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PROBLEM 18.56
Determine the rate of change
G
H
of the angular momentum
G
H
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
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
′
′
= = ⋅
= = ⋅
From Eqs. (18.10):
2
2
22
(0.021836)(4) 0.087345 slug ft /s
(0.038820)(3) 0.11646 slug ft /s
0
(0.087345 slug ft /s) (0.11646 slug ft /s)
x xx
y yy
z zz
G
HI
HI
HI
ω
ω
ω
′ ′′
′ ′′
= = =
= = =
= =
′′
= +H ij
PROBLEM 18.56 (Continued)
Components along x and y axes:
34
55
34
(0.087345) (0.11646)
55
0.040761
43
55
43
(0.087345) (0.11646)
55
0.13975
xx y
yxy
HH H
HHH
′′
′′
= −
= −
= −
= +
= +
=
22
( 0.040761 slug ft /s) (0.13975 slug ft /s)
G
=− ⋅+ ⋅H ij
G Axyz
( ) 0 5 ( 0.040761 0.13975 )
(0.20380 ft/lb)
G G Axyz G
= + × = + ×− +
=
HH ΩH j i j
k
PROBLEM 18.57
Determine the rate of change
D
H
of the angular momentum
D
H
of the assembly of Problem 18.3.
PROBLEM 18.3 Two uniform rods AB and CE, each of weight
3 lb and length 2 ft, are welded to each other at their midpoints.
Knowing that this assembly has an angular velocity of constant
magnitude
ω
= 12 rad/s, determine the magnitude and direction
of the angular momentum HD of the assembly about D.
SOLUTION
2
30.093168 lb s /ft, 24 in. 2 ft,
32.2
W
ml
g
=== ⋅==
(12 rad/s)=iω
For rod
ADB,
0, since 0.
Dx x
II
ω
=≈≈Hi
For rod
CDE, use principal axes
, xy
′′
as shown.
9
cos , 41.410
12
θθ
= = °
2
cos 9 rad/s
x
ω ωθ
′
= =
2
sin 7.93725 rad/s
y
ω ωθ
′= =
0
z
ω
′=
0
x
I
′
≈
22
11
(0.093168)(2)
12 12
y
I ml
′
= =
2
0.0310559 lb s ft= ⋅⋅
0 (0.0310559)(7.93725) 0
′
=++j
0.246498 ′
=j
0.246498(sin cos ) 0.163045 0.184874
D
θθ
= += +H ij i j
Let the frame of reference
Dxyz be rotating with angular velocity
(12 rad/s)= = iΩω
Then,
() 0
D D Dxyz D D
= +× =+×HH ΗΗ
Ωω
12 (0.163045 0.184874 )
D
=×+Hi i j
(2.22 lb ft)
D
= ⋅Hk
PROBLEM 18.58
Determine the rate of change
A
H
of the angular momentum
HA of the disk of Problem 18.4.
PROBLEM 18.4 A homogeneous disk of weight
6 lbW=
rotates at the constant rate
1
16
ω
=
rad/s with respect to arm
ABC, which is welded to a shaft DCE rotating at the constant
rate
2
8 rad/s.
ω
=
Determine the angular momentum
A
H
of the
disk about its center A.
PROBLEM 18.59
Determine the rate of change
C
H
of the angular momentum
C
H
of
the disk of Problem 18.5.
PROBLEM 18.5 A thin disk of mass
4 kgm=
rotates at the constant
rate
2
15 rad/s
ω
=
with respect to arm ABC, which itself rotates at the
constant rate
15 rad/s
ω
=
about the y axis. Determine the angular
momentum of the disk about its center C.
PROBLEM 18.60
Determine the rate of change
G
H
of the angular momentum
HG of the disk of Prob. 18.8 for an arbitrary value of
β
,
knowing that its angular velocity
ω
remains constant.
4
G
PROBLEM 18.61
Determine the rate of change
D
H
of the angular momentum HD
of the assembly of Problem 18.3, assuming that at the instant
considered the assembly has an angular velocity
(12=ω
rad/s)i
and an angular acceleration
2
(96 rad/s ) .
α
= − i
PROBLEM 18.62
Determine the rate of change
D
H
of the angular momentum HD
of the assembly of Problem 18.3, assuming that at the instant
considered the assembly has an angular velocity
(12=ω
rad/s)i
and an angular acceleration
2
(96 rad/s ) .=iα
PROBLEM 18.63
A thin homogeneous square of mass m and side a is welded to a vertical
shaft AB with which it forms an angle of
45 .°
Knowing that the shaft
rotates with an angular velocity
ω
=jω
and an angular acceleration
,
α
=jα
determine the rate of change
A
H
of the angular momentum HA of
the plate assembly.
22
2
15
0 ( cos 45 ) ( sin 45 )
12 12
(3 2 )
12
A Axyz x x y y z z
ma ma
ma
αα
α
′′
=+ °+ °
= +
ik
jk
PROBLEM 18.63 (Continued)
With respect to the fixed reference frame,
()
A A Axyz A
= +×
HH H
Ω
PROBLEM 18.64
Determine the rate of change
G
H
of the angular momentum
HG of the disk of Prob. 18.8 for an arbitrary value of
β
,
knowing that the disk has an angular velocity
ω
=
ω
i and an
angular acceleration
α
=
α
i.
24
PROBLEM 18.64 (Continued)
( )( )
2
1cos cos sin
2mr
αβ β β
= +
ij
2
1sin sin cos 0
4
G
44
PROBLEM 18.65
A slender, uniform rod AB of mass m and a vertical shaft CD, each
of length 2b, are welded together at their midpoints G. Knowing that
the shaft rotates at the constant rate
ω
, determine the dynamic
reactions at C and D.
2
22
1cos (sin cos )
3
1sin cos
3
G G Gxyz G B
mb
mb
ω ωββ β
ω ββ
=×+
= −
j ij
k
PROBLEM 18.65 (Continued)
Equations of motion.
We equate the systems of external and effective forces
eff
22
( ):2 ( )
1
2 2 sin cos
3
D D xz G
xz
bC C
bC bC mb
ω ββ
Σ=Σ × + =
− +=−
M M j i kH
ki k
eff :0Σ=Σ + =F F CD
6mb
PROBLEM 18.66
A thin homogeneous triangular plate of weight 10 pounds is welded to a light
vertical axle supported by bearings at A and B. Knowing that the plate rotates
at the constant rate
8 rad/s,
ω
=
determine the dynamic reactions at A and B.
PROBLEM 18.66 (Continued)
Equations of motion. (Weight is omitted for dynamic reactions.)
Eq. (18.28),
() .
A A Axyz A
Σ = +×MH ΩH
2
2
( )0 12 6
1
12
xz x z
mbh mb
h B B hB h
mbh
ω ωω
ω
× + = + ×− + − +
= +
j i k j i j k Bi
k
2
1,0
22
22 2
:3
11 1
,
3 12 4
b
m mx m
mb mb mb
ωω
ωω ω
Σ= + =− =−
=− −− =−
F a AB i i
A i iA i
PROBLEM 18.67
The assembly shown consists of pieces of sheet aluminum of
uniform thickness and of total weight 2.7 lb welded to a light
axle supported by bearings at A and B. Knowing that the
assembly rotates at the constant rate
240
ω
=
rpm, determine
the dynamic reactions at A and B.
PROBLEM 18.67 (Continued)
For calculation of
,
xz
I
use pairs of elements
1
dA
and
2
:dA
21
.dA dA=
2
b
zz
PROBLEM 18.67 (Continued)
32
0 ( 3.4938 10 )(25.133) 1.10346 lb.
−
−− ×
PROBLEM 18.68
The 8-kg shaft shown has a uniform cross section.
Knowing that the shaft rotates at the constant rate
ω
12 rad/s,=
determine the dynamic reactions at A and B.
22
0( )
x xy xz
xz xy
II I
II
ωωωω
ωω
=+× − −
= −
ii jk
jk
Since the shaft lies in the xz plane,
0.
xy
I=
By symmetry, the mass center lies on line AB.
0m=a
eff
: 0 and form a couple.FmΣ= + = =F AB a A BΣ
xz
