown in
Fig. P5-2. The collar slides without friction along a horizontal track while the
rod is free to rotate about the pivot point Q located at the collar. Knowing that
the angle θ describes the orientation of the rod with the vertical, that x is the
horizontal position of the cart, and that gravity acts downward, determine a
system of two differential equations for the collar and the rod in terms of x and
θ.
x
m1
Q
l
θ
g
m2
Figure P 5-2
Solution to Question 5–2
Preliminaries
For this problem it is convenient to apply the following balance laws:
• Newton’s 2nd law to the collar
• Euler’s 1st law to the center of mass of the rod
• Euler’s 2nd law about the center of mass of the rod
In order to use the aforementioned balance laws, we will need the following
kinematic quantities in an inertial reference frame:
• The acceleration of the collar
• The acceleration of the center of mass of the rod
• The rate of change of angular momentum of the rod relative to the center
of mass of the rod
186
Chapter 5. Kinetics of Rigid Bodies
Kinematics
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in F :
Ex
Ez
Ey
Origin at Collar When x = 0
=
To The Right
=
Out of Page
=
Ez × Ex
Next, let R be a reference frame fixed to the rod. Then, choose the following
coordinate system fixed in R:
er
Ez
eθ
Origin at Collar
=
Along Rod
=
Out of Page
=
Ez × er
The relationship between the bases {Ex , Ey , Ez } and {er , eθ , Ez } is shown in
Fig. 5-2 Using Fig. 5-2, we have that
Ey
eθ
θ
Ez , ez
Ex
θ
er
Figure 5-2
tion 6.2.
Relationship Between Bases {Ex , Ey , Ez } and {er , eθ , Ez } for Ques-
er
eθ
Ex
Ey
=
sin θ Ex − cos θ Ey
=
cos θ Ex + sin θ Ey
=
sin θ er + cos θ eθ
= − cos θ er + sin θ eθ
(5.46)
In terms of the basis {Ex , Ey , Ez }, the position of the collar is given as
r = xEx
(5.47)
187
Therefore, the velocity of the collar in reference frame F is given as
F
v = ẋEx
(5.48)
Furthermore, the acceleration of the collar in reference frame F is given as
F
a = ẍEx
(5.49)
Next, the position of the center of mass of the rod relative to the collar is given
as
l
r̄ − r = er
(5.50)
2
In addition, the angular velocity of R in reference frame F is given as
F
ωR = θ̇Ez
(5.51)
Differentiating F ωR in Eq. (5.51), the angular acceleration of reference frame R
in reference frame F is given as
F
αR = θ̈Ez
(5.52)
Then, since the location of the collar is also a point on the rod, the acceleration
of the center of mass of the rod relative to the collar is obtained from rigid body
kinematics as
F
ā − F a = F αR × (r̄ − r) + F ωR × F ωR × (r̄ − r)
(5.53)
Using the expression for r̄ − r from Eq. (5.50), the expression for
Eq. (5.51), and the expression for Eq. (5.52), we obtain
F
+ θ̇Ez × θ̇Ez ×
l
er
ā − a = θ̈Ez ×
2
F
l
er
2
F ωR
from
(5.54)
Simplifying Eq. (5.54), we obtain
F
l
l
ā − F a = − θ̇ 2 er + θ̈eθ
2
2
(5.55)
Then, using the fact that F a = ẍEx , we obtain
F
Finally, we need
F
l 2
l
θ̇ er + θ̈eθ
2
2
ā = ẍEx −
(5.56)
F
d H̄/dt. We have that
F
R
H̄ = Ī
· F ωR
(5.57)
188
Chapter 5. Kinetics of Rigid Bodies
R
where Ī is the moment of inertia tensor of the rod relative to the center of
mass and F ωR is the angular velocity of the rod in reference frame F . Now
since the {er , eθ , Ez } is a principle-axis basis, we have that
Ī
R
= Īr r er ⊗ er + Īθθ eθ ⊗ eθ + Īzz Ez ⊗ Ez
Furthermore, using the expression for
F
F ωR
(5.58)
as given in Eq. (5.51), we obtain
H̄ = Īzz θ̇Ez
(5.59)
Now, for a slender rod of mass M and length l we have that
Ml2
12
(5.60)
Ml2
θ̇Ez
12
(5.61)
Īz z =
Therefore,
F
H̄ =
Differentiating the expression in Eq. (5.61), we obtain
d F Ml2
θ̈Ez
H̄ =
dt
12
F
(5.62)
Kinetics
As stated earlier, to solve this problem we will use the following balance laws:
• Newton’s 2nd law to the collar
• Euler’s 1st law to the center of mass of the rod
• Euler’s 2nd law about the center of mass of the rod
Application of Newton’s 2nd Law to Collar
The free body diagram of the collar is shown in Fig. 5-3. where
N
R
mg
Figure 5-3
Free Body Diagram of Collar for Question 6.2.
189
N
= Force of Track on Collar
R
= Reaction Force of Hinge Due to Rod
mg = Force of Gravity
Then, from the geometry we have that1
N
=
NEy
R
= Rr er + Rθ eθ
mg =
−mgEy
(5.63)
Using Eq. (5.46), the force R can be written as
R = Rr (sin θ Ex − cos θ Ey ) + Rθ (cos θ Ex + sin θ Ey )
(5.64)
R = (Rr sin θ + Rθ cos θ )Ex + (−Rr cos θ + Rθ sin θ )Ey
(5.65)
which gives
The total force on the collar is then given as
F = N + R + mg
= NEy + (Rr sin θ + Rθ cos θ )Ex + (−Rr cos θ + Rθ sin θ )Ey − mgEy
(5.66)
This gives
F = (Rr sin θ + Rθ cos θ )Ex + (N − mg − Rr cos θ + Rθ sin θ )Ey
(5.67)
Setting F equal to mF a using the expression for F a from Eq. (5.49), we obtain
(Rr sin θ + Rθ cos θ )Ex + (N − mg − Rr cos θ + Rθ sin θ )Ey = mẍEx
(5.68)
Equating components, we obtain the following two scalar equations:
mẍ
N − mg − Rr cos θ + Rθ sin θ
= Rr sin θ + Rθ cos θ
(5.69)
= 0
(5.70)
Application of Euler’s 1st Law to Rod
Using the free body diagram of the rod as shown in Fig. 5-4, we have that
F = −R + Mg = −Rr er − Rθ eθ − MgEy
(5.71)
Also, equating F and mF ā using F ā from Eq. (5.56), we have that
−Rr er − Rθ eθ − MgEy = M ẍEx +
1
l
l
θ̈eθ − θ̇ 2 er
2
2
(5.72)
It is noted that, because the rod is a distributed mass, the reaction force between the collar
and the rod is not purely along the direction of the rod. Instead, the reaction force between the
collar and the rod has a component orthogonal to the rod.
190
Chapter 5. Kinetics of Rigid Bodies
Eq. (5.72) can be split into components in the er and eθ directions by taking dot
products. We note that Furthermore, we note that
Ex · e r
Ey · er
Ex · e θ
Ey · eθ
=
sin θ
= − cos θ
=
cos θ
=
sin θ
(5.73)
Taking dot products in the er and eθ directions, respectively, we obtain the
following two scalar equations:
−Rr + Mg cos θ
−Rθ − Mg sin θ
Ml 2
θ̇
2
Ml
θ̈
= M ẍ cos θ +
2
= M ẍ sin θ −
(5.74)
(5.75)
Application of Euler’s 2nd Law to Rod
Referring again to the free body diagram of the rod as shown in Fig. 5-4, we have
that where
−R
Mg
Figure 5-4
Free Body Diagram of Rod for Question 6.4.
−R = Reaction Force of Cart on Rod
Mg = Force of Gravity
Now since gravity passes through the center of mass of the rod, the only moment about the center of mass is due to −R. Consequently,
M̄ = (rR − r̄) × (−R)
(5.76)
Also,
rR
=
xEx
r̄
= xEx +
l
er
2
(5.77)
191
Consequently,
l
rR − r̄ = − er
2
(5.78)
l
M̄ = − er × (−Rr er − Rθ eθ )
2
(5.79)
Then,
This gives
M̄ =
Equating M̄ and
obtain
F
l
Rθ Ez
2
F
(5.80)
d H̄/dt using the expression for
F
F
d H̄/dt from Eq. (5.62), we
l
Ml2
θ̈ = Rθ
12
2
(5.81)
This gives
Rθ =
Ml
θ̈
6
(5.82)
System of Two Differential Equations
The system of two differential equations can be obtained from Eq. (5.69), Eq. (5.70),
Eq. (5.82), Eq. (5.74), and Eq. (5.75). Substituting Eq. (5.82) into Eq. (5.75), we obtain
Ml
Ml
−Mg sin θ −
θ̈ = M ẍ cos θ +
θ̈
(5.83)
6
2
Simplifying this last equation yields the first differential equation as
lθ̈ +
3g
3
ẍ cos θ +
sin θ = 0
2
2
(5.84)
Next, multiplying Eq. (5.74) by sin θ and Eq. (5.75) by cos θ , we have the following two equations:
−Rr sin θ + Mg cos θ sin θ
−Rθ cos θ − Mg sin θ cos θ
= M ẍ sin2 θ −
=
Ml 2
θ̇ sin θ
2
(5.85)
Ml
θ̈ cos θ
M ẍ cos2 θ +
2
Adding these last two equations gives
−Rr sin θ − Rθ cos θ = M ẍ(sin2 θ + cos2 θ) −
Ml 2
Ml
θ̇ sin θ +
θ̈ cos θ
2
2
(5.86)
We then obtain
−Rr sin θ − Rθ cos θ = M ẍ −
Ml 2
Ml
θ̇ sin θ +
θ̈ cos θ
2
2
(5.87)
192
Chapter 5. Kinetics of Rigid Bodies
Now substitute the expression for −Rr sin θ − Rθ cos θ from Eq. (5.69) into this
Eq. (5.87). This gives
−mẍ = M ẍ −
Ml
Ml 2
θ̇ sin θ +
θ̈ cos θ
2
2
(5.88)
Rearranging this last equation, we obtain the second differential equation as
(M + m)ẍ −
Ml 2
Ml
θ̇ sin θ +
θ̈ cos θ = 0
2
2
(5.89)
The system of differential equations is then given as
lθ̈ +
3
3g
sin θ
ẍ cos θ +
2
2
= 0
(5.90)
Ml 2
Ml
(M + m)ẍ −
θ̇ sin θ +
θ̈ cos θ
2
2
= 0
193
Question 5–3
A bulldozer pushes a boulder of mass m with a known force P up a hill inclined
at a constant inclination angle β as shown in Fig. P5-3. For simplicity, the boulder is modeled as a uniform sphere of mass m and radius r . Assuming that
the boulder rolls without slip along the surface of the hill, that the coefficient
of dynamic Coulomb friction between the bulldozer and the boulder is μ, that
the force P is along the direction of the incline and passes through the center of
mass of the boulder, and that gravity acts downward, determine the differential
equation of motion of the boulder in terms of the variable x.
Friction (μ)
No Slip
P O
r
x
g
β
Figure P 5-3
Solution to Question 5–3
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in F :
Ex
Ez
Ey
Origin at Point O when x = 0
=
Up Incline
=
Into Page
=
Ez × E x
Now for this problem it is helpful to do some of the kinetics before proceeding
with the remainder of the kinematics. First, the free body diagram of the sphere
is shown in Fig. 5-5 where P is the point on the sphere that instantaneously is
sliding on the plate and Q is the point on the sphere that is instantaneously in
contact with the incline.
194
Chapter 5. Kinetics of Rigid Bodies
Ff
P
mg
Q
P
R
Figure 5-5
N
Free Body Diagram of Sphere for Question 5–3.
Using Fig. 5-5, the forces acting on the sphere are given as
N
R
P
mg
Ff
=
=
=
=
=
Force
Force
Force
Force
Force
of
of
of
of
of
Incline on Sphere
Rolling
Plate on Sphere
Gravity
Friction Due to Contact of Sphere with Bulldozer
From the geometry we have that
N = NEy
(5.91)
R = REx
(5.92)
P = P Ex
(5.93)
mg = mguv
(5.94)
Ff
vrel
= −μP
vrel (5.95)
where uv is the unit vector in the vertically downward direction. Now uv is
shown in Fig. 5-6.
Using Fig. 5-6, we have that
uv = − sin βEx + cos βEy
(5.96)
Therefore, the force of gravity is obtained as
mg = −mg sin βEx + mg cos βEy
(5.97)
Next, in order to obtain the correct direction for the friction force Ff , we need
to determine vrel . We note that
plate
F
vrel = F vR
P − vP
(5.98)
195
Ex
⊗
β
Ey
uv
Figure 5-6
where
F vplate
P
F vR
P
Unit Vector in Vertically Downward Direction for Question 5–3.
is the velocity of point P on the sphere in reference frame F and
is the of point P on the plate velocity of the plate in reference frame F
(we note in this case that the plate is the surface on which the sphere slides).
Now we have from the geometry of the problem that
rP = xEx − r Ex = (x − r )Ex
Consequently,
F plate
vP
Next, we need to determine
incline, we have that
Also,
F
F R
vP .
=
(5.99)
F
d
(rP ) = ẋEx
dt
(5.100)
Since the sphere rolls without slip along the
F
vQ = 0
(5.101)
vO = F vQ + F ωR × (rO − rQ )
(5.102)
where R is the reference frame of the sphere and F ωR is the angular velocity of
the sphere in reference frame F . Because for this problem the motion is planar,
we have that
F R
ω = ωEz
(5.103)
Furthermore, from the definition of the coordinate system above we have that
rQ
= xEx + r Ey
(5.104)
rO
= xEx
(5.105)
Consequently,
rO − rQ = −r Ey
(5.106)
vO = ωEz × (−r Ey ) = r ωEx
(5.107)
We then have that
196
Chapter 5. Kinetics of Rigid Bodies
Furthermore, a second expression for rO is given as
rO = xEx
Therefore,
(5.108)
F
d
(5.109)
(rO ) = ẋEx
dt
Differentiating F vO in Eq. (5.109), the acceleration of point O in reference frame
F is given as
F
aO = ẍEx
(5.110)
F
vO =
Setting the result of Eq. (5.107) equal to the result of Eq.(5.109), we obtain
ẋ = r ω
(5.111)
which gives
ω=
ẋ
r
(5.112)
Differentiating this last result, we obtain
ẍ
r
The angular velocity of the sphere is then given as
ω̇ =
(5.113)
ẋ
Ez
(5.114)
r
Furthermore, the angular acceleration of the sphere in reference frame F is
given as
F
d F R ẍ
F R
α =
α
(5.115)
= Ez
dt
r
We can then use the expression for ω from Eq. (5.112) to determine F vP . We
have that
F
vP = F vQ + F ωR × (rP − rQ )
(5.116)
F
ωR =
Then, using Eq. (5.99) and Eq. (5.104), we have that
rP − rQ = −r Ex − r Ey
Consequently,
F R
vP
= ωEz × (−r Ex − r Ey ) = ẋEx − ẋEy
(5.117)
(5.118)
Then vrel is obtained from Eq. (5.100) and Eq. (5.118) as
plate
F
vrel = F vR
P − vP
= −ẋEy
(5.119)
The force of friction Ff is then given as
Ff = −μP
−ẋEy
= μP Ey
ẋ
(5.120)
We now have expressions for all of the forces acting the sphere and can proceed
to solving parts (a) and (b).
197
Determination of Differential Equation of Motion
The differential equation of motion can be obtained using Euler’s law about the
point of contact of the sphere with the incline (i.e. about point Q), i.e. we can
apply
F
d F
HQ
(5.121)
MQ − (r̄ − rQ ) × mF aQ =
dt
Noting that r̄ = rO , we have that
MQ − (rO − rQ ) × mF aQ = ḢQ
(5.122)
Now the acceleration of the contact point Q is obtained as
F
aQ = F aO + F αR × (rQ − rO ) + F ωR × ω × (rQ − rO )
Using F aO from Eq. (5.110), F ωR from Eq. (5.114),
the fact that rQ − rO = r Ey , we obtain F aQ as
F
aQ
F αR
(5.123)
from Eq. (5.115), and
ẍ
ẋ
ẋ
Ez × r Ey
= ẍEx + Ez × r Ey + Ez ×
r
r
r
ẋ 2
Ey
= ẍEx − ẍEx −
r
ẋ 2
Ey
=−
r
(5.124)
Consequently, the inertial moment −(rO − rQ ) × mF aQ is given as
−r Ey × m(−
ẋ 2
Ey ) = 0
r
(5.125)
Since the inertial moment is zero, for this problem Eq. (5.121) reduces to
MQ =
d F
HQ
dt
F
(5.126)
Next, looking at the free body diagram above, it can be seen that the forces
N and R both pass through point Q. Therefore, the moment relative to point Q
is due to only the forces P, Ff , and mg. We then have that
MQ = (rP − rQ ) × P + (rP − rQ ) × Ff + (rO − rQ ) × mg
(5.127)
where
rP − rQ
= −r Ex − r Ey
(5.128)
rO − rQ
= −r Ey
(5.129)
198
Chapter 5. Kinetics of Rigid Bodies
Therefore,
MQ = (−r Ex − r Ey ) × F Ex + (−r Ex − r Ey ) × μP Ey
+ (−r Ey ) × (−mg sin βEx + mg cos βEy )
(5.130)
This last expression simplifies to
MQ = r P Ez − r μP Ez = r P (1 − μ)Ez − mgr sin βEz
(5.131)
Furthermore, the angular momentum relative to the contact point is given as
F
F
HQ = H̄ + (rQ − rO ) × m(F vQ − F vO )
Now we have that
F
R
H̄ = Ī
(5.132)
· F ωR
(5.133)
Now since {Ex , Ey , Ez } is a principle-axis basis, we have that
R
Ī
Then, substituting Ī
= Īxx Ex ⊗ Ex + Īyy Ey ⊗ Ey + Īzz Ez ⊗ Ez
R
(5.134)
from Eq. (5.134) into Eq. (5.133), we obtain
F
H̄ = Īzz ωEz = Īzz
ẋ
Ez
r
(5.135)
Now we have for a uniform sphere we have that Īzz = 2mr 2 /5. Consequently,
F
H̄ =
2mr ẋ
2mr 2 ẋ
Ez =
Ez
5 r
5
(5.136)
Next, since F vQ = 0, we have that
(rQ − rO ) × m(F vQ − F vO ) = mr Ey × (−ẋEx ) = mr ẋEz
(5.137)
Consequently,
F
HQ =
2mr ẋ
7mr ẋ
Ez + mr ẋEz =
Ez
5
5
F
Then, differentiating F HQ in reference frame F , we obtain d
(5.138)
F
HQ /dt as
7mr ẍ
d F
Ez
HQ =
(5.139)
dt
5
F from Eq. (5.131) with d F HQ /dt from Eq. (5.139), we obtain
F
Equating MQ
r P (1 − μ) − mgr sin β =
7
mr ẍ
5
(5.140)
Simplifying Eq. (5.140), we obtain the differential equation of motion as
7
mẍ + mg sin β = P (1 − μ)
5
(5.141)
199
Question 5–4
A uniform slender rod of mass m and length l pivots about its center at the
fixed point O as shown in Fig. P5-4. A torsional spring with spring constant
K is attached to the rod at the pivot point. The rod is initially at rest and the
spring is uncoiled when a linear impulse F̂ is applied transversely at the lower
end of the rod. Determine (a) the angular velocity of the rod immediately after
the impulse F̂ is applied and (b) the maximum angle θmax attained by the rod
after the impulse is applied.
m
K
O
l
F̂
Figure P 5-4
Solution to Question 5–4
Preliminaries
We note for this problem that the fixed point O is the center of mass. Furthermore, since the rod is constrained to rotate about point O, we need not consider
the translational motion of the center of mass of the rod. Then, since the only
motion that needs to be considered is the rotational motion of the rod about
its center of mass, the only kinematic quantity of interest in this problem is the
angular momentum of the rod.
Kinematics
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Ex
Ez
Ey
Origin at Point O
=
=
=
Along Rod (Down)
Out of Page
Ez × Ex
200
Chapter 5. Kinetics of Rigid Bodies
Next, let R be a reference frame fixed to the rod. Then, choose the following
coordinate system fixed in reference frame R:
er
ez
eθ = ez × er
Origin at Point O
=
=
Along Rod (Down)
Ez
Then, since the rod rotates about the ez -direction, the angular velocity of the
rod in reference frame F can then be expressed as
F
ωR = ωez
(5.142)
Furthermore, the angular momentum of the rod in reference frame F relative
to the point O is given as
F
F R
HO = IR
(5.143)
O · ω
Now since {er , eθ , ez } is a principle-axis basis, the inertia tensor can be expressed as
O
O
O
IR
(5.144)
O = Ir r er ⊗ er + Iθθ eθ ⊗ eθ + Izz ez ⊗ ez
Substituting the results of Eq. (5.144) and Eq. (5.142) into Eq. (5.143), we obtain
F
HO as
O
F
O
O
HO = IrOr er ⊗ er + Iθθ
eθ ⊗ eθ + Izz
ez ⊗ ez · ωez = Izz
ωez
(5.145)
Finally, we have for a slender uniform rod that
O
=
Izz
ml2
12
(5.146)
Substituting the result of Eq. (5.146) into Eq. (5.145), we obtain
F
HO =
ml2
ωez
12
(5.147)
Kinetics
Since the center of mass of the rod is the fixed point O, in order to solve this
problem we only need to apply angular impulse and angular momentum relative
to the center of mass. From this point forward we will use the general notation
for the center of mass rather than using the subscript O. Then, applying angular
impulse and angular momentum relative to point O, we have that
ˆ = F H̄ − F H̄
M̄
(5.148)
The free body diagram of the rod during the application of F̂ is shown in Fig. 57. Now, we first note that the impulse due to the torsional spring, τs , is zero
201
τ̂
R̂
F̂
Figure 5-7
5–4.
Free Body Diagram of Rod During Application of F̂ for Problem
because F̂ is assumed to be applied instantaneously and, thus, the orientation
of the rod does not change during the application of F̂. Next, we see that the
reaction impulse at point O, R̂, is inconsequential because R̂ passes through
point O. Therefore, the impulse applied to the rod about point O is given as
ˆ = (r − r̄) × F̂
M̄
F̂
Now we see that
rF̂ =
l
er
2
(5.149)
(5.150)
Furthermore, since F̂ is applied horizontally, we have that
F̂ = F̂ eθ
(5.151)
Substituting Eq. (5.150) and Eq. (5.151) into Eq. (5.149), we obtain
ˆ = l e × F̂ e = lF̂ E
M̄
r
z
θ
2
2
(5.152)
Next, since the rod is initially at rest, we have that
F
H̄ = 0
(5.153)
Then, substituting ω2 into the expression for HO from Eq. (5.143), we have that
F
H̄ =
ml2 ω Ez
12
(5.154)
F
Setting M̂ from Eq. (5.152) equal to H̄ from Eq. (5.154, we obtain
ml2 lF̂
Ez =
ω Ez
2
12
(5.155)
Dropping Ez and solving for ω , we obtain
ω =
6F̂
ml
(5.156)
202
Chapter 5. Kinetics of Rigid Bodies
The angular velocity of the rod the instant after the impulse F̂ is applied is then
given as
6F̂
F R
ez
ω
=
(5.157)
ml
Now, after F̂ has been applied, the rod starts to rotate. Therefore, the only
forces and torques acting on the rod after the application of F̂ are the reaction
force, R, at point O and the spring torque, τs . Since F v̄ = 0, we see that R · vO =
0 which implies that R does no work. Furthermore, the spring torque τs is
conservative with potential energy
F
U = F Us =
1
Kθ 2
2
(5.158)
Since the only forces or torques acting on the rod after F̂ is applied are conservative or do no work, energy is conserved. The total energy of the rod is then
given as
F
E = FT + FU
(5.159)
The kinetic energy is given as
F
T =
1F
1 F
H̄ · F ωR
m v̄ · F v̄ +
2
2
(5.160)
since F v̄ = 0, the kinetic energy reduces to
T =
Substituting
1F
2 H̄
1F
H̄ · F ωR
2
(5.161)
from Eq. (5.143) into Eq. (5.161), we obtain
F
T =
ml2 ω2
24
(5.162)
ml2 θ̇ 2
24
(5.163)
Observing that ω = θ̇, we have that
F
T =
Now the point where θ attains its maximum value is where θ̇ = 0. Applying
conservation of energy, we have that
F
T1 + F U 1 = F T2 + F U 2
(5.164)
where point “1” is immediately after F̂ is applied and point “2” is when θ̇ = 0.
We then have that
ml2 ω21
F
(5.165)
T1 =
24
203
Now we note that ω1 = ω from Eq. (5.156). Consequently, we obtain F T1 as
F
#
ml2
T1 =
24
6F̂
ml
$2
(5.166)
Furthermore, we have that U1 = 0 since the spring is initially uncoiled. Next, we
see that T2 = 0 since ω2 = θ̇2 = 0, i.e.
F
T2 =
ml2 θ̇22
ml2 ω22
=
=0
24
24
(5.167)
1
2
Kθmax
2
(5.168)
Last, we have that
F
U2 =
Then, applying Eq. (5.164), we obtain
ml2
24
#
6F̂
ml
$2
=
Solving for θmax gives
1
2
Kθmax
2
θmax
6F̂
=
ml
ml2
12K
(5.169)
(5.170)
204
Chapter 5. Kinetics of Rigid Bodies
Question 5–5
A homogeneous cylinder of mass m and radius r moves along a surface inclined
at a constant inclination angle β as shown in Fig. P5-5. The surface of the incline
is composed of a frictionless segment of known length x between points A and
B and a segment with a coefficient of friction μ from point B onwards. Knowing
that the cylinder is released from rest at point A and that gravity acts vertically
downward, determine (a) the velocity of the center of mass and the angular
velocity when the disk reaches point B, (b) the time (measured from point B)
when sliding stops and rolling begins, and (c) the velocity of the center of mass
and the angular velocity of the disk when sliding stops and rolling begins.
m
O
r
g
P
A
Frictionless
x
Friction (μ)
B
β
Figure P 5-5
Solution to Question 5–5
Kinematics
Let F be a fixed reference frame. Then, choose the following coordinate system
fixed in reference frame F :
Ex
Ez
Ey
Origin at O
att = 0
=
=
=
Down Incline
Into Page
Ez × Ex
In terms of the basis {Ex , Ey , Ez }, the position of the center of mass of the
cylinder is given as
r̄ = rO = xEx
(5.171)
205
Now, since the basis {Ex , Ey , Ez } is fixed, the velocity of the center of mass of
the cylinder is given as
F
v̄ = ẋEx = v̄Ex
(5.172)
Finally, the acceleration of the center of mass of the cylinder is given as
F
ā =
d F ˙ = āEx
v̄ = ẍEx = v̄
dt
F
(5.173)
Next, since the cylinder rotates about the Ez -direction, the angular velocity of
the cylinder in the fixed reference frame F is given as
F
ωR = ωEz
(5.174)
Finally, the velocity of the instantaneous point of contact, P , in reference frame
F is given as
F
vP = F v̄ + F ωR × (rP − r̄) = v̄Ex + ωEz × r Ey = (v̄ − r ω)Ex
(5.175)
where we note that
rP − r̄ = r Ey
(5.176)
Kinetics
The kinetics of this problem are divided into the following two distinct segments:
(a) the frictionless segment
(b) the segment with friction
We will analyze each of these segments separately.
Kinetics During Frictionless Segment
The free body diagram of the cylinder during the frictionless segment is shown
in Fig. 5-8. It can be seen from Fig. 5-8 that the following forces act on the
cylinder during the frictionless segment:
N
= Normal Force of Incline on Cylinder
mg = Force of Gravity
Now from the geometry of the problem we have that
N = NEy
(5.177)
mg = mguv
(5.178)
(5.179)
206
Chapter 5. Kinetics of Rigid Bodies
mg
N
Figure 5-8
Free Body Diagram of Cylinder During Frictionless Segment for
Question 5–5.
.
Ez⊗
Ex
β
Ey
Figure 5-9
5–5.
uv
Direction of Unit Vertical uv in Terms of Ex and Ey for Question
where uv is the unit vector in the vertically downward direction as shown in
Fig. 5–9.
Using Fig. 5–9, we have
(5.180)
uv = sin βEx + cos βEy
Consequently, the force of gravity is given as
mg = mg sin βEx + mg cos βEy
(5.181)
Now we know that the force of gravity is conservative. Furthermore, because
the normal force N acts at point P , we have that
N · F vP = NEy · (v̄ − r ω)Ex = 0
(5.182)
Eq. (5.182) implies that the power of N is zero which implies that N does no
work. Then, since gravity is the only force other than N and is conservative,
energy is conserved during the frictionless segment, i.e.,
F
E = F T + F U = constant
(5.183)
Now the kinetic energy in reference frame F is given as
F
T =
1 F
1 F
m v̄ · F v̄ + m H̄ · F ωR
2
2
(5.184)
207
Now we are given that the disk is released from rest which implies that
F
H̄(t = 0) = 0
(5.185)
Furthermore, from the free body diagram of Fig. 5-8 is is seen that both of the
forces that act on the cylinder during the frictionless segment pass through the
center of mass of the cylinder. Consequently, the resultant moment about the
center of mass of the cylinder during the frictionless segment is zero, i.e.,
M̄ = 0
(5.186)
Then, since M̄ ≡ 0 during the frictionless segment, we have that
d F H̄ = 0
dt
(5.187)
H̄ = constant
(5.188)
F
Eq. (5.187) implies that
F
during the frictionless segment. Then, Eq. (5.185), together with Eq. (5.187)
implies that
F
H̄ = 0
(5.189)
during the frictionless segment. Consequently, the kinetic energy of the cylinder
during the frictionless segment reduces to
F
T =
1 F
m v̄ · F v̄
2
(5.190)
Substituting the expression for F v̄ from Eq. (5.172) into Eq. (5.190), we obtain
the kinetic energy as
1
1
F
T = mv̄Ex · v̄Ex = mv̄ 2
(5.191)
2
2
Next, since the only conservative force acting on the cylinder is that due to
gravity, the potential energy in reference frame F is given as
F
U = F Ug = −mg · r̄
(5.192)
Substituting the results of Eq. (5.181) and Eq. (5.171) into Eq. (5.192), we obtain
the potential energy in reference frame F as
F
U = F Ug = −(mg sin βEx + mg cos βEy ) · xEx = −mgx sin β
(5.193)
Then, substituting the results of Eq. (5.191) and Eq. (5.193) into Eq. (5.183), we
obtain the total energy of the cylinder during the frictionless segment as
F
E=
1
mv̄ 2 − mgx sin β = constant
2
(5.194)
208
Chapter 5. Kinetics of Rigid Bodies
Then, using the principle of work and energy for a rigid body, we have that
F
E0 = F E1
(5.195)
where F E0 and F E1 are the total energies of the cylinder at the beginning and
end of the frictionless segment. Eq. (5.195) implies that
1
1
mv̄02 − mgx0 sin β = mv̄12 − mgx1 sin β
2
2
(5.196)
Now we know that v̄0 = 0 and x0 are both zero. Therefore,
1
mv̄12 − mgx1 sin β = 0
2
Then, knowing that x1 = x, we can solve Eq. (5.197) for v̄1 to give
v̄1 = 2gx sin β
(5.197)
(5.198)
Eq. (5.198) implies that the velocity of the center of mass of the cylinder at the
end of the frictionless segment is given as
F
v̄(t1 ) = 2gx sin βEx
(5.199)
F
Finally, since H̄ ≡ 0 during the frictionless segment, the angular velocity of the
cylinder during the frictionless segment is also zero which implies that
F
ωR (t1 ) = 0
(5.200)
Segment with Friction
The free body diagram of the cylinder during the segment with friction is shown
in Fig. 5-10.
Ez⊗
Ex
β
Ey
uv
Figure 5-10 Free Body Diagram of Cylinder During Segment with Friction for
Question 5–5.
It can be seen that the key difference between the segment with friction and
the frictionless segment is that a friction force, Ff , acts at the instantaneous
209
point of contact. Recalling that the expression for the force of sliding Coulomb
friction is given as
vrel
Ff = −μN
(5.201)
vrel Now we know that
vrel = F vP R − F vP S
(5.202)
where S denotes the inclined surface. Now since the incline is fixed, we have
that
F
vP S = 0
(5.203)
which implies that
vrel = F vP R
(5.204)
Then, using the result of Eq. (5.175), we have that
vrel = (v̄ − r ω)Ex
(5.205)
Now we know that, at the beginning of the segment with friction that
v̄(t1 ) − r ω(t1 ) = v̄(t1 ) > 0
(5.206)
Therefore, during the period when the cylinder is sliding, we must have that
v̄ − r ω > 0
(5.207)
|v̄ − r ω| = v̄ − r ω
(5.208)
(v̄ − r ω)Ex
vrel
=
= Ex
vrel |v̄ − r ω|
(5.209)
which implies that
Therefore, we have that
The force of sliding Coulomb friction is then given as
Ff = −μNEx
(5.210)
Now, in order to solve for the velocity of the center of mass of the cylinder
and the angular velocity of the cylinder at the instant that sliding stops and
rolling begins, we need to apply both the principle of linear impulse and linear
momentum and the principle of angular impulse and angular momentum. First,
we can apply the principle of linear impulse and linear momentum by applying
Euler’s 1st law to the cylinder during the segment with friction as
F = F ā
(5.211)
First, the resultant force acting on the cylinder during the segment with friction
is given as
F = N + mg + Ff
(5.212)
210
Chapter 5. Kinetics of Rigid Bodies
Using the expressions for N, mg, and Ff from Eq. (5.177), Eq. (5.181), and
Eq. (5.210), we obtain the resultant force acting on the cylinder as
F = NEy + mg sin βEx + mg cos βEy − μNEx
Then, using F from Eq. (5.213) and the expression for
Eq. (5.211), we have that
F
(5.213)
ā from Eq. (5.173) in
NEy + mg sin βEx + mg cos βEy − μNEx = mẍEx
(5.214)
Simplifying Eq. (5.214), we obtain
(mg sin β − μN)Ex + (N + mg cos β)Ey − μNEx = mẍEx
(5.215)
Equating components in Eq. (5.215) yields the following two scalar equations:
mg sin β − μN = mẍ
(5.216)
N + mg cos β = 0
(5.217)
N = −mg cos β
(5.218)
Eq. (5.217) implies that
Consequently, the magnitude of the normal force, N, is given as
N = mg cos β
(5.219)
Substituting N from Eq. (5.219) into Eq. (5.216) gives
mg sin β − μmg cos β = mẍ
(5.220)
g(sin β − μ cos β) = ẍ
(5.221)
Eq. (5.220) simplifies to
Integrating Eq. (5.221) from t = t1 to t = t2 where t2 is the time when sliding
stops and rolling begins, we have that
t2
t1
g(sin β − μ cos β)dt =
t2
t1
ẍdt = ẋ(t2 ) − ẋ(t1 ) = v̄(t2 ) − v̄(t1 )
(5.222)
Now since the quantity g(sin β − μ cos β) is constant, we have that
t2
t1
g(sin β − μ cos β)dt = g(sin β − μ cos β)(t2 − t1 )
(5.223)
Substituting the result of Eq. (5.223) into Eq. (5.222) gives
g(sin β − μ cos β)(t2 − t1 ) = v̄(t2 ) − v̄(t1 )
(5.224)
Next, we can apply the principle of angular impulse and angular momentum
to the cylinder during the segment with friction indirectly by applying Euler’s
211
2nd law. Using the center of mass of the cylinder as the reference point, we have
Euler’s 2nd law as
F
d F H̄
(5.225)
M̄ =
dt
Now since N and mg pass through the center of mass, the only moment acting
on the cylinder during the friction segment is that due to friction and is given
as
(5.226)
M̄ = (rP − r̄) × Ff
Recalling that rP − r̄ = r Ey and using the expression for Ff from Eq. (5.210), we
have that
M̄ = r Ey × (−μNEx ) = r Ey × (−μmg cos βEx ) = r μmg cos βEz
(5.227)
Furthermore, the angular momentum of the cylinder relative to the center of
mass in reference frame F is given as
F
R
H̄ = Ī
· F ωR
(5.228)
Since {Ex , Ey , Ez } is a principle-axis basis, we can write the moment of inertia
tensor of the cylinder as
R
Ī
= Īxx Ex × Ex + Īyy Ey × Ey + Īzz Ez × Ez
Then, using the expression for
F ωR
F
(5.229)
from Eq. (5.174), we have that
H̄ = Īzz ωEz
(5.230)
Now we have for a uniform circular cylinder that
mr 2
2
(5.231)
mr 2
ωEz
2
(5.232)
Īzz =
We then have that
F
H̄ =
Then, integrating Eq. (5.225) from t1 to t2 , we have that
t2
t1
M̄dt =
d F F
F
H̄ dt = H̄(t2 ) − H̄(t1 )
dt
t2 F
t1
(5.233)
Now, noting that r μmg cos β and the vector Ez are constant, we have from
Eq. (5.227) that
t2
t1
M̄dt =
t2
t1
r μmg cos βEz dt = r μmg cos β(t2 − t1 )Ez
(5.234)
212
Chapter 5. Kinetics of Rigid Bodies
Substituting the result of Eq. (5.234) and the result of Eq. (5.232) into Eq. (5.233),
we obtain
mr 2
mr 2
ω(t2 )Ez −
ω(t1 )Ez
2
2
r μmg cos β(t2 − t1 )Ez =
(5.235)
Now we recall the cylinder is not rotating at the beginning of the friction segment. Consequently, ω(t1 ) = 0. Using this last fact, dropping the dependence
on Ez and solving Eq. (5.235) for ω(t2 ), we obtain
2gμ(t2 − t1 ) cos β
r
ω(t2 ) =
(5.236)
Lastly, we know that, at time t2 , when sliding stops and rolling begins, we
have the following kinematic constraint:
F
vP (t2 ) = 0
(5.237)
Using the expression for F vP from Eq. (5.175), we have that
v̄(t2 ) − r ω(t2 ) = 0
(5.238)
Solving Eq. (5.238) for ω(t2 ), we obtain
ω(t2 ) =
v̄(t2 )
r
(5.239)
Using Eq. (5.224), Eq. (5.236), and Eq. (5.239), we can solve for the velocity
of the center of mass of the cylinder and the angular velocity of the cylinder at
time t2 when sliding stops and rolling begins. First, substituting the result of
Eq. (5.239) into Eq. (5.236), we have that
v̄(t2 )
2gμ(t2 − t1 ) cos β
=
r
r
(5.240)
Solving Eq. (5.240) for v̄(t2 ), we obtain
v̄(t2 ) = 2gμ(t2 − t1 ) cos β
(5.241)
Then, substituting the result of Eq. (5.241) into Eq. (5.224), we have that
g(sin β − μ cos β)(t2 − t1 ) = 2gμ(t2 − t1 ) cos β − v̄(t1 )
(5.242)
Solving Eq. (5.242) for t2 − t1 gives
t2 − t 1 =
v̄(t1 )
g(3μ cos β − sin β)
(5.243)
Substituting the result of Eq. (5.243) into Eq. (5.241), we have that
v̄(t2 ) =
2μ cos βv̄(t1 )
(3μ cos β − sin β)
(5.244)
213
Now we have from the end of the frictionless segment that v̄(t1 ) =
Consequently, we have from Eq. (5.244) that
2μ cos β 2gx sin β
v̄(t2 ) =
(3μ cos β − sin β)
2gx sin β.
(5.245)
The velocity of the center of mass of the cylinder at the instant when sliding
stops and rolling begins is then given as
2μ cos β 2gx sin β
F
Ex
v̄(t2 ) =
(5.246)
(3μ cos β − sin β)
Finally, substituting the result of Eq. (5.245) into Eq. (5.239), we have that
2μ cos β 2gx sin β
(5.247)
ω(t2 ) =
r (3μ cos β − sin β)
The angular velocity of the cylinder at the instant when sliding stops and rolling
begins is then given as
2μ cos β 2gx sin β
F R
Ez
ω (t2 ) =
(5.248)
r (3μ cos β − sin β)
214
Chapter 5. Kinetics of Rigid Bodies
Question 5–6
One end of a uniform slender rod of mass m and length l slides along a frictionless vertical surface while the other end of the rod slides along a frictionless horizontal surface as shown in Fig. P5-6. The angle θ formed by the rod
is measured from the vertical. Knowing that gravity acts vertically downward,
determine (a) the differential equation of motion for the rod while it maintains
contact with both surfaces and (b) the value of the angle θ at which the rod loses
contact with the vertical surface. In obtaining your answers, you may assume
that the initial conditions are θ(t = 0) = 0 and θ̇(t = 0) = 0.
g
A
θ
C
m, l
B
Figure P 5-6
Solution to Question 5–6
Kinematics:
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Ex
Ez
Ey
Origin at C
=
=
=
To The Right
Out of Page
Ez × Ex
Now we need the acceleration of the center of mass of the rod and the angular
acceleration of the rod. The position of the center of mass is given as
rC =
l
l
sin θ Ex + cos θ Ey
2
2
(5.249)
Differentiating rC with respect to time, we obtain
F
vC =
F
l
l
d
(rC ) = θ̇ cos θ Ex − θ̇ sin θ Ey
dt
2
2
(5.250)
215
Then, differentiating F vC with respect to time in reference frame F , we obtain
the acceleration of the center of mass of the rod in reference frame F as
F
aC =
l
d F l
aC =
θ̈ cos θ − θ̇ 2 sin θ Ex −
θ̈ sin θ + θ̇ 2 cos θ Ey (5.251)
dt
2
2
F
Also, the angular velocity of the rod in reference frame F is given as
F
ωR = θ̇Ez
(5.252)
which implies that the angular acceleration of the rod is
F
αR = θ̈Ez
(5.253)
Kinetics
For this problem it is most convenient to apply Euler’s laws using the center
of mass of the rod as the reference point. The free body diagram of the rod is
given in Fig. 5-11 where
Nx
A
C
mg
B
Ny
Figure 5-11
Free Body Diagram of Rod for Question 5.7.
Nx = Reaction Force of Vertical Wall on Rod
Ny = Reaction Force of Floor on Rod
mg = Force of Gravity
From the geometry we have that
Nx =
Nx Ex
Ny =
Ny Ey
mg = −mgEy
(5.254)
Therefore, the resultant force on the rod is given as
F = Nx + Ny + mg = Nx Ex + Ny Ey − mgEy = Nx Ex + (Ny − mg)Ey
(5.255)
216
Chapter 5. Kinetics of Rigid Bodies
Then, applying Euler’s 1st law by setting F equal to mF aC where F aC is obtained
from Eq. (5.251), we have that
ml ml θ̈ cos θ − θ̇ 2 sin θ Ex −
θ̈ sin θ + θ̇ 2 cos θ Ey
2
2
(5.256)
Equating components in Eq. (5.256), we obtain the following two scalar equations:
Nx Ex + (Ny − mg)Ey =
Nx
Ny − mg
ml θ̈ cos θ − θ̇ 2 sin θ
2
ml = −
θ̈ sin θ + θ̇ 2 cos θ
2
=
(5.257)
(5.258)
Next, we apply Euler’s 2nd law relative to the center of mass. Since the force
of gravity passes through the center of mass, the resultant moment about the
center of mass is due to only Nx and Ny and is given as
MC = (rx − rC ) × Nx + (ry − rC ) × Ny
We note that
rx
ry
= l cos θ Ey
= l sin θ Ex
(5.259)
(5.260)
Consequently,
rx − rC
ry − rC
l
l
l
l
sin θ Ex + cos θ Ey = − sin θ Ex + cos θ Ey
= l cos θ Ey −
2
2
2
2
l
l
l
l
sin θ Ex + cos θ Ey = sin θ Ex − cos θ Ey
=
l sin θ Ex −
2
2
2
2
(5.261)
The moment MC is then given as
l
l
l
l
sin θ Ex − cos θ Ey × Ny Ey
MC = − sin θ Ex + cos θ Ey × Nx Ex +
2
2
2
2
(5.262)
which simplifies to
l
l
l
MC = − Nx cos θ Ez + Ny sin θ Ez =
Ny sin θ − Nx cos θ Ez
2
2
2
(5.263)
Next, we have that
F
HC = IC · F ωR
(5.264)
Now since {Ex , Ey , Ez } is a principle-axis basis, we have that
C
C
C
Ex ⊗ Ex + Iyy
Ey ⊗ Ey + Izz
Ez ⊗ Ez
IC = Ixx
(5.265)
217
Then, we obtain F HC as
F
C F R
C
θ̇Ez
HC = Izz
ω = Izz
(5.266)
Now, we know that
C
Izz
=
F
ml2
12
(5.267)
C
HC = Izz
θ̇Ez
(5.268)
Then, differentiating F HC in reference frame F , we obtain
d F C
θ̈Ez
HC = Izz
dt
F Then, setting MC equal to d F HC /dt, we obtain
F
ml2
l
θ̈
Ny sin θ − Nx cos θ =
2
12
(5.269)
(5.270)
Simplifying this last result gives
Ny sin θ − Nx cos θ =
ml
θ̈
6
(5.271)
(a) Differential Equation of Motion While Rod Maintains Contact With Wall and
Floor
Now we can determine the differential equation using the results from Eq. (5.257),
Eq. (5.258), and Eq. (5.271). First we solve Eq. (5.258) for Ny which gives
Ny = mg −
ml θ̈ sin θ + θ̇ 2 cos θ
2
(5.272)
Next we substitute Nx from Eq. (5.257) and Ny from the last expression into
Eq. (5.271). This gives
,
ml ml
ml θ̈
sin θ −
mg −
θ̈ sin θ + θ̇ 2 cos θ
θ̈ cos θ − θ̇ 2 sin θ cos θ =
2
2
6
(5.273)
Multiplying out this last expression gives
ml
ml 2
ml
ml 2
ml
θ̈ sin2 θ−
θ̇ cos θ sin θ −
θ̈ cos2 θ+
θ̇ sin θ cos θ =
θ̈
2
2
2
2
6
(5.274)
2
2
Noting that sin θ + cos θ = 1, we obtain
mg sin θ −
mg sin θ −
ml
ml
θ̈ =
θ̈
2
6
(5.275)
218
Chapter 5. Kinetics of Rigid Bodies
from which we obtain
2ml
θ̈ = mg sin θ
3
(5.276)
The differential equation of motion while the rod maintains contact with the
wall and the floor is given as
3g
θ̈ =
sin θ
(5.277)
2l
(b) Value of θ When Rod Loses Contact With Vertical Wall
The rod will lose contact with the vertical wall when Nx = 0, i.e. when Nx = 0.
Consequently, we need to determine the value of θ such that
Nx =
ml θ̈ cos θ − θ̇ 2 sin θ = 0
2
(5.278)
It is seen from this last expression that we need to find expressions for θ̈ and
θ̇ 2 in terms of θ. We have an expression for θ̈ in terms of θ from the result of
part (a), i.e.
3g
sin θ
(5.279)
θ̈ =
2l
Now we note that
θ̈ =
dθ̇
dθ̇ dθ
dθ̇
=
= θ̇
dt
dθ dt
dθ
(5.280)
Therefore,
θ̇
3g
dθ̇
=
sin θ
dθ
2l
(5.281)
Separating variables in this last expression, we obtain
θ̇dθ̇ =
3g
sin θ dθ
2l
(5.282)
Integrating both sides of this last equation gives
θ̇
θ̇0
which gives
#
θ̇ 2
2
$θ̇
θ̇0
θ
3g
sin θ dθ
2l
(5.283)
θ
3g
cos θ
= −
2l
θ0
(5.284)
θ̇dθ̇ =
θ0
Noting that θ(t = 0) = 0 and θ̇(t = 0) = 0, we obtain
3g
θ̇ 2
=
(1 − cos θ )
2
2l
(5.285)
219
Consequently,
θ̇ 2 =
3g
(1 − cos θ )
l
(5.286)
Substituting this last result for θ̇ 2 and the original differential equation from
Eq. (5.277) into Eq. (5.278), we obtain
3g
ml 3g
sin θ cos θ −
(1 − cos θ ) sin θ = 0
(5.287)
2
2l
l
Simplifying this last expression, we obtain
1
sin θ cos θ − (1 − cos θ ) sin θ = 0
2
(5.288)
sin θ [cos θ − 2(1 − cos θ )] = 0
(5.289)
sin θ [3 cos θ − 2] = 0
(5.290)
sin θ = 0 or 3 cos θ − 2 = 0
(5.291)
θ = 0 or θ = cos−1 (2/3)
(5.292)
This gives
which simplifies to
Therefore, we have that
This implies that
Since θ = 0 occurs before the motion starts, we reject this solution. Then the
angle θ at which the rod loses contact with the vertical wall is
θ = cos−1 (2/3)
(5.293)
220
Chapter 5. Kinetics of Rigid Bodies
Question 5–7
A homogeneous semi-circular cylinder of mass m and radius r rolls without
slip along a horizontal surface as shown in Fig. P5-7. The center of mass of
the cylinder is located at point C while point O is located at the center of the
main diameter of the cylinder. Knowing that the angle θ is measured from
the vertical and that gravity acts downward, determine the differential equation
of motion for the cylinder. In obtaining your answers, you may assume that
4r /(3π ) ≈ 0.42r .
4r
3π
g
O
C
r
θ
P
Figure P 5-7
Solution to Question 5–7
Kinematics
First, let F be a fixed reference frame. Then choose the following coordinate
system fixed in F :
Ex
Ez
Ey
Origin at O When θ = 0
=
To The Right
=
Into Page
=
Ez × Ex
Next, let R be a reference frame fixed to the cylinder. Then choose the following
coordinate system fixed in R:
er
ez
eθ
Origin at O Moving with Cylinder
=
Along OC
=
Into Page(= Ez )
=
Ez × er
The relationship between the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in
Fig. 5-12. Using Fig. 5-12, we have that
er
eθ
= − sin θ Ex + cos θ Ey
= − cos θ Ex − sin θ Ey
(5.294)
221
eθ
θ
e , Ez
⊗z
Ex
θ
er
Ey
Figure 5-12
Geometry of Coordinate Systems for Question 5–7.
Now since θ is the angle measured from the vertically downward direction,
the angular velocity of the cylinder in the fixed reference frame is given as
F
ωR = θ̇Ez
(5.295)
Next, since the cylinder rolls without slip along a fixed surface, we have that
F
vP = 0
(5.296)
Then, we have from kinematics of rigid bodies that
F
vO − F vP = F vO = F ωR × (rO − rP )
(5.297)
Now, from the geometry of the problem we have that
rO − rP = −r Ey
(5.298)
Consequently, we obtain the velocity of point O in reference frame F as
F
vO = θ̇Ez × (−r Ey ) = r θ̇Ex
(5.299)
Furthermore, from kinematics of rigid bodies we have that
F
vC − F vO = F ωR × (rC − rO )
(5.300)
Now, the position of point O is given as
rO = xEx
(5.301)
Furthermore, the position of the center of mass of the cylinder is given as
rC = xEx + 0.42r er
(5.302)
Consequently, we have the position of C relative to O as
rC − rO = 0.42r er
Therefore,
Then,
F
F
(5.303)
vC − F vO = θ̇Ez × 0.42r er = 0.42r θ̇eθ
(5.304)
vC = F vO + 0.42r θ̇eθ = r θ̇Ex + 0.42r θ̇eθ
(5.305)
222
Chapter 5. Kinetics of Rigid Bodies
Kinetics
The free body diagram of the cylinder is shown in Fig. 5-13. The forces acting
mg
R
N
Figure 5-13
Free Body Diagram for Question 5–7.
on the cylinder are
N = Reaction Force of Surface on Disk
R = Rolling Force
mg = Force of Gravity
Now it is important to notice that the forces N and R act point P and F vP = 0.
Consequently, neither N nor R do any work. Since the only other force acting on
the cylinder is the conservative force of gravity, we have that
F
Consequently,
Now we know that
E = constant
(5.306)
d F E =0
dt
(5.307)
F
E = FT + FU
(5.308)
Since point C is the center of mass of the cylinder, the kinetic energy is given as
F
T =
1 F
1
m vC · F vC + F HC · F ωR
2
2
(5.309)
Using the expression for F vC from Eq. (5.305), we have that
1 1 F
m vC · F vC = m r θ̇Ex + 0.42r θ̇eθ · r θ̇Ex + 0.42r θ̇eθ
2
2
(5.310)
Simplifying this expression, we obtain
1 1 F
m vC · F vC = m r 2 θ̇ 2 + (0.42r θ̇)2 + 0.84r 2 θ̇ 2 Ex · eθ
2
2
(5.311)
223
Noting that 0.422 = 0.18 and that
Ex · eθ = Ex · − cos θ Ex − sin θ Ey = − cos θ
(5.312)
we obtain
1 1 F
m vC · F vC = m r 2 θ̇ 2 + 0.18r 2 θ̇ 2 − 0.84r 2 θ̇ 2 cos θ
2
2
(5.313)
Simplifying further, we have that
1 F
1
m vC · F vC = mr 2 θ̇ 2 [1.18 − 0.84 cos θ ]
2
2
(5.314)
Next, the angular momentum of the cylinder relative to the center of mass C is
given as
F
F R
HC = IR
(5.315)
C · ω
Now since the cylinder is symmetric about the er -direction, we have that {er , eθ , ez }
is a principle-axis basis. Consequently, the moment of inertia tensor IR
C can be
expressed as
C
C
C
IR
(5.316)
C = Ir r er ⊗ er + Iθθ eθ ⊗ eθ + Izz ez ⊗ ez
Then, using the expression for F ωR from Eq. (5.295), we obtain F HC as
C
F
C
C
HC = IrCr er ⊗ er + Iθθ
eθ ⊗ eθ + Izz
ez ⊗ ez · θ̇ez = Izz
(5.317)
θ̇ez
Now for a semicircular cylinder we have that
C
Izz
= 0.32mr 2
(5.318)
Consequently, the angular momentum of the cylinder relative to the center of
mass of the cylinder is given as
F
HC = 0.32mr 2 θ̇Ez
(5.319)
Using the expression for F HC from Eq. (5.319), we obtain
1
1F
HC · F ωR =
0.32mr 2 θ̇Ez · θ̇Ez = 0.16mr 2 θ̇ 2
2
2
(5.320)
The kinetic energy of the cylinder in reference frame F is then given as
F
T =
1
mr 2 θ̇ 2 [1.18 − 0.84 cos θ ] + 0.16mr 2 θ̇ 2
2
(5.321)
Eq. (5.321) simplifies to
F
T =
1
mr 2 θ̇ 2 [1.5 − 0.84 cos θ ]
2
(5.322)
224
Chapter 5. Kinetics of Rigid Bodies
Now since the only conservative force acting on the cylinder is that due to gravity, the potential energy of the cylinder in reference frame F is given as
F
U = −mg · rC
(5.323)
Now since the force of gravity acts vertically downward, we have that
mg = mgEy
(5.324)
Then, using the expression for rC from Eq. (5.302), we obtain F U as
F
U = −mgEy · (xEx + 0.42r er ) = −0.42mgr Ey · er
(5.325)
Now, using the expression for er from Eq. (5.294), we have that
Ey · er = Ey · (− sin θ Ex + cos θ Ey ) = cos θ
(5.326)
Consequently, the potential energy in reference frame F is given as
F
U = −0.42mgr cos θ
(5.327)
The total energy of the system in reference frame F is then given as
F
E = FT + FU =
1
mr 2 θ̇ 2 [1.5 − 0.84 cos θ ] − 0.42mgr cos θ
2
(5.328)
Differentiating F E with respect to time, we obtain
d F 1
E = mr 2 θ̇ θ̈ [1.5 − 0.84 cos θ ]+ mr 2 θ̇ 2 0.84θ̇ sin θ +0.42mgr θ̇ sin θ = 0
dt
2
(5.329)
Simplifying this last expression gives
mr 2 θ̇ θ̈ [1.5 − 0.84 cos θ ] + 0.42mr 2 θ̇ 2 θ̇ sin θ + 0.42mgr θ̇ sin θ = 0 (5.330)
We then obtain
θ̇ mr 2 θ̈ [1.5 − 0.84 cos θ ] + 0.42mr 2 θ̇ 2 sin θ + 0.42mgr sin θ = 0
(5.331)
Noting that θ̇ ≠ 0 as a function of time, we obtain the differential equation of
motion as
mr 2 θ̈ [1.5 − 0.84 cos θ ] + 0.42mr 2 θ̇ 2 sin θ + 0.42mgr sin θ = 0
(5.332)
225
Question 5–8
A homogeneous sphere of radius r rolls without slip along a fixed spherical
surface of radius R as shown in Fig. P5-8. The angle θ measures the amount by
which the sphere has rotated from the vertical direction. Knowing that gravity
acts downward and assuming the initial conditions θ(0) = 0 and θ̇(0) = 0, determine the differential equation of motion while the sphere maintains contact
with the spherical surface.
θ
m
C
g
O
r
R
Figure P 5-12
Solution to Question 5–8
Preliminaries
For this problem it is convenient to apply the following balance laws:
• Euler’s 1st law to the center of mass of the rolling sphere
• Euler’s 2nd law about the center of mass of the rolling sphere
In order to use the aforementioned balance laws, we will need the following
kinematic quantities in an inertial reference frame:
• The acceleration of the center of mass of the rolling sphere
• The rate of change of angular momentum of the rolling sphere about the
center of mass of the rolling sphere
226
Chapter 5. Kinetics of Rigid Bodies
Kinematics
Let F be a fixed reference frame. Then, choose the following coordinate system
fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
Along OC When θ = 0
Into Page
Ez × Ex
Next, let A be a reference frame fixed to the direction OC. Then, choose the
following coordinate system fixed in reference frame A:
er
ez
eφ
Origin at O
=
=
=
Along OC
Into Page (= Ez )
ez × er
The geometry of the bases {Ex , Ey , Ez } and {er , eφ , ez } is shown in Fig. 5-14
where φ be the angle between the direction Ex and the direction er . Using the
Ex
er
φ
e z , Ez ⊗
φ
Ey
eθ
Figure 5-14
tion 5–8.
Relationship Between Bases {Ex , Ey , Ez } and {er , eφ , ez } for Ques-
geometry in Fig. 5-15, we have the following relationship between {Ex , Ey , Ez }
and {er , eφ , ez }:
cos φEx + sin φEy
er =
eφ = − sin φEx + cos φEy
(5.333)
Ex =
cos φer − sin φeφ
Ey =
sin φer + cos φeφ
Next, the angular velocity of reference frame A in reference frame F is given as
F
ωA = φ̇ez = φ̇Ez
(5.334)
227
Furthermore, denoting the reference frame of the rolling sphere by R and observing that θ describes the rotation of the rolling sphere relative to the fixed
vertical direction, we have that
F
ωR = θ̇ez = θ̇Ez
(5.335)
The position of the center of mass of the rolling sphere is then given as
r̄ = (R + r )er
(5.336)
Differentiating r̄ in reference frame F using the transport theorem, we have that
F
v̄ =
F
A
d
d
(r̄) =
(r̄) + F ωA × r̄
dt
dt
(5.337)
Now we have that
Ad
dt
=
(r̄)
F ωA
0
× r̄ = θ̇ez × (R + r )er = (R + r )φ̇eφ
(5.338)
Consequently,
F
Differentiating
that
F
v̄ = (R + r )φ̇eφ
(5.339)
v̄ in reference frame F using the transport theorem, we have
F
ā =
d F Ad F F A F
v̄ =
v̄ + ω × v̄
dt
dt
F
(5.340)
Now we have that
Ad
dt
F
F ωA
v̄
=
(R + r )φ̈eφ
× r̄ = θ̇ez × (R + r )φ̇eφ = −(R + r )φ̇2 er
(5.341)
Consequently,
F
ā = −(R + r )φ̇2 er + (R + r )φ̈eφ
(5.342)
Now we see that Eq. (5.342) is an expression for F ā in terms of the derivatives
of φ. However, in this problem we are interested in obtaining the differential
equation of motion in terms of θ. Therefore, we need to eliminate φ in favor of
θ.
Eliminating φ is accomplished as follows. First, we know that the sphere
rolls without slip along the fixed sphere. Denoting Q as the instantaneous point
of contact between the two spheres, we have that
F
F
vR
Q ≡ vQ = 0
(5.343)
228
Chapter 5. Kinetics of Rigid Bodies
Then, applying the relative velocity property for two points on a rigid body, we
have that
F
v̄ − F vQ = F ωR × (r̄ − fQ )
(5.344)
Now we have that rQ = Rer which implies that r̄ − rQ = r er . Then, substituting
from Eq. (5.335) into Eq. (5.344), we obtain
F ωR
F
v̄ − F vQ = θ̇ez × r er = r θ̇eφ
(5.345)
Finally, observing from Eq. (5.343) that F vQ = 0, we obtain F v̄ as
F
v̄ = r θ̇eφ
(5.346)
Then, setting the result of Eq. (5.346) equal to the result of Eq. (5.339), we obtain
r θ̇ = (R + r )φ̇
(5.347)
Solving Eq. (5.347) for φ̇, we obtain
φ̇ =
r
θ̇
R+r
(5.348)
Differentiating φ̇ in Eq. (5.348), we obtain
φ̈ =
r
θ̈
R+r
(5.349)
The acceleration of the center of mass of the rolling sphere is then given as
2
r
r
F
θ̇ er + (R + r )
θ̈ eφ
ā = −(R + r )
(5.350)
R+r
R+r
Simplifying Eq. (5.350), we obtain
F
ā = −
r 2 θ̇ 2
er + r θ̈eφ
R+r
(5.351)
Next, we need to compute the rate of change of the angular momentum of
the sphere relative to the center of mass of the sphere in reference frame F . We
have that
F
R
(5.352)
H̄ = Ī · F ωR
R
where Ī is the moment of inertia tensor relative to the center of mass and F ωR
is the angular velocity of the sphere in reference frame F . Now since {er , eφ , ez }
is a principle-axis basis, the moment of inertia tensor Ī
Ī
R
R
can be written as
= Īr r er ⊗ er + Īφφ eφ ⊗ eφ + Īzz ez ⊗ ez
(5.353)
Then, substituting the expression for F ωR as given in Eq. (5.335) and the expression for the moment of inertia tensor from Eq. (5.353) into Eq. (5.352), we
F
obtain H̄ as
F
H̄ = Īzz θ̇ez
(5.354)
229
Now we have for a sphere that
2
mr 2
5
(5.355)
2
mr 2 θ̇ez
5
(5.356)
Īzz =
Consequently, Eq. (5.354) simplifies to
F
H̄ =
F
Differentiating H̄ in reference frame F , we obtain
d F 2
H̄ = mr 2 θ̈ez
dt
5
F
(5.357)
Kinetics
As stated earlier, this problem will be solved using the following balance laws:
• Euler’s 1st law to the rolling sphere
• Euler’s 2nd law about the center of mass of the rolling sphere
The free body diagram of the rolling sphere is shown in Fig. 5-15. It can be seen
that the following forces act on the sphere:
Fr
= Force of Rolling
N
= Reaction Force of Fixed Sphere on Rolling Sphere
mg = Force of Gravity
Given the geometry of the problem, we have that
N
= Ner
Fr
= Fr e φ
mg = −mgEx
(5.358)
Then, substituting the expression for Ex in terms of er and eφ from Eq. (5.333),
the force of gravity can be written as
mg = −mg cos φer + mg sin φeφ
(5.359)
Then, using the fact that θ is zero when the sphere is at the top of the fixed
sphere, we have that
θ(t = 0) = φ(t = 0) = 0
(5.360)
Consequently, integrating Eq. (5.348), we obtain
φ=
rθ
R+r
(5.361)
Substituting the expression for φ into Eq. (5.359), we obtain
mg = −mg cos
rθ
R+r
+ mg sin
rθ
R+r
eφ
(5.362)
230
Chapter 5. Kinetics of Rigid Bodies
Fr
mg
N
Figure 5-15
Free Body Diagram of Rolling Sphere for Question 5–8.
Application of Euler’s 1st To the Rolling Sphere
Euler’s 1st law states that
F = mF ā
(5.363)
For this problem, the resultant force acting on the sphere is given as
F = N + Fr + mg
(5.364)
Using the expressions for the forces as given in Eq. (5.358) and Eq. (5.362), we
have the resultant force as
r
r
θ + mg sin
θ eφ
R+r
R+r
F = Ner + Fr eφ − mg cos
Then, setting F from Eq. (5.365) equal to mF ā using
obtain
rθ
Ner + Fr eφ − mg cos
R+r
rθ
+ mg sin
R+r
F
(5.365)
ā from Eq. (5.351), we
r 2 θ̇ 2
er + r θ̈eφ
eφ = m −
R+r
(5.366)
Simplifying Eq. (5.366), we obtain
r 2 θ̇ 2
rθ
rθ
er + Fr + mg sin
= −m
er + mr θ̈eφ
N − mg cos
R+r
R+r
R+r
(5.367)
Eq. (5.367) yields the following two scalar equations:
rθ
R+r
rθ
Fr + mg sin
R+r
N − mg cos
= −m
r 2 θ̇ 2
R+r
= mr θ̈
(5.368)
(5.369)
Application of Euler’s 2nd About the Center of Mass of the Rolling Sphere
We recall Euler’s 2nd law relative to the center of mass of a rigid body as
M̄ =
d F H̄
dt
F
(5.370)
231
F F We already have d H̄ from Eq. (5.357). Next, observing that mg and N both
pass through the center of mass of the sphere, we obtain M̄ as
M̄ = (rQ − r̄) × Fr
(5.371)
Then, using the fact that rQ −r̄ = −r er and the expression for Fr from Eq. (5.358),
we obtain M̄ as
(5.372)
M̄ = −r er × Fr eφ = −r Fr ez
F F Then, setting M̄ from Eq. (5.372) equal to d H̄ /dt using the expression for
F F d H̄ /dt from Eq. (5.357), we obtain
2
mr 2 θ̈
5
(5.373)
2
Fr ez = − mr θ̈
5
(5.374)
−r Fr =
Solving Eq. (5.373) for Fr , we obtain
Differential Equation of Motion
The differential equation of motion can now be found using Eq. (5.368) and
Eq. (5.374). In particular, substituting Fr from Eq. (5.374) into Eq. (5.368), we
obtain
2
rθ
= mr θ̈
(5.375)
− mr θ̈ + mg sin
5
R+r
Rearranging Eq. (5.375), we obtain
rθ
7
mr θ̈ − mg sin
5
R+r
=0
(5.376)
Simplifying Eq. (5.376), we obtain
θ̈ −
rθ
5g
sin
7r
R+r
=0
(5.377)
232
Chapter 5. Kinetics of Rigid Bodies
Question 5–10
A uniform circular disk of mass m and radius r rolls without slip along a plane
inclined at a constant angle β with horizontal as shown in Fig. P5-10. Attached
at the center of the disk is a linear spring with spring constant K. Knowing
that the spring is unstretched when the angle θ is zero and that gravity acts
downward, determine the differential equation of motion for the disk in terms
of the angle θ.
g
K
θ
O
r
P
β
Figure P 5-13
Solution to Question 5–10
Kinematics
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in F :
Ex
Ez
Ey
Origin at Point O when t = 0
=
Down Incline
=
Into Page
=
Ez × Ex
Now since the disk rolls without slip along the incline, we have that
F
vP = 0
(5.378)
Then, using the velocity property for two points on a rigid body, the velocity of
the center of mass of the disk is obtained as
F
v̄ = F vP + F ωR × (r̄ − rP )
(5.379)
233
In terms of the basis {Ex , Ey , Ez }, the angular velocity of the disk in reference
frame F is given as
F
ωR
r̄ − rP
= θ̇Ez
(5.380)
= −r Ey
(5.381)
where R denotes the reference frame of the disk. Consequently, the velocity of
the center of mass of the disk in reference frame F is given as
F
v̄ = θ̇Ez × (−r Ey ) = r θ̇Ex
(5.382)
Computing the rate of change of F v̄ in reference frame F , we obtain the acceleration of the center of mass of the disk as
F
ā =
d F v̄ r θ̈Ex
dt
F
(5.383)
Next, the angular momentum relative to the instantaneous point of contact
is given as
F
F
(5.384)
HP = H̄ + (rQ − r̄) × m(F vQ − F v̄)
Now we have that
F
R
H̄ = Ī
· F ωR
(5.385)
Now since {Ex , Ey , Ez } is a principle-axis basis, we have that
R
= Īxx Ex ⊗ Ex + Īyy Ey ⊗ Ey + Īzz Ez ⊗ Ez
Ī
Then, substituting Ī
R
(5.386)
from Eq. (5.386) into Eq. (5.385), we obtain
F
H̄ = Īzz θ̇Ez
(5.387)
Now we have for a uniform circular disk that Īzz = mr 2 /2. Consequently,
F
H̄ =
mr 2
θ̇Ez
2
(5.388)
Next, since F vQ = 0, we have that
(rQ − rO ) × m(F vQ − F vO ) = mr Ey × (−r θ̇Ex ) = mr 2 θ̇Ez
(5.389)
Consequently,
mr 2
3
θ̇Ez + mr 2 θ̇Ez = mr 2 θ̇Ez
(5.390)
2
2
The rate of change of angular momentum relative to the point of contact is then
given as
F
3
d F
HQ = mr 2 θ̈Ez
(5.391)
dt
2
F
HQ =
234
Chapter 5. Kinetics of Rigid Bodies
Fs
mg
R
N
Figure 5-16
Free Body Diagram of Cylinder for Question 5–10.
Kinetics
The free body diagram of the cylinder is shown in Fig. 5-16.
Using Fig. 5-16, the forces acting on the cylinder are given as
N
R
Fs
mg
=
=
=
=
Force of Incline on Cylinder
Force of Rolling
Spring Force
Force of Gravity
From the geometry we have that
N = NEy
(5.392)
R = REx
(5.393)
Fs
= −K( −
0 )us
(5.394)
mg = mguv
(5.395)
(5.396)
where uv is the unit vector in the vertically downward direction and us is the
direction of the spring force. Now uv is shown in Fig. 5-17.
⊗
β
Ey
Figure 5-17
Ex
uv
Unit Vector in Vertically Downward Direction for Question 5–10.
235
Using Fig. 5-17, we have that
uv = sin βEx + cos βEy
(5.397)
Therefore, the force of gravity is obtained as
mg = mg sin βEx + mg cos βEy
(5.398)
Next, we have for the spring force that
= r̄ − rA (5.399)
where A is the attachment point of the spring. Now suppose we say that the
spring is attached a distance L from where O is located when θ = 0. Then, we
have that
(5.400)
rA = −LEx
Then we have that
r̄ − rA = xEx − (−LEx ) = (x + L)Ex
(5.401)
Furthermore, the unstretched length of the spring is given as
0
=L
(5.402)
Consequently, we obtain
r̄ − rA − L = x
(5.403)
Now since the disk rolls without slip, we have that
x = rθ
(5.404)
r̄ − rA − L = r θ
(5.405)
which implies that
Finally, the direction us is given as
us =
r̄ − rA
= Ex
r̄ − rA (5.406)
The spring force is then given as
Fs = −Kr θEx
(5.407)
Using the forces above, we can now apply the general form of Euler’s 2nd
law to the disk, i.e.,
MQ − (r̄ − rQ ) × mF aQ =
d F
HQ
dt
F
(5.408)
236
Chapter 5. Kinetics of Rigid Bodies
First, since N and R pass through point Q, the moment relative to point Q is
given as
(5.409)
MQ = (r̄ − rQ ) × (mg + Fs )
where we note that both the force of gravity and the spring force act at the
center of mass. Then, using the expressions for mg and Fs from above, we
obtain
MQ = (−r Ey ) × (mg sin βEx + mg cos βEy − Kr θEx )
= mgr sin βEz − Kr 2 θEz
(5.410)
= (mgr sin β − Kr 2 θ)Ez
Next, we have that
F
aQ = F ā + F αR × (rQ − rO ) + F ωR × ω × (rQ − rO )
(5.411)
Using F ā from Eq. (5.383) and the fact that rQ − r̄ = r Ey , we obtain F aQ as
F
aQ = r θ̈Ex + θ̈Ez × r Ey + θ̇Ez × θ̇Ez × r Ey
= r θ̈Ex − r θ̈Ex − r θ̇ 2 Ey
(5.412)
= −r θ̇ 2 Ey
Consequently, the inertial moment −(r̄ − rQ ) × mF aQ is given as
−r Ey × m(−r θ̇ 2 Ey ) = 0
(5.413)
Then, since the inertial moment is zero, we can set the moment relative to Q
equal to the rate of change of angular momentum relative to Q to obtain
mgr sin β − Kr 2 θ =
3
mr 2 θ̈
2
(5.414)
Rearranging, we obtain the differential equation of motion as
3
mr 2 θ̈ + Kr 2 θ − mgr sin β = 0
2
(5.415)
237
Solution to Question 5–11
A slender rod of mass m and length l is suspended from a massless collar at
point O as shown in Fig. P5-11. The collar in turn slides without friction along
a horizontal track. The position of the collar is denoted as x while the angle
formed by the rod with the vertical is denoted θ. Given that a known horizontal
force P is applied to the rod at the point O and that gravity acts downward,
determine a system of two differential equations describing the motion of the
rod in terms of the variables x and θ.
x(t)
P
O
m
θ
g
l
Figure P 5-16
Solution to Question 5–11
Kinematics
First, let F be a reference frame fixed to the track. Then, choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O When x = 0
=
To The Right
=
Out of Page
=
Ez × Ex
Next, let R be a reference frame fixed to the rod. Then, choose the following
coordinate system fixed in reference frame R:
er
ez
eθ
Origin at O
=
=
=
Along Rod
Out of Page
ez × er
238
Chapter 5. Kinetics of Rigid Bodies
Ey
eθ
θ
ez , Ez
Ex
θ
er
Figure 5-18
Geometry of Bases {Ex , Ey , Ez } and {er , eθ , ez } for Question 5–11.
The geometry of the bases {Ex , Ey , Ez } and {er , eθ , ez } is given in Fig. 5-18.
Using Fig. 5-18, we have that
er
= sin θ Ex − cos θ Ey
(5.416)
eθ
= cos θ Ex + sin θ Ey
(5.417)
Ex
= sin θ er + cos θ eθ
(5.418)
Ey
= − cos θ er + sin θ eθ
(5.419)
Furthermore, we have that
The position of the center of mass of the rod is then given as
r̄ = rO + rC/O
(5.420)
where C is the center of mass of the rod. Now we have that
rO
rC/O
= xEx
l
er
=
2
(5.421)
(5.422)
Therefore,
l
(5.423)
er
2
Furthermore, the angular velocity of the rod in reference frame F is given as
r̄ = xEx +
F
ωR = θ̇ez
(5.424)
239
Then, the velocity of the center of mass of the rod in reference frame F is given
as
F
dr̄ FdrO FdrC/O
F
v̄ =
=
+
= F vO + F vC/O
(5.425)
dt
dt
dt
Now we have that
F
vO =
F
drO
= ẋEx
dt
(5.426)
Furthermore, using the rate of change transport theorem, we have that
F
R
drC/O
drC/O F R
=
+ ω × rC/O
dt
dt
(5.427)
Now we have that
R
drC/O
dt
F
= 0
ωR × rC/O
(5.428)
= θ̇ez ×
l
lθ̇
er =
eθ
2
2
(5.429)
Therefore, we obtain F vC/O as
F
vC/O =
lθ̇
eθ
2
(5.430)
which implies that
F
v̄ = ẋEx +
lθ̇
eθ
2
(5.431)
The acceleration of the center of mass is then obtained as
F
ā =
d F Fd F Fd F
v̄ =
vO +
vC/O = F aO + F aC/O
dt
dt
dt
F
First, we obtain F aO as
F
aO = ẍEx
(5.432)
(5.433)
Furthermore, using the rate of change transport theorem, we have F aC/O as
F
aC/O =
Rd d F
F
vC/O =
vC/O + F ωR × F vC/O
dt
dt
F
(5.434)
Now we know that
d F
vC/O
=
dt
R
F
ωR × F vC/O
lθ̈
eθ
2
= θ̇ez ×
(5.435)
lθ̇
lθ̇ 2
eθ = −
er
2
2
(5.436)
240
Chapter 5. Kinetics of Rigid Bodies
Therefore,
F
aC/O = −
lθ̇ 2
lθ̈
er +
eθ
2
2
(5.437)
which implies that
lθ̇ 2
lθ̈
er +
eθ
(5.438)
2
2
Next, the angular momentum of the rod relative to the center of mass is
given as
F
R
(5.439)
H̄ = Ī · F ωR
F
ā = ẍEx −
Now since {er , eθ , ez } is a principle-axis basis, we have that
R
Ī
= Īr r er ⊗ er + Īθθ eθ ⊗ eθ + Īzz ez ⊗ ez
(5.440)
F
Then, using the expresion for F ωR from Eq. (5.424), we obtain H̄ as
F
H̄ = Īr r er ⊗ er + Īθθ eθ ⊗ eθ + Īzz ez ⊗ ez · θ̇ez = Īzz θ̇ez
(5.441)
F
Computing the rate of change of H̄ in reference frame F , we obtain
d F H̄ == Īzz θ̈ez
dt
F
(5.442)
Now for a slender uniform rod we have that Īzz = ml2 /12. Consequently, we
have that
F
d F ml2
θ̈ez
(5.443)
H̄ ==
dt
12
Kinetics
This problem will be solved by applying the following two balance laws: (1)
Euler’s 1st law and (2) Euler’s 2nd law relative to the center of mass of the rod.
The free body diagram of the rod is shown in Fig. 5-19.
Using Fig. 5-19, we see that the following three forces act on the rod
N
= Reaction Force of Track
P
= Known Horizontal Force
mg = Force of Gravity
Now we have that
N = NEy
(5.444)
P = P Ex
(5.445)
mg = −mgEy
(5.446)
The resultant force acting on the particle is then given as
F = N + P + mg = NEy + P Ex − mgEy = P Ex + (N − mg)Ey
(5.447)
241
N
P
mg
Figure 5-19
Free Body Diagram of Rod for Question 5–11.
Application of Euler’s 1st Law to Rod
Then, applying Euler’s 1st law, we have that
F = mF ā
(5.448)
Using the expression for F ā from Eq. (5.438) and the resultant force from Eq. (5.447),
we obtain
mlθ̇ 2
mlθ̈
er +
eθ
P Ex + (N − mg)Ey = mẍEx −
(5.449)
2
2
Then, computing the projection of Eq. (5.449) in the Ex -direction, we have that
P = mẍ −
mlθ̇ 2
mlθ̈
er · Ex +
e θ · Ex
2
2
(5.450)
Now from Eq. (5.416) and Eq. (5.417) we have that
er · Ex
= sin θ
(5.451)
e θ · Ex
= cos θ
(5.452)
Therefore,
P = mẍ −
mlθ̈
mlθ̇ 2
sin θ +
cos θ
2
2
(5.453)
Next, computing the projection of Eq. (5.449) in the Ey -direction, we have that
N − mg = −
mlθ̈
mlθ̇ 2
er · Ey +
eθ · Ey
2
2
(5.454)
Now from Eq. (5.416) and Eq. (5.417) we have that
er · Ey
= − cos θ
(5.455)
eθ · Ey
= sin θ
(5.456)
242
Chapter 5. Kinetics of Rigid Bodies
Therefore,
N − mg =
mlθ̈
mlθ̇ 2
cos θ +
sin θ
2
2
(5.457)
The two equations that result from the application of Euler’s 1st law to the rod
are then given from Eq. (5.453) and Eq. (5.457) as
mlθ̈
mlθ̇ 2
sin θ +
cos θ
2
2
mlθ̇ 2
mlθ̈
N − mg =
cos θ +
sin θ
2
2
P = mẍ −
(5.458)
(5.459)
Application of Euler’s 2nd Law to Rod
Applying Euler’s 2nd law to the center of mass of the rod, we have that
M̄ =
d F H̄
dt
F
(5.460)
Now we already have the rate of change of angular momentum relative to the
center of mass of the rod from Eq. (5.442). Next, since gravity passes through
the center of mass of the rod and the forces N and P both act at point O, the
moment relative to the center of mass is given as
M̄ = (rO − r̄) × (N + P)
(5.461)
Now we have that
l
(5.462)
rO − r̄ = − er
2
Furthermore, using the expressions for N and P from Eq. (5.444) and Eq. (5.445),
respectively, we obtain
l
l
M̄ = − er × NEy + P Ex = −
(5.463)
Ner × Ey + P er × Ex
2
2
Now we have that
e r × Ey
= sin θ Ez = sin θ ez
(5.464)
er × Ex
= cos θ Ez = cos θ ez
(5.465)
Consequently,
l
l
M̄ = − er × NEy + P Ex = − (N sin θ + P cos θ ) ez
2
2
F
(5.466)
F
Then, setting M̄ in Eq. (5.466) equal to d( H̄)/dt using the expression for
F
F
d( H̄)/dt from Eq. (5.443), we obtain
−
l
ml2
θ̈ez
(N sin θ + P cos θ ) ez =
2
12
(5.467)
243
Eq. (5.467) simplifies to
N sin θ + P cos θ = −
ml
θ̈
6
(5.468)
Determination of System of Two Differential Equations of Motion
The first differential equation of motion is obtained directly from Eq. (5.458),
i.e.,
mlθ̇ 2
mlθ̈
cos θ −
sin θ = P
(5.469)
mẍ +
2
2
The second differential equation is obtained using Eq. (5.459) and Eq. (5.468).
First, multiplying Eq. (5.459) by sin θ , we have that
N sin θ − mg sin θ =
mlθ̈
mlθ̇ 2
cos θ sin θ +
sin2 θ
2
2
(5.470)
Then, subtracting Eq. (5.468) from Eq. (5.470), we obtain
P cos θ + mg sin θ = −
mlθ̇ 2
mlθ̈
ml
θ̈ −
cos θ sin θ −
sin2 θ
6
2
2
(5.471)
Simplifying Eq. (5.470), we obtain the second differential equation of motion as
mlθ̇ 2
ml cos θ sin θ + mg sin θ = −P cos θ
1 + 3 sin2 θ θ̈ +
6
2
(5.472)
A system of two differential describing the motion of the rod is then given as
mlθ̈
mlθ̇ 2
cos θ −
sin θ
2
2
mlθ̇ 2
ml cos θ sin θ + mg sin θ
1 + 3 sin2 θ θ̈ +
6
2
mẍ +
= P
(5.473)
= −P cos θ (5.474)
Alternate System of Two Differential Equations
Now while Eqs. (5.473) and (5.474) are perfectly valid, a more elegant system
of differential equations is obtained by manipulating Eqs. (5.473) and (5.474).
First, multiplying Eq. (5.473) by cos θ , we obtain
mẍ cos θ +
mlθ̇ 2
mlθ̈
cos2 θ −
sin θ cos θ = P cos θ
2
2
(5.475)
Then, adding Eqs. (5.475) and (5.474) gives
mẍ cos θ +
ml mlθ̈
mlθ̇ 2
cos2 θ −
sin θ cos θ +
1 + 3 sin2 θ θ̈
2
2
6
mlθ̇ 2
cos θ sin θ + mg sin θ = 0
+
2
(5.476)
244
Chapter 5. Kinetics of Rigid Bodies
Now it is seen that the second and fourth terms in Eq. (5.476) cancel. Consequently, Eq. (5.476) simplifies to
mẍ cos θ +
mlθ̈
ml cos2 θ +
1 + 3 sin2 θ θ̈ + mg sin θ = 0
2
6
(5.477)
Eq. (5.477) can be further simplified to give
mẍ cos θ +
2ml
θ̈ + mg sin θ = 0
3
(5.478)
An alternate system of differential equations is then obtained using Eq. (5.473)
and (5.478) as
mlθ̇ 2
mlθ̈
cos θ −
sin θ
2
2
2mlθ̈
+ mg sin θ
mẍ cos θ +
3
mẍ +
= P
(5.479)
= 0
(5.480)
Derivation of 2nd Differential Equation Using Point O As Reference Point
It is ntoed that one of the differential equations can be obtained by using point
O as the reference point. In particular, we know from Eq. (5.433 that the acceleration of point O in reference frame F is F aO = ẍEx ≠ 0. Consequently, point
O is not inertially fixed and it is necessary to apply the general form of Euler’s
2nd law relative to point O, i.e.,
MO − (r̄ − rO ) × mF aO =
d F
HO
dt
F
(5.481)
Now examining the free body diagram of Fig. (5-19), we see that the forces P and
N pass through point O. Consequently, the moment relative to point O is due
entirely to the force of gravity and is given as
MO = (r̄ − rO ) × mg
(5.482)
Using the expressions for rO − r̄ from Eq. (5.462), we have that
r̄ − rO =
l
er
2
(5.483)
Next the force of gravity is given as
mg = −mgEy = −mg(− cos θ er +sin θ eθ ) = mg cos θ er −mg sin θ eθ (5.484)
where we have used the expression for Ey from Eq. (5.419). The moment relative
to point O is then given as
MO =
l
mgl
er × (mg cos θ er − mg sin θ eθ ) = −
sin θ ez
2
2
(5.485)
245
Next, the inertial moment due to the acceleration of point O is then given as
−(r̄ − rO ) × mF aO =
l
er × mẍEx
2
(5.486)
Using the expression for Ex from Eq. (5.418), we obtain
l
mlẍ
cos θ ez (5.487)
−(r̄ − rO ) × mF aO = − er × mẍ(sin θ er + cos θ eθ ) = −
2
2
Then, the angular momentum relative to point O is obtained as
ml2
l
lθ̇
F
F
F
F
θ̇ez + − er m
eθ
HO = H̄ + (rO − r̄) × m( vO − v̄) =
(5.488)
12
2
2
Simplifying Eq. (5.488) gives
F
HO =
ml2
ml2
ml2
θ̇ez +
θ̇ez =
θ̇ez
12
4
3
(5.489)
The rate of change of F HO in reference frame F is then given as
ml2
d F
θ̈ez
HO =
dt
3
F
(5.490)
Substituting the results of Eqs. (5.485), (5.487), and (5.490) into (5.481), we obtain
mgl
mlẍ
ml2
sin θ ez −
cos θ ez =
θ̈ez
−
(5.491)
2
2
3
Dropping the dependence in Eq. (5.491) and simplifying, we obtain
mẍ cos θ +
2ml
θ̈ + mg sin θ = 0
3
It is seen that Eq. (5.492) is identical to Eq. (5.478).
(5.492)
246
Chapter 5. Kinetics of Rigid Bodies
Question 5–12
A uniform slender rod of mass m and length 2l slides without friction along a
fixed circular track of radius R as shown in Fig. P5-12. Knowing that θ is the
angle from the vertical to the center of the rod and that gravity acts downward,
determine the differential equation of motion for the rod.
g
O
R
a
m
θ
B
C
A
2l
Figure P 5-18
Solution to Question 5–12
Kinematics
First, let F be a reference frame fixed to the track. Then, choose the following
coordinate system fixed in reference frame F :
Ex
Ez
Ey
Origin at O
=
=
=
along OC When θ = 0
Out of Page
Ez × Ex
Next, let R be a reference frame fixed to the rod. Then, choose the following
coordinate system fixed in reference frame R:
er
ez
eθ
Origin at O
=
=
=
Along OC
Out of Page
ez × er
247
The geometry of the coordinate systems {Ex , Ey , Ez } and {er , eθ , ez } is shown
in Fig. 5-20.
eθ
θ
e z , Ez
Ey
θ
er
Ex
Figure 5-20
Unit Vertical Direction for Question 5–12.
Using Fig. 5-20, we have that
Ex
= cos θ er − sin θ eθ
(5.493)
Ey
= sin θ er + cos θ eθ
(5.494)
Next, the position of the center of mass of the rod is given as
r̄ = rC = aer
(5.495)
Furthermore, the angular velocity of reference frame R in reference frame F is
given as
F R
ω = θ̇Ez
(5.496)
Then, applying the rate of change transport theorem between reference frame
R and reference frame F , we obtain the velocity of the center of mass of the
rod as
F
dr̄ Rdr̄ F R
F
=
+ ω × r̄
v̄ =
(5.497)
dt
dt
Now we have that
R
dr̄
= 0
dt
F R
ω × r̄ = θ̇ez × (aer ) = aθ̇eθ
(5.498)
(5.499)
Adding Eq. (5.498) and Eq. (5.499), we obtain the velocity of the center of mass
of the rod in reference frame F as Differentiating vC in Eq. (5.500), we obtain
F
v̄ = aθ̇eθ
(5.500)
248
Chapter 5. Kinetics of Rigid Bodies
Then, applying the rate of change transport theorem to F v̄, we obtain the acceleration of the center of mass of the rod in reference frame F as
d F R d F F R F
ā =
v̄ + ω × v̄
dt
dt
(5.501)
d F v̄
= aθ̈eθ
dt
F R
ω × F v̄ = θ̇ez × (aθ̇eθ ) = −aθ̇ 2 er
(5.502)
F
ā =
F
Now we have that
R
(5.503)
Adding Eq. (5.502) and Eq. (5.503), we obtain the acceleration of the center of
mass of the rod in reference frame F as
F
ā = −aθ̇ 2 er + aθ̈eθ
(5.504)
Next, from the kinematic properties of a rigid body, we have that
F
vA
F
vB
=
F
=
F
v̄ + F ωR (rA − rC )
F
(5.505)
R
(5.506)
rA − rC
= −leθ
(5.507)
rB − rC
= leθ
(5.508)
v̄ + ω (rB − rC )
Now we note that
Substituting rA − rC and rB − rC into Eq. (5.505) and Eq. (5.506), respectively, we
obtain
vA
= aθ̇eθ + θ̇Ez × (−leθ ) = lθ̇er + aθ̇eθ
(5.509)
vB
= aθ̇eθ + θ̇Ez × (leθ ) = −lθ̇er + aθ̇eθ
(5.510)
Kinetics
The free body diagram of the rod is shown in Fig. 5-21. It can be seen that the
following three forces act on the rod:
NA = Reaction Force of Track at Point A
NB = Reaction Force of Track at Point B
mg = Force of Gravity
From the geometry of the problem, it is seen that the reaction forces NA and
NB are in the directions orthogonal to the track at points A and B, respectively.
249
B
NB
C
A
mg
NA
Figure 5-21
Free Body Diagram for Question 4.
Suppose we let uA and uB be the directions of NA and NB , respectively. Then we
can write
NA
= NA uA
(5.511)
NB
= NB uB
(5.512)
Now, it can be seen that uA and uB must lie along the line segments from O to
A and O to B, respectively. Therefore,
uA
= rA /rA (5.513)
uB
= rB /rB (5.514)
Noting that rA = aer − leθ and rB = aer + leθ , we obtain
uA
=
uB
=
aer − leθ
√
a2 + l2
aer + leθ
√
a2 + l2
(5.515)
(5.516)
which gives
NA
NB
aer − leθ
= NA √
a2 + l2
aer + leθ
= NB √
a2 + l2
(5.517)
(5.518)
Next, the force of gravity is given as
mg = mgEx
(5.519)
Using the expression for Ex from Eq. (5.493), the force of gravity is obtained as
mg = mgEx = mg(cos θ er − sin θ eθ ) = mg cos θ er − mg sin θ eθ
Now we will solve this problem by the following two methods:
• Euler’s Laws Relative to the Center of Mass of the Rod
• The Work-Energy Theorem for a Rigid Body
(5.520)
250
Chapter 5. Kinetics of Rigid Bodies
Method 1: Euler’s Laws Using Center of Mass of Rod as Reference Point
Application of Euler’s 1st Law to Rod
Using the free body diagram of Fig. 5-21, the resultant force acting on the rod is
given as
(5.521)
F = NA + NB + mg
Substituting NA , NB , and mg from Eq. (5.517), Eq. (5.518), and Eq. (5.520) and
aC from Eq. (5.504), we obtain
aer − leθ
aer + leθ
+ NB √
+ mg cos θ er − mg sin θ eθ
F = NA √
a2 + l2
a2 + l2
Then, setting F equal to mF ā using the expression for
obtain
F
(5.522)
ā from Eq. (5.504), we
aer − leθ
aer + leθ
F = NA √
+NB √
+mg cos θ er −mg sin θ eθ = −maθ̇ 2 er +maθ̈eθ
2
2
a +l
a2 + l2
(5.523)
Rearranging Eq. (5.523) gives
a
l
√
(NA + NB ) + mg cos θ er − √
(NA − NB ) + mg sin θ
a2 + l2
a2 + l2
= −maθ̇ 2 er + maθ̈eθ
(5.524)
Equating components, we obtain the following two scalar equations:
−maθ̇ 2
maθ̈
a
√
(NA + NB ) + mg cos θ
2
a + l2
l
(NA − NB ) − mg sin θ
= −√
2
a + l2
=
(5.525)
(5.526)
Application of Euler’s 2nd Law Relative to Center of Mass ofRod
Observing that mg passes through the center of mass, the moment relative to
the center of mass of the rod is given as
M̄ = MC = (rA − rC ) × NA + (rB − rC ) × NB
(5.527)
Substituting the expressions for rA −rC and rB −rC from Eq. (5.507 and Eq. (5.508,
respectively, and the expressions for NA and NB from Eq. (5.517) and Eq. (5.518),
respectively, into Eq. (5.527), we obtain M̄ as
aer − leθ
aer + leθ
+ (leθ ) × NB √
M̄ = (−leθ ) × NA √
2
2
a +l
a2 + l2
(5.528)
251
Eq. (5.528) simplifies to
M̄ = √
al
a2
+ l2
(NA − NB ) ez
(5.529)
Next, the angular momentum of the rod relative to the center of mass of the rod
is given as
F
R
H̄ = Ī · F ωR
(5.530)
Now since {er , eθ , ez } is a principle-axis basis, we have that
R
Ī
= Īr r er ⊗ er + Īθθ eθ ⊗ eθ + Īzz ez ⊗ ez
Consequently,
F
H̄ = Īzz θ̇ez
(5.531)
(5.532)
Now for a slender uniform rod we have that
Īzz =
m(2l)2
ml2
mL2
=
=
12
12
3
(5.533)
F
Where L = 2l is the length of the rod. Therefore, we obtain H̄ as
F
H̄ =
ml2
θ̇ez
3
(5.534)
F
The rate of change of H̄ in reference frame F is then obtained as
d F ml2
θ̈ez
H̄ =
dt
3
F
(5.535)
˙ from Eq. (5.535), we obtain
Setting M̄ from Eq. (5.529) equal to H̄
al
ml2
θ̈ = √
(NA − NB )
2
3
a + l2
(5.536)
The differential equation of motion is obtained using Eq. (5.526) and Eq. (5.536).
Dividing Eq. (5.526) by a gives
l
ml2
θ̈ = √
(NA − NB )
2
3a
a + l2
(5.537)
Then, adding Eq. (5.536) and Eq. (5.537), we obtain
maθ̈ +
ml2
θ̈ = −mg sin θ
3a
(5.538)
Rearranging Eq. (5.538), we obtain the differential equation of motion as
(3ma2 + ml2 )θ̈ + 3mga sin θ = 0
(5.539)
252
Chapter 5. Kinetics of Rigid Bodies
Method 2: Work-Energy Theorem
From the work-energy theorem for a rigid body we have that
n
d F . nc F
E =
Fi · vi + τnc · F ωR
dt
i=1
(5.540)
Now since the only non-conservative forces acting on the rod are NA and NB , the
first term in Eq. (5.540) is given as
n
.
F
F
F
Fnc
i · v i = N A · v A + NB · v B
(5.541)
i=1
Using
lsupFvA and F vB from Eq. (5.509) and Eq. (5.510), respectively, and NA and NB
from Eq. (5.517) and Eq. (5.518), respectively, we obtain
n
.
aer − leθ
aer + leθ
F
√
Fnc
· (lθ̇er + aθ̇eθ ) + NB √
· (lθ̇er − aθ̇eθ ) = 0
i · vi = NA
2
2
a +l
a2 + l2
i=1
(5.542)
Furthermore, since no pure torques act on the rod, we have that
τnc · F ωR = 0
(5.543)
Consequently,
d F E =0
dt
which implies that the total energy is a constant, i.e.
F
E = F T + F U = constant
(5.544)
(5.545)
Now the kinetic energy in reference frame F is given as
F
T =
1F
1 F
m v̄ · F v̄ +
H̄ · F ωR
2
2
(5.546)
F
Substituting F v̄ from Eq. (5.500) and H̄ from Eq. (5.530), we obtain
1
1 ml2
3ma2 + ml2 2
F
θ̇Ez · θ̇Ez =
θ̇
T = (aθ̇eθ ) · (aθ̇eθ ) +
2
2
3
6
(5.547)
Next, the potential energy is due entirely to gravity and is given as
F
U = −mg · r̄
(5.548)
Using the expression for mg from Eq. (5.520) and r̄ = rC from Eq. (5.495), we
obtain
F
U = −(mg cos θ er − mg sin θ eθ ) · (aer ) = −mga cos θ
(5.549)
253
Adding the kinetic energy in Eq. (5.547), and the potential energy in Eq. (5.549),
we obtain
3ma2 + ml2 2
F
θ̇ − mga cos θ = constant
E=
(5.550)
6
Computing the rate of change of F E in Eq. (5.550), we obtain
d F 3ma2 + ml2
θ̇ θ̈ + mgaθ̇ sin θ = 0
E =
dt
3
(5.551)
Noting that θ̇ ≠ 0 as a function of time, we obtain
3ma2 + ml2
θ̈ + mga sin θ = 0
3
(5.552)
Multiplying this last equation through by three, we obtain
(3ma2 + ml2 )θ̈ + 3mga sin θ = 0
(5.553)
254
Chapter 5. Kinetics of Rigid Bodies
Question 5–17
A uniform disk of mass m and radius r rolls without slip along the inside of
a fixed circular track of radius R as shown in Fig. P5-17. The angles θ and φ
measure the position of the center of the disk and the angle of rotation of the
disk, respectively, relative to the vertically downward direction. Knowing that
the angles θ and φ are simultaneously zero and that gravity acts downward,
determine the differential equation of motion for the disk in terms of the angle
θ.
g
R
φ
O
m
θ
r
C
Figure P 5-19
Solution to Question 5–17
Kinematics
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in F :
Ex
Ez
Ey
Origin at O
=
=
=
Along OC When θ = 0
Out of Page
Ez × Ex
255
Next, let R be a reference frame fixed to the direction of OC. Then, choose the
following coordinate system fixed in reference frame R:
Origin at O
=
=
=
er
ez
eθ
Along OC
Out Page
ez × er
The geometry of the bases {Ex , Ey , Ez } and {er , eθ , ez } is shown in Fig. 5-22.
eθ
θ
e z , Ez
Ey
θ
er
Ex
Figure 5-22
Unit Vertical Direction for Question 5–17.
Using Fig. 5-22, we have that
Ex
= cos θ er − sin θ eθ
(5.554)
Ey
= sin θ er + cos θ eθ
(5.555)
Next, the position of the center of mass of the disk is given as
r̄ = rC = (R − r )er
(5.556)
Furthermore, the angular velocity of reference frame R in reference frame F is
given as
F R
ω = θ̇Ez
(5.557)
Then, applying the rate of change transport theorem between reference frame
R and reference frame F , we obtain the velocity of the center of mass of the
disk as
F
dr̄ Rdr̄ F R
F
=
+ ω × r̄
v̄ =
(5.558)
dt
dt
Now we have that
R
dr̄
= 0
dt
F R
ω × r̄ = θ̇ez × (R − r )er = (R − r )θ̇eθ
(5.559)
(5.560)
256
Chapter 5. Kinetics of Rigid Bodies
Adding Eq. (5.559) and Eq. (5.560), we obtain the velocity of the center of mass
of the rod in reference frame F as Differentiating vC in Eq. (5.558), we obtain
F
v̄ = (R − r )θ̇eθ
(5.561)
We can obtain a second expression for F v̄ as follows. Since point P and point C
are both fixed in the disk, we have that
F
v̄ − F vP = F ωD × (r̄ − rP )
(5.562)
where F ωD is the angular velocity of the disk in reference frame F . Now, since
φ describes the rotation of the disk relative to the vertical and φ is measured
in the direction opposite the angle θ, we have that
F
ωD = −φ̇ez
(5.563)
Then, noting that r̄ − rP = −r er , we have that
F
v̄ − F vP = −φ̇ez × (−r er ) = r φ̇eθ
(5.564)
Furthermore, since the disk rolls without slip, we have that
F
Therefore,
F
vP = 0
(5.565)
v̄ = r φ̇eθ
(5.566)
Then, setting the result of Eq. (5.566) equal to the result of Eq. (5.561), we obtain
r φ̇ = (R − r )θ̇
(5.567)
Solving Eq. (5.567) for φ̇, we obtain
φ̇ =
R−r
θ̇
r
(5.568)
Next, applying the rate of change transport theorem to F v̄ using the expression
for F v̄ in Eq. (5.561, we obtain the acceleration of the center of mass of the disk
in reference frame F as
F
ā =
d F R d F F R F
ā =
v̄ + ω × v̄
dt
dt
F
(5.569)
Now we have that
d F v̄
= (R − r )θ̈eθ
dt
F R
ω × F v̄ = θ̇ez × (aθ̇eθ ) = −(R − r )θ̇ 2 er
R
(5.570)
(5.571)
257
Adding Eq. (5.570) and Eq. (5.571), we obtain the acceleration of the center of
mass of the rod in reference frame F as
F
ā = −(R − r )θ̇ 2 er + (R − r )θ̈eθ
(5.572)
Finally, since point C and point P are both fixed to the disk, the acceleration of
point P in reference frame F is obtained as
F
aP = F ā + F αD × (rP − r̄) + F ωD F ωD × (rP − r̄)
(5.573)
Now we have that
F
αD =
R−r
d F D θ̈ez
ω
= −φ̈ez = −
dt
r
F
(5.574)
Therefore
F
R−r
R−r
R−r
θ̈ez × (r er ) −
θ̇ez × −
θ̇ez × (r er )
aP = ā −
r
r
r
F
(5.575)
Simplifying Eq. (5.575), we obtain
F
aP = F ā − (R − r )θ̈eθ −
Then, substituting the expression for
obtain
F
F
(R − r )2 θ̇ 2
er
r
ā from Eq. (5.572) into Eq. (5.576), we
aP = −(R − r )θ̇ 2 er −
(R − r )2 θ̇ 2
er
r
Simplifying Eq. (5.577), we obtain F aP as
R−r
F
θ̇ 2 er
aP = −(R − r ) 1 +
r
Kinetics
The free body diagram of the disk is shown in Fig. 5-23.
R
mg
N
Figure 5-23
(5.576)
Free Body Diagram for Question 5–17.
(5.577)
(5.578)
258
Chapter 5. Kinetics of Rigid Bodies
Using Fig. 5-23, it is seen that the following forces act on the disk
N
= Reaction Force of Track on Disk
FR
= Rolling Force of Track on Disk
mg = Force of Gravity
Now from the geometry we have that
N = Ner
FR
(5.579)
= FR eθ
(5.580)
mg = mgEx
(5.581)
Using the expression for Ex from Eq. (5.554), the force of gravity can be written
in terms of the basis {er , eθ , ez } as
mg = mg cos θ er − mg sin θ eθ
(5.582)
Now we will use the following three methods to determine the differential equation of motion: (1) Euler’s 1st law and Euler’s 2nd law relative to the center of
mass of the disk, (2) Euler’s 2nd law relative to the instantaneous point of contact of the disk with the track and (3) the alternate form of the work-energy
theorem for a rigid body.
Method 1: Euler’s 1st Law and Euler’s 2nd Law Relative to Center of Mass of
Disk
Applying Euler’s 1st law, we have that
F = mF ā
(5.583)
Now, using the forces as given in Eq. (5.579), Eq. (5.580), and Eq. (5.582), the
resultant force acting on the disk is given as
F = N + FR + mg = Ner + FR eθ + mg cos θ er − mg sin θ eθ
(5.584)
Simplifying Eq. (5.584), we obtain
F = (N + mg cos θ )er + (FR − mg sin θ )eθ
Then, setting F in Eq. (5.585) equal to mF ā using the expression for
Eq. (5.572), we obtain
(5.585)
F
ā from
(N + mg cos θ )er + (FR − mg sin θ )eθ = −m(R − r )θ̇ 2 er + m(R − r )θ̈eθ (5.586)
Equating components in Eq. (5.586), we obtain
N + mg cos θ
= −m(R − r )θ̇ 2
(5.587)
FR − mg sin θ
= m(R − r )θ̈
(5.588)
259
Applying Euler’s 2nd law relative to the center of mass of the disk, we have
that
F
d F H̄
(5.589)
M̄ =
dt
Now we have that
F
D
(5.590)
H̄ = Ī · F ωD
Now since {er , eθ , ez } is a principle-axis basis, we have that
D
Ī
= Īr r er ⊗ er + Īθθ eθ ⊗ eθ + Īzz ez ⊗ ez
Then, using the expression for
from Eq. (5.568), we obtain
F ωD
from Eq. (5.563) and the expression for φ̇
R−r
θ̇ez
r
Now we have for a uniform circular disk that
F
(5.591)
H̄ = −Īzz
Īzz =
mr 2
2
(5.592)
(5.593)
F
Consequently, H̄ becomes
F
H̄ = −
mr (R − r )
mr 2 R − r
θ̇ez = −
θ̇ez
2
r
2
(5.594)
F
Computing the rate of change of H̄ in reference frame F , we obtain
F
d F mr (R − r )
θ̈ez
(5.595)
H̄ = −
dt
2
Next, since the forces N and mg pass through the center of mass of the disk,
the moment applied to the disk relative to the center of mass of the disk is due
entirely to bf FR and is given as
M̄ = (rP − r̄) × FR = r er × FR eθ = r FR ez
F
(5.596)
F
Then, setting M̄ from Eq. (5.596) equal to d( H̄)/dt, we obtain
mr (R − r )
θ̈ = r FR
2
Simplifying Eq. (5.597), we obtain
−
(5.597)
m(R − r )
θ̈
(5.598)
2
The differential equation of motion can now be obtained using Eq. (5.588)
and Eq. (5.598). Substituting the expression for FR from Eq. (5.598) into Eq. (5.588),
we obtain
m(R − r )
θ̈ − mg sin θ = m(R − r )θ̈
(5.599)
−
2
Simplifying Eq. (5.599), we obtain the differential equation of motion as
FR = −
3
m(R − r )θ̈ + mg sin θ = 0
2
(5.600)
260
Chapter 5. Kinetics of Rigid Bodies
Method 2: Euler’s 2nd Law Relative to Instantaneous Point of Contact
Since the point of contact is an arbitrary point, we have that
F
MP − (r̄ − rP ) × m aP =
d F HP
dt
F
(5.601)
First, the angular momentum relative to point P is given as
F
F
HP = H̄ + (r̄ − rP ) × m(F v̄ − F vP )
(5.602)
Then, using the fact that r̄−rP = −r er , the fact that F v̄− F vP = F v̄ = (R −r )θ̇eθ ,
F
and the expression for H̄ from Eq. (5.594), we obtain F HP as
F
HP = −
mr (R − r )
3mr (R − r )
θ̈ez + (−r er ) × m(R − r )θ̇eθ = −
θ̇ez (5.603)
2
2
Then, computing the rate of change of F HP in reference frame F , we obtain
3mr (R − r )
d F θ̈ez
HP = −
dt
2
F
(5.604)
Next, since the forces N and FR pass through point P , the moment acting on the
disk relative to point P is due entirely to gravity and is given as
MP = (r̄ − rP ) × mg = −r er × (mg cos θ er − mg sin θ eθ ) = mgr sin θ ez (5.605)
Finally, the inertial moment is given as
R−r
F
2
−(r̄ − rP ) × m aP = −(−r er ) × m −(R − r ) 1 +
θ̇ er = 0
r
(5.606)
Substituting the results of Eq. (5.604), Eq. (5.605), and Eq. (5.606) into Eq. (5.601),
we obtain
3mr (R − r )
θ̈ez
(5.607)
mgr sin θ = −
2
Simplifying Eq. (5.607), we obtain the differential equation of motion as
3
m(R − r )θ̈ + mg sin θ = 0
2
(5.608)
Method 3: Alternate Form of Work-Energy Theorem for a Rigid Body
Of the three forces acting on the disk, we know that gravity is conservative.
Furthermore, since the velocity of point P in reference frame F is zero, we
know that neither N nor bf FR does work. Finally, since no pure torques act
on the disk, we know that the work done by all non-conservative forces and
non-conservative pure torques is zero. Therefore, we have that
d F E =0
dt
(5.609)
261
Now the total energy in reference frame F is given as
F
E = FT + FU
(5.610)
First, the kinetic energy of the disk in reference frame F is given as
F
T =
1F
1 F
m v̄ · F v̄ +
H̄ · F ωD
2
2
(5.611)
Using the expression for F v̄ from Eq. (5.561), we have that
1 F
1
m v̄ · F v̄ = m(R − r )2 θ̇ 2
2
2
(5.612)
F
Next, using the expression for H̄, the expression for F ωD from Eq. (5.563), and
the expression for φ̇ from Eq. (5.568), we have that
#
$ mr 2 R − r
1
R−r
1
1F
F D
−
θ̇ez · −
θ̇ez = m(R − r )2 θ̇ 2 (5.613)
H̄ · ω =
2
2
2
r
r
4
Adding Eq. (5.612) and Eq. (5.613), we obtain the kinetic energy of the disk in
reference frame F as
3
F
T = m(R − r )2 θ̇ 2
(5.614)
4
Next, the potential energy of the disk in reference frame F is given as
F
U = F Ug = −mg · r̄
= −(mg cos θ er − mg sin θ eθ ) · (R − r )er = −mg(R − r ) cos θ
(5.615)
The total energy in reference frame F is then given as
F
E = FT + FU =
3
m(R − r )2 θ̇ 2 − mg(R − r ) cos θ
4
(5.616)
Computing the rate of change of F E and setting the result equal to zero, we
obtain
d F 3
E = m(R − r )2 θ̇ θ̈ + mg(R − r )θ̇ sin θ = 0
(5.617)
dt
2
Eq. (5.617) can be re-written as
3
m(R − r )θ̈ + mg sin θ = 0
(5.618)
(R − r )θ̇
2
Observing that θ̇ is not zero as a function of time, we obtain the differential
equation of motion as
3
m(R − r )θ̈ + mg sin θ = 0
2
(5.619)
262
Chapter 5. Kinetics of Rigid Bodies
Question 5–20
A uniform circular disk of mass m and radius r rolls without slip along a plane
inclined at a constant angle β with horizontal as shown in Fig. P5-20. Attached
to the disk at the point A (where A lies in the direction of P O) is a linear spring
with spring constant K and a nonlinear damper with damping constant c. The
damper exerts a force of the form
Fd = −cvA 3
vA
vA where vA is the velocity of point A. Knowing that the spring is unstretched when
the angle θ is zero and that gravity acts downward, determine the differential
equation of motion for the disk.
K
g
c
θ
A
O
r
P
β
Figure P 5-24
Solution to Question 5–20
Kinematics
First, let F be a fixed reference frame. Then, choose the following coordinate
system fixed in reference frame F :
Ex
Ez
Ey
Origin at Point O
When t = 0
=
=
=
Along Incline
Into Page
E z × Ex
263
Now, since the disk rolls without slip along a fixed surface, we have that
F
vP = 0
(5.620)
Then, from kinematics of rigid bodies, the velocity of point O is given as
F
vO = F vP + F ωR × (rO − rP )
(5.621)
where R is the reference frame of the disk. Now, since the angle θ describes the
amount that the disk has rotated since time t = 0, we have that
F
ωR = θ̇Ez
(5.622)
Furthermore, noting that rO − rP = −r Ey , we obtain vO as
F
vO = θ̇Ez × (−r Ey ) = r θ̇Ex
(5.623)
Differentiating F vO in reference frame F , we obtain the acceleration of point O
in reference frame F as
F
aO =
d F vO = r θ̈Ex
dt
F
(5.624)
Finally, we obtain the velocity of point A in reference frame F as
F
vA = F vP + F ωR × (rA − rP )
(5.625)
Noting that rA − rP = −2r Ey , we obtain F vA as
F
vA = θ̇Ez × (−2r Ey ) = 2r θ̇Ex
(5.626)
Kinetics
The free body diagram of the disk is shown in Fig. 5-24. It can be seen that the
forces acting on the disk are as follows:
mg
R
N
Fs
Fd
=
=
=
=
=
Force of Gravity
Rolling Force
Normal Force Applied by Incline on Disk
Spring Force
Damping Force
Now we have the following
R
=
REx
N
=
NEy
mg = mguv
(5.627)
264
Chapter 5. Kinetics of Rigid Bodies
Fs
Fd
mg
R
N
Figure 5-24
Free Body Diagram for Question 6.11
Ex
Ey
Figure 5-25
6.11
β
uv
Downward Direction uV in Terms of Basis {Ex , Ey } for Question
where uv is the unit vector in the vertically downward direction. Using the
geometry shown in Fig. 5-25, we have that
uv = sin θ Ex + cos θ Ey
(5.628)
The force of gravity is then given as
mg = mg sin βEx + mg cos βEy
(5.629)
Next, we need the spring force. We know that the spring force has the form
Fs = −K [ρ − L]
ρ
ρ
(5.630)
where ρ is the position of the spring measured from its unstretched length L.
Now for this problem we have that
ρ = rA − rA (t = 0)
(5.631)
We can get rA using vA from Eq. (5.626). Noting that rA (t = 0) = −r Ey , we have
that
rA = 2r θEx − r Ey
(5.632)
Therefore, ρ is given as
ρ = 2r θEx − r Ey − (−r Ey ) = 2r θEx
(5.633)
265
Furthermore, the spring is unstretched when θ = 0 which implies that L = 0.
Therefore, we obtain the spring force as
Fs = −2Kr θEx
(5.634)
Finally, the damping force is given as
Fd = −cFvA 3
Fv
F v
A
(5.635)
A
Using F vA from Eq. (5.626), we obtain Fd as
Fd = −c(2r θ̇)3 Ex = −8cr 3 θ̇ 3 Ex
(5.636)
Now that we have all of the forces, this problem can be solved by performing
a moment balance about the point of contact, P . Since P is not a fixed point, we
have that
F
d F HP
(5.637)
MP − (r̄ − rP ) × mF aP =
dt
Noting that r̄ = rO , Eq. (5.637) can be written as
MP − (rO − rP ) × mF aP = ḢP
(5.638)
rO − rP = −r Ey
(5.639)
Now we have that
Furthermore, we have aP as
F
aP = F aO + α × (rP − rO ) + F ωR ×
F
ωR × (rP − rO )
(5.640)
Differentiating ω in Eq. 5.622), we obtain the angular acceleration of the disk,
α, as
F
d F R F R
α =
ω
(5.641)
= θ̈Ez
dt
Furthermore, using F aO from Eq. (5.624), we obtain F aP as
F
aP = r θ̈Ex + θ̈Ez × r Ey + θ̇Ez × θ̇Ez × r Ey
(5.642)
Simplifying Eq. (5.642) gives
F
aP = r θ̈Ex − r θ̈Ex − r θ̇ 2 Ey = −r θ̇ 2 Ey
(5.643)
−(rO − rP ) × mF aP = r Ey × m(−r θ̇ 2 Ey ) = 0
(5.644)
We then obtain
266
Chapter 5. Kinetics of Rigid Bodies
Therefore, Eq. (5.637) reduces to
MP =
d F HP
dt
F
(5.645)
Examining the free body diagram in Fig. 5-24), we see that the forces R and N
pass through point P . Consequently, the moment relative to point P is given as
MP = (rA − rP ) × (Fs + Fd ) + (rO − rP ) × mg
(5.646)
Now we note that rA − rP = −2r Ey . Then, substituting the expressions for Fs
from Eq. (5.634), Fd from Eq. (5.636), and mg from Eq. (5.629), we obtain MP as
MP = (−2r Ey ) × (−2Kr θEx − 8cr 3 θ̇ 3 Ex ) + (−r Ey ) × (mg sin βEx + mg cos βEy )
(5.647)
Eq. (5.647) simplifies to
MP = (−4Kr θ − 16cr 4 θ̇ 3 + mgr sin β)Ez
(5.648)
Next, the angular momentum relative to point P is given as
F
F R
HP = IR
P · ω
(5.649)
Now since {Ex , Ey , Ez } is a principle-axis basis, we have that
P
P
P
IR
P = Ixx Ex ⊗ Ex + Iyy Ey ⊗ Ey + Izz Ez ⊗ Ez
(5.650)
Consequently, we obtain F HP as
F
P
HP = Izz
θ̇Ez
(5.651)
Now from the parallel-axis theorem we have that
P
Izz
= Īzz + md2
(5.652)
Noting that Īzz = mr 2 and that d = r , we obtain
P
Izz
=
mr 2
3
+ mr 2 = mr 2
2
2
We then obtain F HP as
F
HP =
3
mr 2 θ̇Ez
2
(5.653)
(5.654)
Differentiating F HP in reference frame F , we obtain
d F 3
HP = mr 2 θ̈Ez
dt
2
F
(5.655)
267
Then, setting
F
d
F
HP /dt in Eq. (5.655) equal to MP from Eq. (5.648), we obtain
3mr 2
θ̈Ez = (−4Kr θ − 16cr 4 θ̇ 3 + mgr sin β)Ez
2
(5.656)
Dropping Ez from both sides and rearranging, we obtain the differential equation of motion as
3
mr 2 θ̈ + 16cr 4 θ̇ 3 + 4Kr θ − mgr sin β = 0
2
(5.657)