*19–40.
l
ω
0
P
SOLUTION
Ans.
(b) From part (a) Ans.v=1
4v0=1
4(4) =1 rad s
v=1
4v0
c1
12 ml2dv0=c1
3ml2dv
vl=1.5 m.
0=4 rad>s,m=2 kg,
V0
A
thin rod of mass m has an angular velocity while
rotating
on a smooth surface. Determine its new angular
velocity
just after its end strikes and hooks onto the peg and
the
rod starts to rotate about P without rebounding. Solve
the
problem (a) using the parameters given, (b) setting
19–41.
Tests of impact on the fixed crash dummy are conducted
using the 300-lb ram that is released from rest at
and allowed to fall and strike the dummy at If the
coefficient of restitution between the dummy and the ram is
determine the angle to which the ram will
rebound before momentarily coming to rest.
ue=0.4,
u=90°.
u=30°,
SOLUTION
Ans. u=66.9°
1
2a300
32.2 b(7.178)2–300(10) =0 – 300(10 sin u)
T
2+V
2=T
3+V
3
v¿=7.178 ft>s
1:
+2
e=0.4 =v¿-0
0 – (–17.944)
v=17.944 ft>s
0–300(10 sin 30°) =1
2a300
32.2 b(v)2–300(10)
u
10 ft 10 ft
19–42.
z
A
G
D
E
C
B
I
0.3 m 0.3 m
0.3 m 0.3 m
0.1 m
0.1 m
uu
The vertical shaft is rotating with an angular velocity of
when .If a force Fis applied to the collar so
that ,determine the angular velocity of the shaft.
Also, find the work done by force F.Neglect the mass of
rods GH and EF and the collars Iand J.The rods AB and
CD each have a mass of 10 kg.
u=90°
u=0°3 rad>s
SOLUTION
19–43.
SOLUTION
Applying Eq. 19–17, we have
(1)
Coefficient of Restitution:Applying Eq. 19–20, we have
(2)
Solving Eqs. (1) and (2) yields
Thus, the angular velocity of the slender rod is given by
Ans.v2=(yB)2
2
=6.943
2
=3.47 rad
>
s
(yG)2=2.143 ft>s(yB)2=6.943 ft>s
0.8 =(yB)2–(yG)2
6–0
e=(yB)2–(yG)2
(yG)1–(yB)1
a3
32.2 b(6)(2) =0.2070c(yB)2
2d+a3
32.2 b(yG)2(2)
C
mb(yG)1
D
(rb)=Izv2+
C
mb(yG)2
D
(rb)
(Hz)1=(Hz)2
2
12 a5
32.2 b
The mass center of the 3-lb ball has a velocity of
when it strikes the end of the smooth 5-lb
slender bar which is at rest. Determine the angular velocity
of the bar about the zaxis just after impact if .e=0.8
(vG)1=6ft>s
(vG)16ft/s
2ft
z
2ft
O
B
A
*19–44.
SOLUTION
Ans.v=6.45 rad>s
+2(9.81)(–0.25) +5(9.81)(–0.7)
=1
2c1
12(2)(0.5)2+2(0.25)2+1
2(5)(0.2)2+5(0.7)2dv2
1
2c1
12(2)(0.5)2+2(0.25) +1
2(5)(0.2)2+5(0.7)2d(3.572)2+0
T
2+V
2=T
3+V
3
v=3.572 rad>s
2(2.4261)(0.25) +5(2.4261)(0.7) =c1
12(2)(0.5)2+2(0.25)2+1
2(5)(0.2)2+5(0.7)2dv
©(Hs)1=©(Hs)2
(vG)1=2.4261 m>s
0 +2(9.81)(0.3) +5(9.81)(0.3) =1
2 (2)(vG)2
1+1
2(5)(vG)2
1
T
0+V
0=T
1+V
1
The pendulum consists of a slender 2-kg rod AB and 5-kg
disk. It is released from rest without rotating.When it falls
0.3 m, the end Astrikes the hook S, which provides a
permanent connection. Determine the angular velocity of
the pendulum after it has rotated .Treat the pendulum’s
weight during impact as a nonimpulsive force.
90°
AB0.2 m 0.3 m
0.5 m
S
19–45.
SOLUTION
.The mass moment of inertia of the block about point Dis
. si )tcapmi eht retfa( kcolb eht fo ygrene citenik laitini ehT
Applying Eq. 18–18, we have
Conservation of Angular Momentum:Since the weight of the block and the normal
reaction Nare nonimpulsive forces, the angular momentum is conserves about
point D. Applying Eq. 19–17, we have
Ans.y
=
5.96 ft
>
s
ca 10
32.2 byd(0.5) =0.2070(4.472)
(myG)(r¿)=IDv2
(HD)1=(HD)2
v2=4.472 rad>s
1
2(0.2070) v2
2+5.00 =0+7.071
T2+V2=T3+V3
1
2IDv2
2=1
2(0.2070) v2
2
ID=1
12 a10
32.2 b
A
12+12
B
+a10
32.2 b
A
20.52+0.52
B
2=0.2070 slug #ft2
10(0.7071) =7.071 ft #lb
The 10-lb block slides on the smooth surface when the
corner Dhits a stop block S. Determine the minimum
velocity vthe block should have which would allow it to tip
over on its side and land in the position shown. Neglect the
size of S.Hint: During impact consider the weight of the
block to be nonimpulsive.
1ft
v
A
AB
CDS
BC
D
1ft
1031
19–46.
SOLUTION
(1)
a
(2)
(3)
Solving Eqs. (1)–(3) for hyields
Ans.h=7
5r
Require n2=v2r
0+(P)¢t(h)=c2
5mr
2+mr
2dv2
+(HA)1+©
1MAdt=(HA)2
0+P(¢t)=mn2
A
P
h
C
r
Determine the height hat which a billiard ball of mass m
must be struck so that no frictional force develops between
it and the table at A. Assume that the cue Conly exerts a
horizontal force Pon the ball.
5
19–47.
The pendulum consists of a 15-kg solid ball and 6-kg rod. If it
is released from rest when u
1=90°,
determine the angle u
2
after the ball strikes the wall, rebounds, and the pendulum
swings up to the point of momentary rest. Take
e=0.6.
100 mm
2 m
A
u
SOLUTION
1033
*19–48.
The 4-lb rod AB is hanging in the vertical position. A 2-lb
block, sliding on a smooth horizontal surface with a velocity
of 12 ft/s, strikes the rod at its end B. Determine the velocity of
the block immediately after the collision. The coefficient
of restitution between the block and the rod at Bis e=0.8.
SOLUTION
Here,.Applying Eq. 19–17, we have
[1]
Coefficient of Restitution:Applying Eq. 19–20, we have
[2]
Solving Eqs. [1] and [2] yields
Ans.
(vB)2=12.96 ft s:
(vb)2=3.36 ft s:
(:
+) 0.8 =(vB)2–(vb)2
12 –0
e=(vB)2–(vb)2
(vb)1–(vB)1
a2
32.2 b(12)(3) =0.3727c(vB)2
3d+a2
32.2 b(vb)2(3)
[mb(vb)1](rb)=IAv2+[mb(vb)2](rb)
(HA)1=(HA)2
v2=(vB)2
3
B
A
3ft
12 ft/s
Ans:
(vb)2=3.36 ft>sS
1034
19–49.
A
B
500 mm
100 mm
150 mm
u
The hammer consists of a 10-kg solid cylinder
C
and 6-kg
uniform slender rod AB.If the hammer is released from rest
when and strikes the 30-kg block Dwhen ,
determine the velocity of block Dand the angular velocity of
the hammer immediately after the impact.The coefficient of
restitution between the hammer and the block is .e=0.6
u=0°u=90°
SOLUTION
and
.Initially,.Since the hammer
rotates about the fixed axis,and
.The mass moment of inertia of rod AB and cylinder C
about their mass centers is and
.Thus,
Then,
Conservation of Angular Momentum: The angular momentum of the system is
conserved point A.Then,
(1)16.5vD–3.55v3=22.08
=30vD(0.55) –0.125v3–6[v3(0.25)](0.25) –0.025v3–10[v3(0.55)](0.55)
0.125(6.220) +6[6.220(0.25)](0.25) +0.025(6.220) +10[6.220(0.55)](0.55)
(HA)1=(HA)2
v2=6.220 rad>s
0+0=1.775v2
2+(–68.67)
T
1+V
1=T
2+V
2
=1.775 v22
= 1
2 (0.125)v22+1
2 (6)
C
v2(0.25)
D
2+1
2 (0.025)v22+1
2 (10)
C
v2(0.55)
D
2
T2=1
2IGABv2
2+1
2 mAB(vGAB)2
2+1
2 IGC v2
2+1
2 mC(vGC)2
2
IC=1
12 m(3r2+h2)=1
12(10)
C
3(0.052)+0.152
D
=0.025 kg #m2
IGAB =1
12 ml2=1
12 (6)(0.52)=0.125 kg #m2
(vGC)2=v2rGC =v2(0.55)
(vGAB)2=v2rGAB =v2(0.25)
T
1=0–6(9.81)(0.25) –10(9.81)(0.55) =-68.67 J
V
2=(V
g)2=-W
AB(yGAB)2–W
C(yGC)2=W
AB(yGAB)1+W
C(yGC)1=0
1035
.Thus,
(2)
S
olving Eqs. (1) and (2),
Ans.v3=0.934 rad>s(vD)3=1.54 m>s
(vD)3+0.55v3=2.053
0.6=(vD)3–
C
–v3(0.55)
D
3.421 –0
e=(vD)3–
C
(vP)x
D
3
C
(vP)x
D
2–(vD)2
:
+
v
3rGC =v3 (0.55);
Ans:
(vD)3=1.54
m>s
v
3=0.934
rad>s
1036
19–50.
The 20-kg disk strikes the step without rebounding.
Determine the largest angular velocity
v1
the disk can have
and not lose contact with the step, A.
200 mm
◊1
30 mm
A
SOLUTION
Ans:
v
1=7.17 rad>s
1037
19–51.
SOLUTION
. Applying Eq. 19–17, we have
(1)
Coefficient of Restitution:Applying Eq. 19–20, we have
(2)
Equating Eqs. (1) and (2) yields
Ans.u=tan–1
¢
A
7
5e
≤
tan2u=7
5e
5
7tan u=ecos u
sin u
y2
y1
=ecos u
sin u
e=–(y2sin u)
–y1cos u
e=0–(yb)2
(yb)1–0
y2
y1
=5
7tan u
(my1)(rsin u)=a2
5mr2bay2cos u
rb+(my2)(rcos u)
C
mb(yb)1
D
(r¿)=IGv2+
C
mb(yb)2
D
(r–)
(HA)1=(HA)2
v2=y2cos u
r
The solid ball of mass mis dropped with a velocity onto
the edge of the rough step. If it rebounds horizontally off
the step with a velocity , determine the angle at which
contact occurs. Assume no slipping when the ball strikes the
step.The coefficient of restitution is e.
uv2
v
1
r
v1
v2
u
Ans:
u=
tan–1
aA7
5
e
b
1038
*19–52.
25 mm
150 mm
A
G
1
The wheel has a mass of 50 kg and a radius of gyration of
125 mm about its center of mass G. Determine the
minimum value of the angular velocity of the wheel, so
that it strikes the step at Awithout rebounding and then
rolls over it without slipping.
1
SOLUTION
.The mass moment of inertia of the wheel about its mass
center is .Thus,
(1)
Conservation of Energy: With reference to the datum in Fig. a,
and .Since the
wheel is required to be at rest in the final position, .The initial kinetic energy
of the wheel is
.Then
Substituting this result into Eq.(1),we obtain
Ans.v1=3.98 rad>s
v2=3.587 rad>s
0.953125v2
2+0=0+12.2625
T
2+V
2=T
3+V
3
0.953125v2
2
T
2=1
2 m(vG)2
2+1
2IGv2
2=1
2(50)[v2(0.15)]2+1
2(0.78125)(v2
2) =
T
3=0
V
3=(V
g)3=W(yG)3=50(9.81)(0.025) =12.2625 JW(yG)2=0
V
2=(V
g)2 =
v1=1.109v2
50[v1(0.15)](0.125) +0.78125v1=50[v2(0.15)](0.15) +0.78125v2
(HA)1=(HA)2
IG=mkG
2=50(0.1252)=0.78125 kg #m2
(vG)2=v2r=v2(0.15)
V
V
1039
19–53.
25 mm
150 mm
A
G
1
The wheel has a mass of 50 kg and a radius of gyration of
125 mm about its center of mass G.If it rolls without
slipping with an angular velocity of before it
strikes the step at A,determine its angular velocity after it
rolls over the step.The wheel does not loose contact with
the step when it strikes it.
1=5 rad>s
SOLUTION
and .The mass moment of inertia of the wheel about its
mass center is .Thus,
(1)
Conservation of Energy: With reference to the datum in Fig. a,
and .The initial
kinetic energy of the wheel is
.Thus,
and .
Ans.v3=2.73 rad>s
19.37 +0=0.953125v3
2+12.2625
T
2+V
2=T
3+V
3
T
3=0.953125v3
2
19.37 JT
2=0.953125v2
2=0.953125(4.5082) =
1
2 (0.78125)v2=0.953125v2
T=1
2 mvG
2+1
2IGv2=1
2(50)[v(0.15)]2 +
V
3=(V
g)3=W(yG)3=50(9.81)(0.025) =12.2625 JW(yG)2=0
V
2=(V
g)2 =
v2=4.508 rad>s
50(0.75)(0.125) +0.78125(5) =50[v2(0.15)](0.15) +0.78125v2
(HA)1=(HA)2
IG=mkG
2=50(0.1252)=0.78125 kg #m2
v2=v2r=v2(0.15)0.75 m>s
V
V
v
3=2.73
rad>s
1040
19–54.
The rod of mass m and length L is released from rest
without rotating. When it falls a distance L, the end A strikes
the hook S, which provides a permanent connection.
Determine the angular velocity
v
of the rod after it has
rotated
90°.
Treat the rod’s weight during impact as a
nonimpulsive force.
A
L
L
S
SOLUTION
T1+V1=T2+V2
0+mgL =
1
2
mvG
2+0
v
G=2
2gL
H1=H2
m
2
2gL
aL
2b
=
1
3
mL
2(v2)
v2=
3
2
2
2gL
L
T2+V2=T3+V3
1
2a
1
3
mL
2
b9(2gL)
4L
2+0=
1
2a
1
3
mL
2
b
v2–mg
a
L
2b
3
4
gL =
1
6
L
2v2–g
a
L
2b
v=
A
7.5
g
L
Ans.
Ans:
v=
A
7.5
g
L
19–55.
SOLUTION
Ans.hB=2hG=0.980 ft
hG=0.490 ft
1
2c1
3a15
32.2 b (2)2d(4.865)2=0+15(hG)
T
3+V
3=T
4+V
4
v3=(vB)3
2=9.730
2=4.865 rad>s
(vB)3=-9.730 ft>s=9.730 ft>s
c
A
+T
B
e=0–(v
B
)3
(v
B
)2–0 ;
0.7 =0–(v
B
)3
13.90
v2=6.950 rad>s
Hence (v
B)2=6.950(2) =13.90 rad>s
0+15(1) =1
2c1
3a15
32.2 b(2)2dv2
2
T
1=V
1=T
2+V
2
T
he 1
5
-lb rod AB is released from rest in the vertical
position.
If the coefficient of restitution between the floor
and
the cushion at Bis determine how high the end
of the rod rebounds after impact with the floor
.
e=0.7,
2 ft
B
*19–56.
Aball having a mass of 8 kg and initial speed of
rolls over a 30-mm-long depression.Assuming
that the ball rolls off the edges of contact first A,then B,
without slipping,determine its final velocity when it
reaches the other side.
v2
v1=0.2 m>s
B
v2
v10.2 m/s
SOLUTION
1043
19–57.
A
so
lid
b
a
ll
w
i
t
h
a mass m
i
s t
h
rown on t
h
e groun
d
suc
h
t
h
at
at the instant of contact it has an angular velocity and
velocity components and as shown. If the
ground is rough so no slipping occurs, determine the
components of the velocity of its mass center just after
impact. The coefficient of restitution is e.
1vG2y1
1vG2x1
V1
SOLUTION
Coefficient of Restitution
(ydirection):
Ans.
Conservation of angular momentum about point on the ground:
c
Since no slipping
, then,
T
herefore
Ans.(yG)x 2 =5
7a(yG)x 1 –2
5 v1 rb
v2=
5a(vG)x 1 –2
5v1rb
7r
(vG)x2=v2 r
–2
5 mr2v1+m(vG)x 1r=2
5mr2v2+m(vG)x 2 r
+)
(HA)1=(HA)2
(
A
+T
B
e=0–(yG)y 2
(yG)y1–0
(yG)y2=-e(yG)y1=e(yG)y1
c
(v
G
)
y1
(v
G
)
x1
r
G
V
1
7a
5
1044
19–58.
The pendulum consists of a 10-lb solid ball and 4-lb rod. If
it is released from rest when determine the angle
of rebound after the ball strikes the wall and the
pendulum swings up to the point of momentary rest. Take
e=0.6.
u1
u0=0°,
0.3 ft
0.3 ft
2ft
A
θ
SOLUTION
Just before impact:
Since the wall does not move,
Ans.u1=39.8°
1
2(1.8197)(3.2685)2–4(1) –10(2.3) =0–4(1) sin u1–10(2.3 sin u1)
T
3+V
3=T
4+V
4
v¿=7.518
2.3 =3.2685 rad>s
(vP)=7.518 ft>s
a:
+be=0.6 =(vP)–0
0–(–12.529)
vP=2.3(5.4475) =12.529 ft>s
v=5.4475 rad>s
0+0+1
2(1.8197)v2–4(1) –10(2.3)
T
1+V
1=T
2+V
2
IA=1
3a4
32.2 b(2)2+2
5a10
32.2 b(0.3)2+a10
32.2 b(2.3)2=1.8197 slug #ft2
Ans:
u
1=
39.8°