978-0133915426 Chapter 20 Part 1

subject Type Homework Help
subject Pages 14
subject Words 1692
subject Authors Russell C. Hibbeler

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page-pf1
20–1.
The propeller of an airplane is rotating at a constant speed
while the plane is undergoing a turn at a constant rate
Determine the angular acceleration of the propeller if
(a) the turn is horizontal, i.e., and (b) the turn is
vertical, downward, i.e., vtj.
vtk,
vt.
vsi,
SOLUTION
For ,.
Ans.
(b) For ,.
For ,.
Ans.a=-vsvtk+0=-vsvtk
a=v
#=(v
#
s)XYZ +(v
#
t)XYZ
(v
#
t)XYZ =(v
#
t)xyz +Æ*vt=0+0=0
Æ=0vt
=0+(vtj)*(vsi)=-vsvtk
(v
#
s)XYZ =(v
#
s)xyz +Æ*vs
Æ=vtjvs
a=vsvtj+0=vsvtj
a=v
#=(v
#
s)XYZ +(v
#
t)XYZ
(v
#
t)XYZ =(v
#
t)xyz +Æ*vtk=0+0=0
Æ=0vt
=0+(vtk)*(vsi)=vsvtj
(v
#
s)XYZ =(v
#
s)xyz +Æ*vs
z
V
s
page-pf2
20–2.
SOLUTION
velocity (yaxis).Thus,
Equating kand jcomponents, we have
Angular Acceleration: The angular acceleration will be determined by
investigating the time rate of change of angular velocity with respect to the fixed
XYZ frame. Since always lies in the fixed X–Y plane, then is
observed to have a constant direction from the rotating xyz frame if this frame is
evah ew, htiw 602.qE gniylppA. ta gnitator
Velocity and Acceleration: Applying Eqs.20–3 and 20–4 with the and obtained above
and ,we have
Ans.
Ans.={-0.1125j-0.130k}ms2
+(-0.8660j)*[(-0.8660j)*(0.15j+0.2598k)]
=(0.4330i)*(0.15j+0.2598k)
aA=a*rA+v*(v*rA)
vA=v*rA=(-0.8660j)*(0.15j+0.2598k)={-0.225i}m>s
rA={(0.3 -0.3 cos 60°)j+0.3 sin 60°k}m={0.15j+0.2598k}m
av
a=v
#=(v
#)xyz +vz*v=0+0.5k*(-0.8660j)={0.4330i} rad>s2
(v
#)xyz =0Æ=vz={0.5k} rad>s
v={-0.8660j} rad>sv
a
-v=-1.00 cos 30°
v=0.8660 rad>s
0=-vssin 30° +0.5 vs=1.00 rad>s
-vj=-vscos 30°j-vssin 30°k+0.5k
v=vs+vz
The disk rotates about the zaxis at a constant rate
without slipping on the horizontal plane.
Determine the velocity and the acceleration of point Aon
the disk.
vz=0.5 rad>s
z
A
x
= 0.5 rad/s
z
V
page-pf3
20–3.
The ladder of the fire truck rotates around the zaxis with an
angular velocity which is increasing at
At the same instant it is rotating upward at a
constant rate Determine the velocity and
acceleration of point Alocated at the top of the ladder at
this instant.
v2=0.6 rad>s.
0.8rad>s2.
v1=0.15 rad>s,
z
y
A
40 ft
1
30
Angular acceleration: For ,.
#
#
v=v1={0.15k}rad>sv1
v
page-pf4
*20–4.
The ladder of the fire truck rotates around the zaxis with
an angular velocity of which is increasing
at
while increasing at Determine the
velocity and acceleration of point Alocated at the top of
the ladder at this instant.
0.4 rad>s2.v2=0.6 rad>s
0.2 rad>s2.
v1=0.15 rads,
Ans.
Ans.aA={-3.33i-21.3j+6.66k}ft>s2
+(0.4i+0.09j+0.2k)*(34.641j+20k)
aA=(0.6i+0.15k)*[(0.6i+0.15k)*(34.641j+20k)]
aA=Æ*(Æ*rA/O)+v
#*rA/O
vA={-5.20i-12j+20.8k}ft>s
vA=Æ*rA/O=(0.6i+0.15k)*(34.641j+20k)
Æ
#
=0.2k+0.4i+0.15k*0.6i={0.4i+0.09j+0.2k}rad>s2
v
#=v
#
1k+v
#
2i+v1k*v2i
rA/O={34.641j+20k}ft
y
A
40 ft
30
z
1
v
At the same instant it is rotating upward at
page-pf5
20–5.
If the plate gears A and B are rotating with the angular
velocities shown, determine the angular velocity of gear C
about the shaft DE. What is the angular velocity of DE
about the y axis?
8
100 mm
A
B
DE
25 mm
y
x
vA5 rad/s
C
page-pf6
20–6.
T
h
e con
i
ca
l
spoo
l
ro
ll
s on t
h
e p
l
ane w
i
t
h
out s
li
pp
i
ng.If t
h
e
axle has an angular velocity of and an
angular acceleration of at the instant shown,
determine the angular velocity and angular acceleration of
the spool at this instant.
a1=2 rad>s2
v1=3 rad>s
SOLUTION
Ans.
Ans.a={24.7i-5.49j} rad>s2
=2k+0+(-5.8476 cos 20°j-5.8476 sin 20°k)+(3k)*(-8.7714 cos 20°j-8.7714 sin 20°k)
a=v
#=
A
v
#
1
B
xyz +v1*v1+
A
v
#
2
B
xyz +v1*v2
A
v
#
2
B
xyz =-
2
sin 20° =-5.8476 rad>s2
A
v
#
1
B
xyz =2 rad>s2
={-8.24j} rad>s
v=v1+v2=3k-8.7714 cos 20°j-8.7714 sin 20°k
v2=-
3
sin 20° =-8.7714 rad>s
20
v
1
3rad/s
a
1
2rad/s
2
y
z
page-pf7
20–7.
SOLUTION
Ans.
Ans.aA={8.30i-35.2j+7.02k}ft>s2
aA=(4i+4.5j+2k)*(2.598j+1.5k)+(3k+1.5i)*(-7.794i-2.25j+3.879k)
aA=v
#*rA+Æ*vA
=4i+4.5j+2k
=(2k+0) +(4i+3k*1.5i)
Æ
#
=v
#
1+v
#
2
={-7.79i-2.25j+3.90k}ft>s
=-7.794i+3.897k-2.25j
vA=(3k+1.5i)*(2.598j+1.5k)
vA=Æ*rA
Æ=v1+v2=3k+1.5i
rA=3 cos 30°j+3 sin 30°k={2.598j+1.5k}ft
At a given instant, the antenna has an angular motion
and about the zaxis. At this
same instant the angular motion about the xaxis is
and Determine the velocity
and acceleration of the signal horn Aat this instant. The
distance from Oto Ais d=3ft.
v
#
2=4 rad>s2.v2=1.5 rad>s,
u=30°,
v
#
1=2 rad>s2
v1=3 rad>s
z
xy
u30
O
A
·
·
d
v
1
v
2
v
2
v
1
page-pf8
*20–8.
The disk rotates about the shaft S, while the shaft is
turning about the z axis at a rate of
v
z = 4 rad
>
s, which is
increasing at 2 rad
>
s2. Determine the velocity and
acceleration of point A on the disk at the instant shown.
No slipping occurs.
SOLUTION
z
y
A
B
S
2 rad/s2
4 rad/s
0.1 m
page-pf9
*20–8. Continued
page-pfa
20–9.
The disk rotates about the shaft S, while the shaft is
turning about the z axis at a rate of
v
z = 4 rad
>
s, which is
increasing at 2 rad
>
s2. Determine the velocity and
acceleration of point B on the disk at the instant shown.
No slipping occurs.
SOLUTION
z
y
A
B
S
2 rad/s2
4 rad/s
0.1 m
page-pfb
20–9. Continued
page-pfc
20–10.
The electric fan is mounted on a swivel support such
that the fan rotates about the z axis at a constant rate
of
v
z = 1 rad
>
s and the fan blade is spinning at a
constant rate
vs
= 60 rad
>
s. If f = 45° for the motion,
determine the angular velocity and the angular
acceleration of the blade.
SOLUTION
v=vz+vs
=1k+60 cos 45°j+60 sin 45°k
=42.426j+43.426k
=542.4j+43.4k6 rad>s
Ans.
v
#
=
v
#
z+
v
#
s
=0+0+
v
z*
v
s
=1k*42.426j+43.426k
Ans.
=5-42.4i6 rad>s2
Ans.
\
x
z
Vz
Vs
f
page-pfd
1057
20–11.
The electric fan is mounted on a swivel support such thatthe
fan rotates about the z axis at a constant rate of
v
z = 1 rad
>
s
and the fan blade is spinning at a constant rate
vs
= 60 rad
>
s.
If at the instant f = 45°, f
#
= 2 rad
>
s for the motion,
determine the angular velocity and the angular acceleration
of the blade.
\
x
z
Vz
Vs
f
Ans:
V
={2i+42.4j+43.4k} rad>s
A
=
{
-
42.4i
-
82.9j
+
84.9k} rad
>
s
2
SOLUTION
v=vz+vs+vx
=1k+60
cos
45°j+60
sin
45°k+2i
=2i+42.426j+43.426k
=52i+42.4j+43.4k6 rad>s
Ans.
v
#
=
v
#
z+
v
#
s+
v
#
x
=0+(
v
z+
v
x)*
v
s+
v
z*
v
x
=0+(1k+2i)*(42.426j+43.426k)+1k*(2i)
=-42.426i+84.853k-84.853j+2j
=5-42.4i-82.9j+84.9k6 rad>s2
Ans.
page-pfe
*20–12.
The drill pipe P turns at a constant angular rate
vP
= 4 rad
>
s. Determine the angular velocity and
angular acceleration of the conical rock bit, which rolls
without slipping. Also, what are the velocity and
acceleration of point A?
P
v
P
4 rad/s
SOLUTION
page-pff
1059
20–13.
T
h
e r
i
g
h
t c
i
rcu
l
ar cone rotates a
b
out t
h
e zax
i
s at a constant
rate of without slipping on the horizontal
plane. Determine the magnitudes of the velocity and
acceleration of points Band C.
v1=4 rad>s
SOLUTION
Equating components,
Thus,
Ans.
Ans.
Ans.
Ans.aC=1.60 m>s2
aC={-1.131j-1.131k}m>s2
aC=a*rC+v*vC=16i*(0.1)(0.707)k+(-4j)*(-0.2828i)
aB=1.13 m>s2
aB={1.131k}m>s
aB=a*rB+v*vB=16i*(0.1)(0.707)j+0
vC=0.283 m>s
vC=v*rC=(-4j)*(0.1(0.707)k)={-0.2828i}m>s
vB=0
vB=v*rB=(-4j)*(0.1(0.707)j)=0
a=v
#={16i} rad>s2
=0+(4k)*(-5.66 cos 45°j-5.66 sin 45°k)
v
#=v
#
1+v
#
2=0+v1*v2
Æ=v1
v={-4j} rad>s
v=-4 rad>s,
v2=-5.66 rad>s
0=4+0.707 v2
v=0.707 v2
vj=4k+v2cos 45° j+v2sin 45°k.
v=v1+v2
C
50 mm
z
v
1
4 rad/s
Ans:
vB=0
vC=0.283
m>s
a
B=
1.13
m
>
s
2
a
C=
1.60
m
>
s
2
page-pf10
20–14.
z
x
vs 10 rad/s
vs 6 rad/s2
A
B
C
0.15 m
y
The wheel is spinning about shaft AB with an angular
velocity of = 10 rad>s, which is increasing at a constant
rate of = 6 rad>s , while the frame precesses about the
z axis with an angular velocity of v = 12 rad>s, which is
increasing at a constant rate of v = 3 rad>s . Determine
the velocity and acceleration of point C located on the rim
of the wheel at this instant.
#
p2
p
v
#
s2
vs
SOLUTION
page-pf11
20–15.
At the instant shown, the tower crane rotates about the
zaxis with an angular velocity , which is
increasing at .The boom OA rotates downward
with an angular velocity , which is increasing
at . Determine the velocity and acceleration of
point Alocated at the end of the boom at this instant.
0.8 rad>s2
v2=0.4 rad>s
0.6 rad>s2
v1=0.25 rad>s
v10.25 rad/s
40 ft
z
y
A
O
30
SOLUTION
vA=v*rA=(1 -0.4 i+0.25 k)*(34.64j+20k)
rA=40 cos 30°j+40 sin 30°k={34.64j+20k}ft
={-0.8i-0.1j+0.6k} rad>s2
=xyz +Æ*v=(-0.8 i+0.6k)+(0.25k)*(-0.4 i+0.25k)
Æ={0.25 k} rad>s
v
#(v)
#
page-pf12
*20–16.
Gear Ais fixed while gear Bis free to rotate on the shaft S.
If the shaft is turning about the zaxis at
while increasing at determine the velocity and
acceleration of point Pat the instant shown. The face of
gear Blies in a vertical plane.
2 rad>s2,
vz=5 rad>s,
y
z
AB
S
P
80 mm
80 mm
160 mm
V
z
SOLUTION
vP=Æ*rP
#
Æ={5k-10j} rad>s
vP={-1600i} mm>s
(5k-10j*(160j+80k)vP
page-pf13
20–17.
T
h
e truncate
d
d
ou
bl
e cone rotates a
b
out t
h
e zax
i
s at
without slipping on the horizontal plane.If
at this same instant is increasing at
determine the velocity and acceleration of point Aon
the cone.
v
#
z=0.5 rad>s2,vz
vz=0.4 rad>s
1.5 ft
0.5ft
30
A
z
v
z
0.4rad/s
page-pf14
20–18.
The IA is located along the points of contant of Band C
Ans.
Thus,
Let the x,y,z axes have an angular velocity of , then
Ans.a={-6400i} rad>s2
a=(-40j)*(160k-40j)
a=v
#=v
#
P+v
#
s=0+vP*(vs+vP)
Æ*vP
v=vP+vs
vs=4(40) k={160k} rad>s
vP={-40j} rad>s
vP=40 rad>s
-32i=-0.8 vPi
-32i=3ij k
0-vP4vP
0 0.1 0.4 3
vP=v*rP/O
vP=-32i
rP>O=0.1j=0.4k
=-vPj+4vPk
v=-vPj+vsk
vs=4vP
vP
0.1 =vs
0.4
0.1 ft
B
S
y
z
80 rad/s
Gear A is fixed to the crankshaft S, while gear C is fixed.
Gear B and the propeller are free to rotate. The crankshaft is
turning at 80 rad
>
s about its axis. Determine the magnitudes
of the angular velocity of the propeller and the angular
acceleration of gear B.

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