16–41.
Here , and ,,,.
Ans.
Ans.–3=-0.3[sin 50°(a)+cos 50°(8.70)2]a=-50.5 rad>s2
–2=-0.3 sin 50°(v)v=8.70 rad>s
u=50°u
$=au
#=vu
#=vvy=-2m>s, ay=-3m>s2
y
## =ay=-0.3
A
sin uu
$+cos uu
#2
B
#=vy=-0.3 sin uu
#
At the instant the slotted guide is moving upward
with an acceleration of and a velocity of .
Determine the angular acceleration and angular velocity of
link AB at this instant. Note: The upward motion of the
guide is in the negative ydirection.
2m>s3m>s2
u
=50°,
300 mm
y
a3m/s
A
B
V,A
U
16–42.
At the instant shown,
u=60°
, and rod AB is subjected to a
deceleration of 16 m
>
s
2
when the velocity is
10 m>s
.
Determine the angular velocity and angular acceleration of
link CD at this instant.
SOLUTION
u
x
=–
0.6 sin u
(
u
#
)
(1)
x
$=–
0.6 cos u
(
u
#
)
2–
0.6 sin u
(
u
$
)
(2)
u=60°
#=10 m>s
$=–16 m>s2
300 mm300 mm
D
B
x
A
v 10 m/s
a 16 m/s2
uu
16–43.
The crank AB is rotating with a constant angular velocity of
4 rad
>
s. Determine the angular velocity of the connecting
rod CD at the instant
u=30°.
SOLUTION
0.3 sec230
°
(2
–
cos 60.0
°
)
u
C
B
600 mm
300 mm
16–45.
Time Derivatives: Taking the time derivative of Eq. (1).we have
(2)
However and .From Eq.(2),
(3)
However, the positive root of Eq.(1) is
Substitute into Eq.(3),we have
Ans.
Note: Negative sign indicates that vis directed in the opposite direction to that of
positive x.
v=- r2
1vsin 2u
22r2
1cos2u+r2
2+2r
#1r2
+r1vsin u
x=r1cos u+2r2
1cos 2u+r2
2+2r1r2
v=r1xvsin u
r1cos u–x
0=xv –r1(vcos u–xvsin u)
v=du
dt
v=dx
dt
0=2xdx
dt –2r1
¢
–xsin udu
dt +cos udx
dt
≤
Determine the velocity of rod Rfor any angle of the cam
Cif the cam rotates with a constant angular velocity The
pin connection at Odoes not cause an interference with the
motion of Aon C.
V.
u
C
r
2
r
1
O
V
u
16–46.
The circular cam rotates about the fixed point O with a
constant angular velocity
V.
Determine the velocity v of the
follower rod AB as a function of
u
.
SOLUTION
u
2
2
(R
+
r)2
–
d2 sin2u
–v=–d sin u(v)–
d2 sin 2
u
2
2
(R
+
r)2
–
d2 sin2 u
v
v=vd
°
sin u+d sin 2u
2
2
(R+r)2–d2 sin2 u
¢
Ans.
AB
R
d
r
v
u
O
v
16–47.
Determine the velocity of the rod Rfor any angle of cam
Cas the cam rotates with a constant angular velocity The
pin connection at Odoes not cause an interference with the
motion of plate Aon C.
V.
u
SOLUTION
Ans.v=-rvsin u
x=-rsin uu
x=r+rcos u
u
R
C
r
O
V
678
16–49.
SOLUTION
(1)
(2)
When ,,
thus, (from Eq. (1))
(from Eq.(2))
Also,
At ,
Ans.
Ans.a
C
=-0.577 Lv2=0.577 Lv2c
aC=-Lsin 60°(–v)2+Lcos 60°(–1.155v2)+0+Lsin 60°(v)2
vC=L(cos 60°)(–v)–Lcos 60°(v)=-Lv=Lvc
sC=0
f=60°u=60°
aC=-Ls i n f(f
#)2+Lcos f(f
$)–Lc o s u(u
$)+Lsin u(u
#)2
vC=Lcos ff
#–Lcos uu
#
sC=Ls i n f–Ls i n u
f
$=-1.155v2
u
$=0
u
#=-f
#=v
f=60°u=60°
cos u(u
#)2+s i n uu
$+sinff
$+c o s f(f
#)2=0
s i n uu
#]+sinff
#=0
B a r AB r o t a t e s u n i f o r m l y a b o u t t h e f i x e d p i n Aw i t h a
consta nt a ng ula r velocity Determine the velocity and
acceleration of block C, at the instant u=60°.
V.
A
L
L
B
u
V
Ans:
v
C=L
v
c
a
C=
0.577
Lv
2
c
679
16–50.
The center of the cylinder is moving to the left with a
constant velocity v0. Determine the angular velocity
V
and
angular acceleration
A
of the bar. Neglect the thickness of
the bar.
SOLUTION
r
u
V
Ar
vOO
681
16–51.
SOLUTION
Time derivatives:
Ans.vB=h
dvA
x
#=h
dy
#
x=
¢
h
d
≤
y
The pins at Aand Bare confined to move in the vertical
and horizontal tracks. If the slotted arm is causing Ato move
downward at determine the velocity of Bat the instant
shown.
vA,
90°
y
d
h
B
A
θ
vA
Ans:
vB=
¢
h
d
≤
vA
682
*16–52.
or
(1)
(2)
Since
then,
(3)
Ans.
From Eq. (1) and (2):
(4)
(5)
Substituting Eqs. (1), (2), (3) and (5) into Eq. (4) and simplifying yields
f
$=
f
#2sin f–b
lv2sin u
cos
f
–sin ff
#2+cos ff
$=-ab
lbsin uu
#2
aC=v
#
C=l
#f
#sin f+l
#fcos ff
#+bcos uau
#b2
vC=bvDab
lbsin ucos u
D1–ab
lb2
sin2uT+bvsin u
f
#=ab
lbcos uv
D1–ab
lb2
sin2u
cos f=21–sin2f=A1–ab
lb2sin2u
cos ff
#=b
lcos uu
#
vC=x
#=lsin ff
#+bsin uu
#
sin f=b
lsin ulsin f=bsin u
The crank AB has a constant angular velocity . Determine
the velocity and acceleration of the slider at Casafunction
of . Suggestion: Use the xcoordinate to express the motion
of Cand the coordinate for CB.x 0 when .f=0°f
u
V
C
B
bl
x
x
y
A
θ
φ
ω
a
1–
a
b
lb
2
sin2 u
b
2
683
16–53.
If the wedge moves to the left with a constant velocity v,
determine the angular velocity of the rod as a function of .u
SOLUTION
However,.Therefore,
Time Derivative:Taking the time derivative,
(1)
Since point Ais on the wedge, its velocity is .The negative sign indicates
that vAis directed towards the negative sense of xA.Thus,Eq. (1) gives
Ans.u
#=vsin f
Lcos (f–u)
vA=-v
vA=x
#
A=-
Lcos (f–u)u
#
sin f
x
#
A=Lcos (f–u)(–u
#)
sin f
xA=Lsin (f–u)
sin f
sin
A
180° –f
B
=sinf
xA=Lsin(f–u)
sin
A
180° –f
B
L
v
f
u
Ans:
u
#
=
v
sin
f
L
cos
(f
–
u)
684
16–54.
The crate is transported on a platform which rests on
rollers, each having a radius r. If the rollers do not slip,
determine their angular velocity if the platform moves
forward with a velocity v.
SOLUTION
Time derivatives:
Ans.v=v
2r
sA=v=2ru
#Where u
#=v
sA=2sG=2ru
v
rv
Ans:
v=
v
2r
685
16–55.
Arm AB has an angular velocity of and an angular
acceleration of . If no slipping occurs between the disk D
and the fixed curved surface, determine the angular velocity
and angular acceleration of the disk.
A
V
SOLUTION
Ans.
Ans.a¿=(R+r)a
r
v¿=(R+r)v
r
(R+r)adu
dt b=radf
dt b
ds =(R+r)du=rdf
B
R
A
D
C
r
ωα
‘,
‘
Ans:
v′=
(R+r)
v
r
a′=
(R+r)
a
r
686
*16–56.
SOLUTION
Ans.
Ans.=15 (v2cos u+asin u)
(34 –30 cos u)1
2
–225 v2sin2u
(34 –30 cos u)3
2
aB=s
#=15 vcos uu
#+15v
#sin u
234 –30 cos u
+a–1
2b(15vsin u)a30 sin uu
#b
(34 –30 cos u)
3
2
vB=15 vsin u
(34 –30 cos u)1
2
vB=s
#=1
2(34 –30 cos u)–1
2(30 sin u)u
#
s=232+52–2(3)(5) cos u
At the instant shown, the disk is rotating with an angular
velocity of and has an angular acceleration of .
Determine the velocity and acceleration of cylinder Bat
this instant. Neglect the size of the pulley at C.
AV 3ft
5ft
A
V,AC
u
B
687
16–57.
SOLUTION
Ans.
Also;
Ans.
u=tan–111.485
5.196 =65.7°
vB=2(5.196)2+(11.485)2=12.6 in.>s
(vB)y=6 sin 30° +8.4853 =11.485 in.>s
(vB)x=-6 cos 30° =-5.196 in.>s
(vB)xi+(vB)yj=(–6 cos 30°i+6 sin 30°j)+(4k)*(1.5>sin 45°)i
vB=vG+v*rB>G
u=tan–111.485
5.196 =65.7°
vB=2(5.196)2+(11.485)2=12.3 in.>s
(+c)(vB)y=6 sin 30° +8.4852 =11.485 in.>s
(;
+)(vB)x=6 cos 30° +0=5.196 in.>s
vB=
6+[4(1.5
>sin 45°) =8.4852]
vB=vG+vB>G
At the instant shown the boomerang has an angular velocity
and its mass center Ghas a velocity
Determine the velocity of point Bat this
instant.
vG=6 in.>s.
v=4 rad>s,
45°
30°
1.5 in.
GB
=4rad/s
ω
v
G=6in./s
Ans:
vB=12.6 in.>s
65.7° b
16–58.
SOLUTION
the instant shown, vBis directed towards the negative y axis. Also, block Cis moving
downward vertically due to the constraint of the guide.Then vcis directed toward
negative y axis.
Velocity Equation:.,ereH
Applying Eq. 16–16, we have
Equating iand jcomponents gives
Ans.–4=2.598(0) –2vAB vAB =2.00 rad s
0=-1.50vBC
vBC =0
–4j=-1.50vBCi+(2.598vBC –2vAB)j
–4j=-2vAB j+(vBCk)*(2.598i+1.50j)
vC=vB+vBC *rC>B
rC>B={3 cos 30°i+3 sin 30°j}ft={2.598i+1.50j}ft
If the block at Cis moving downward at 4 ft
/
s, determine
the angular velocity of bar AB at the instant shown.
A
B
AB
C
2ft
3ftvC=4ft/s
30°
ω
689
16–59.
The link AB has an angular velocity of 3 rad
>
s. Determine
the velocity of block C and the angular velocity of link BC
at the instant
u=45°
. Also, sketch the position of link BC
when u
=60°, 45°,
and 30° to show its general plane motion.
SOLUTION
u=30°
1.5 m
0.5 m
vAB 3 rad/s
45
u
A
BC