PROBLEM 18.29
A circular plate of mass m is falling with a velocity
0
v
and no angular
velocity when its edge C strikes an obstruction. Assuming the impact to be
perfectly plastic
( 0),e=
determine the angular velocity of the plate
immediately after the impact.
/
2
CG R= −r ik
Condition of impact:
( 0) ( ) 0
Cy
ev= =
Kinematics:
/C CG
=+×vvωr
() () ( ) ( )
2
()
22 2
Cx Cz y x y z
yxy z
R
vvv
RR R
v
ωωω
ωω ω
+ =+ ++ × −
= + − ++
i k j i j k ik
j j ki j
:0 ( )
2
y xz
R
v
ωω
=++j
()
2
y xz
R
v
ωω
=−+
PROBLEM 18.29 (Continued)
Moments about G:
/
22 2
0 () () ()
( )( )
2
111
2
CG Gx Gy Gz
xx yy zz
xyz
Ct H H H
RCt I I I
RC t mR mR mR
ωωω
+ ×∆ = + +
−×∆ = + +
r ji jk
ik j i j k
i:
2
1
4
2
x
RC t mR
ω
∆=
(3)
j:
2
1
02y
mR
ω
=
0
y
ω
=
2
1
RC t mR
0
2
xz
0
2 2 2( )
x xz
v
R
ω ωω
= −+
0
3 2 22
xz
v
R
ωω
+=
(5)
0
2 2 2( )
z xz
v
R
ω ωω
= −+
0
2 3 22
xz
v
R
ωω
+=
(6)
xyz
5
R
PROBLEM 18.30
For the plate of Problem 18.29, determine (a) the velocity of its mass center
G immediately after the impact, (b) the impulse exerted on the plate by the
obstruction during the impact.
/
2
CG R= −r ik
Condition of impact:
( 0) ( ) 0
Cy
ev= =
Kinematics:
/C CG
=+×vvωr
() () ( ) ( )
2
()
22 2
Cx Cz y x y z
yxy z
R
vvv
RR R
v
ωωω
ωω ω
+ =+ ++ × −
= + − ++
i k j i j k ik
j j ki j
:0 ( )
2
y xz
R
v
ωω
=++j
()
2
y xz
R
v
ωω
=−+
PROBLEM 18.30 (Continued)
Moments about G:
/
22 2
0 () () ()
( )( )
2
111
2
CG Gx Gy Gz
xx yy zz
xyz
Ct H H H
RCt I I I
RC t mR mR mR
ωωω
+ ×∆ = + +
−×∆ = + +
r ji jk
ik j i j k
i:
2
1
4
2
x
RC t mR
ω
∆=
(3)
j:
2
1
02y
mR
ω
=
0
y
ω
=
2
1
RC t mR
0
2
xz
0
2 2 2( )
x xz
v
R
ω ωω
= −+
0
3 2 22
xz
v
R
ωω
+=
(5)
0
2 2 2( )
z xz
v
R
ω ωω
= −+
0
2 3 22
xz
v
R
ωω
+=
(6)
0
5
PROBLEM 18.31
A square plate of side a and mass m supported by a ball-and–socket
joint at A is rotating about the y axis with a constant angular
velocity
0
ω
=j
ω
when an obstruction is suddenly introduced at B
in the xy plane. Assuming the impact at B to be perfectly plastic
(e
0),=
determine immediately after impact (a) the angular velocity
of the plate, (b) the velocity of its mass center G.
1()
2
z zy
a
ω ωω
′′
′′
= −+
i jk
Also,
2
/
1( 2)
4
GA y y z
m ma
ωω ω
′′ ′
′′ ′
×= − + +r v ij k
22
11
12 6
G xx yy zz y z
PROBLEM 18.31 (Continued)
Principle of impulse–momentum.
Moments about A:
0
( ) ( )( )
AA
a Ft+− × ∆ =H j kH
0/ 0 /
() ( )
G GA G GA
m aF t m+× −∆=+×Hrv iHrv
0
24 4 y
2 22
00
1 11 2
:2
24 12 4 8
y yy
ma ma ma
ω ω ωω ω
′ ′′
′=+=j
22
11
0
2 16
y
0
PROBLEM 18.32
Determine the impulse exerted on the plate of Problem 18.31 during
the impact by (a) the obstruction at B, (b) the support at A.
PROBLEM 18.31 A square plate of side a and mass m supported
by a ball–and–socket joint at A is rotating about the y axis with a
constant angular velocity
0
ω
=j
ω
when an obstruction is suddenly
introduced at B in the xy plane. Assuming the impact at B to be
perfectly plastic (e
0),=
determine immediately after impact (a) the
angular velocity of the plate, (b) the velocity of its mass center G.
1()
2
z zy
a
ω ωω
′′
′′
= −+
i jk
Also,
2
/
1( 2)
4
GA y y z
m ma
ωω ω
′′ ′
′′ ′
×= − + +r v ij k
22
11
12 6
G xx yy zz
yz
ma ma
ωω
′′
′′
= +
jk
PROBLEM 18.32 (Continued)
Principle of impulse–momentum.
Moments about A:
0
( ) ( )( )
AA
a Ft
′
+− × ∆ =H j kH
0/ 0 /
()
G GA G GA
m aF t m
′
+× +∆=+×Hrv iHrv
PROBLEM 18.33
The coordinate axes shown represent the principal centroidal axes of inertia of a 3000–lb space probe whose radii
of gyration are
1.375 ft,
x
k=
1.425 ft,
y
k=
and
1.250 ft.
z
k=
The probe has no angular velocity when a 5-oz
meteorite strikes one of its solar panels at Point A with a velocity
0
(2400 ft/s) (3000 ft/s)=−+v ij
(3200 ft/s)k
relative to the probe. Knowing that the meteorite emerges on the other side of the panel with no change in the
direction of its velocity, but with a speed reduced by 20 percent, determine the final angular velocity of the
probe.
SOLUTION
2
3000 93.17 lb s /ft
PROBLEM 18.33 (Continued)
Let
A
H
be the angular momentum of the probe and
m′
be its mass. Conservation of angular momentum
about the origin for a system of particles consisting of the probe plus the meteorite:
PROBLEM 18.34
The coordinate axes shown represent the principal centroidal axes of inertia of a 3000-lb space probe whose
radii of gyration are
1.375 ft,
x
k=
1.425 ft,
y
k=
and
1.250 ft.
z
k=
The probe has no angular velocity when a
5-oz meteorite strikes one of its solar panels at Point A and emerges on the other side of the panel with no
change in the direction of its velocity, but with a speed reduced by 25 percent. Knowing that the final angular
velocity of the probe is
(0.05 rad/s) (0.12 rad/s)
z
ω
=−+i jkω
and that the x component of the resulting
change in the velocity of the mass center of the probe is
0.675 in./s,−
determine (a) the component
z
ω
of the
final angular velocity of the probe, (b) the relative velocity
0
v
with which the meteorite strikes the panel.
SOLUTION
Masses: Space probe:
2
3000 93.17 lb s /ft
32.2
m′= = ⋅
2
50.009705 lb s /ft
PROBLEM 18.34 (Continued)
Initial linear momentum of the space probe,
(lb s):⋅
0
0m′′=v
Final linear momentum of the space probe,
(lb s):⋅
0.675
xyz yz
(lb s):⋅
PROBLEM 18.35
A 1200-kg satellite designed to study the sun has an angular velocity
0
ω
= (0.050 rad/s)i + (0.075 rad/s)k when two small jets are activated
at A and B in a direction parallel to the y axis. Knowing that the
coordinate axes are principal centroidal axes, that the radii of gyration
of the satellite are
1.120 m,
x
k=
1.200 m,
y
k=
0.900 m,
z
k=
and that
each jet produces a 50–N thrust, determine (a) the required operating
time of each jet if the angular velocity of the satellite is to be reduced
to zero, (b) the resulting change in the velocity of the mass center G.
PROBLEM 18.35 (Continued)
Solving (1) and (2) simultaneously,
( )
6.4518 N s,∆=− ⋅
A
Ft
( )
54.298 N s∆=− ⋅
B
Ft
Since both impulse components are negative, the 50-N thrusts of jets A and B act in the negative y direction.
PROBLEM 18.36
If jet A in Prob. 18.35 is inoperative, determine (a) the required operating
time of jet B to reduce to zero the x component of the angular velocity of
the satellite, (b) the resulting final angular velocity, (c) the resulting
change in the velocity of the mass center G.
0
Principal radii of gyration:
1.120 m, 1.200 m, 0.900 m
x yz
k kk= = =
Jet thrust
50 N=
parallel to the y axis.
Principal moments of inertia.
2
22
Final angular momentum if
( )
2
0.
x
=ω
( ) ( )
( )
( )
2
22
ωωω
=++
G xx yy zz
II IH i jk
PROBLEM 18.36 (Continued)
Principle of impulse and momentum. Moments about G:
( )
( )
( )
/G
12
G BG B
Ft+ ×∆ =Hr jΗ
2
972
z
Since
( )
B
Ft∆
is negative, the 50-N thrust of jet B acts in the negative y direction.
(a) Operating time of jet B.
( )
47.04
∆−
B
Ft
PROBLEM 18.37
Denoting, respectively, by
,ω
,
O
H
and T the angular velocity, the angular momentum, and the kinetic energy
of a rigid body with a fixed Point O, (a) prove that
2;
O
T⋅=Hω
(b) show that the angle
θ
between
ω
and
HO will always be acute.
PROBLEM 18.38
Show that the kinetic energy of a rigid body with a fixed Point O can be
expressed as
2
1
2
,
OL
TI
ω
=
where
ω
is the instantaneous angular velocity of
the body and
OL
I
is its moment of inertia about the line of action OL of
.
ω
Derive this expression (a) from Eqs. (9.46) (or Eq. B.19 in the Appendix) and
(18.19), (b) by considering T as the sum of the kinetic energies of particles
i
P
describing circles of radius
i
r
about line OL.
2OL
PROBLEM 18.39
Determine the kinetic energy of the disk of Problem 18.1.
PROBLEM 18.1 A thin, homogeneous disk of mass m and radius r
spins at the constant rate
1
ω
about an axle held by a fork-ended vertical
rod, which rotates at the constant rate
2
.
ω
Determine the angular
momentum
G
H
of the disk about its mass center G.
21
8
PROBLEM 18.40
Determine the kinetic energy of the plate of Problem 18.2.
PROBLEM 18.2 A thin rectangular plate of weight 15 lb rotates about
its vertical diagonal AB with an angular velocity
ω
. Knowing that the
z axis is perpendicular to the plate and that
ω
is constant and equal to
5 rad/s, determine the angular momentum of the plate about its mass
center G.
12 0.8(5 rad/s) 4 rad/s
15
90.6(5 rad/s) 3 rad/s
15
0
x
y
z
ωω
ωω
ω
′
′
= = =
= = =
=
2
2
2
1 15 lb 9
ft 0.021836 slug ft
12 32.2 12
1 15 lb 12 ft 0.038820 slug ft
12 32.2 12
x
y
I
I
′
′
= = ⋅
= = ⋅