PROBLEM 18.151
A four-bladed airplane propeller has a mass of 160 kg and a radius of gyration of
800 mm. Knowing that the propeller rotates at 1600 rpm as the airplane is
traveling in a circular path of 600-m radius at 540 km/h, determine the magnitude
of the couple exerted by the propeller on its shaft due to the rotation of the
airplane.
SOLUTION
We assume senses shown for
,
,
xy
ωω
and
.v
PROBLEM 18.152
A 2.4–kg piece of sheet steel with dimensions
160 640 mm×
was
bent to form the component shown. The component is at rest
( 0)
ω
=
when a couple
0
(0.8 N m)= ⋅Mk
is applied to it.
Determine (a) the angular acceleration of the component,
(b) the dynamic reactions at A and B immediately after the
couple is applied.
SOLUTION
2
0.1024
23.4375 kg/m
mass per unit area
A
=
=
Angular velocity and angular acceleration:
ω
αα
=
=
k
k
ω
Angular momentum about G:
G xz yz z
yz z
III
II
ω ωω
ωω
=−− +
=−+
H i jk
jk
PROBLEM 18.152 (Continued)
Since the mass center lies on the rotation axis,
0=a
m
Σ= + =
= −
FAB a
BA
PROBLEM 18.153
A homogeneous disk of weight
6 lbW=
rotates at the constant
rate
1
16 rad/s
ω
=
with respect to arm ABC, which is welded to
a shaft DCE rotating at the constant rate
2
8 rad/s.
ω
=
Determine the dynamic reactions at D and E.
y z yz A
DD EE m
+ ++ =
j k j ka
PROBLEM 18.153 (Continued)
Resolve into components.
2
22
2
22
/
0
()
()
( )2 ( ) ( )
yy
zz
D D A AD A
yz A A
D E mb c
D E mc b
m
M l E E lcb m
ωω
ωω
+= −
+= +
Σ==+×
+ × + = + +− ×
M H Hr a
i i j kH ijk a
222 2 2
0 2 12 2
22
12 2 2
11
( )2 2 42
1
2
zy
M lE lE m r b c m r lc lb
m r lb lc
ω ωω ω
ωω ω ω
− + = ++ + − −
+ +−
ijk i j
k
i:
222
02
1
4
M mrbc
ω
= ++
k:
22
12 2 2
22
12 2 2
1
22
1
22
y
y
m
E r lb lc
l
m
D r lb lc
l
ωω ω ω
ωω ω ω
= +−
=− +−
j:
22
22 1
22
22 1
1
22
1
22
z
z
m
E lc lb r
l
m
D lc lb r
l
ωω ω
ωω ω
= +−
= ++
6
PROBLEM 18.154
A 48-kg advertising panel of length
2 2.4 ma=
and width 2b
1.6 m=
is
kept rotating at a constant rate
1
ω
about its horizontal axis by a small
electric motor attached at A to frame ACB. This frame itself is kept
rotating at a constant rate
ω
2 about a vertical axis by a second motor
attached at C to the column CD. Knowing that the panel and the frame
complete a full revolution in 6 s and 12 s, respectively, express, as a
function of the angle
,
θ
the dynamic reaction exerted on column CD by
its support at D.
2
2
22
2
0 ( ) sin cos
1sin cos
3
z zz x y xy
xy
II
mb
ω θθ
ω θθ
′ ′′ ′ ′′
′
=−−
= −
PROBLEM 18.154 (Continued)
2 22
12 2
21
cos sin cos
33
mb mb
ωω θ ω θ θ
′
Σ= −Mi k
2 22
21
PROBLEM 18.155
A 2500-kg satellite is 2.4 m high and has octagonal bases of sides
1.2 m. The coordinate axes shown are the principal centroidal axes of
inertia of the satellite, and its radii of gyration are
0.90 m
xz
kk= =
and
0.98 m.
y
k=
The satellite is equipped with a main 500–N thruster
E and four 20-N thrusters A, B, C, and D, which can expel fuel in the
positive y direction. The satellite is spinning at the rate of 36 rev/h
about its axis of symmetry
,
y
G
which maintains a fixed direction in
space, when thrusters A and B are activated for 2 s. Determine (a) the
precession axis of the satellite, (b) its rate of precession, (c) its rate of
spin.
0
2 22
2
2 22
2
00
2 rad 1 h
(36 rev/h) rev 3600 s
0.062832 rad/s
2500 kg
(2500)(0.90) 2025 kg m
2025 kg m
(2500)(0.98) 2401 kg m
( ) (2401)(0.062832) (150.86 kg m /s)
0.6
xx
zx
yy
Gy
m
I mk
II
I mk
I
a
π
ω
ω
=
=
=
= = = ⋅
= = ⋅
= = = ⋅
= = = ⋅
=
Hj j j
m, 0.6 1.2 cos 45° 1.4485b=+=
0
22
( ) ()
150.86 ( 57.941)(2)
(115.88 kg m /s) (150.86 kg m /s)
115.88 0.057225 rad/s 32.788 rev/h
2025
36 rev/h
GG G
x
x
x
y
y
y
t
H
I
H
I
H
ω
ω
= +∆
= +−
=− ⋅+ ⋅
==−=− =−
= =
HH M
ji
ij
(a) Precession axis:
115.88
x
H
22
42.327
4.798
48.693 rev/h
xy
γ
γθ
ω ωω
= °
−= °
= +
=
Law of sines.
sin sin( ) sin
ϕψ ω
γ γθ θ
= =
−
sin 37.529
PROBLEM 18.156
The space capsule has no angular velocity when the jet at A is activated
for 1 s in a direction parallel to the x axis. Knowing that the capsule
has a mass of 1000 kg, that its radii of gyration are
1.00 m
zy
kk= =
and
1.25 m,
z
k=
and that the jet at A produces a thrust of 50 N,
determine the axis of precession and the rates of precession and spin
after the jet has stopped.
Gx
j:
2
62.5 N m s ( ) 62.5 kg m /s
Gy
H− ⋅ ⋅= =− ⋅
2
PROBLEM 18.156 (Continued)
Moments of inertia:
2 22
2 22
2 22
(1000 kg)(1 m) 1000 kg m
(1000 kg)(1.25 m) 1562.5 kg m
xk
yy
zz
I mk
I mk
= = = ⋅
= = = ⋅
Angular velocity vector components:
22
() 0
() 62.5 0.0625 rad/s
1000
() 100 0.0640 rad/s
1562.5
tan 0.97656 44.321
0.089455 rad/s.
Gx
x
x
Gy
y
y
Gz
z
z
y
z
yz
H
I
H
I
H
I
ω
ω
ω
ω
γγ
ω
ω ωω
= =
−
= = = −
−
= = = −
= = = °
= +=
Rates of precession and spin.
sin sin 32.005
sin( ) 0.089455 sin12.316
sin sin 32.005
ω γθ
ψθ
−°
= = °
Since
,
γθ
>
the precession is retrograde.
PROBLEM 18.157
A homogeneous disk of mass m connected at A and B to a fork–ended shaft of
negligible mass which is supported by a bearing at C. The disk is free to rotate
about its horizontal diameter AB and the shaft is free to rotate about a vertical
axis through C. Initially, the disk lies in a vertical plane
0
( 90 )
θ
= °
and the shaft
has an angular velocity
0
8
φ
=
rad/s. If the disk is slightly disturbed, determine
for the ensuring motion (a) the minimum value of
,
φ
(b) the maximum value of
.
θ
2
2
21
2
2 0 10
y
CC
CC
θϕ
θϕ
= −
= +
PROBLEM 18.157 (Continued)
Data:
1
2
22 2
12
0
0
0
1
2
sin cos 1 cos
90
0
16 rad/s
e
e
ee
θθ θ
θ
θ
ϕ
=
=
+=+
= °
=
=
2
1
(1 cos 90 )(8)
8 rad/s
8
1 cos
C
ϕθ
=+°
=
=+
84
2
21
CC
θϕ
= −
(b)
2
max 2 1 min
()
64 (8)(4)
CC
θϕ
= −
= −
2
max
PROBLEM 18.158
The essential features of the gyrocompass are shown. The rotor
spins at the rate
ψ
about an axis mounted in a single gimbal,
which may rotate freely about the vertical axis AB. The angle
formed by the axis of the rotor and the plane of the meridian is
denoted by
,
θ
and the latitude of the position on the earth is
denoted by
.
λ
We note that line OC is parallel to the axis of the
earth, and we denote by
e
ω
the angular velocity of the earth about
its axis.
(a) Show that the equations of motion of the gyrocompass are
22
cos sin cos sin cos 0
ze e
II I
θωω λθ ω λθ θ
′′
+− =
0
z
I
ω
=
where
z
ω
is the rectangular component of the total angular
velocity
ω
along the axis of the rotor, and I and I′ are the
moments of inertia of the rotor with respect to its axis of
symmetry and a transverse axis through O, respectively.
(b) Neglecting the term containing
2
,
e
ω
show that for small values
of
θ
, we have
cos 0
ze
I
I
ωω λ
θθ
+=
′
and that the axis of the gyrocompass oscillates about the
north-south direction.
e ez
PROBLEM 18.158 (Continued)
The angular momentum
O
H
of the rotor is
O xz yy zz
II I
ωωω
=++H i jk
where
and .
xy z
III II
′
= = =
Recalling Eq. (2), we write
O O Oxyz O
cos cos
cos sin sin cos cos
cos sin ( sin )
Oe z
e ee
e ez
I II
II I
ω λ θθ θ ω
ωλθθωλωλθ
ω λ θ θω λ ω
′′
Σ=− + +
+− +
′′
−+
M ijk
i jk
(4)
We observe that the rotor is free to spin about the z axis and free to rotate about the y axis. Therefore,
ze e
Q. E. D
Setting the coefficient of k equal to zero,
( cos sin ) ( sin ) ( cos sin )( sin ) 0
ze e e e
I II
ω ω λ θ θω λ ω λ θθω λ
′′
+− + −− + =
Observing that the last two terms cancel out, we have
PROBLEM 18.158 (Continued)
where the coefficient of sin
θ
is a constant. The rotor, therefore, oscillates about the line NS as a
simple pendulum. For small oscillations,
sin ,
θθ
≈
and Eq. (8) yields
cos 0
ze
I
ωω λ