Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
SOLUTION
*12–172.
The rod OA rotates clockwise with a constant angular velocity
of 6 rad
>
s. Two pin-connected slider blocks, located at B, move
freely on OA and the curved rod whose shape is a limaçon
described by the equation r = 200(2 − cos
u
) mm. Determine
the speed of the slider blocks at the instant
u=150°
.
O
400 mm
200 mm
600 mm
r
6 rad/s
B
A
u
SOLUTION
12–173.
Determine the magnitude of the acceleration of the slider
blocks in Prob. 12–172 when
u
= 150°.
O
400 mm
200 mm
600 mm
r
6 rad/s
B
A
u
SOLUTION
r2=4 cos 2u
rr
#
=-4 sin 2u u
#
rr
$
=r
#2=-4 sin 2u u
$
-8 cos 2u u
#2
when
u=0
,
u
#
=6
,
u
$
=0
r=2
,
r=0
,
r
$
=-144
v
r=
r
#
=
0 Ans.
v
u=
ru
#
=
2(6)
=
12 ft
>
s Ans.
a
r=
r
$
-
ru
#2=-
144
-
2(6)
2=-
216 ft
>
s
2
Ans.
a
u=
ru
$
+2
r
#
u
#
=2(0) +2(0)(6) =0
Ans.
12–174.
A double collar C is pin connected together such that one
collar slides over a fixed rod and the other slides over a
rotating rod. If the geometry of the fixed rod for a short
distance can be defined by a lemniscate, r2 = (4 cos 2
u
) ft2,
determine the collar’s radial and transverse components of
velocity and acceleration at the instant
u
= 0° as shown. Rod
OA is rotating at a constant rate of
u
#
= 6 rad
>
s.
r2 fi 4 cos 2 u
u fi 6 rad/s
O
rC
A
·
12–175.
SOLUTION
A block moves outward along the slot in the platform with
a speed of where tis in seconds.The platform
rotates at a constant rate of 6 rad/s. If the block starts from
rest at the center, determine the magnitudes of its velocity
and acceleration when t=1s.
r
#
=14t2m>s,
θ
θ
·=6rad/s
r
SOLUTION
*12–176.
The car travels around the circular track with a constant
speed of 20 m
>
s. Determine the car’s radial and transverse
components of velocity and acceleration at the instant
u
=
p
>4 rad
.
r (400 cos u) m
r
u
SOLUTION
2
12–177.
The car travels around the circular track such that its
transverse component is
u
= (0.006t2) rad, where t is in
seconds. Determine the car’s radial and transverse
components of velocity and acceleration at the instant t
=
4 s.
r (400 cos u) m
r
u
12–178.
SOLUTION
200
600
p
The car travels along a road which for a short distance
is defi ned by r=(200>u) ft, where u is in radians. If it
maintains a constant speed of v=35 ft>s, determine the
radial and transverse components of its velocity when
u=p>3 rad. r
θ
SOLUTION
r=8
u=0.6 t
r
#
=0
u
#
=0.6
r
$
=0
u
$
=0
z=1.5 sin u
z
#
=1.5 cos u u
#
z
$
=-1.5 sin
u
(
u
#
)2+1.5 cos
u
u
$
At
t=4 s
u=2.4
z
#
=-0.6637
z
$
=-0.3648
vr=0
Ans.
vu=4.80 ft>s
Ans.
vz=-0.664 ft>s
Ans.
a
r=0-8(0.6)2=-2.88 ft>s2
Ans.
au=0+0=0
Ans.
a
z=-0.365 ft>s2
Ans.
12–179.
A horse on the merry-go-round moves according to the
equations r
=
8 ft,
u
=
(0.6t) rad, and z
=
(1.5 sin
u
) ft,
where t is in seconds. Determine the cylindrical components
of the velocity and acceleration of the horse when t
=
4 s.
z
z
r
u
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
r=8
r
#
=0
u
#
=2
r
$
=0
u
$
=0
z=1.5 sin u
z
#
=1.5 cos u u
#
z
$
=-1.5 sin
u
(
u
#
)2+1.5 cos
u
u
$
v
r
=r
#
=0
v
u=
r u
#
=
8(2)
=
16 ft
>
s
(v
z
)
max
=z
#
=1.5(cos 0°)(2) =3 ft>s
(v
z
)
min
=z
#
=1.5(cos 90°)(2) =0
v
max =2
(16)2
+
(3)2
=
16.3 ft
>
s Ans.
v
min =2
(16)2
+
(0)2
=
16 ft
>
s Ans.
a
r=
r
$
-
r(u
#
)
2=
0
-
8(2)
2=-
32 ft
>
s
2
a
u=
r u
$
+
2 r
#
u
#
=
0
+
0
=
0
(
a
z)max =
z
$
=-1.5(sin 90°)(2)2=-6
(
a
z)min =
z
$
=-1.5(sin 0°)(2)2=0
a
max =2
(
-
32)2
+
(0)2
+
(
-
6)2
=
32.6 ft
>
s2 Ans.
a
min =2
(
-
32)2
+
(0)2
+
(0)2
=
32 ft
>
s2 Ans.
*12–180.
A horse on the merry-go-round moves according to the
equations r
=
8 ft,
u
#
=
2 rad
>
s and z
=
(1.5 sin
u
) ft, where
t is in seconds. Determine the maximum and minimum
magnitudes of the velocity and acceleration of the horse
during the motion.
z
z
r
u
12–181.
If the slotted arm rotates counterclockwise with a
constant angular velocity of , determine the
magnitudes of the velocity and acceleration of peg at
.The peg is constrained to move in the slots of the
fixed bar and rotating bar .ABCD
u=30°
P
u
#
=2 rad>s
AB
SOLUTION
When ,
Velocity:
Thus, the magnitude of the peg’s velocity is
Ans.
Acceleration:
Thus, the magnitude of the peg’s acceleration is
Ans.a=2ar2+au2=212.322+21.232=24.6 ft>s2
au=ru
$
+2r
#u
#
=0+2(5.333)(2) =21.23 ft>s2
ar=r
$
-ru2
#
=30.79 -4.619(22)=12.32 ft>s2
v=2vr2+vu2=25.3332+9.2382=10.7 ft>s
vu=ru
#
=4.619(2) =9.238 ft>svr=r
#
=5.333 ft>s
r
$|u=30° =4[0 +22(sec3 30° +tan2 30° sec 30°)] =30.79 ft>s2
r
#|u=30° =(4 sec30° tan30°)(2) =5.333 ft>s
r|u=30° =4 sec 30° =4.619 ft
u=30°
=4[secu(tanu)u
#
+u
#2(sec3u+tan2u secu)] ft>s2
u
$
=0 r
$
=4 [secu(tanu)u
$
+ u
#
(sec u (sec2u)u
#
+tan u secu(tan u)u
#
)]
u
#
=2 rad>s
r
#
=(4 secu(tanu)u
#
) ft>s
A
D
P
C
r (4 sec ) ft
u
u
4 ft
12–182.
The peg is constrained to move in the slots of the fixed bar
and rotating bar .When , the angular
velocity and angular acceleration of arm are
and ,respectively.Determine
the magnitudes of the velocity and acceleration of the peg
at this instant.P
u
$
=3 rad>s2
u
#
=2 rad>s
AB
u=30°ABCD
SOLUTION
When ,
Velocity:
Thus, the magnitude of the peg’s velocity is
Ans.
Acceleration:
Thus, the magnitude of the peg’s acceleration is
Ans.a=2ar2+au2=220.322+35.192=40.6 ft>s2
au=ru
$
+2r
#u
#
=4.619(3) +2(5.333)(2) =35.19 ft>s2
ar=r
$
-ru2
#
=38.79 -4.619(22)=20.32 ft>s2
v=2vr2+vu2=25.3332+9.2382=10.7 ft>s
vu=ru
#
=4.619(2) =9.238 ft>svr=r
#
=5.333 ft>s
r
$|u=30° =4[(sec 30° tan 30°)(3) +22(sec3 30° +tan2 30° sec 30°)] =38.79 ft>s2
#
r|u=30° =(4 sec 30° tan 30°)(2) =5.333 ft>s
r|u=30° =4 sec 30° =4.619 ft
u=30°
=4[secu(tanu)u
$
+ u
# 2(sec3u°+tan2u°secu°)] ft>s2
u
$
=3 rad>s2
r
$
=4[secu(tanu)u
$
+ u
#
(secusec2uu
#
+tanu secu(tanu)u
#
)]
u
#
=2 rad>sr
#
=(4 secu(tanu)u
#
) ft>s
A
D
B
P
C
r
fi
(4 sec
)
ft
u
u
4 ft
12–183.
SOLUTION
Since
Thus,
Ans.a=
`
ar
`
=6.67 m>s2
au=0
u
$
=0
0=0+60 u
$
v
#
=r
#u
#
+ru
$
v=ru
#
=60u
$
au=ru
$
+2r
#u
#
=-6.67 m>s2
=0-60(0.333)2
ar=r
$
-r(u
#
)2
#
20 =60 u
#
n=2(nr)2+(nu)2
nu=ru
#
=60 u
#
nr=r
#
=0
n=20
r
$
=0
r
#
=0
r=60
A truck is traveling along the horizontal circular curve of
radius with a constant speed
Determine the angular rate of rotation of the radial line r
and the magnitude of the truck’s acceleration.
u
#v=20 m>s.
r=60 m
rfi60 m
·
u
u
*12–184.
A truck is traveling along the horizontal circular curve of
r
adius with a speed of which is increasing
at
Determine the truck’s radial and transverse
components of acceleration.
3m>s2.
20 m>sr=60 m
SOLUTION
Ans.
Ans.
au=
a
t=
3m
>
s2
a
r=-an=-6.67 m>s2
a
n=
n2
r=(20)2
60 =6.67 m>s2
at=3m>s2
r=60
rfi60 m
·
u
u
SOLUTION
#
u=5
r=100(2 -cos
u
)
r
=100 sin uu
#
=500 sin u
r
$
=500 cos uu
#
=2500 cos u
At u
=120°,
vr=r
#
=500 sin 120°=433.013
v
u=
ru
#
=
100 (2
-
cos 120
°
)(5)
=
1250
v
=2
(433.013)2
+
(1250)2
=
1322.9 mm
>
s
=
1.32 m
>
s Ans.
12–185.
The rod OA rotates counterclockwise with a constant
angular velocity of
u
#
= 5 rad
>
s. Two pin-connected slider
blocks, located at B, move freely on OA and the curved rod
whose shape is a limaçon described by the equation
r = 100(2 − cos
u
) mm. Determine the speed of the slider
blocks at the instant
u
= 120°.
A
u
O
B
r
y
x
·
u fi 5 rad/s
SOLUTION
u
#
=5
u
#
=0
r=100(2 -cos
u
)
r
#
=100 sin uu
#
=500 sin u
r
$
=500 cos uu
#
=2500 cos u
a
r=
r
$
-
ru
#2=
2500 cos u
-
100(2
-
cos u)(5)
2=
5000(cos 120
°-
1)
=-
7500 mm
>
s
2
a
s=
r
#
u
+
2ru
#
=
0
+
2(500 sin u)(5)
=
5000 sin 120
°=
4330.1 mm
>
s
2
a
=2
(
-
7500)2
+
(4330.1)2
=
8660.3 mm
>
s2
=
8.66 m
>
s2 Ans.
12–186.
Determine the magnitude of the acceleration of the slider
blocks in Prob. 12–185 when
u
= 120°.
A
u
O
B
r
y
x
·
u fi 5 rad/s
196
12–187.
The searchlight on the boat anchored 2000 ft from shore is
turned on the automobile, which is traveling along the
straight road at a constant speed of Determine the
angular rate of rotation of the light when the automobile is
from the boat.r=3000 ft
80 ft>s.
80 ft/s
ru
u
SOLUTION
r
#
=-2000 csc uctn u
#
r=2000 csc u
*12–188.
$
required angular acceleration of the light at this instant.u
a velocity of at the instant determine ther=3000 ft,80 ft>s
If the car
>
is accelerating at and has 15 ft s2
80 ft/s
ru
u
SOLUTION
At ,
Since
Then,
Ans.u
$
=0.00404 rad>s2
au=15 sin 41.8103° =10 m>s
au=3000 u
$
+2(-3354.102)(0.0177778)2
au=ru
$
+2r
#u
#
r
#
=-3354.102 u
#
u=41.8103°r=3000 ft
r
#
=-2000 csc uctn uu
#
r=2000 csc u
in Prob. 12–187
12–189.
A particle moves along an Archimedean spiral ,
where is given in radians. If (constant),
determine the radial and transverse components of the
particle’s velocity and acceleration at the instant
. Sketch the curve and show the components on
the curve.
u=p>2 rad
u
#
=4 rad>su
r
=(8
u
)ft
SOLUTION
Time Derivatives: Since is constant, .
Velocity:Applying Eq. 12–25, we have
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
Ans.
Ans.au=ru
$
+2r
#u
#
=0+2(32.0)(4) =256 ft>s2
ar=r
$
-ru
#2=0-4p
A
42
B
=-201 ft>s2
vu=ru
#
=4p(4) =50.3 ft>s
vr=r
#
=32.0 ft>s
r=8u=8ap
2b=4pft
r
#
=8u
#
=8(4) =32.0 ft>sr
$
=8u
$
=0
u
=0u
y
u
r(8 u)ft
r
12–190.
SOLUTION
Velocity: Applying Eq. 12–25, we have
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
Ans.
Ans.au=ru
$
+2r
#u
#
=4p(5) +2(32.0)(4) =319 ft>s2
ar=r
$
-ru
#2=40 -4p
A
42
B
=-161 ft>s2
vu=ru
#
=4p(4) =50.3 ft>s
vr=r
#
=32.0 ft>s
r
$
=8u
$
=8(5) =40 ft>s2
2b=4pft
Solve Prob. 12–189 i f the particle has an angular
acceleration when at rad.u=p2u
#
=4 rad>su
$
=5 rad>s2
y
u
r(8 u)f
t
r
>
Ans:
12–191.
SOLUTION
At ,
Ans.
Ans.a=2(-0.75)2+(0)2+(3.353)2=3.44 ft
>
s2
az=3.353
au=0+0=0
ar=0-3(0.5)2=-0.75
v=2(0)2+(1.5)2+(5.761)2=5.95 ft>s
vz=5.761
vu=3(0.5) =1.5
vr=0
z
$
=3.353
z
#
=5.761
z=-0.8382
t=3s
u
$
=0r
$
=0z
$
=-12 sin 2t
u
#
=0.5 r
#
=0z
#
=6 cos 2t
u=0.5 tr=3z=3 sin 2t
r
z
A
u
The arm of the robot moves so that r = 3 ft is constant, and
4
1
its grip
A moves along the path z = 3 sin u2ft, where u is in
radians. If ut= 10.5 2rad, where t is in seconds, determine the
magnitudes
of the grip’s
velocity and acceleration when t = 3 s.
