978-0133915426 Chapter 15 Part 6

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

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

1.5 m

1 m

3 kg

6 m/s

z

*15–100.

Each ball has a negligible size and a mass of 10kgand is

attached to the end of a rod whose mass maybeneglected.

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) Nm

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

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

90

v

A

70 km/h

v

B

r

B

57 m

r

A

60 m

y

z

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