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574
15–95.
Determine the angular momentum
H
p
of the 6-lb particle
about point P.
SOLUTION
y
z
8 ft
12 ft
P
4 ft/s
6 lb
B
A
575
*15–96.
Determine the angular momentum
Ho
of each of the two
particles about point O.
SOLUTION
y
O
B
P
1.5 m
5 m
1 m
4 m
4 kg
4 m
576
15–97.
Determine the angular momentum
H
p
of each of the two
particles about point P.
SOLUTION
y
O
B
P
1.5 m
5 m
1 m
4 m
4 kg
4 m
577
15–98.
Determine the angular momentum
HO
of the 3-kg particle
about point O.
SOLUTION
x
P
O
A
2 m
2 m
2 m
1.5 m
1.5 m
1 m
3 kg
6 m/s
z
578
15–99.
Determine the angular momentum
HP
of the 3-kg particle
about point P.
SOLUTION
x
P
O
A
2 m
2 m
2 m
1.5 m
1.5 m
1 m
3 kg
6 m/s
z
*15–100.
Each ball has a negligible size and a mass of 10kgand is
attached to the end of a rod whose mass maybeneglected.
If the rod is subjected to a torque M
=
(t
2+
2) N
#
m,
where t is in seconds, determine the speed of each ball when
t=3
s. Each ball has a speed v = 2 m
>
s when
t=0
.
SOLUTION
M (t2 2) N m
0.5 m
v
15–101.
The 800-lb roller-coaster car starts from rest on the track
having the shape of a cylindrical helix. If the helix descends
8 ft for every one revolution, determine the speed of the car
when
t=4 s
. Also, how far has the car descended in this
time? Neglect friction and the size of the car.
SOLUTION
u
8 ft
r 8 ft
15–102.
The 800-lb roller-coaster car starts from rest on the track
having the shape of a cylindrical helix. If the helix descends
8 ft for every one revolution, determine the time required
for the car to attain a speed of 60 ft
>
s. Neglect friction and
the size of the car.
8 ft
r 8 ft
15–103.
A4-lb ball Bis traveling around in a circle of radius
with a speed If the attached cord is pulled down
through the hole with a constant speed determine
the ball’s speed at the instant How much work has
to be done to pull down the cord? Neglect friction and the size
of the ball.
r2=2 ft.
vr=2ft>s,
1vB21=6ft>s.
r1=3ft
SOLUTION
Ans.
Ans.©U1-2=3.04 ft #lb
1
2(4
32.2)(6)2+©U1-2=1
2(4
32.2)(9.22)2
T1+©U1-2=T2
v2=292+22=9.22 ft>s
vu=9ft>s
4
32.2 (6)(3) =4
32.2 vu(2)
H1=H2
B
(vB)1=6ft>s
r1=3ft
*15–104.
Ans.
Ans.t=0.739 s
(3 -1.5213) =2t
¢r=vrt
r2=1.5213 =1.52 ft
4
32.2(6)(3) =4
32.2(11.832)(r2)
H1=H2
vu=11.832 ft>s
12 =2(vu)2+(2)2
A 4-lb ball Bis traveling around in a circle of radius
with a speed If the attached cord is pulled
down through the hole with a constant speed
determine how much time is required for the ball to reach
a speed of 12 ft/s. How far is the ball from the hole when
this occurs? Neglect friction and the size of the ball.
r2
vr=2ft>s,
1vB21=6ft>s.
r1=3ft
B
(vB)1=6ft>s
r1=3ft
584
15–105.
The two blocks A and B each have a mass of 400 g. The
blocks are fixed to the horizontal rods, and their initial
velocity along the circular path is 2 m
>
s. If a couple moment
of
M=(0.6) N #m
is applied about CD of the frame,
determine the speed of the blocks when
t=3 s.
The mass
of the frame is negligible, and it is free to rotate about CD.
Neglect the size of the blocks.
SOLUTION
M 0.6 N
m
C
A
B
0.3 m 0.3 m
Ans:
v=9.50 m>s
585
15–106.
SOLUTION
But,
Thus,
or, Q.E.D.u
$
+ag
Rbsin u=0
gsin u=-Ru
$
s=Ru
gsin u=-
dv
dt =-
d2s
dt2
Asmall particle having amass mis placed inside the
semicircular tube.The particle is placed at the position
shown and released. Apply the principle of angular
momentum about point and show that
the motion of the particle is governed by the differential
equation u
$
+1g>R2sin u=0.
O1©MO=H
#
O2,
R
O
u
Rb
586
15–107.
M (30t2) Nm
F 15t N
4 m
SOLUTION
Ans.v=3.33 m>s
0+L5 s
0
30 t2 dt +L5 s
0
15t(4)dt =150v(4)
Lt1
t2
If the rod of negligible mass is subjected to a couple
moment of and the engine of the car
supplies a traction force of to the wheels,
where tis in seconds, determine the speed of the car at the
instant .The car starts from rest. The total mass of
the car and rider is 150 kg. Neglect the size of the car.
t=5 s
F=(15t) N
M=(30t2) N #m
Ans:
v=3.33
m>s
587
*15–108.
When the 2-kg bob is given a horizontal speed of ,it
begins to rotate around the horizontal circular path A. If the
force Fon the cord is increased, the bob rises and then
rotates around the horizontal circular path B. Determine
the speed of the bob around path B. Also, find the work
done by force F.
1.5 m>s
A
B
300 mm
600 mm
SOLUTION
(2)
Eliminating Ffrom Eqs. (1) and (2) yields
(3)
When ,.Using Eq. (3), we obtain
Solving for the root 1, we obtain
Conservation of Angular Momentum: By observing the free-body diagram of
the system shown in Fig. b,notice that Wand Fare parallel to the zaxis,MShas
no zcomponent, and FSacts through the zaxis.Thus,they produce no angular
impulse about the zaxis.As a result, the angular momentum of the system is
conserved about the zaxis.When and ,
and .Thus,
(4)v2 sin u2=1.6867
0.3373(2)(1.5) =0.3 sin u2 (2)v2
r1mv1=r2mv2
A
Hz
B
1=
A
Hz
B
2
r=r2=0.3 sin u2
r=r1=0.6 sin 34.21° =0.3373 m
u=u2
u=u1=34.21°
u1=34.21°
6
cos2 u1+0.3823 cos u1-1=0
1-cos2 u1
cos u1
=1.52
9.81(0.6)
v=v1=5 m>sl=0.6 m
1-cos2 u
cos u=v2
9.81l
sin2 u
cos u=v2
9.81l
;©Fn=ma n;
F sin u=2av2
l sin ub
588
Ans.UF=8.32 N #m
1
2 (2)(1.52)+UF-2(9.81)(0.3366) =1
2 (2)(1.992)2
1
2 mv1 2+UF+
A
-Wh
B
=1
2 mv2 2
T1+©U1-2=T2
Ans:
v2=1.99 m>s
UF=8.32 N #m
589
15–109.
The elastic cord has an unstretched length
l0=1.5 ft
and a
stiffness
k=12 lb>ft
. It is attached to a fixed point at A and
a block at B, which has a weight of 2 lb. If the block is
released from rest from the position shown, determine its
speed when it reaches point C after it slides along the
smooth guide. After leaving the guide, it is launched onto
the smooth horizontal plane. Determine if the cord becomes
unstretched. Also, calculate the angular momentum of the
block about point A, at any instant after it passes point C.
SOLUTION
TB+VB=TC+VC
0+1
2
(12)(5 -1.5)2=
1
2a
2
32.2 b
v2
C+1
2
(12)(3 -1.5)2
vC=43.95 =44.0 ft>s
Ans.
There is a central force about A, and angular momentum about A is conserved.
HA=
2
32.2
(43.95)(3) =8.19 slug
#
ft2
>
s Ans.
If cord is slack
AD =1.5 ft
(HA)1=(HA)2
8.19 =
2
32.2
(vu)D(1.5)
(v
u
)
D
=88 ft>s
But
TC+VC=TD+VD
1
2a2
32.2 b
(43.95)2+
1
2
(12)(3 -1.5)2=
1
2a2
32.2 b
(vD)2+0
vD=48.6 ft>s
Since
vD 6 (vu)D
cord will not unstretch. Ans.
Ans:
vC=44.0 ft>s
H
A=
8.19 slug
#
ft
2>
s.
The cord will not unstretch.
BC
A
4 ft
3 ft
k 12 lb/ft
15–110.
The amusement park ride consists of a 200-kg car and
passenger that are traveling at 3 m
>
s along a circular path
having a radius of 8 m. If at
t=0,
the cable OA is pulled in
toward O at 0.5 m
>
s, determine the speed of the car when
t=4 s
. Also, determine the work done to pull in the cable.
SOLUTION
A
O
r
15–111.
r
A
2ft
B
v
r
4ft/s
r
B
1f
t
A
A
b
ox
h
av
i
ng a we
i
g
h
t of 8
lb
i
s mov
i
ng aroun
d
i
n a c
i
rc
l
e of
radius with a speed of while
connected to the end of a rope. If the rope is pulled inward
with a constant speed of determine the speed of
the box at the instant How much work is done
after pulling in the rope from Ato B? Neglect friction and
the size of the box.
rB=1ft.
vr=4ft>s,
1vA21=5ft>srA=2ft
SOLUTION
(nB)tangent =10 ft>s
(Hz)A=(Hz)B;a8
32.2 b(2)(5) =a8
32.2 b(1)(vB)tangent
592
*15–112.
60
60
90
v
A
70 km/h
v
B
r
B
57 m
r
A
60 m
y
z
55 m
55 m
A
B
u
A toboggan and rider, having a total mass of 1
5
0 kg, enter
horizontally tangent to a 90° circular curve with a velocity
of If the track is flat and banked at an angle
of 60°, determine the speed and the angle of “descent,”
measured from the horizontal in a vertical x–zplane,at
which the toboggan exists at B. Neglect friction in the
calculation.
uvB
vA=70 km>h.
SOLUTION
(1)
Datum at B:
(2)
Since
Solving Eq. (1) and Eq (2):
Ans.
Ans.u=20.9
vB=21.9 m>s
h=(rA-rB) tan 60° =(60 -57) tan 60° =5.196
1
2(150)(19.44)2+150(9.81)h=1
2(150)(nB)2+0
T
A+V
A=T
B+V
B
150(19.44)(60) =150(nB) cos u(57)
vA=70 km>h=19.44 m>s
Ans:
v
B=21.9 m>s
u=20.9
593
15–113.
SOLUTION
(1)
(2)
Solving,
Ans.
Ans.rB=13.8 Mm
v
B=10.2 km>s
-66.73(10-12)(5.976)(1024)(700)
rB
1
2 (700)[10(103)]2-66.73(10-12)(5.976)(1024)(700)
[15(106)]
=1
2 (700)(vB)2
1
2 ms (vA)2-GMe ms
rA
=1
2 ms (vB)2-GMems
rB
TA+VA=TB+VB
700[10(103) sin 70°](15)(106)=700(vB)(rB)
ms (vA sin fA)rA=ms (vB)rB
(HO)1=(HO)2
An earth satellite of mass 700 kg is launched into a free-
flight trajectory about the earth with an initial speed of
when the distance from the center of the
earth is If the launch angle at this position is
determine the speed of the satellite and its
closest distance from the center of the earth. The earth
has a mass Hint: Under these
conditions, the satellite is subjected only to the earth’s
gravitational force, Eq. 13–1. For part of
the solution, use the conservation of energy.
F=GMems>r2,
Me=5.976110242 kg.
rB
vB
fA=70°,
rA=15 Mm.
vA=10 km>s
A
rA
rB
vB
vA
f
Ans:
v
B=10.2 km>s
rB=13.8 Mm
