Ans.
and
Ans.=[3.05i–30.9j+1.10k]m>s2
+2(1.5k)*(–3j+5.196k)+(–4.098j+1.098k)
=(–0.75i–3.375j)+0.5k*(12 cos 30°j+12 sin 30° k)+(1.5k)*
C
(1.5k)*(12 cos 30° j+12 sin 30° k)
D
aB=aA=Æ
#
*rB>A+Æ*(Æ*rB>A)+2Æ*(vB>A)xyz +(aB>A)xyz
=[–17.8i–3j+5.20k]m>s
=(–2.25i)+1.5k*(12 cos 30° j+12 sin 30°k)+(–3j+5.196k)
*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
z
y
5 in.
A
B
v1 4 rad/s
fi
v1 3 rad/s2
v2 2 rad/s
20–45.
SOLUTION
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
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
Ω=55k6 rad>s
Ω
#
=0
rB=1.2 sin 30°j+1.2 cos 30°k=50.6j+1.0392k6 m
v
B
=r
#
B
=(r
#
B
)
xyz
+Ω*r
B
=0+(5k)*(0.6j+1.0392k)=5–3i6 m>s
aB=r
$
B=
3
(r
$
B)xyz +Ω*(r
#
B)xyz
4
+Ω
#
*rB+Ω*r
#
B
=[0+0]+0+[(5k)*(–3i)]
=5–15j6 m>s2
Ω
xyz
=52i6 rad>s
Ω
#
xyz =0
r
A
>
B
=1.6 cos 45°j–1.6 sin 45°k=51.1314j–1.1314k6 m
(v
A
>
B
)
xyz
=r
#
A
>
B
=(r
#
A
>
B
)
xyz
+Ω
xyz
*r
A
>
B
=0+(2i)*(1.1314j–1.1314k)
=52.2627j+2.2627k6 m>s
(aA
>
B)xyz =r
$
A
>
B=
3
(
r
$
A
>
B)xyz +Ωxyz *(r
#
A
>
B)xyz
4
+
3
Ω
#
xyz *rA
>
B
4
+
3
Ωxyz *r
#
A
>
B
4
(aA
>
B)xyz =[0+0]+0+
3
(2i)*(2.2627j+2.2627k)
4
=5–4.5255j+4.5255k6 m>s2
v
A
=v
B
+Ω*r
A
>
B
+(v
A
>
B
)
xyz
=(–3i)+
3
(5k)*(1.1314j–1.1314k)
4
+(2.2627j+2.2627k)
=5–8.66i+2.26j+2.26k6 m>s
Ans.
aA=aB+Ω
#
*rA
>
B+Ω*(Ω*rA
>
B)+2Ω*(vA
>
B)xyz +(aA
>
B)xyz
=(–15j)+0+(5k)*[(5k)*(1.1314j–1.1314k)] +[2(5k)*(2.2627j+2.2627k)] +(–4.5255j+4.5255k)
=5–22.6i–47.8j+4.53k6 m>s2
Ans.
v
2
v
1
u
f
1.2 m
z
A
r
x
y
B
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
Ω=55k6 rad>s
Ω
#
=
{2k} rad
>
s
2
rB=1.2 sin 30°j+1.2 cos 30°k=50.6j+1.0392k6 m
v
B
= r
#
B
=(r
#
B
)
xyz
+Ω*r
B
=0+(5k)*(0.6j+1.0392k)=5–3i6 m>s
aB=r
$
B=
3
(r
$
B)xyz +Ω*(r
#
B)xyz
4
+Ω
#
*rB+Ω* r
#
B
=[0+0]+(2k)*(0.6j+1.0392k)+[(5k)*(–3i)]
=5–1.2i–15j6 m>s2
Ω
xyz
=52i6 rad>s
Ω
#
xyz =
5
3i
6
rad
>
s
2
r
A
>
B
=1.6 cos 45°j–1.6 sin 45°k=51.1314j–1.1314k6 m
(v
A
>
B
)
xyz
=r
#
A
>
B
=(r
#
A
>
B
)
xyz
+Ω
xyz
*r
A
>
B
=0+(2i)*(1.1314j–1.1314k)
=52.2627j+2.2627k6 m>s
(aA
>
B)xyz =r
$
A
>
B=
3
( r
$
A
>
B)xyz +Ωxyz *( r
#
A
>
B)xyz
4
+
3
Ω
#
xyz *rA
>
B
4
+
3
Ωxyz *r
#
A
>
B
4
(a
A
>
B
)
xyz
=[0+0]+(3i)*(1.1314j–1.1314k)+[(2i)*(2.2627j+2.2627k)]
=5 –1.1313j+7.9197k6 m>s2
v
A
=v
B
+Ω*r
A
>
B
+(v
A
>
B
)
xyz
=(–3i)+[(5k)*(1.1314j–1.1314k)] + (2.2627j+2.2627k)
=5–8.66i+2.26j+2.26k6 m>s
Ans.
aA=aB+Ω
#
*rA
>
B+Ω*(Ω*rA
>
B)+2Ω*(vA
>
B)xyz +(aA
>
B)xyz
=(–1.2i–15j)+(2k)*(1.1314j–1.1314k)+(5k)*[(5k)*(1.1314j–1.1314k)]
+[2(5k)*(2.2627j+2.2627k)] +(–1.1313j+7.9197k)
=5–26.1i–44.4j+7.92k6 m>s2
Ans.
v
2
v
1
u
f
1.2 m
z
A
r
x
B
y
5 ft
P
B
C
A
u
15 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
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.
20–50.
SOLUTION
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
r6ft
P
v
1
6 rad/s
=
=
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,
z
A
r6ft
P
v
1
6rad/s
=
=
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
rO=0
vO=0
aO=0
𝛀
A
>
O
={–0.4i} rad>s
𝛀
#
A
>
O=0
r
A
>
O
=4 cos 30°j+40 sin 30°k
=34.64j+20k
(v
A
>
O
)
xyz
=(r
#
A
>
O
)
xyz
+𝛀
A
>
O
*r
A
>
O
=0+(–0.4i)*(34.64j+20k)
=8j–13.856k
(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
=0+0+0+(–4i)*(8j–13.86k)
=–5.542j–3.2k
v
A
=v
O
+𝛀*r
A
>
O
+(v
A
>
O
)
xyz
=0+0.25k*(34.64j+20k)+(8j–13.856k)
={–8.66i+8j–13.9k} ft>s
Ans.
aA=aO+𝛀
#
*rA
>
O+𝛀*(𝛀*rA
>
O)+2𝛀*(vA
>
O)xyz +(aA
>
O)xyz
=0+0+(0.25k)*(0.25k)*(34.64j+20k)
+2(0.25k)*(8j–13.856k)–5.542j–3.2k
aA={–4i–7.71j–3.20k} ft>s2
Ans.
Ans:
vA=5–8.66i+8j–13.9k6 ft>s
a
A=
{
–
4i
–
7.71j
–
3.20k} ft
>
s
2
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
𝛀
#
=5
0.4k
6
rad
>
s
2
rO=0
vO=0
aO=0
𝛀
A
>
O
=5–0.4i6 rad>s
𝛀
#
A
>
O=
5
–0.8i
6
rad
>
s
2
r
A
>
O
=4
cos
30°j+40
sin
30°k
=34.64j+20k
(v
A
>
O
)
xyz
=(r
#
A
>
O
)
xyz
+Ω
A
>
O
*r
A
>
O
=0+(–0.4i)*(34.64
j+20k)
=8j–13.856k
(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
=0+0+(–0.8i)*(34.64j+20k)+(–4i)*(8j–13.86k)
=10.457j–30.913k
v
A
=v
O
+𝛀*r
A
>
O
+(v
A
>
O
)
xyz
=0+0.25k*(34.64j+20k)+(8j–13.856k)
=5–8.66i+8j–13.9k6 ft>s
Ans.
aA= aO+𝛀
#
*rA
>
O+𝛀*(𝛀*rA
>
O)+2𝛀*(vA
>
O)xyz +(aA
>
O)xyz
=0+(0.4k)*(34.64j+20k)
+(0.25k)*[(0.25k)*(34.64j+20k)] +2(0.25k)*(8j–13.856k)+10.457j–30.913k
aA=5–17.9i+8.29j–30.9k6 ft>s2
Ans.
30
A
x
O
z
v1 fi 0.25 rad/s
v2 fi 0.4 rad/s
40 ft
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
rB=5–2i+1.5k6 ft
v
B
=r
#
B
=(r
#
B
)
xyz
+𝛀*r
B
=(2k)*(–2i+1.5k)
=5–4j6 ft>s
aB=r
$
B=
3
(r
$
B)xyz +𝛀*(r
#
B)xyz
4
+𝛀
#
*rB+𝛀*r
#
B
=(6k)*(–2i+1.5k)+(2k)*(–4j)
=58i–12j6 ft>s2
𝛀
C
>
B
=57i6 rad>s
𝛀
#
C
>
B=
5
1i
6
rad
>
s
2
r
C
>
B
=1 cos 30°j+1 sin 30°k=50.866j+0.5k6 ft
(v
C
>
B
)
xyz
=(r
#
C
>
B
)
xyz
+𝛀
C
>
B
*r
C
>
B
=(2 cos 30°j+2 sin 30°k)+(7i)*(0.866j+0.5k)
=5–1.768j+7.062k6 ft>s
(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
=(–0.5 cos 30°j–0.5 sin 30°k)+(7i)*(1.732j+1k)+(1i)*(0.866j+0.5k)
+(7i)*(–1.768j+7.06k)
=5–57.37j+0.3640k6 ft>s2
v
C
=v
B
+𝛀*r
C
>
B
+(v
C
>
B
)
xyz
=(–4j)+(2k)*(0.866j+0.5k)+(–1.768j+7.06k)
vC=5–1.73i–5.77j+7.06k6 ft>s
Ans.
aC=aB+𝛀
#
*rC
>
B+𝛀*(𝛀*rC
>
B)+2𝛀*(vC
>
B)xyz +(aC
>
B)xyz
=(8i–12j)+(6k)*(0.866j+0.5k)+(2k)*
3
(2k)*(0.866j+0.5k)
4
+2(2k)*(–1.768j+7.062k)+(–57.37j+0.364k)
aC=59.88i–72.8j+0.365k6 ft>s2
Ans.
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