PROBLEM 18.68 (Continued)
2
aa a a
PROBLEM 18.69
After attaching the 18–kg wheel shown to a balancing machine and
making it spin at the rate of 15 rev/s, a mechanic has found that to balance
the wheel both statically and dynamically, he should use two corrective
masses, a 170–g mass placed at B and a 56-g mass placed at D. Using a
right-handed frame of reference rotating with the wheel (with the z axis
perpendicular to the plane of the figure), determine before the corrective
masses have been attached (a) the distance from the axis of rotation to the
mass center of the wheel and the products of inertia
xy
I
and
,
zx
I
(b) the
force–couple system at C equivalent to the forces exerted by the wheel on
the machine.
zx
PROBLEM 18.69 (Continued)
2 32
(18)( 1.15267 10 )(94.248)
(184.3 N)
n
mm
ω
−
′= =− =−− ×
=
Fa r j
j
C x xy xz
II I
ωωω
=−−H i jk
22
C C C xz xy
II
ωω
′==×= −
MH H j k
ω
CC
C
PROBLEM 18.70
When the 18-kg wheel shown is attached to a balancing machine and
made to spin at a rate of 12.5 rev/s, it is found that the forces exerted by
the wheel on the machine are equivalent to a force–couple system
consisting of a force
(160 N)=Fj
applied at C and a couple
(14.7 N m) ,
C
= ⋅Mk
where the unit vectors form a triad which rotates
with the wheel. (a) Determine the distance from the axis of rotation to the
mass center of the wheel and the products of inertia
xy
I
and
.
zx
I
(b) If
only two corrective masses are to be used to balance the wheel statically
and dynamically, what should these masses be and at which of the Points
A, B, D, or E should they be placed?
PROBLEM 18.70 (Continued)
(b) Positions for the balance masses.
182 mm 0.182 m
AB E D
yy y y= =−=−= =
75 mm 0.075 m
ABE D
xxx x−= = =−= =
B
AD A D
Then
0.1425
AD E
mm−=−
AD E
Solving,
0.01605 kg, 0.1585 kg
AD E
mm= =
A
PROBLEM 18.71
Knowing that the assembly of Problem 18.65 is initially at rest
(
ω
0)=
when a couple of moment
00
=M Mj
is applied to shaft CD,
determine (a) the resulting angular acceleration of the assembly,
(b) the dynamic reactions at C and D immediately after the couple is
applied.
3
G G Gxyz mb
PROBLEM 18.71 (Continued)
Equations of motion: Equivalence of applied and effective forces.
2 22
eff 0
11
( ) : 2 ( ) sin cos cos
33
D D xz
b C C M mb mb
αβ β α β
Σ=Σ × + + = +M M j ik j i j
2 22
11
i:
2
1
2 sin cos
3
z
bC mb
αβ β
=
(3)
j:
22
0
1cos
3
M mb
αβ
=
(4)
x
0
22
3
11
sin cos sin cos
66 cos
z
M
C mb mb mb
αβ β β β β
= =
0
M
x
0
tan
2
M
b
β
=
Ck
eff
0
2
M
b
PROBLEM 18.72
Knowing that the plate of Problem 18.66 is initially at rest
( 0)
ω
=
when a couple
of moment
0
(0.75 ft= ⋅M
lb)j is applied to it, determine (a) the resulting
angular acceleration of the plate, (b) the dynamic reactions A and B immediately
after the couple has been applied.
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 18.72 (Continued)
Equations of motion. We first use Eq. (18.28):
M()
A A Axyz A
Σ = +×HH
Ω
A Axyz mbh mb
Substituting for
and ( )
A A Axyz
HH
into Eq. (18.28), noting that
0,
ω
=
and computing
A
ΣM
from diagram:
2
0
11
12 6
xz
2
0
11
12 6
xz
M hB hB mbh mb
αα
− +=− +jk i i j
Equating the coefficients of the unit vectors:
j:
2
0
1
6
M mb
α
=
0
2
6M
mb
α
=
(a)
2
10 lb 2
32.2
6(0.75 lb ft) 14.49 rad/s
( )(1 ft)
α
⋅
= =
2
(14.49 rad/s )=j
α
k:
0
x
hB−=
0
x
B=
i:
1 1 1 10 lb
, (1)(14.49)
12 12 12 32.2
zz
hB mbh B mb
αα
=− =−=−
4 4 32.2
PROBLEM 18.73
The assembly of Problem 18.67 is initially at rest (
ω
0)=
when a couple
0
M
is applied to axle AB. Knowing that the
resulting angular acceleration of the assembly is
2
(150 rad/s ) ,=αi
determine (a) the couple
0
,M
(b) the
dynamic reactions at A and B immediately after the couple is
applied.
SOLUTION
2
2.7 0.08385 lb s /ft
0.75
A
Moments and products of inertia:
mass (area)
()II
r
=
44 4
44
32
112
333
22
(0.1118)(0.5)
33
4.658 10 lb s ft
x
x
Ibb b
Ib
r
−
=+=
= =
= × ⋅⋅
22 4
44
31 5 1 1
() ()
22 2 2 2
11
xy
Ibbbbb b b
= + −=−
PROBLEM 18.73 (Continued)
For calculation of
,
xz
I
use pairs of elements
1
dA
and
2
:dA
21
dA dA=
2
1 21
0
22
b
xz
zz
But
.zx=
23 4 4 4
0
21 5
a
xz
4 4 32
55
(0.1118)(0.5) 2.9115 10 lb s ft
12 12
xz
Ib
r
−
=− =− =− × ⋅⋅
3 3 32
− −−
A A A Axyz A
0
4( ) ( )
y z x xy xz x xy xz
M bB B I I I I I I
α α αω ω ω ω
+ × + = − − +× − −i i j k i j ki i j k
22
0
4 4 ( )( )
z y x xy xz xz xy
M bB bB I I I I I
α αω αω
− + =−− −+i j ki j k
PROBLEM 18.74
The shaft of Problem 18.68 is initially at rest
( 0)
ω
=
when a couple M0 is applied to it. Knowing that the
resulting angular acceleration of the shaft is
2
(20 rad/s ) ,=αi
determine (a) the couple M0, (b) the
dynamic reactions at A and B immediately after the
couple is applied.
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.
x xy xz
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=Σ + = =F AB a A B
PROBLEM 18.74 (Continued)
Calculation of
x
I
and
.
xz
I
Divide the shaft into eight segments, each of length
200 mm 0.2 ma= =
Let
m′
be the mass of one segment.
8 kg 1 kg
x
Total:
2
10
3
x
I ma
′
=
For , , , and ,
0
xz
I=
2
aa a a
PROBLEM 18.75
The assembly shown weighs 12 lb and consists of 4 thin 16-in.–
diameter semicircular aluminum plates welded to a light 40–in.–
long shaft AB. The assembly is at rest
( 0)
ω
=
at time t = 0
when a couple
0
M
is applied to it as shown, causing the
assembly to complete one full revolution in 2 s. Determine
(a) the couple M0, (b) the dynamic reactions at A and B at
0.t=
PROBLEM 18.75 (Continued)
23
22
42
42
42
0.37267 0.13345 lb s /ft
2 2 (0.66667)
1
(0.13345) (0.66667) 0.041407 lb s ft
2
4
(0.13345) (0.66667) 0.035147 lb s ft
3
4
(0.13345) (0.66667) 0.035147 lb s ft
3
x
xy
xz
m
r
I
I
I
rππ
π
= = = ⋅
= = ⋅⋅
= − =− ⋅⋅
= − =− ⋅⋅
Since the mass center lies on the rotation axis,
0=a
eff
0mΣ= + =Σ = =FAB F a
= −AB
00
() 2 ( )
C yz
M b b M bB BΣ = +− × + × = + × +M i i A iB i i j k
0
22
zy
M bB bB=−+ijk
where
2 40 in. 3.3333 ftb= =
CC
MΣ=H
Resolve into components.
(a) i:
0
(0.041407)(3.1416)
x
MI
α
= =
0
(0.1301 lb ft)= ⋅Mi
(b) j:
2
2
z xy xz
bB I I
αω
− =−+
( 0.035147)(3.1416) 0 0.0331 lb
3.3333
z
B−− +
=−=−
0.0331 lb
z
A=
k:
2
2
y xz xy
bB I I
αω
=−−
( 0.035147)(3.1416) 0 0.0331 lb
3.3333
y
B−+
=−=
0.0331 lb
y
A= −
(0.0331 lb) (0.0331 lb)=−+A jk
(0.0331 lb) (0.0331 lb)= −B jk
PROBLEM 18.76
For the assembly of Problem 18.75, determine the dynamic
reactions at A and B at
2 s.t=
PROBLEM 18.75 The assembly shown weighs 12 lb and
consists of four thin 16-in.–diameter semicircular aluminum
plates welded to a light 40–in.-long shaft AB. The assembly is at
rest
( 0)
ω
=
at time t
0=
when a couple
0
M
is applied to it as
shown, causing the assembly to complete one full revolution in
2 s. Determine (a) the couple M0, (b) the dynamic reactions at
A and B at
0.t=
PROBLEM 18.76 (Continued)
23
22
0.37267 0.13345 lb s /ft
2 2 (0.66667)
m
r
rππ
= = = ⋅
42
1
PROBLEM 18.77
The sheet–metal component shown is of uniform thickness
and has a mass of 600 g. It is attached to a light axle
supported by bearings at A and B located 150 mm apart. The
component is at rest when it is subjected to a couple M0 as
shown. If the resulting angular acceleration is
α
2
(12 rad/s ) ,=k
determine (a) the couple M0, (b) the
dynamic reactions at A and B immediately after the couple
has been applied.
y yz xz
zz
MI
α
Σ=
(3)
Calculation of the required moment and products of inertia.
22
Let
r
be the mass per unit area.
2
35.556 kg/m
m
A
r
= =
The component is comprised of 3 parts: triangle , triangle , and rectangle as shown. Let the lengths of
75 mm be labeled b.
el el
xz yz
PROBLEM 18.77 (Continued)
Coordinates of the element mass center:
el el el
1
,,
2
x b y y zz=−= =
PROBLEM 18.77 (Continued)
Triangle . The equation of the lower edge is
.
2
b
yz
=−−
xz yz z
Rectangle . Area:
32
(0.150)(0.075) 11.25 10 mA
−
= = ×
Mass:
3
400 10 kgmA
r
−
= = ×
12 12
z
Totals.
62
187.5 10 kg m
xz xz
II
−
=Σ= × ⋅
62
93.75 10 kg m
yz yz
II
−
=Σ=− × ⋅
62
0
zz
0