PROBLEM 15.176
Knowing that at the instant shown the rod attached at A has an
angular velocity of 5 rad/s counterclockwise and an angular
acceleration of 2 rad/s2 clockwise, determine the angular velocity and
the angular acceleration of the rod attached at B.
SOLUTION
Geometry: Apply the law of sines to the triangle ABP to determine the lengths
AP
and
.BP
rel
PROBLEM 15.176 (Continued)
and a relative acceleration
BP =α
PROBLEM 15.177
The Geneva mechanism shown is used to provide an
intermittent rotary motion of disk S. Disk D rotates with a
constant counterclockwise angular velocity
D
ω
of 8 rad/s. A
pin P is attached to disk D and can slide in one of the six
equally spaced slots cut in disk S. It is desirable that the
angular velocity of disk S be zero as the pin enters and leaves
each of the six slots; this will occur if the distance between the
centers of the disks and the radii of the disks are related as
shown. Determine the angular velocity and angular
acceleration of disk S at the instant when
150 .
f
= °
SOLUTION
Geometry:
222
PROBLEM 15.177 (Continued)
Equate the two expressions for
P
v
and resolve into components.
: 1.54914 10cos(30 )
βω β
= °+
S=α
PROBLEM 15.178
In Problem 15.177, determine the angular velocity and
angular acceleration of disk S at the instant when
135 .
f
= °
PROBLEM 15.177 The Geneva mechanism shown is used
to provide an intermittent rotary motion of disk S. Disk D
rotates with a constant counterclockwise angular velocity
D
ω
of 8 rad/s. A pin P is attached to disk D and can slide in
one of the six equally spaced slots cut in disk S. It is
desirable that the angular velocity of disk S be zero as the pin
enters and leaves each of the six slots; this will occur if the
distance between the centers of the disks and the radii of the
disks are related as shown. Determine the angular velocity
and angular acceleration of disk S at the instant when
150 .
f
= °
SOLUTION
Geometry:
222
S
S
PROBLEM 15.178 (Continued)
Motion of disk D. (Rotation about B)
( ) (1.25)(8) 10 in./s
BP
ω
= = =v
30°
S=α
PROBLEM 15.179
At the instant shown, bar BC has an angular velocity of 3 rad/s
and an angular acceleration of
2
2 rad/s ,
both counterclockwise.
Determine the angular acceleration of the plate.
PROBLEM 15.179 (Continued)
Relative to the frame (plate), the acceleration of pin B is
22
rel
/ rel rel
30
() () 4
BF t t
v
aa
= −= −
a ji j i
PROBLEM 15.180
At the instant shown bar BC has an angular velocity of 3 rad/s and
an angular acceleration of
2
2 rad/s ,
both clockwise. Determine
the angular acceleration of the plate.
2
//
2
22
(2 (6 3) (3)(6 3)
12 6 54 27
(60 in./s ) (15 in./s )
B BC B C BC B C
ω
= ×−
=− ×−+ − −+
= ++ −
= −
a rr
k) ij ij
ji i j
ij
α
PROBLEM 15.180 (Continued)
PROBLEM 15.181*
Rod AB passes through a collar that is welded to link DE. Knowing that
at the instant shown block A moves to the right at a constant speed of 75
in./s, determine (a) the angular velocity of rod AB, (b) the velocity
relative to the collar of the point of the rod in contact with the collar, (c)
the acceleration of the point of the rod in contact with the collar. (Hint:
Rod AB and link DE have the same
ω
and the same
.)α
PROBLEM 15.181* (Continued)
PROBLEM 15.182*
Solve Problem 15.181, assuming that block A moves to the left at a
constant speed of
75 in./s.
PROBLEM 15.181 Rod AB passes through a collar which is welded
to link DE. Knowing that at the instant shown block A moves to the
right at a constant speed of 75 in./s, determine (a) the angular
velocity of rod AB, (b) the velocity relative to the collar of the point
of the rod in contact with the collar, (c) the acceleration of the point
of the rod in contact with the collar. (Hint: Rod AB and link DE have
the same
ω
and the same
.)α
SOLUTION
Let
ω
=ω
and
α
=α
be the angular velocity and angular acceleration of the link DE and collar rigid
body. Let F be a frame of reference moving with this body. The rod AB slides in the collar relative to the
frame of reference with relative velocity
u=u
30°
and relative acceleration
u=u
30 .°
Note that this
relative motion is a translation that applies to all points along the rod. Let Point A be moving with the end of
the rod and
A′
be moving with the frame. Point E is a fixed point.
6 in. 12 in.
60°
PROBLEM 15.182* (Continued)
PROBLEM 15.183*
In Problem 15.157, determine the acceleration of pin P.
PROBLEM 15.157 The motion of pin P is guided by slots cut in
rods AD and BE. Knowing that bar AD has a constant angular
velocity of 4 rad/s clockwise and bar BE has an angular velocity
of 5 rad/s counterclockwise and is slowing down at a rate of
2 rad/s2, determine the velocity of P for the position shown.
PROBLEM 15.183* (Continued)
Acceleration of Point P″ on rod BE coinciding with the pin
2
//
2
P BE P B BE P E
ω
′′
=×−
aαr r
P−+=a ij
PROBLEM 15.184
The bowling ball shown rolls without slipping on the horizontal xz plane with an
angular velocity
xyz
ωωω
=++i jk
ω
. Knowing that
(4.8 m / s) (4.8 m / s) (3.6 m / s)
A
=−+v i jk
and
(9.6 m / s)
D
=vi
(7.2 m / s) ,+k
determine (a) the angular velocity of the bowling ball, (b) the
velocity of its center C.
SOLUTION
At the given instant, the origin is not moving.
: 4.8 4.8 3.6
0.109 0.109 0
A A x yz
ω ωω
=× −+ =
i jk
v r i jkω
( )
4.8 4.8 3.6 0.109 0.109 0.109
z z xy
ω ω ωω
−+ =− + + −i jk i j k
( )
: 0.109 4.8 44.037 rad/s
: 0.109 4.8 44.037 rad/s
: 0.109 3.6 33.028 rad/s
zz
zz
xy xy
ωω
ωω
ωω ωω
−= =−
=−=−
− = −=
i
j
k
: 9.6 7.2
0 0.218 0
D D xyz
ωωω
=× +=
i jk
v r ikω
9.6 7.2 0.218 0.218
zx
ωω
+=− +ik i k
: 0.218 9.6 44.037 rad/s
: 0.218 7.2 33.028 rad/s
zz
xx
ωω
ωω
−= =−
= =
i
j
33.028 0
yx
ωω
=−=
( ) .a Angular velocity
( ) ( )
33.0 rad/s 44.0 rad/s = −ikω
( ) b Velocity of point C.
( )
33.028 44.037 0.109
4.8 3.6
CC
=×= − ×
= +
vr i k j
ik
ω
( ) ( )
4.80 m/s 3.60 m/s
C
= +v ik
PROBLEM 15.185
The bowling ball shown rolls without slipping on the horizontal xz plane with an
angular velocity
xyz
ωωω
=++i jk
ω
. Knowing that
(3.6 m / s) (4.8 m / s) (4.8 m / s)
B
=−+v i jk
and
(7.2 m / s)
D
=vi
(9.6 m / s) ,+k
determine (a) the angular velocity of the bowling ball, (b) the
velocity of its center C.
SOLUTION
At the given instant, the origin is not moving.
: 3.6 4.8 4.8
0 0.109 0.109
B B xy z
ωω ω
=× −+ =
ij k
v r i jkω
( )
3.6 4.8 4.8 0.109 0.109 0.109
yz x x
ωω ω ω
−+ = − − +i jk i j k
( )
: 0.109 3.6 33.028 rad/s
: 0.109 4.8 44.037 rad/s
: 0.109 4.8 44.037 rad/s
yz yz
xx
xx
ωω ωω
ωω
ωω
− = −=
−=− =
= =
i
j
k
: 7.2 9.6
0 0.218 0
D D xyz
ωωω
=× +=
i jk
v r ikω
7.2 9.6 0.218 0.218
zx
ωω
+=− +ik i k
: 0.218 7.2 33.028 rad/s
: 0.218 9.6 44.037 rad/s
zz
xx
ωω
ωω
−= =−
= =
i
k
33.028 0
yz
ωω
= +=
( ) .a Angular velocity
( ) ( )
44.0 rad/s 33.0 rad/s = −ikω
( ) b Velocity of point C.
( ) ( )
44.037 33.028 0.109
3.6 4.8
CC
=×= − ×
= +
vr i k j
ik
ω
( ) ( )
3.60 m/s 4.80 m/s
C
= +v ik
PROBLEM 15.186
Plate ABD and rod OB are rigidly connected and rotate about
the ball–and–socket joint O with an angular velocity
ω
=
ω
x
i +
ω
x
j +
ω
z k. Knowing that vA = (80 mm/s)i + (360 mm/s)j +
(vA)z k and
1.5 rad/s,
x
ω
=
determine (a) the angular velocity
of the assembly, (b) the velocity of Point D.
SOLUTION
1.5 rad/s (1.5 rad/s)
x yz
ω ωω
= = ++
i jk
ω
PROBLEM 15.186 (Continued)
(b) Velocity of D.
D