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1180
21–69.
The top has a mass of 90 g, a center of mass at G,and a radius
of gyration about its axis of symmetry.About
any transverse axis acting through point Othe radius of
gyration is .If the top is connected to a ball-and-
socket joint at Oand the precession is ,
determine the spin .Vs
vp=0.5 rad>s
kt=35 mm
k=18 mm
SOLUTION
Ans.vs=c=3.63
A
103
B
rad>s
+0.090(0.018)2(0.5)(0.7071)
C
0.5(0.7071) +c
#
D
0.090(9.81)(0.06) sin 45° =-0.090(0.035)2(0.5)2(0.7071)2
©Mx=-If
#2sin ucos u+Izf
#
sin uaf
#
cos u+c
#b
vp=0.5 rad>s
Vp
Vs
60 mm
G
O
45
21–70.
SOLUTION
undergoes steady precession.
and
.
Thus,
©Mx=-If
#2sin ucos u+Izf
#
sin u
A
f
#
cos +c
#
B
=3.4507
A
10-3
B
slug #ft2
I=Ix=Iy=a1
32.2 ba4
12 b2
Iz=a1
32.2 ba1
12 b2
=215.67
A
10-6
B
slug #ft2
#
#
The 1-lb top has a center of gravity at point G. If it spins
about its axis of symmetry and precesses about the vertical
axis at constant rates of and ,
respectively, determine the steady state angle .The radius
of gyration of the top about the zaxis is ., and
about the xand yaxes it is .kx=ky=4in
kz=1in
u
vp=10 rad>svs=60 rad>s
y
x
O
z
3 in.
vp10 rad/s
u
G
vs60 rad/s
-1 sin u(0.25) =-3.4507
A
10-3
B
(-10)2sin ucos u+215.67
A
10-6
B
(-10) sin u[(-10) cos u+60]
1182
21–71.
The space capsule has a mass of 2 Mg, center of mass at G,
and radii of gyration about its axis of symmetry (zaxis) and
its transverse axes (xor yaxis) of and
, respectively. If the capsule has the angular
velocity shown, determine its precession and spin .
Indicate whether the precession is regular or retrograde.
Also, draw the space cone and body cone for the motion.
c
#
f
#
kx=ky=5.5 m
kz=2.75 m
SOLUTION
.Thus,
(1)
(2)
Solving Eqs. (1) and (2),
Using these results,
Ans.
Ans.
Since , the motion is regular precession.Ans.I7Iz
=212 rad>s
c
#
=I-Iz
IIz
HGcos u=
B
60 500 -15 125
60 500(15125)
R
4.9446
A
106
B
cos 30°
f
#
=HG
I=HG
60 500 =
4.9446
A
106
B
60 500 =81.7 rad>s
HG=4.9446
A
106
B
kg #m2>su=66.59°
HGcos u=1 964 795.13
150 cos 30° =HGcos u
15 125
vz=HGcos u
Iz
HGsin u=4 537 500
150 sin 30° =HGsin u
60 500
vy=HGsin u
I
=60 500 kg #m2
y
x
G
30
v150 rad/s
*21–72.
The 0.25 kg football is spinning at vz
=15 rad>s
as shown.
If
u=40°
, determine the precession about the zaxis. The
radius of gyration about the spin axis is
k
z
=0.042 m
, and
about a transverse axis is
k
y
=0.13 m
.
SOLUTION
#
2
2
2
Z
G
z
vz 15 rad/s
1184
21–73.
S
OLUTION
F
rom Eq. 21–34 and Hence
H
owever, and
Q.E.D.tan u=I
Iz
tan b
vy
vz
=tan b=Iz
Itan u
vz=vcos bvy=vsin b
vy
vz
=Iz
Itan uvz=HGcos u
Iz
vy=HGsin u
I
T
he projectile shown is subjected to torque-free motion.
T
he transverse and axial moments of inertia are Iand ,
r
espectively.If represents the angle between the
p
recessional axis Zand the axis of symmetry z,and
i
s the angle between the angular velocity and the
z
axis,show that and are related by the equation
.tan u=(I>Iz) tan b
ub
V
b
u
Iz
G
Z
u
V
y
xz
b
I
z
1185
21–74.
SOLUTION
Use the result of Prob. 21–75.
Using the law of sines:
Ans.c
#
=
2.35 rev
>
h
sin 9.189°
2=sin (20° -9.189°)
c
#
b=9.189°
tan 20° =a1600(1.8)2
1600(1.2)2btan b
tan u=aI
Izbtan b
I=1600(1.8)2,Iz=1600(1.2)2
The radius of gyration about an axis passing through the
axis of symmetry of the 1.6-Mg space capsule is
and about any transverse axis passing through the center of
mass G, If the capsule has a known steady-state
precession of two revolutions per hour about the Zaxis,
determine the rate of spin about the zaxis.
kt=1.8 m.
kz=1.2 m,
Gz
Z
20
21–75.
The rocket has a mass of 4 Mg and radii of gyration
and It is initially spinning about the
zaxis at when a meteoroid Mstrikes it at A
and creates an impulse Determine the axis
of precession after the impact.
I=5300i6N#s.
vz=0.05 rad>s
ky=2.3 m.kz=0.85 m
SOLUTION
Since
then
The axis of precession is defined by HG.
Thus,
Ans.
Ans.
Ans.g=cos-1(0.159) =80.9°
b=cos-1(0.9874) =9.12°
a=cos-1(0) =90°
uHG=900j+144.5k
911.53 =0.9874j+0.159k
HG=900j+[4000(0.85)2](0.05)k=900j+144.5k
vz=0.05 rad>s
GA
M
z
x
y
3m
z
ω
*21–76.
SOLUTION
, and .
Thus,
Ans.
Ans.
Also,
Thus,
Ans.b=tan-1
¢
vy
vz
≤
=tan-1a12.57
34.92 b=19.8°
vz=HGcos u
Iz
=0.02 cos 45°
0.405
A
10-3
B
=34.92 rad>s
vy=HGsin u
I=0.02 sin 45°
1.125
A
10-3
B
=12.57 rad>s
=22.35 rad>s=22.3 rad>s
c
#
=I-Iz
IIz
HGcos u=
1.125
A
10-3
B
-0.405
A
10-3
B
1.125
A
10-3
B
(0.405)
A
10-3
B
(0.02) cos 45°
f
#
=HG
I=0.02
1.125
A
10-3
B
=17.78 rad>s=17.8 rad>s
u=45°=1.125
A
10-3
B
kg #m2
z
y
x
45
V
B
HG0.02 kg m2/s
vector kes
k#2f
b
c. HG = 0.02 g m >s, determine its precession and spin
Also, find the angle that the angular velocity ma
with the z axis.
k
The football has a mass of 450 g and radii of gyration
about its axis of symmetry (z axis) and its transverse axes
(x or y axis) of
respectively. If the football has an angular momentum of
kz = 30mm and kx = y = 50mm,
1188
21–77.
The satellite has a mass of 1.8 Mg, and about axes passing
through the mass center G the axial and transverse radii of
gyration are
k
z
=0.8 m
and
kt=1.2 m,
respectively. If it is
spinning at vs
=6 rad>s
when it is launched, determine its
angular momentum. Precession occurs about the Z axis.
SOLUTION
I=1800(1.2)2=2592 kg #m2
Iz=1800(0.8)2=1152 kg #m2
Applying the third of Eqs. 21–36 with
u=5°
c
#
=6 rad>s
c
#
=
I-I
z
Hz
HG cos u
0=
2592 -1152
2592(1152) HG cos 5°
HG=12.5 Mg #m2>s
Ans.
5
vs
z
G
Z
1189
21–78.
The radius of gyration about an axis passing through the
axis of symmetry of the 1.2 Mg satellite is ,
known spin of about the zaxis, determine the
steady-state precession about the zaxis.
2700 rev>h
kz=1.4 m
SOLUTION
The moment inertia of the satelite about the zeht dna si sixa
. si sixa esrevsnart sti tuoba etiletas eht fo aitreni tnemom
Applying the third of Eq.21–36 with ,we have
Applying the second of Eq. 21–36, we have
Ans.f
#
=HG
I=19.28(103)
5808 =3.32 rad s
HG=19.28
A
103
B
kg #m2>s
1.5p=c5808 -2352
5808(2352) dHGcos 15°
c
#
=I-Iz
II
z
HGcos u
u=15°
I=1200
A
2.202
B
=5808 kg #m2
Iz=1200
A
1.42
B
=2352 kg #m2
#
15°
z
Z
G
and about any transverse axis passing through the
center of mass G, If the satelite has a
kt=2.20 m.
