82
Chapter 3. Kinetics of Particles
Then, the principle unit normal vector is obtained as
F
en =
F
det /dt
(3.158)
det /dt
Now we have from the basic kinematic equation that
F
where
F
det
dt
A
det
det F A
=
+ et
dt
dt
(3.159)
= 0
aer + e
er + ae
et = Ez

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 = ( r 2 )ur + (2 + r )u
r
r
(3.343)
Then, adding Eq. (3.340) and Eq. (3.343), we obtain the acceleration of the pa

110
Chapter 3. Kinetics of Particles
Finally, we have that
us =
r rA
r rA
=
r rP
r rP
=
r ur
= ur
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)
T

108
Chapter 3. Kinetics of Particles
Now we have that
B
d
rm/P
dt
F
=
B rm/P
r ur
(3.334)
= uz r ur = r u
Adding the two expressions in Eq. (3.334), we have that
F
vm/P = r ur + r u
(3.335)
Then, adding Eq. (3.332) and Eq. (3.335), we obtain the velocity

107
reference frame F as
F
B = uz
(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 dened above, we have tha

106
Chapter 3. Kinetics of Particles
in reference frame B:
ur
uz
u
Origin at O
=
=
=
Along P m
Out of Page (= Ez = ez )
uz ur
The geometry of the bases cfw_Ex , Ey , Ez , cfw_ex , ey , ez , and cfw_ur , u , uz is shown in
Fig. 3-10. Using Fig. 3-10, we h

102
Chapter 3. Kinetics of Particles
Question 311
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 othe

105
Question 312
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
i

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
r rA
r rA
us =
=
r er
= er
r
(3.314)
Finally, the unstretched lengt

111
Question 313
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
f

112
Chapter 3. Kinetics of Particles
Next, let A be a reference frame xed to the plane formed by the vectors Ez and
OQ. Then, choose the following coordinate system xed in reference frame A:
er
Ez
e
Origin at O
=
=
=
Along OQ
Up
Ez e r
The position of the

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

119
Determination of Dierential Equation Using Newtons 2nd Law
Setting F from Eq. (3.401) equal to mF a using F a from Eq. (3.397), we have that
mg cos er + (T + mg sin )e = m R 2 (l R) er m(l R) 2 e
(3.402)
Equating components in Eq. (3.402) results in t

117
Ex
er
Ey
e z , Ez
e
Figure 3-13
Geometry of Question 317.
Now we note that the rope has a xed length l. Since the length of the portion
of the rope wrapped around the cylinder is R, the exposed portion of the rope
must have length l R. Furthermore, si

118
Chapter 3. Kinetics of Particles
Now we have that
A
d
dt
F
F
v = (R ) + (l R) er
(3.394)
A F v = ez (l R)er = (l R) 2 e
(3.395)
Adding Eq. (3.394) and Eq. (3.395), we obtain
F
a = R 2 + (l R) er (l R) 2 e
(3.396)
Eq. (3.396) simplies to
F
a = R 2 (l

115
Then, dividing the rst equation in Eq. (3.375) by this last result, we obtain
r
r r 2
= 2
R
r + rr
+g
R
(3.379)
Rearranging and simplifying this last equation, we obtain the second dierential
equation as
r
gr
r 2
r + 2 r 2 r 2 +
=0
(3.380)
1+
R
R
R
Th

116
Chapter 3. Kinetics of Particles
Question 317
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 xed cylinder of radius R. As t

114
Chapter 3. Kinetics of Particles
The resultant force on the particle is then given as
r
er + Ez
mgEz
F = N + mg = N R
2
r
1+
R
which can be re-written as
F = N
r
R
er + N
r 2
1+
R
(3.372)
1
r
1+
R
2
mg Ez
(3.373)
Setting F = mF a using F a from pa

113
Kinetics
We need to apply Newtons 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 313.
force of the parabolo

103
The velocity in reference frame F is computed from the rate of change transport
theorem as
F
dr Adr F A
F
=
+ r
v=
(3.306)
dt
dt
where
Adr
dt
= r er
F A
(3.307)
r = Ez r er = r e
Adding the two expressions in Eq. (3.307), we obtain F v as
F
v = r er

101
Now, using F v from Eq. (3.276), we obtain d(F v)/dt as
d
dt
F
v = x sec x + x 2 sec x tan x = sec x x + x 2 tan x
(3.294)
Then, using from Eq. (3.290) we obtain
F
v = cos x(x sec x)2 = x 2 sec x
(3.295)
The acceleration of the particle in reference

91
Now, it is seen that Eq. (3.224) has no unknown reaction forces. Consequently,
Eq. (3.224) is one of the dierential equations of motion. Dropping m from
Eq. (3.224), the rst dierential equation of motion can be written as
r + 2 = 0
r
(3.226)
The second

87
Question 37
A particle of mass m slides without friction along the inner surface of a xed
cone of semi-vertex angle as shown in the Fig. P3-7. The equation for the cone
is given in cylindrical coordinates as
z = r cot
Knowing that the basis cfw_Ex , E

85
Using the expression for
reference frame F as
F
Fv
from Eq. (3.154), we obtain the kinetic energy in
1
m r0 ea [aer + e ] r0 ea [aer + e ]
2
1
2
= mr0 (a2 + 1) 2 e2a
2
T =
(3.186)
Next, since gravity is the only conservative force acting on the particl

86
Chapter 3. Kinetics of Particles
Simplifying Eq. (3.194) and setting the result equal to zero, we obtain
d
dt
F
2
E = mr0 (a2 + 1)e2a ( a 2 ) + mgr0 ea (cos a sin ) = 0
(3.195)
0 as a function of time (otherwise the particle would not be
Now since
mo

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
e et
a
aer + e
=
= er
2+1
2+1
a
a
1
aer + e
=
= e
2+1
2+1
a
a
(3.177)
Substituting the expressions in E

89
N
m
mg
Figure 3-6
Free Body Diagram of Particle for Question 37.
Kinetics
In order to determine the dierential equations of motion, we need to apply
Newtons 2nd Law, i.e., we need to apply F = mF a. The free body diagram of
the particle is shown in Fig

88
Chapter 3. Kinetics of Particles
A:
er
ez
e
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 ez
(3.198)
Furthermore, the angular velocity of reference frame A in reference frame F is
given

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 35.
the following two forces act on the collar: (1) the reaction force of the track, N,
and (2) gravity, mg. Since N

90
Chapter 3. Kinetics of Particles
where
f
r
f
f
z
= cot
(3.213)
= 0
(3.214)
= 1
(3.215)
f = cot er + ez
(3.216)
We then obtain f as
The unit normal to the surface of the cone is then given as
n=
f
f
=
cot er + ez
1 + cot2
(3.217)
Now from trigonometr