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PROBLEM 15.51
In the simplified sketch of a ball bearing shown, the diameter of the inner
race A is 60 mm and the diameter of each ball is 12 mm. The outer race B
is stationary while the inner race has an angular velocity of 3600 rpm.
Determine (a) the speed of the center of each ball, (b) the angular velocity
of each ball, (c) the number of times per minute each ball describes a
complete circle.
SOLUTION
2 226.19
π
PROBLEM 15.52
A simplified gear system for a mechanical watch is
shown. Knowing that gear A has a constant angular
velocity of 1 rev/h and gear C has a constant angular
velocity of 1 rpm, determine (a) the radius r, (b) the
magnitudes of the accelerations of the points on gear B
that are in contact with gears A and C.
SOLUTION
Point where A contacts B:
AA
r
ω
n
PROBLEM 15.53
Arm ACB rotates about Point C with an angular velocity of
40 rad/s counterclockwise. Two friction disks A and B are
pinned at their centers to arm ACB as shown. Knowing that the
disks roll without slipping at surfaces of contact, determine the
angular velocity of (a) disk A, (b) disk B.
SOLUTION
Disk A: Plane motion = Translation with A + Rotation about A.
/
12 mm,
A E A EA
r= = −v vv
1440
960=
12 A
ω
+
1440 960 200 rad/s
12
A
ω
+
= =
(a)
200 rad/s
A
=ω
(b)
24.0 rad/s
B
=ω
PROBLEM 15.54
Arm ACB rotates about Point C with an angular velocity of
40 rad/s counterclockwise. Two friction disks A and B are
pinned at their centers to arm ACB as shown. Knowing that the
disks roll without slipping at surfaces of contact, determine the
angular velocity of (a) disk A, (b) disk B.
SOLUTION
Arm : Fixed axis rotation.ACB
//
0.3 in.,
AC A AC AB
rr
ω
= =v
(0.3)(40)=
12 in./s=
//
1.8 in.,
BC B BC AB
rr
ω
= =v
(1.8)(40)=
72 in./s=
Disk B: Plane motion = Translation with B + Rotation about B.
/
0.6 in.,
B D B BA
r= = −v vv
0 72=
0.6 B
ω
+
72 120 rad/s
0.6
B
ω
= =
(a)
104.0 rad/s
A
=ω
(b)
120.0 rad/s
B
=ω
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.55
Knowing that at the instant shown the velocity of collar A is
900 mm/s to the left, determine (a) the angular velocity of rod ADB,
(b) the velocity of Point B.
SOLUTION
B
PROBLEM 15.56
Knowing that at the instant shown the angular velocity of rod DE is
2.4 rad/s clockwise, determine (a) the velocity A, (b) the velocity of
Point B.
SOLUTION
B
PROBLEM 15.57
Knowing that the disk has a constant angular velocity of 15 rad/s clockwise,
determine the angular velocity of bar BD and the velocity of collar D when
(a)
0,
θ
=
(b)
90 ,
θ
= °
(c)
180 .
θ
= °
SOLUTION
PROBLEM 15.57 (
Continued)
()c
180 .
θ
= °
42 in./s
B
v=
2.8
sin , 16.26
10
ββ
= = °
Bar :BD
/D B DB
= +v vv
D
v
[
42=
]
/DB
v
+
β
10
DB
DB
4.38 rad/s
DB
=ω
tan
DB
vv
β
=
12.25 in./s
D
=v
PROBLEM 15.58
The disk has a constant angular velocity of 20 rad/s clockwise.
(a) Determine the two values of the angle
θ
for which the velocity of collar D is
zero. (b) For each of these values of
,
θ
determine the corresponding value of the
angular velocity of bar BD.
SOLUTION
2.8 in., 10 in.
A BD
rl= =
From geometry, sin sin
BD A A
l rr
βθ
= +
(1)
is tangent to the circular path of ,
B
Bv
thus
B AA
r
ω
=v
θ
For rod BD
/B D BD BD
l
ω
=v
β
//
0
B D BD BD
=+=+vvv v
/BD B
=vv
For
,
βθ
=
sin sin so that sin sin
BD A A
l rr
βθ β β
= = +
2.8
sin , 22.9 ,
10 2.8
A
BD A
r
lr
ββ
= = = °
−−
22.9
θ
= °
For
180 , sin sin ,
θ βθ β
= °+ =−
sin sin
BD A A
l rr
ββ
= −
2.8
sin , 12.6
10 2.8
A
BD A
r
lr
ββ
= = = °
++
192.6
θ
= °
( ) For matching magnitudesb
/DB B
vv=
( )( )
2.8 20
, 5.6 rad/s
10
AA
BD BD A A BD
BD
r
lr l
ω
ω ωω
= = = =
For 22.9 ,
θ
= °
5.60 rad/s
BD =ω
For 192.6 ,
θ
= °
5.60 rad/s
BD
=ω
PROBLEM 15.59
The test rig shown was developed to perform fatigue testing
on fitness trampolines. A motor drives the 9-in radius
flywheel AB, which is pinned at its center point A, in a
counterclockwise direction. The flywheel is attached to slider
CD by the 18-in connecting rod BC. Knowing that the “feet”
at D should hit the trampoline twice every second, at the
instant when
θ
= 30°, determine (a) the angular velocity of the
connecting rod BC, (b) the velocity of D, (c) the velocity of
midpoint CB.
SOLUTION
E
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.60
In the eccentric shown, a disk of 2-in.-radius revolves about shaft O
that is located 0.5 in. from the center A of the disk. The distance
between the center A of the disk and the pin at B is 8 in. Knowing that
the angular velocity of the disk is 900 rpm clockwise, determine the
velocity of the block when
30 .
θ
= °
SOLUTION
B=v
PROBLEM 15.61
In the engine system shown,
160 mml=
and
60 mm.b=
Knowing that the crank
AB rotates with a constant angular velocity of 1000 rpm clockwise, determine the
velocity of the piston P and the angular velocity of the connecting rod when
(a)
0,
θ
=
(b)
90 .
θ
= °
SOLUTION
AB
(1000)(2 ) 104.72 rad/s
60
π
(a)
0.
θ
= °
Crank AB. (Rotation about A)
/0.06 m
BA =r
.Rod BD
(Plane motion = Translation with B + Rotation about B)
/D B DB
v= +vv
D
v
[6.2832=
]
/
[
DB
v+
]
/
0
6.2832 m/s
D
DB
v
v
=
=
vv
PD
=
0
P=v
6.2832
0.16
B
BD
v
l
ω
= =
39.3 rad/s
BD
=ω
PD
P=v
/(0.06)(104.72) 6.2832 m/s
B B A AB
r
ω
= = =v
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.62
In the engine system shown
160 mml=
and b
60 mm.=
Knowing that crank AB
rotates with a constant angular velocity of 1000 rpm clockwise, determine the
velocity of the piston P and the angular velocity of the connecting rod when
60 .
θ
= °
SOLUTION
(1000)(2 )
1000 rpm 104.72 rad/s
60
AB
π
ω
= = =
.
60
θ
= °
Crank AB. (Rotation about A)
/
3 in.
BA
=r
30°
/(0.06)(104.72) 6.2832 m/s
B B A AB
r
ω
= = =v
60°
Rod BD. (Plane motion = Translation with B + Rotation about B.)
Geometry.
sin sinlr
βθ
=
0.06
sin sin sin 60
0.16
18.95
r
l
βθ
β
= = °
= °
/D B DB
= +v vv
[D
v
] [314.16=
/
60 ] [ DB
v°+
]
β
Draw velocity vector diagram.
180 30 (90 ) 78.95
ϕβ
= °− °− °− = °
Law of sines.
/
sin sin 30 sin (90 )
DB
DB
v
vv
ϕβ
= =
° °−
sin
cos
6.2832 sin 78.95
cos18.95
6.52 m/s
B
D
v
v
ϕ
β
=
°
=°
=
PD
vv=
6.52 m/s
P=v
PROBLEM 15.62 (Continued)
/
sin 30
cos
6.2832sin 30
B
DB
v
v
β
°
=
°
0.16
BD
l
BD
PROBLEM 15.63
Knowing that at the instant shown the angular velocity of rod AB
is 15 rad/s clockwise, determine (a) the angular velocity of rod
BD, (b) the velocity of the midpoint of rod BD.
SOLUTION
M=v
PROBLEM 15.64
In the position shown, bar AB has an angular velocity of 4 rad/s clockwise.
Determine the angular velocity of bars BD and DE.
8
BD
BD =ω
PROBLEM 15.65
Linkage DBEF is part of a windshield
wiper mechanism, where points O, F and D
are fixed pinned connections. At the
position shown,
θ
= 60° and link EB is
horizontal. Knowing that link EF has a
counterclockwise angular velocity of
4 rad/s at the instant shown, determine the
angular velocity of links EB and DB.
SOLUTION
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.66
Robert’s linkage is named after Richard Robert (1789–1864) and can be used
to draw a close approximation to a straight line by locating a pen at Point F.
The distance AB is the same as BF, DF and DE. Knowing that the angular
velocity of bar AB is 5 rad/s clockwise in the position shown, determine
(a) the angular velocity of bar DE, (b) the velocity of Point F.
66
BD DE
BD
PROBLEM 15.66 (Continued)
F
