Problem 20.60* By substituting the components of
HOfrom Eqs. (20.9) into the equation
MO=dHOx
dt i+dHOy
dt j+dHOz
dt k+||×HO
derive Eqs. (20.12).
Problem 20.61 A ship has a turbine engine. The spin
axis of the axisymmetric turbine is horizontal and
aligned with the ship’s longitudinal axis. The turbine
rotates at 10,000 rpm. Its moment of inertia about its
spin axis is 1000 kg-m2. If the ship turns at a constant
rate of 20 degrees per minute, what is the magnitude of
the moment exerted on the ship by the turbine?
where Ixx =1
2Izz =500 kg-m2
˙
ψ=20 π
180 1
60 =0.005818 (rad/s),
˙
φ=10000(2π/60)=1047.2 rad/s,
θ=90◦.
Mx=6092 N-m
(top view
of turbine)
ψ
= 20°/min
φ
= 10,000 rpm
z
Problem 20.62 The center of the car’s wheel Atravels
in a circular path about Oat 15 mi/h. The wheel’s radius
is 1 ft, and the moment of inertia of the wheel about its
axis of rotation is 0.8 slug-ft2. What is the magnitude
of the total external moment about the wheel’s center of
Strategy: Treat the wheel’s motion as steady preces-
sion with nutation angle θ=90◦.
O
680
Problem 20.63 The radius of the 5-kg disk is R=
0.2 m. The disk is pinned to the horizontal shaft and
rotates with constant angular velocity ωd=6 rad/s rela-
tive to the shaft. The vertical shaft rotates with constant
angular velocity ω0=2 rad/s. By treating the motion of
the disk as steady precession, determine the magnitude
of the couple exerted on the disk by the horizontal shaft.
R
vd
v0
Problem 20.64 The helicopter is stationary. The zaxis
of the body-fixed coordinate system points downward
and is coincident with the axis of the helicopter’s rotor.
The moment of inertia of the rotor about the zaxis
is 8600 kg-m2. Its angular velocity is −258k(rpm). If
the helicopter begins a pitch maneuver during which its
angular velocity is 0.02j(rad/s), what is the magnitude
of the gyroscopic moment exerted on the helicopter by
the rotor? Does the moment tend to cause the helicopter
to roll about the xaxis in the clockwise direction (as seen
in the photograph) or the counterclockwise direction?
y
x
z
Problem 20.65 The bent bar is rigidly attached to the
vertical shaft, which rotates with constant angular veloc-
ity ω0. The disk of mass mand radius Ris pinned to
the bent bar and rotates with constant angular velocity
ωdrelative to the bar. Determine the magnitudes of the
force and couple exerted on the disk by the bar.
b
h
0
β
ω
d
ω
R
Solution:
(a) The center of mass of the disk moves in a horizontal circular path
(b) By orienting a coordinate system as shown, with the z axis nor-
mal to the disk and the x axis horizontal, the disk is in steady
+1
Problem 20.66 The bent bar is rigidly attached to the
vertical shaft, which rotates with constant angular veloc-
ity ω0. The disk of mass mand radius Ris pinned to the
bent bar and rotates with constant angular velocity ωd
relative to the bar. Determine the value of ωdfor which
no couple is exerted on the disk by the bar.
682
Problem 20.67 A thin circular disk undergoes
moment-free steady precession. The zaxis is perpen-
dicular to the disk. Show that the disk’s precession rate
is ψ=−2φ/cos θ. (Notice that when the nutation angle
is small, the precession rate is approximately two times
Z
y
z
θ
Problem 20.68 The rocket is in moment-free steady
precession with nutation angle θ=40◦and spin rate
˙
φ=4 revolutions per second. Its moments of inertia are
Ixx =10,000 kg-m2and Izz =2000 kg-m2. What is the
rocket’s precession rate ˙
ψin revolutions per second?
z
X
x
Y
y
Z
φ
θ
ψ
Problem 20.69 Sketch the body and space cones for
the motion of the rocket in Problem 20.68.
Solution: The angle θ=40◦. The angle βdefined by
X
cone
x
Z
ψ
Problem 20.70 The top is in steady precession with
nutation angle θ=15◦and precession rate ˙
ψ=1
revolution per second. The mass of the top is 8 ×
10−4slug, its center of mass is 1 in. from the point,
and its moments of inertia are Ixx =6×10−6slug-ft2
and Izz =2×10−6slug-ft2. What is the spin rate ˙
φof
the top in revolutions per second?
Y
Z
X
y
x
z
ψ
φ
θ
684
Problem 20.71 Suppose that the top described in
Problem 20.70 has a spin rate ˙
φ=15 revolutions per
second. Draw a graph of the precession rate (in revolu-
tions per second) as a function of the nutation angle θ
for values of θfrom zero to 45◦.
Solution: The behavior of the top is described in Eq. (20.32),
˙
φ=v
R=−ω0h
4R.
The solution, ˙
ψ1,2=−b±√b2−c.
The two solutions, which are real over the interval, are graphed as a
function of θover the range 0 ≤θ≤45◦. The graph is shown.
Precession rate vs Nutation angle
10
Problem 20.72 The rotor of a tumbling gyroscope can
be modeled as being in moment-free steady precession.
The moments of inertia of the gyroscope are Ixx =Iyy =
0.04 kg-m2and Izz =0.18 kg-m2. The gyroscope’s spin
rate is ˙
φ=1500 rpm and its nutation angle is θ=20◦.
(a) What is the precession rate of the gyroscope in
rpm?
Problem 20.73 A satellite can be modeled as an 800-
kg cylinder 4 m in length and 2 m in diameter. If the
nutation angle is θ=20◦and the spin rate ˙
φis one
revolution per second, what is the satellite’s precession
rate ˙
ψin revolutions per second?
z
Z
θ
y
φ
of a homogenous cylinder are
Problem 20.74 The top consists of a thin disk bonded
to a slender bar. The radius of the disk is 30 mm and its
mass is 0.008 kg. The length of the bar is 80 mm and
its mass is negligible compared to the disk. When the
top is in steady precession with a nutation angle of 10◦,
the precession rate is observed to be 2 revolutions per
second in the same direction the top is spinning. What
is the top’s spin rate?
10⬚
2(0.008 kg)(0.03 m)2
Eq. 20.32 is
Mx=(Izz −Ixx)˙
ψ2cos 10◦+Izz ˙
ψ˙
φ
Solving we find
10°
•
686
Problem 20.75 Solve Problem 20.58 by treating the
motion as steady precession.
45 deg/s
Solution: The view of an airplane’s landing gear looking from
behind the airplane is shown in Fig. (a). The radius of the wheel is
45◦/s=0.7853 rad/s. The spin vector is aligned with the zaxis, from
which
˙
φ=v
R=30
0.3=100 rad/s
The moments and products of inertia of the wheel are Izz =mR2/2=
2 kg-m2. The moment is
Mx=Izz ˙
ψ˙
φsin 90◦=2(0.7854)(100)=157 N-m.
Problem 20.76* The two thin disks are rigidly con-
nected by a slender bar. The radius of the large disk is
200 mm and its mass is 4 kg. The radius of the small
v0
y
Problem 20.77* Suppose that you are testing a car and
use accelerometers and gyroscopes to measure its Euler
angles and their derivatives relative to a reference coor-
dinate system. At a particular instant, ψ=15◦,θ=4◦,
φ=15◦, the rates of change of the Euler angles are
zero, and their second derivatives with respect to time
are ¨
ψ=0, ¨
θ=1 rad/s2, and ¨
φ=−0.5 rad/s2. The car’s
principal moments of inertia, in kg-m2, are Ixx =2200,
Iyy =480, and Izz =2600. What are the components of
the total moment about the car’s center of mass?
y
zZ
Y
X
x
Problem 20.78* If the Euler angles and their second
derivatives for the car described in Problem 20.77
have the given values, but their rates of change are
˙
ψ=0.2 rad/s, ˙
θ=−2 rad/s, and ˙
φ=0, what are the
components of the total moment about the car’s center
of mass?
688
Problem 20.79* Suppose that the Euler angles of the
car described in Problem 20.77 are ψ=40◦,θ=20◦,
and φ=5◦, their rates of change are zero, and the com-
ponents of the total moment about the car’s center of
mass are
Mx=−400 N-m,
My=200 N-m,
Mz=0.
What are the x,y, and zcomponents of the car’s angular
acceleration?
Problem 20.80 The mass of the bar is 6 kg. Determine
the moments and products of inertia of the bar in terms
of the coordinate system shown.
x
y
1 m
from which the inertia matrix for the element Oy about the x′axis is
0.1667 0 0
000.1667
(b) The horizontal element Ox of the bar. The mass of the horizontal
690
The inertia matrix is
00 0
v
and Izz =Ixx +Iyy , where the subscripts vand hdenote the vertical
and horizontal bars respectively. Noting that the masses are
00
3(3)
Problem 20.81 The object consists of two 1-kg verti-
cal slender bars welded to a 4-kg horizontal slender bar.
0.1 m0.1 m
0.2 m
Solution:
Ixy =0.01 kg-m2,I
xz =0,I
yz =0.
Problem 20.82 The 4-kg thin rectangular plate lies in
the x-y plane. Determine the moments and products of
Solution: From Appendix B, the moments of inertia of the plate’s
area are
Problem 20.83 If the 4-kg plate is rotating with angu-
lar velocity ω=6i+4j−2k(rad/s), what is its angular
momentum about its center of mass?
Solution: Angular momentum is
0.12 0 0
6
692
Problem 20.84 The 30-lb triangular plate lies in the
x–yplane. Determine the moments and products of iner-
z
IA
xy =1
8(6)2(4)2=72 ft4.
The plate’s area and mass are
A=1
2(6)(4)=12 ft2
AIA
Iyz =Izx =O,
Problem 20.85 The 30-lb triangular plate lies in the
x–yplane.
(a) Determine its moments and products of inertia in
terms of a parallel coordinate system x′y′z′with its
origin at the plate’s center of mass.
(b) If the plate is rotating with angular velocity ω=
20i−12j+16k(rad/s), what is its angular momen-
tum about its center of mass?
3(6)=4ft,
Problem 20.86 Determine the inertia matrix of the
2.4-kg steel plate in terms of the coordinate system
shown.
y
220 mm
Solution: Equation (20.39) gives the plate’s moments and
8(0.22)4−1
4(0.05)2(0.15)2=0.000279 m4.
The area is
AIy=0.0357 kg-m2,
AIA
Problem 20.87 The mass of the steel plate is 2.4 kg.
(a) Determine its moments and products of inertia in
terms of a parallel coordinate system x′y′z′with its
origin at the plate’s center of mass.
(b) If the plate is rotating with angular velocity
ω=20i+10j−10k(rad/s), what is its angular
momentum about its center of mass?
Using the results of the solution of Problem 20.86 and the parallel
0.215
694
Problem 20.88 The slender bar of mass mrotates
about the fixed point Owith angular velocity ω=
ωyj+ωzk. Determine its angular momentum (a) about
its center of mass and (b) about O.
x
z
y
O
l
Solution:
12 ,
Ixy =Ixz =Iyz =0.
The angular momentum about Ois
=
0mL2
30
00 mL2
3
ωy
ωz
=
3
mL2ωz
3
,
Problem 20.89 The slender bar of mass mis parallel
to the xaxis. If the coordinate system is body fixed and
its angular velocity about the fixed point Ois ω=ωyj,
what is the bar’s angular momentum about O?
y
h
l
696
Problem 20.90 In Example 20.8, the moments and
products of inertia of the object consisting of the booms
AB and BC were determined in terms of the coordinate
system shown in Fig. 20.34. Determine the moments and
products of inertia of the object in terms of a parallel
x
B
6 m
001,056,000
The mass of the boom AB is mAB =4800 kg. The mass of the boom
rearrange Eq. (20.42) to obtain the moments and products of inertia
about the parallel axis passing through the center of mass of the two
booms when the moments and products of inertia in the x,y,zsystem
are known:
I(G)
x′x′=I(0)
xx −(d2
y+d2
x)m =15600 kg-m2.
I(G)
y′y′=I(o)
yy −(d2
x+d2
z)m =226800 kg-m2.
I(G)
z′z′=I(o)
zz −(d2
x+d2
y)m =242400 kg-m2
I(G)
x′y′=I(0)
xy −dxdym=−32400 kg-m2,
I(G)
x′z′=0,I(G)
y′z′=0 .
The inertia matrix for the x′,y′,z′system is
[I]G=
15,600 32,400 0
32,400 226,800 0
0 0 242,400
kg-m2.
y
y
A
18 m
C
Problem 20.91 Suppose that the crane described in
Example 20.8 undergoes a rigid-body rotation about the
vertical axis at 0.1 rad/s in the counterclockwise direc-
tion when viewed from above.
(a) What is the crane’s angular velocity vector ωin
terms of the body-fixed coordinate system shown
in Fig. 20.34?
(b) What is the angular momentum of the object con-
sisting of the booms AB and BC about its center
of mass?
698