Thus,
and
aB=aA
#
*rB>A+Æ*(Æ*rB>A)+2Æ*(vB>A)xyz +(aB>A)xyz
20–43. Continued
Ans:
1098
*20–44.
At the instant shown, the rod AB is rotating about the
z axis with an angular velocity
v
1 = 4 rad
>
s and an
angular acceleration
v
#
1 = 3 rad
>
s2. At this same instant,
the circular rod has an angular motion relative to the
rod as shown. If the collar C is moving down around
the circular rod with a speed of 3 in.
>
s, which is
increasing at 8 in.
>
s2, both measured relative to the rod,
determine the collar’s velocity and acceleration at this
instant.
SOLUTION
C
>
B
=52i6 rad>s
#
C
>
B=
5
8i
6
rad
>
s
2
>
>
z
x
y
5 in.
A
B
v1 4 rad/s
fi
v1 3 rad/s2
v2 2 rad/s
fi
v2 8 rad/s2
Ans:
1099
20–45.
SOLUTION
Relative to XYZ, let xyz have
Relative to xyz, let coincident ,‚ have
((rP/O)xyz changes direction relative to XYZ.)
(rP>O)xyz
z¿y¿x¿
The particle Pslides around the circular hoop with a
constant angular velocity of while the hoop
rotates about the xaxis at a constant rate of If
at the instant shown the hoop is in the x–y plane and the
angle determine the velocity and acceleration of
the particle at this instant.
u=45°,
v=4 rad>s.
u
#
=6 rad>s,
200 mm
z
P
y
O
θ
=4rad/s
V
Ans:
1100
20–46.
At the instant shown, the industrial manipulator is
rotating about the z axis at
v
1 = 5 rad
>
s, and about joint
B at
v
2 = 2 rad
>
s. Determine the velocity and
acceleration of the grip A at this instant, when f = 30°,
u
= 45°, and r = 1.6 m.
SOLUTION
#
=0
rB=1.2 sin 30°j+1.2 cos 30°k=50.6j+1.0392k6 m
xyz
=52i6 rad>s
#
xyz =0
v
2
v
1
u
f
1.2 m
z
A
r
x
B
1101
20–47.
At the instant shown, the industrial manipulator is rotating
about the z axis at
v
1 = 5 rad
>
s, and v
#
1=
2 rad
>
s
2
; and
about joint B at
v
2 = 2 rad
>
s and v
#
2=
3 rad
>
s
2
. Determine
the velocity and acceleration of the grip A at this instant,
when f = 30°,
u
= 45°, and r = 1.6 m.
SOLUTION
=55k6 rad>s
#
=
{2k} rad
>
s
2
v
2
v
1
u
f
1.2 m
z
A
r
x
B
Ans:
1102
Ans:
P=5
*20–48.
At the given instant, the rod is turning about the z axis
with a constant angular velocity
v
1 = 3 rad
>
s. At this
same instant, the disk is spinning at
v
2 = 6 rad
>
s when
v
#
2 = 4 rad
>
s2, both measured relative to the rod.
Determine the velocity and acceleration of point P on
the disk at this instant.
y
z
v2 6 rad/s
v2 4 rad/s2
O
1.5 m
0.5 m
0.5 m
x
2 m
v1 3 rad/s
P
fi
SOLUTION
Motion of point A. Point A is located at the center of the disk. At the instant
consider, the fixed XYZ frame and rotating
x y z
frame are set coincident with
origin at Point O. The
x y z
frame is set rotate with constant angular velocity
𝛀
Motion of P with Respect to A. The xyz and
x y z
frame are set coincident with
origin at A. Here
x y z
frame is set to rotate at
𝛀 ″ =
V
x y z
of which
Motion of P. Here,
𝛀=
V
1=53
k6 rad>s
and
𝛀
#
=0
.
1103
y
5 ft
x
P
B
C
A
u
15 ft
3 ft
2 ft
4 ft
v1 0.8 rad/s
v1 1.30 rad/s2
v2 3 rad/s
v2 2 rad/s2
z
vO 2 ft/s
O
SOLUTION
𝛀=50.8k6 rad>s
𝛀
#
=5
1.3k
6
rad
>
s
2
𝛀
P
>
A
={3j} rad>s
>
20–49.
At the instant shown, the backhoe is traveling forward
at a constant speed vO = 2 ft
>
s, and the boom ABC is
rotating about the z axis with an angular velocity
v
1 = 0.8 rad
>
s and an angular acceleration
v
#
1=
1.30 rad
>
s
2
. At this same instant the boom is
rotating with
v
2 = 3 rad
>
s when v
#
2=
2 rad
>
s
2
, both
measured relative to the frame. Determine the velocity
and acceleration of point P on the bucket at this instant.
1104
20–50.
SOLUTION
vP=vO+Æ*rP/O+(rP/O)xyz
Æ
#
=0
Æ=v1={6k} rad>s
At t
h
e
i
nstant s
h
own, t
h
e arm OA of t
h
e conveyor
b
e
l
t
i
s
rotating about the zaxis with a constant angular velocity
while at the same instant the arm is rotating
upward at a constant rate If the conveyor is
running at a constant rate determine the velocity
and acceleration of the package Pat the instant shown.
Neglect the size of the package.
r
#=5ft>s,
v2=4 rad>s.
v1=6 rad>s,
z
A
O
r6ft
u45
P
v
1
6 rad/s
=
=
=
Ans:
1105
20–51.
At t
h
e
i
nstant s
h
own, t
h
e arm OA of t
h
e conveyor
b
e
l
t
i
s
rotating about the zaxis with a constant angular velocity
while at the same instant the arm is rotating
upward at a constant rate If the conveyor is
running at a rate which is increasing at
determine the velocity and acceleration of the package Pat
the instant shown. Neglect the size of the package.
r
$=8ft>s2,r
#=5ft>s,
v2=4 rad>s.
v1=6 rad>s,
SOLUTION
(aP/O)xyz =8cos 45j+8 sin 45°k96.18j39.60k
Æ=v1={6k} rad>s
z
A
O
r6ft
u45
P
v
1
6rad/s
=
=
=
Ans:
1106
30
y
A
x
O
z
v1 fi 0.25 rad/s
v2 fi 0.4 rad/s
40 ft
*20–52.
The crane is rotating about the z axis with a constant rate
v
1 = 0.25 rad
>
s, while the boom OA is rotating downward
with a constant rate
v
2 = 0.4 rad
>
s. Compute the velocity
and acceleration of point A located at the top of the boom
at the instant shown.
SOLUTION
𝛀={0.25k} rad>s
𝛀
#
=0
r
(v
A
>
O
)
xyz
=(r
#
A
>
O
)
xyz
+𝛀
A
>
O
*r
A
>
O
v
Ans:
1107
20–53.
Solve Prob. 20–52 if the angular motions are increasing
at v
#
1=
0.4 rad
>
s
2
and v
#
2=
0.8 rad
>
s
2
at the instant
shown.
SOLUTION
𝛀=50.25k6 rad>s
(aA
>
O)xyz =
3
(r
$
A
>
O)xyz +𝛀A
>
O*(r
#
A
>
O)xyz
4
+𝛀
#
A
>
O*rA
>
O+𝛀A
>
O*r
#
A
>
O
30
y
A
x
O
z
v1 fi 0.25 rad/s
v2 fi 0.4 rad/s
40 ft
Ans:
1108
20–54.
At the instant shown, the arm AB is rotating about the fixed
bearing with an angular velocity
v
1 = 2 rad
>
s and angular
acceleration v
#
1=
6 rad
>
s
2
. At the same instant, rod BD is
rotating relative to rod AB at
v
2 = 7 rad
>
s, which is
increasing at v
#
2=
1 rad
>
s
2
. Also, the collar C is moving
along rod BD with a velocity
r
#
=2 ft>s
and a deceleration
r
$
=
0.5 ft
>
s
2
, both measured relative to the rod.
Determine the velocity and acceleration of the collar at
this instant.
SOLUTION
𝛀=52k6 rad>s
𝛀
#
=5
6k
6
rad
>
s
2
aB=r
$
B=
3
(r
$
B)xyz +𝛀*(r
#
B)xyz
4
+𝛀
#
*rB+𝛀*r
#
B
>
(aC
>
B)xyz =
3
(r
$
C
>
B)xyz +𝛀C
>
B*(r
#
C
>
B)xyz
4
+𝛀
#
C
>
B*rC
>
B+𝛀C
>
B*r
#
C
>
B
u 30
A
C
B
v2 7 rad/s
v2 1 rad/s2
v1 2 rad/s
v1 6 rad/s2
r 1 ft
D
y
x2 ft
1.5 ft
z