16–21.
The motor turns the disk with an angular velocity of
v
=
(
5t
2+
3t
)
rad
>
s, where t is in seconds. Determine the
magnitudes of the velocity and the n and t components of
acceleration of the point A on the disk when
t=3 s
.
SOLUTION
t=3 s
=
t=3 s
=
u
150 mm
16–22.
SOLUTION
If the motor turns gear Awith an angular acceleration of
when the angular velocity is ,
determine the angular acceleration and angular velocity of
gear D.
vA=20 rad>saA=2 rad>s2
B
C
D
50 mm
100 mm
16–23.
If the motor turns gear Awith an angular acceleration of
when the angular velocity is ,
determine the angular acceleration and angular velocity of
gear D.
vA=60 rad>saA=3 rad>s2
SOLUTION
A
B
C
D
50 mm
100 mm
*16–24.
SOLUTION
first. Applying Eq. 16–2, we have
However, where is the angular velocity of propeller.Then,
Ans.vB=rA
r
B
vA=a0.5
1.2 b(506.25) =211 rad>s
vB
vArA=vBrB
vA=100t4|1.5 s
0=506.25 rad>s
LvA
0
dv=L1.5 s
0
400t3dt
dv=adt
The gear Aon the drive shaft of the outboard motor has a
radius .and the meshed pinion gear Bon the
propeller shaft has a radius .Determine the
angular velocity of the propeller in ,if the drive shaft
rotates with an angular acceleration ,
where tis in seconds.The propeller is originally at rest and
the motor frame does not move.
a=(400t3) rad>s2
t=1.5 s
rB=1.2 in
rA=0.5 in
2.20 in.
P
B
A
16–25.
SOLUTION
first. Applying Eq. 16–2, we have
The angular acceleration of gear Aat is given by
However, and where and are the angular
velocity and acceleration of propeller.Then,
Motion of P:The magnitude of the velocity of point Pcan be determined using
Eq. 16–8.
Ans.
The tangential and normal components of the acceleration of point Pcan be
determined using Eqs. 16–11 and 16–12, respectively.
The magnitude of the acceleration of point Pis
Ans.a
P=2
a2
r+
a2
n=2
12.892
+
31.862
=
34.4 ft
>
s2
an=v2
BrP=
A
13.182
B
a2.20
12 b=31.86 ft>s2
ar=aBrP=70.31a2.20
12 b=12.89 ft>s2
vP=vBrP=13.18a2.20
12 b=2.42 ft>s
aB=rA
rB
aA=a0.5
1.2 b(168.75) =70.31 rad>s2
vB=rA
rB
vA=a0.5
1.2 b(31.64) =13.18 rad>s
aB
vB
aArA=aBrB
vArA=vBrB
aA=400
A
0.753
B
=168.75 rad>s2
t=0.75 s
vA=100t4|0.75 s
0=31.64 rad>s
LvA
0
dv=L0.75 s
0
400t3dt
dv=adt
2.20 in.
P
B
A
For the outboard motor in Prob. 16–24, determine the
magnitude of the velocity and acceleration of point P
located on the tip of the propeller at the instant t = 0.75 s.
16–26.
The pinion gear Aon the motor shaft is given a constant
angular acceleration If the gears Aand B
have the dimensions shown, determine the angular velocity
and angular displacement of the output shaft C, when
starting from rest. The shaft is fixed to Band turns
with it.
t=2s
a=3 rad>s2.
SOLUTION
Ans.
Ans.uC=uB=1.68 rad
6(35) =uB(125)
uArA=uBrB
vC=vB=1.68 rad>s
6(35) =vB(125)
vArA=vBrB
uA=6 rad
uA=0+0+1
2(3)(2)2
u=u0+v0t+1
2act2
vA=0+3(2) =6 rad>s
v=v0+act
C
125 mm
35 mm
A
B
16–27.
SOLUTION
Ans.v=100t32|t=1.3 =148 rad s
Lv
0
dv=Lt
0
1502tdt
dv=adt
aP=1502t
(3002t)(0.7) =ap(1.4)
aArA=aBrB
The gear Aon the drive shaft of the outboard motor has a
radius and the meshed pinion gear Bon the
propeller shaft has a radius Determine the
angular velocity of the propeller in if the drive
shaft rotates with an angular acceleration
where tis in seconds.The propeller is
originally at rest and the motor frame does not move.
a=13002t2rad>s2,
t=1.3 s
rB=1.4 in.
rA=0.7 in.
2.2 in.
B
A
*16–28.
The gear A on the drive shaft of the outboard motor has a
radius rA
=
0.7 in. and the meshed pinion gear B on the
propeller shaft has a radius rB
=
1.4 in. Determine the
magnitudes of the velocity and acceleration of a point P
located on the tip of the propeller at the instant t
=
0.75 s.
the drive shaft rotates with an angular acceleration
a
=
(300
1t
) rad
>
s2, where t is in seconds. The propeller is
originally at rest and the motor frame does not move.
SOLUTION
2.2 in.
P
B
A
658
16–29.
A stamp S, located on the revolving drum, is used to label
canisters. If the canisters are centered 200 mm apart on the
conveyor, determine the radius of the driving wheel A
and the radius of the conveyor belt drum so that for each
revolution of the stamp it marks the top of a canister. How
many canisters are marked per minute if the drum at Bis
rotating at ? Note that the driving belt is
twisted as it passes between the wheels.
vB=0.2 rad>s
rB
rA
A
r
A
r
B
S
SOLUTION
l=2p(rA)
16–30.
SOLUTION
At the instant shown, gear A is rotating with a constant
angular velocity of vA=6 rad>s. Determine the largest
angular velocity of gear B and the maximum speed of
point C.
100 mm
B
C
A
100 mm
100 mm
100 mm
B
ω
A= 6 rad/s
ω
660
16–31.
Determine the distance the load Wis lifted in using
the hoist. The shaft of the motor Mturns with an angular
velocity , where tis in seconds.v=100(4 +t) rad>s
t
=5s
300 mm
30 mm
50 mm 225 mm
B
A
E
C
W
D
40 mm
M
SOLUTION
first. Applying Eq. 16–1, we have
Here,.Then, the angular displacement of gear Bis given by
Since gear Cis attached to the same shaft as gear B, then .
Also,,then, the angular displacement of gear Dis given by
Since shaft Eis attached to gear D,.The distance at which the
load Wis lifted is
Ans.sW=rEuE=(0.05)(57.78) =2.89 m
uE=uD=57.78 rad
uD=rC
rD
uC=a30
300 b(577.78) =57.78 rad
rDuD=rCuC
uC=uB=577.78 rad
uB=rA
rB
uA=a40
225 b(3250) =577.78 rad
rAuA=rBuB
uA=3250 rad
LuA
0
du=L5s
0
100(4 +t)dt
du=vdt
Ans:
sW=2.89 m
16–33.
A
B
125 mm
200 mm
vA
vB
The driving belt is twisted so that pulley Brotates in the
opposite direction to that of drive wheel A. If the angular
displacement of Ais rad, where tis in
seconds, determine the angular velocity and angular
acceleration of Bwhen t=3 s.
uA=(5t3+10t2)
SOLUTION
663
16–34.
Forashort time amotorofthe random-orbit sander drives
the
gear Awith an angular velocity of
where tis in seconds.This gear is
connected
to gear B,which is fixed connected to the shaft
CD
.The end of this shaft is connected to the eccentric
spindle
EF and pad P,which causes the pad to orbit around
shaft
CD at aradius of 15 mm. Determine the magnitudes
of
the velocity and the tangential and normal components
of
acceleration of the spindle EF when after
starting
from rest.
t=2s
vA=401t3+6t2rad>s,
SOLUTION
Ans.
Ans.
Ans.(aE)n=600 m>s2
(aE)n=v2
BrE=c1
4(40)
A
t3+6t
B
d2
(0.015) 2t=2
(aE)t=2.70 m>s2
(aE)t=aBrE=1
4
A
120t2+240
B
(0.015) 2t=2
aB=1
4aA
aA(10) =aB(40)
aArA=aBrB
aA=dvA
dt =d
dt
C
40
A
t3+6t
BD
=120t2+240
vE=3m>s
vE=vBrE=1
4vA(0.015) =1
4(40)
A
t3+6t
B
(0.015) 2t=2
vB=1
4vA
vArA=vBrB
40 mm
10 mm
15 mm
A
A
B
C
C
D
E
V
Ans:
vE=3
m>s
(a
E
)
t=
2.70
m
>
s
2
(a
E
)
n=
600
m
>
s
2
16–35.
If the shaft and plate rotates with a constant angular velocity
of ,determine the velocity and acceleration of
point Clocated on the corner of the plate at the instant
shown. Express the result in Cartesian vector form.
v=14 rad>s
SOLUTION
Thus,
Since is constant
fo noitarelecca dna yticolev ehT.nesohc si,ecneinevnoc roF
point Ccan be determined from
Ans.
and
Ans. =[38.4i–64.8j+40.8k]m s2
=0+(–6i+4j+12k)*[(–6i+4j+12k)*(–0.3i+0.4j)]
aC=a*rC+V*(V*rc)
=[–4.8i–3.6j–1.2k] m>s
=(–6i+4j+12k)*(–0.3i+0.4j)
vC=v*rC
rC=[–0.3i+0.4j] m
a=0
v
v=vuOA =14a–3
7 i+2
7 j+6
7kb=[–6i+4j+12k] rad>s
uOA =
–0.3i+0.2j+0.6k
2(–0.3)2+0.22+0.62
=-
7 i+2
7 j+6
7 k
C
O
D
z
0.2 m
0.4 m
0.6 m
A
v
a
16–37.
The rod assembly is supported by ball-and-socket joints at
Aand B. At the instant shown it is rotating about the yaxis
with an angular velocity and has an angular
acceleration Determine the magnitudes of
the velocity and acceleration of point Cat this instant.
Solve the problem using Cartesian vectors and
and 16–13.
a=8 rad>s2.
v=5 rad>s
SOLUTION
Ans.
Ans.aC=212.42+(–4.3)2=13.1 m>s2
={12.4i–4.3k}m>s2
=8j*(–0.4i+0.3k)–52(–0.4i+0.3k)
aC=a*r–v2r
vC=21.52+22=2.50 m>s
vC=5j*(–0.4i+0.3k)={1.5i+2k}m>s
vC=v*r
0.3 m
z
x
y
A
C
B
0.4 m
0.4 m
AV
Eqs. 16–9
16–38.
The sphere starts from rest at
u=0°
and rotates with an
angular acceleration of a
=
(4u
+
1) rad
>
s
2
, where
u
is in
radians. Determine the magnitudes of the velocity and
acceleration of point P on the sphere at the instant
u=6
rad.
P
r 8 in.
30fi
SOLUTION
v dv=a du
L
v
0
v dv=
L
u
0
(4u+1) du
v=2u
u=6 rad
=
=
668
16–39.
SOLUTION
Time derivatives:
however, and
Ans.
Ans.a=ry2
A(2y2–r2)
y2(y2–r2)3
2
a=v
#=ryA
C
–y–2y
#
A
y2–r2
B
–1
2+
A
y–1
BA
–1
2
BA
y2–r2
B
–3
2(2yy
#)
D
¢
2y2–r2
y
≤
v=r
y
#
=-yA,u
#
=vcos u=
2y2–r2
y
cos uu
#
=-
r
y2y
#
T
h
e en
d
Aof t
h
e
b
ar
i
s mov
i
ng
d
ownwar
d
a
l
ong t
h
e s
l
otte
d
guide with a constant velocity Determine the angular
velocity and angular acceleration of the bar as a
function of its position y.
AV
vA.
y
r
v
A
A
V,A
U
y2
(
y2
–
r2
)
669
Ans:
v
=28.9 rad>s
b
a
=
470 rad
>
s
2
d
*16–40.
At the instant
u=60°
, the slotted guide rod is moving
to the left with an acceleration of 2 m
>
s2 and a velocity of
5 m
>
s. Determine the angular acceleration and angular
velocity of link AB at this instant.
shown in Fig. a, which is
x=0.2 cos u m
Time Derivatives. Using the chain rule,
x
#
=–
0.2(sin u)u
#
(1)
x
$
=–
0.2[(cos u)u
#2+
(sin u)u
$
] (2)
Here
x
#
=v
,
x
$
=a
,
u
#
=v
and
u
$
=a
when
u =60°
. Realizing that the velocity
and acceleration of the guide rod are directed toward the negative sense of x,
v=–5 m>s
and a
=–2 m>s2
. Then Eq (1) gives
–s=(–0.2(sin 60°)
v
v
=28.87 rad>s=28.9 rad>s
b Ans.
Subsequently, Eq. (2) gives
–
2
=–
0.2[cos 60°
(
28.87
2
)
+
(sin 60°)a]
a
=–469.57 rad>s2=470 rad>s2
d Ans.
The negative sign indicates that
A
is directed in the negative sense of
u
.
v 5 m/s
a 2 m/s2
A
u