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
=
+ r
dt
dt
(3.5)
Now we have
A
dr
= r0 sin er
dt
F A
r = Ez r0 cos er = r0 cos e
(3.6)
(3.7)
Adding the expressio

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

73
We then obtain
F
s F s0 = R 1 + 2 ( 0 )
(3.82)
Solving Eq. (3.82) for s, the arclength is given as
F
s = F s0 + R 1 + 2 ( 0 )
(3.83)
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 veloc

68
Chapter 3. Kinetics of Particles
Question 32
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, de

72
Chapter 3. Kinetics of Particles
following coordinate system to describe the motion of the particle:
er
ez
e
Origin at O
=
=
=
Along Radial Direction of Circle
Ez from Reference Frame F
ez er
Now, since is the angle formed by the helix with the horizon

79
Now we know that since the maximum distance is obtained when the velocity of
the bead is zero, we must have that 1 = 0. Furthermore, since the initial value
of is zero, we have that 0 = 0. Consequently, Eq. (3.141) reduces to
1
2
mR 2 0 (1 + 2 ) = mgR1

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)
Consequently,
d
d

66
Chapter 3. Kinetics of Particles
Then, using the expression for en from Eq. (3.17), we have that
er en
= er ( cos er sin e ) = cos
(3.36)
e en
= e ( cos er sin e ) = sin
(3.37)
Substituting the results of Eq. (3.36) and Eq. (3.37) into Eq. (3.35) giv

71
Solution to Question 33
A bead of mass m slides along a xed circular helix of radius R and constant
helical inclination angle as shown in Fig. P3-3. The equation for the helix is
given in cylindrical coordinates as
z = R tan
(3.71)
Knowing that gravit

55
Question 221
A slender rod of length l is hinged to a collar as shown in Fig. P2-21. The collar
slides freely along a xed horizontal track. Knowing that x is the horizontal
displacement of the collar and that describes the orientation of the rod relati

57
Now, since F vO is expressed in the basis cfw_Ex , Ey , Ez , we have that
F
aO =
F
d
dt
F
vO = xEx
(2.327)
Furthermore, since F vP /O is expressed in the basis cfw_er , e , ez and cfw_er , e , ez
rotates with angular velocity F A , we can obtain F aP

51
Now we can obtain an expression for F B in terms of the basis cfw_ur , u , u by
expressing ez in terms of ur and u as
ez = sin ur + cos u
(2.286)
Consequently,
F
B = (sin ur + cos u ) u = sin ur u + cos u
(2.287)
Then, the velocity of point P in refer

60
Chapter 2. Kinematics
Dierentiating et in Eq. (2.345), we obtain
F
det
= x sin xEx + x cos xEy
dt
Consequently,
F
det
dt
= x = Fv
(2.347)
(2.348)
which implies that
en =
F
F
det /dt
det /dt
=
x sin xEx + x cos xEy
= sin xEx + cos xEy
x
(2.349)
Then, us

54
Chapter 2. Kinematics
Now the two terms required to obtain F v are given as
F
dr
= r er
(2.305)
dt
F B
r = ( cos er sin e + ez ) r er = r e + r sin (2.306)
ez
Therefore, the velocity of the particle in reference frame F is
F
v = r er + r e + r sin ez

62
Chapter 3. Kinetics of Particles
Solution to Question 31
Kinematics
Let F be a reference frame xed to the track. Then, choose the following coordinate system xed in reference frame F :
Ex
Ez
Ey
Origin at point O
=
=
=
Along OP when = 0
Out of page
Ez E

56
Chapter 2. Kinematics
Next, the angular velocity of reference frame A in reference frame F is given as
F
A = ez
(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

67
Rearranging Eq. (3.48) and separating variables, we obtain
g
d = cos 2d
r0
(3.49)
Integrating this last equation, we obtain
1 2 2
g
0 =
[sin 2]0
2
2r0
(3.50)
Noting that (t = 0) = 0, this last equation simplies to
g
2
2 = 0
sin 2
r0
(3.51)
Solvin

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
e + ez
1 + 2
e ez
=
1 + 2
=
(3.104)
Using Eq. (3.104) we can obtain an ex

64
Chapter 3. Kinetics of Particles
which implies that
2 cos er 2 sin e
2 cos er 2 sin e
en =
= cos er sin e
(3.17)
The principal unit bi-normal vector to the track is then obtained as
eb = et en = ( sin er + cos e ) ( cos er sin e ) = ez
(3.18)
The acce

Chapter 3
Kinetics of Particles
Question 31
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
Knowing that gravity acts downward and assuming the initial

59
Arc-length Parameter as a Function of x
Now we recall the arc-length equation as
d
dt
F
s = F v = x 1 + tan2 x = x sec x
(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

43
Therefore,
F
vP /Q = R( )i R sin( )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 = Luy L sin uz + +R( )i R sin( )iz
(2.249)
It is noted that this last expression can be converted

49
Question 219
A particle P slides without friction along the inside of a xed hemispherical bowl
of radius R as shown in Fig. P2-19. The basis cfw_Ex , Ey , Ez is xed to the bowl.
Furthermore, the angle is measured from the Ex -direction to the directio

45
Determination of Intrinsic Basis
The position of the particle in terms of the basis cfw_er , e , Ez is given as
r = r er = aer
(2.250)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F A
= Ez
(2.251)
The veloci

52
Chapter 2. Kinematics
Question 220
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 about an axis along the direction of the shaft

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 cfw_er , e , ez and
cfw_er , e , ez is xed in reference frame B, the velocity of the particle as viewed by
an observer xed to t

50
Chapter 2. Kinematics
Finally, let B be a reference frame xed to the direction OP . Then choose the
following coordinate system xed in reference frame B:
ur
u
u
Origin at O
=
=
=
Along OP
e
ur u
The relationship between the bases cfw_Ex , Ey , Ez and

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
a = a 1 + 2
1/2
a(2 + 2 ) 2
en
1 + 2 + 2 et +
1 + 2
(2.280)
Simplifying Eq. (2.280) g

76
Chapter 3. Kinetics of Particles
Equating components in Eq. (3.110) yields the following three scalar equations:
mg
1 + 2
Nn
= mR 1 + 2
= mR 2
mg
=
1 + 2
Nb
(3.111)
(3.112)
(3.113)
It is noted that, because it contains no reaction forces, Eq. (3.111)

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
(