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PROBLEM 13.1

A 400-kg satellite is placed in a circular orbit 6394 km above the surface

of the earth. At this elevation the acceleration o f gravity is 4.09 m/s .

Knowing that its orbital speed is 20 000 km/h, determine the kinetic

energy o f the satellite.

OLUTION

iven: Mass of satellite, m= 400 kg

peed of satellite, v = 20.0 x 103 km/h

nd: Kinetic energy, T

v= (20.0 x 103 km/h)^~ ~ (1000 m/km)

= 5555 m/s

T = ~mv2 = |( 4 0 0 kg)(5.555 x 103m/s'f

T= 6.17 x 109 N-m

ote: Acceleration o f gravity has no effect on the mass o f the satellite.

T = 6.17 GJ

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A 1000-lb satellite is placed in a circular orbit 3000 mi above the surface

of the earth. At this elevation the acceleration o f gravity is 8.03 ft/s2.

Knowing that its orbital speed is 14,000 mi/h, determine the kinetic

energy o f the satellite.

PROBLEM 13.2

OLUTION

ven: Weight of satellite, W= 1000 lb

peed of satellite, v = 14,000 mi/h

nd: Kinetic energy, T

v =(14,000 mi/h)(5280 ft/mi)" h '

v3600 s= 20,533 fit/s

Mass of satellite = -X - = 31.0559 lbs2/ft(32.2 ft/s2 j

T = m v 2 = ^(31.0559)(20,533)2 = 6.5466 x 109 lb-ft

T= 6.55 x 109 lb-ft

ote: Acceleration of gravity has no effect on the mass of the satellite.

T = 6.55 x 109 lb-ft

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PROBLEM 13.3

An 8-lb stone is dropped from a height h and strikes the ground with a

velocity o f 85 ft/s. (a)Find the kinetic energy of the stone as it strikes the

ground and the height h from which it was dropped. (b) Solve part a,

assuming that the same stone is dropped on the moon. (Acceleration of

gravity on the moon = 5.31 f t/s ").

SOLUTION

Given: Weight of stone, W - 8 lb

Velocity of stone, v = 85 ft/s

Acceleration of gravity on the Moon, g, = 5.31 ft/s2

Find: (a) Kinetic energy, T

Height h,from which stone was dropped

(b) Tand hon the Moon

(a) On the earth

T = 1 mv2 = 1 f 8 lb , 1(85 ft/s)2 = 897.52 lb-ft2 2 V32.2 ft/s2J V

T = 898 lb-ft 0 + C /,6 - C/frjction = h ~ T, = - ^ [ v , 2 - V2 ]2 g L J

28.0208 - 8.282209 - (/friction = 1f 16 lb

\32.2 ft/s"

C/friction = -28 .0208 + 8.282209 + 17.88820

= -1.8504 ft -lb

C/fric,io = 1.850 ft-lb

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PROBLEM 13.11

Boxes are transported by a conveyor belt with a velocity v() to a fixed

incline atA where they slide and eventually fall off at B. Knowing that

/jk = 0.40, determine the velocity of the conveyor belt if the boxes leave

the incline atBwith a velocity o f 2 m/s.

SOLUTION

Given: AtA, v = v0

\W ForAB, nk= 0.40

A .V AtB, v - 2m/s\

\ Find: v0

y

ta=^ mvo Th=i mv\ =^m(2m/s)2

TB=2 m

N UA_B =(IVsin 15 - nkN)(6 m)

\ S F = 0 N - fFcos 15 = 0

N =IFcos 15

UA_B= FF(sin 15 - 0.4 0cos l5)(6 m)

Va-b = -(0.7653 \)W =-0.76531mg

TA + U A -B = T B

j/nvo - 0.7653\mg= 2 m

v02 = (2)(2 + (0.76531)(9.81 m/s: ))

vj; = 19.0154

v0 = 4.36 m/s A

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Boxes are transported by a conveyor belt with a velocity v0 to a fixed

incline atA where they slide and eventually fall off at B. Knowing that

Hk = 0.40, determine the velocity of the conveyor belt if the boxes are to

have a zero velocity atB.

PROBLEM 13.12

OLUTION

v = v

V= 0

Hk = 0.40

ta = 2 OTVo Tb = 0

UA_B = ( f fs i n l 5 - ^ ) ( 6 m )

\ I F = 0 N - ffcosl5 = 0

N = cos 15

Ua-b = ^ (s inl5-0 .40cos l5) (6m)

UA_B = -(0 .76 53 l)ff = -0.7653 lwg

Ta+ TJ A_ B = Tb

mvs - 0.7653Img = 0

vs =(2)(0.76531)(9.81 m/s2)

vs = 15.015

v0 = 3.87 m/s ^

Down to the left.

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PROBLEM 13.13

A . " x

' " l V * I " T "

830l

LA

In an ore-mining operation, a bucket full of ore is suspended from a

traveling crane which moves slowly along a stationary bridge. The bucket

is to swing no more that 12 ft horizontally when the crane is brought to a

sudden stop. Determine the maximum allowable horizontal speed v of the

crane.

SOLUTION

I

A * T ~ I

Given: Crane moves at velocity, v and stops suddenly.

Bucket is to swing no more than 12 ft horizontally.

Find: Maximum allowable velocity v

v, = v v2 = 0

t * 2

U\_2 = -mgh d = 12 ft

AB2 = (30 ft)2 = d 2 + y 2 = (12 ft)2 + y 2

y 2 = 900 - 144 = 756 y = V756

h =30 - y = 30 - J l5 6 = 2.5045 ft

T\+ Ui_2 =T2

^m v2 - mg(2.5045) - 0

v2 = 2g(2.5045) = 2(32.2)(2.5045)

v2 = 161.289

v = 12.6999

v = 12.70 ft/s

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PROBLEM 13.14

In an ore-mining operation, a bucket full of ore is suspended from a

traveling crane which moves slowly along a stationary bridge. The crane

is traveling at a speed of 10 ft/s when it is brought to a sudden stop.

Determine the maximum horizontal distance through which the bucket

will swing.

SOLUTION

Given: Crane moves at velocity, v = 10 ft/s and stops suddenly.

Find: Maximum horizontal distance, dmoved by the bucket.

N I, Refer to Problem 13.13 free body.1 I il i

t 1 2 1 w tin AV2 caT, = mv = ------- (10 f t ) = 50

' 2 2 g g

T, = 0 f/,_ 2 = -Wh

7] + u x 2 = r, 50 - Wh = 0g

h = = = 1.55279 ftg 32.2

AB~ = (30)2 = d 2 + y 2 = d 2+ (30 - 1.55279)2

d 2 = 90.7562 d = 9.5266

d =9.53 ft

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PROBLEM 13.15

10 m/s CarB is towing carAwith a 5-m cable at a constant speed of 10 m/s on

an uphill grade when the brakes of car B are fully applied causing it to

skid to a stop. Car A, whose driver had not observed that car B was

slowing down, then strikes the rear of car B.Neglecting air resistance

and rolling resistance and assuming a coefficient o f kinetic friction of 0.9,

determine the speed o f carAjus t before the collision.

SOLUTION

CarB:Given: CarBtowing carAuphill at a constant speed of 10 m/s

CarBskids toA stop. p k =0 .9

CarA strikes rear of carB.

Find: Speed of carAbefore collision, vA

Car .4:

VTlCj 0 5*

Let d = Distance traveled by carBafter braking.

; ^ 1 - 2= T2 - 7j - = (-mg sin5 - F ) d

d =

d =

pi gsin5 + to.9pigcos5

50 50

9.81 (sin 5 + 0.9cos5) (9.8l) (0.9837)

d= 5.181m traveled byB

For carA,travel to contact

t/,m m 1 2 1 2

1 - C = T A ~ T \ = - Z m V A -2 * 2

{-pigsm5)(d+ 5) = ^ p i v 2A - ~ p f ( l 0 ) 2

L v2a- 50 = (-9.81sin5)(5.181 + 5)

~ v 2 = 41.2952 A

vA = 9.087

v. = 9.09 m/s M

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10 m s

PROBLEM 13.16

in,nfS_ Car B is towing carA at a constant speed of 10 m/s on an uphill grade

when the brakes of carAare fully applied causing all four wheels to skid.

The driver of carB does not change the throttle setting or change gears.

The masses of the cars A and B are 1400 kg and 1200 kg, respectively,

and the coefficient of kinetic friction is 0.8. Neglecting air resistance and

rolling resistance, determine (a) the distance traveled by the cars before

they come to a stop, (b) the tension in the cable.

SOLUTION

F =0.8Na

Given: CarBtows carAat 10 m/s uphill.

CarAbrakes for 4 wheels skid, n k =0.8

CarBcontinues in same gear and throttle setting.

Find: (a)Distance, d,traveled to stop

(b) Tension in cable

(a) Fj = traction force (from equilibrium)

Fl = (I400g)sin5 + (I200g)sin5

= 2600(9.8l)sin5

For system: A + B

Ux_2 = [(Fj - 1400gsin5 - 1200gsin5) - F~\d

= T2 - T x =0 - ~ m A+Bv2 = -1(2600)(10)2

Since (Fj - 1400gsin5 - 1200gsin5) = 0

- F d = -0 .8 [1400 (9.81) cos 5]d = -130,000 N m

d =11.88 m -^

(b) Cable tension, T

Ux_2 = [ T - 0 .8^](11.88) = T2 -7 i

(T- 0 .8 (l40 0) (9 .8 l)co s5 ) ll .8 8 = - ^ ^ ( l 0 ) :

(F - 10945) = -5892

= 5.053 kN

T =5.05 kN

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fiomtfh A trailer truck has a 4400-lb cab and an 18,000-lb trailer. It is traveling

on level ground at 60 mi/h and must slow down to a stop in 3000 ft.

Determine (a) the average braking force that must be supplied, (b) the

average force in the coupling if 60 percent of the braking force is

supplied by the trailer and 40 percent by the cab.

PROBLEM 13.17

SOLUTION

W t |W ,

B

O M O

Given: 4400 lb cab, 18,000 lb trailer on level ground.

Truck comes to a stop in 3000 ft.

60% of braking force from trailer.

40% o f braking force from cab.

Find: (a) Average braking force

(b)Average force in the coupling

(a) Trailer and cab together

V] = (60 mi/h)5280 ft

f l h 1I mi J 1,3600s j

V, = 88 ft/s v2 = 0

88 ft/s

T\+Ux_2 =T2 Tx= ( m T + mc )(v1)2

1

32.2 ft/s2(4,400 lb + 18,000 lb)(88 ft/s)2

= 2.6936 x 106 ft-lb

T2 = 0

7j + Ux_2 = T2 (2.693 x 106 ft-lb) - (3000 ft)Fb = 0

Fb = 897.86 Fb= 898 lb

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PROBLEM 13.19

The system shown, consisting of a 20-kg collar A and a 10-kg

counterweight B, is at rest when a constant 500-N force is applied to

collarA. (o) Determine the velocity ofA just before it hits the support at

C. (b) Solve part a assuming that the counterweight B is replaced by a

98.1-N downward force. Ignore friction and the mass of the pulleys.

OLUTION

Given: System at rest when 500 N force is applied to collarA.No

friction. Ignore pulleys mass.

Find: (a)Velocity, vA ofAjust before it hits C.

*6

IQfej Kinematics

(b) vA If counter weightBis replaced by a 98.1 N

downward force.

X . = 2 X ,

vH = 2vA

(a) Blocks A and B

7j = 0 T ^ 2 ^ 22 = ~ m H VH + ~ m A VA

72 = i ( l 0 k g ) ( 2 v /1)2 + i (2 0 k g ) (v /12)

T2 =(30 kg)(v 4)2

Ux_2 = (500)*.., + {WA){X A) - [WB)(X B)

Ux_2 =(500 N)(0.6 m) + (20 kg x 9.81 m/s2)(0.6 m)

- (lO kg x 9.81 m/s: )(l.2 m)

/,_2 = 300 + 117.72 - 117.72 = 300 J

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7j + Ux_2 = T2 0 + 300 J = (30 kg)v^

VA = 10

vA = 3.16 m/s A

(b) Since the 10 kg mass atBis replaced by a 98.1 N force, kinetic

energy at is,

T 2 = \ m A = ^ { 2 0 k gy A 7 1 = 0

The work done is the same asin part (a)

U{_2= 300 J

71 + t/,_2 =T2 0 + 300 J = (10 kg)v^

va = 30

Va

= 5.48 m/sM

PROBLEM 13.19 CONTINUED

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PROBLEM 13.20

The 10-kg blockA and the 4-kg block Bare both at a height h = 0.5 m

above the ground when the system is released from rest. After A hits the

ground without rebound it is observed that B reaches a maximum height

of 1.18 m. Determine (a) the speed of A just before impact, (b) the

rhL

amount of energy dissipated by axle friction in the pulley.

OLUTION

@l a !

( I 0 ,1? Y*i BI___

Given: mA = 10 kg; mB = 4 kg; A = 0.5 m

System released from rest.

BlockAhits the ground without rebound.

BlockBreaches a height of 1.18 m.

Find: (a) vAjust before blockAhits the ground.

(b) Energy, Er , dissipated by the pulley friction.

(a) vB at = vA at just before impact,

from to ; BlockB

h = T2 = j i n bvb =^ ( 4)vl = 2vb

Tension in the cord is zero, thus

U2_3 = - (4 kg)(9.81 m/s2 j(0.18 m) = -7.0632 J

T, + U2_3 = Tj; 2v | = 7.0632; v | = 3.5316

v2 = v2 = 3.5316 vB = vA = 1.8793 vA = 1.879 m/s 0.25 The blocks move.

Constraint: vA + 3vs = 0

(a) U\_2 = mAg( sin 30 )(d A) - mAg(cos 30)M k{d A)

- mBg (s in 3 0 ) f^ -] - mBg(co s30 )//ifeV J J \ 3 J

= 10(9.81)(0.5)(0.5) - 10(9.81)(0.866)(0.2)(0.5)

- 8(9.81)(0.5)[ ~ | - 8(9.81)(0.866)(0.2)

= 17.985-10.761 = 7.224 N-m

(Gravity) (Friction)

^0.5^1

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T\ = > T2 = j m Av2A + = \ [ W ) v2a+

= 5.444v> = Ux_2

vA = 1.152 m/s

vA= 1.152 m/s -4

(b) Aalone:

u i- 2 =w^ ( s i n 3 0 ) ( ^ ) - M Ag(co s30o){ik (dA) - T (d A)

= M { va)2

Ux_2 =10(9.81)(0.5)(0.5) - 10(9.81)(0.866)(0.2)(0.5) - T(0.5)

= ^(1 0)(1.152)2 = 6.6348

T= 18.789 N

T =18.79 N ^


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Four 3-kg packages are held in place by friction on a conveyor which is

disengaged from its drive motor. When the system is released from rest,

package 1 leaves the belt atA just as package 4comes onto the inclined

portion of the belt at B. Determine (a) the velocity of package 2 as it

leaves the belt atA,(b) the velocity of package 3as it leaves the belt atA.

Neglect the mass o f the belt and rollers.

PROBLEM 13.23

SOLUTIONGiven: Conveyor is disengaged, packages held by friction and system is released from rest. Neglect mass of

elt and rollers. Package 1 leaves the belt as package 4 comes onto the belt.

ind: (a) Velocity of package 2 as it leaves the belt atA.

(b) Velocity of package 3 as it leaves the belt atA.

a) Package 1 falls of f the belt, and 2, 3, 4 move down.

0

= 0.8 m3

T2 = 3H]

T2 = 2 (3 ks )v2

T> = 4.5v;

Ux_2 =(3)(1F)(0.8) = (3)(3 kg) x 9.81 m/s2(0.8)

t/,_2 = 70.632 J

7] +(/,_ 2 = T2 0 + 70.632 = 4.5v|

v| = 15.696

v2 = 3.9618 Vt = 3.96 m/sA

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) Package 2 falls of f the belt and its energy is lost to the system, and 3 and 4 move down 2 ft.


t b ' i

n = (2) - m v .

T{ = (3 kg)(l5.696)

7| = 47.088 J

T3 = (2)1 2- m v ,2 3

(3kg)(vf)

r3 = 3v|

U2- 3 = 2( F )(0 .8 ) = (2 )(3 kg X9.81 m/s2)(0.8 m)

U2_3 = 47.088 J

2| + U2_3 = r 3 = 47.088 + 47.088 = 3v32

v3 = 31.392

v3 = 5.6029

v3 = 5.60 m/s ^

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PROBLEM 13.24

Two blocksA andB, of mass 8 kg and 10 kg, respectively, are connected

by a cord which passes over pulleys as shown. A 6-kg collar C is placed

on block A and the system is released from rest. After the blocks move

1.8 m, collar C is removed and blocks A and B continue to move.

Determine the speed o f blockAjust before it strikes the ground.

SOLUTIONGiven: mA = 8 kg; mB = 10 kg; mc = 6 kg

System released from rest.

Collar C removed after blocks move 1.8 m.

Find: vA, just before it strikes the ground.

Posit ion 1 to position 2

V| = 0 7j = 0

At 2, before C is removed from the system

T2 = + mB + mc) v2

Ti = ~(24 kg)v; = \2vj

u \-2 = (mA + mc ~ a )* (L 8 m)

t/,_2 = (8 + 6 - 10)g( l.8 m) = 70.632 J

Tx + t/ ,_ 2 = T2; 0 + 70.632 = 12v,2

vf = 5.886

Position 2 to position 3

1 1ftn = ~{ m A + mBy 2 = (5.886) = 52.974

T3 = ^ (mA + Wfl)V3 = 9v3

Ur _3 = (mA - mB)g ( 2 - 0.6) = (-2 kg)(9.81 m/s2)(l.4 m)

U2._3 = -27.468 J

T; + U2._3 = T3 = 52.974 - 27.468 = 9vj

v32 = 2.834 v3 = 1.68345 vA =1.683 m/s 4

0-6

I .*2.

r -

r

s u m

u .Bi

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PROBLEM 13.25

0.71b

0.51b

A 0.7-lb block rests on top of a 0.5-lb block supported by but not attached

to a spring of constant 9 lb/ft. The upper block is suddenly removed.

Determine (a) the maximum velocity reached by the 0.5-lb block, (b)the

maximum height reached by the 0.5-lb block.

LUTIONGiven: A 0.7 lb block rests on a 0.5 lb block which is not attached to a

spring of constant 9 lb/ft; upper block is suddenly removed.

VvJ 0 J k

D 5 - ^ _ a

T T

0 - 5 U , |

%

Find: (a) vmax o f 0.5 lb block

(b) maximum height reached by the 0.5 lb block

At the initial position (1), the force in the spring equals the weight of both

blocks, i.e., 1.2 lb.

Thus at a distance .v, the force in the spring is,

Fs = 1 .2 - k x

Fs = 1.2 - 9 *

Max velocity of the 0.5 lb block occurs while the spring is still in contact

with the block.

CD

7; = o05

g )

2 0.25 2v = v

9 ,t/|_2 = ( l .2-9x) x = 0.077778 ftdx

max 0.7(0.077778) - -(0.07777 8)2

v-^ = 3.5063

vmax = 1.87249

vmax = 1.872 ft/s^

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b)

l_ 0-S4fa

1

I f.

O .S J t

o

x0 = Initial compression

1.21bXn = 0.133333 ft

9 lb/ft

Fs = \ . 2 - 9 x

t x = o, r3= o

t/ ,_3 = Fsdx - 0.5F

I/ ,_3 = ( l.2-9x)fc-0.5A

9 9= 1.2x0 - jc0 - 0.5A

= 1.2(0.133333) - |(0.1 33 33 3)2 - 0.5h

= 0.08 - 0.5/i

7; + (/ l 3 = r3: 0 + (0.08 - 0.5/*) = 0

h =0.16 ft

h =1.920 in. 4

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Solve Prob. 13.25, assuming that the 4-lb block is attached to the spring.

PROBLEM 13.26

OLUTION

) Same as 13.25 solution for Part (a) vmax = 1-872 ft/s ft1 2 ,

(/ ,- 2 = Wx + - k A ' j _ f , 1 2 ,

- k Ai f

UU2 =20*+ - (144)

2 *

12

2

- 2 ( 1 4 4 )12

\ 2

2 ( 1 4 4 )

a jc ^ 2

32.2

cbc 32.2U v

( 1 'I ' x 'X --------- -144

I 127 J ,

= 20 - (144)

- 20 - 144* + 12 - 36*

x = = 0.177778 ft(= 2.1333 in.)180 v '

For * = 0.177778 ft

U!_2 = 20(0.177778) + 0.5 - 72(0.094444)2 - 0.56889

3.5556 + 0.5 - 0.64222 - 0.56889 = -2

20

32.2

vmax = 3-0264

vmax = 3.03 ft/s

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PROBLEM 13.31

An uncontrolled automobile traveling at 100 km/h strikes squarely a

highway crash cushion of the type shown in which the automobile is

" brought to rest by successively crushing steel barrels. The magnitude Fof

the force required to crush the barrels is shown as a function of the distance

xthe automobile has moved into the cushion. Knowing that the automobile

has a mass of 1000 kg and neglecting the effect of friction, determine

(a) the distance the automobile will move into the cushion before it comes

to rest, (b)the maximum deceleration of the automobile.

U,_2 = (80k N )(l.5m) + (l20k N )(rf-1 .5) = 120 + 120

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PROBLEM 13.32

A piston of mass mand cross-sectional areaA is in equilibrium under the

pressurep at the center of a cylinder closed at both ends. Assuming that the

piston is moved to the left a distance a/2 and released, and knowing that the

pressure on each side of the piston varies inversely with the volume,

determine the velocity of the piston as it again reaches the center of the

cylinder. Neglect friction between the piston and the cylinder and express

your answer in terms o f m, a, p,andA.

OLUTIONessures vary inversely as the volume

tially at

t ,

I I i L _

^ =AaP Ax

Aa

a.

Pa

P A ( la - x) f (2a - x)

v = 0 x =

x = a, T2 = mv

i-2 = Jl{P L - P R) A d x = ^ P a A2 2

Wj. 2 = paA lnx + In(2a - x) fa

1 1

x 2a - xdx

i-2 = PaA lna + lna - Ina , 5a

In

2 3a2f 4 lMj_2 =paA lna - In----- = paAIn

4 v3 ,,

T\ + C/j_2 = T2 0 + paA\n.f 4\ 1 2 mv

2

2paA\n

0.5754m

paA

mv = 0.759 . 1 ^ - 1 2--------------------- = - j r fvA - -jrfvo

(b)

GM = 9.81(6,370,000)2, = v02 +2GM

vA = (10,400)2 + 2GM

1000

1 1

1 1

1000

vA =9.6082 km/s

2GM

8623 7360

v., = 9.61 km/s A

n = vo +1000

Vg =(10,400 )2 + 2GM1000

vtt = 7.3536 km/s

J 1_

rB r0

1 1

14720 7360

vB = 7.35 km/s A

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PROBLEM 13.37

A 0.75-lb brass (nonmagnetic) block A and an 0.5-lb steel magnet Bare

in equilibrium in a brass tube under the magnetic repelling force of

another steel magnet C located at a distance jc = 0 .15 in. from B. The

force is inversely proportional to the square of the distance between B

and C. If block A is suddenly removed, determine (a) the maximum

velocity of B, (b) the maximum acceleration of B. Assume that air

resistance and friction are negligible.

SOLUTION

( a )

For x = 0.15 m = 0.0125 ft

F =-------- ----- T = 6400(0.0125 ft)'

= 6400 k/ft2

ZF = -W A - W B + F = 0

= -1.25 lb + 6400 k/ft2 = 0

k = 195.3 x 10'6 lb -ft2

1 2 0.5 2 0.25 i 7] = 0, V, = 0, T2 = mBv = v = v* lb-ft

2 2g g

For max v,d 0.25

v- = 0dv\_ g

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) Max acceleration at

-0 .5 = 0, x2 = =2fl95.3 x 10"6)0 5 V Ix2 0.5

x = 0.01976 ft

.01976( k 0.25 20125 ~~2 I _ Vmax

\X J g

- 0.5xX

100,976 0.25 2

0.0125 8

Vmax = 4g (0.00211) = 0.2718

vmax = 0.5213 ft/s

x = 0.0125 ft

v i t ^ u rI F = mBa : - W B = - a B* g

195.3 x 10' 6 A _ 0.5 = 0.5 = aR

(0.0125) 32.2

aB =48.295 ft/s2

vmax = 0.521 ft/s

W+ / Nc - WcosG = -----^

8 R

Nc = (250 kg x 9.81 m/s2)

/ 2 \

cos# +

SR j

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Nd= (250 x 9.81)


( v2A1+ -

gR

ce

rk and energy fro m A to D

vD > vc and cos# < 1, N D > N c

va = > t a = 0 Td = i m v2D = \ 2 5 v p

Ua - d = w ( 21 + 18) = (25 kg)(9.81 m/s2)(45 m) = 110362.5

7a + U a - d = t d 0 + 110362.5 = 125vq

Nd = 250g 1+72gJ

v2d = 882.90

= 250(9.81) 1+882.90

= 5518.1 N72(9.81)

= b = 73 1N ;iV max = AD =5520 N 4

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PROBLEM 13.45

A small block slides at a speed v = 3 m/s on a horizontal surface at a

height h = 1 m above the ground. Determine (a) the angle 9 at which it

will leave the cylindrical surface BCD, (b) the distancexat which it will

hit the ground. Neglect friction and air resistance.

OLUTION

)

aTh

4- JloCos e

W Irvig

N=o S '>v.4u *

lock leaves surface at C when the normal force N = 0

+/ mgcos9 = man

vrgcos9 =

Work-energy principle

se Equation (1)

v2 = gh cos 9 = gy

t b = mvc UB-c = y) = m g ( h - y c )

7# + Ub-c - Tc

4.5m + mg(h - y ) = /hv2-

14.5 + g (h - y ) = - g y c

(1)

(2)

4.5 +gh = gyc

(4.5 + gh)yc = ------------

(I")

(4.5 + (9.8|)(|)2)

y =0.97248 m (3)

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yc = hcos9 cos# = = _ q 97248h 1m

9 = cos '0.97248 = 13.473


(b)

From Equations (1) and (3)

AtC;

AtE:

AtE:

vc = = ^9.81(0.97248) = 3.0887 m/s

(vc )t = vc cos# = 3.0887cos 13.47

= 3.0037 m/s

(vc ) ( = -v r sin# = 3.0887 sin 13.47

= -0.71947 m/s

y = yc + (vc )) t - I g r = 0.97248 - 0.71947f - ^ (9 .8l) /2

y E = 0: 4.9 05 r + 0.7194r - 0.97248 = 0

t= 0.37793 s

x = /icos# + (vc )xt = l(sin 13.47) + 3.0037(0.37793)

= 0.23294 + 1.3519 = 1.3681m

9 = 13.47 A

x =1.368 m A

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PROBLEM 13.46

A small block slides at a speed V on a horizontal surface. Knowing that

h = 8 ft, determine the required speed o f the block if it is to leave the

cylindrical surfaceBCDwhen 6 = 40.

OLUTION

From Problem 13.45, block leaves the surface when N = 0

Vrgcos# =

h

h= 8 ft, G = 40

Vvc = % co s# = 8(32.2)(cos4 0) = 197.33

^ * B 0 = 4 0 Work-energy principle

T ^ t t - c o s

J F [ X c T 1 2B = - rc = ii m = -im(l97.33) = 98.667m

Ub-c = mgh{l ~c osG)

+ UB -C ~ Tc

jmv2 + mg/j(l - cosG) - ~^mvc

v 2c = 197.33 - 2 g h ( \ - c o s G )

= 197.33 - 2(32.2)(8)(1 - cos40) = 76.796

vc = 8.76 ft/s

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PROBLEM 13.47

(a)A 60-kg woman rides a 7-kg bicycle up a 3 percent slope at a constant

speed of 2 m/s. How much power must be developed by the woman?

(b) A 90-kg man on a 9-kg bicycle starts down the same slope and

maintains a constant speed of 6 m/s by braking. How much power is

dissipated by the brakes? Ignore air resistance and rolling resistance.

SOLUTION

f t )

t o o

6

M U

tan 9 = -----100

9 = 1.718

W = Wg + Ww = (7 + 60) kg X9.81 m/s2

W = 657.3 N

pw = i V v = (IF sin #)(v )

Pw = (657.3)(sinl.718)(2)

Pw = 39.41H

Pw =39.4 W ~ + ^ j = 6(0.5 + h )

(K )2 = 0

T\ + V \ - T 2 + V2: K ) , + K ) , = W 2 +(K.)2

25.6 = 6(0.5 + h)

h= 3.767 ft h =45.2 in. v\e = v\g = 0 constraint: y B( J )= 2xA( )

8 lb

32.2 ft/s'+ -

I 2 J 2

31b

32.2 ft/s2 J B 64.4

y B= 0.5 ft, xA = 0.25 ft, V2e = ^( 2 0 lb/ft)(0.25 ft)2 = 0.625 ft-lb

V2 = -3 lb(0.5 ft) = -1.5 ft-lb

r ..

T. + V, = T, + V2: 0 = + 0.625 - 1.51 1 2 2 64.4

Maximum velocity when acceleration = 0.

vB = 3.36 ft/s A

310 -x L C'oJUo

\ 3M>

B

V O ' 5 F t J V 0-t f ' X s *

v2g = -3 (0.6 ft) = - 1.8

v* = ^(20)(0 .3)2 = 0.9

5vT. +V. = T2 + V2: 0 = - 5 - - 0 .9 ,1 1 64.4

vB =3.40 ft/s =0.5

32.2= 0.01553

A tB Tb ~ = ~ wi(48.30) = 24.15 m

Vg = mg(7.5 + 1.5) = 9mg = 9(0.5) = 4.5 lb-ft

Ta + VA = Tg + Vg\ i(0 .0 1 5 5 3 )v 02 = 24.15(0 .0155 3) + 4.5

v0 = 62 7.82 v0 = 25 .05 6 vn = 25.1 f t/ s

A tC

Tc = - m v l = 0.007765v v c = 7.5mg = 7.5(0.5) = 3.75

Ta +Va =Tc + Vc \ 0.007765vq = 0.007765v^ + 3.75

0.0 07 76 5(2 5.05 6)2 - 3.75 = 0.007765v^-

v2 = 144.87

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(b)

r ^0 .S

r

-* LF = m a - = 0.01553(144.87)

1.5

Loop 2

LTb-O B

N = 1.49989

{Package in tube} Nc = 1.500 lb - A

r- \ S $ t (a) AtB,tube supports the package so,

'fcTUM

VB * 0

vs = 0, r f l = 0 VB = m*(7.5 + 1.5)

= 4.5 lb-ft

ta + va = tb + Vb

-^(0.015 53)^ = 4.5 =>vA = 24.073

vA = 24.1 ft/sA

(b) At C Tc = 0.007765v, Vc = l.Smg = 3.75

Ta + VA = Tc + Vc : 0.007765(24.073)2 = 0.007765v + 3.75

v'c = 96.573

0 . 5

Nc = 0.0155396.573

= 0.99985 1.5

{Package on tube} Nc = 1.000 1b A

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PROBLEM 13.76

If the package of Prob. 13.75is not to hit the horizontal surface at C with

a speed greater than 10 ft/s, (a) show that this requirement can be

satisfied only by the second loop, (b) determine the largest allowable

initial velocity v0 when the second loop is used.

SOLUTION

(a) Loop 1

From 13.75, atB

7- 5

t l f %H M M |

[jJ

Y03 - O S

I'PvTU W

\

(2)

v2g = gr = 48.3 ft/s~

j j

T O .r Vft"

I t

vB = 6.9498 fVs

tr = = i(0.01553)(48 .3) = 0.37505

= m g(7.5 + 1.5) = (0.5)(9) = 4.5 lb-ft

rc = = ^(0 .01553) = 0.007765v

Fc = 7.5(0.5) = 3.75 lb-ft

TB + V = Tc+ Vc : 0.37505 + 4.5 = 0.007765v + 3.75

v = 144.887 => vc = 12.039 ft/s

12.04 ft/s > 10 ft/s => Loop (1) does not work 4

(b) Loop 2 At A1

Ta = pivq = 0.007765Vq

vA - 0

At Cassume vr = 10 ft/s

Tc = i mvl =0.007765 (10) ' = 0.7765

vc = 7.5(0.5) = 3.75

Ta + VA = Tc + Vc : 0.007765vq = 0.7765 + 3.75

v0 = 24.144 vn = 24.1 ft/s 4

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Prove that a force F (x ,y ,z )i s conservative if, and only if, the following

relations are satisfied:

dF\__ Fy_ ^ = dF^ dFz _ dFx

dy dx dz dy dx dz

PROBLEM 13.77

OLUTION

r a conservative force, Equation (13.22) must be satisfied

dV_dx

8V_dy

dV_dz

e now write d F x _ _ d 2 y d Fy _ d V

dy dxdy dx dydx

ncedxdy dydx dy dx

d2V = d2V 3'f dFy

e obtain in a similar way

dz dy dx dz

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The force F = (_yzi + zxj +xyk)/x yz acts on the particle P (x ,y ,z ) which

moves in space, (a)Using the relation derived in Prob. 13.77, show that this

force is a conservative force. (b) Determine the potential function

associated with F.

PROBLEM 13.78

SOLUTION

(a ) Fx =yz

Fy =zx

xyz xyz

_ d t i ) = 0 dJy_d

dy fy dx t

= d_Fy_

dy dxThus

The other two equations derived in Problem 13.80 are checked in a similar way.

(b) Recall that Fx = , Fy = , Fz =

Equating (2) and (3)

From (5)

Thus

F F =dx' y d y ~ dz

Fx = - = ~ V = - lnx + f ( y , z ) (1)X ox

F = = ~ V = - Iny + g ( z ,x ) (2)y fy

Fz = = V = -In z + h( x , y ) (3)z dz

Equating (1) and (2)

- lnx + f ( y , z ) = - In y + g (z ,x )

Thus f { y ^ z ) = _lnT + k { z ) (4)

g (z ,x ) = - ln x + k ( z) (5)

- lnz + h(x,y) = - Iny + g (z ,x j

g ( z ,x ) = -ln z + /(x)

g(z,x) = -lnx + k(z)

k(z ) = - l n z

/(x) = -lnx

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From (4)

f ( y , z) = - lny - lnz

Substitute for f [ y , z) in(l)

V =- h i - Iny - lnz


V = -lnxyz 4

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The force F = (xi + yj+ zk )/ ^x 2 + y 2 + z2j acts on the particle

P (x ,y ,z ) which moves in space. (a) Using the relation derived in Prob.

13.77, prove that F is a conservative force. (b) Determine the potential

function V (x ,y ,z "jassociated with F.

PROBLEM 13.79

OLUTION

a) Fx= -----------

Fy =-----------

? -(x2 + y 2 + z2)2 (x2 + y 2 + z 2j 2

9 fx _ x ( ~2) i2y ) dFy .. y ( ~ \ ) 2y

^ (x 2 + y 2 + z 2f ^ (x2 + y 2 + z 2)

Thus dFx _ 8Fy

dy dx

The other two equations derived in Problem 13.79 are checked in a similar fashion

dZ F = J L F = JO Ldx y dy dz

b) Recalling that Fx = , Fy - ,Fz = -

Fx = ~ V = -{ * T dx

& ^ + y 2 + z 2y

V = (x2 +y 2 +z2) 2 + f( y ,z )

Similarly integrating and shows that the unknown function f { x ,y ) is a constant.dy dz

V =

(x2 + y 2 + z2)2

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PROBLEM 13.80

A force F acts on a particle P(x, y) which moves in the x y plane.

Determine whether F is a conservative force and compute the

work of F when P describes the path ABCA knowing that

(a) F = (far + y)i + (kx +y)j, (b) F = (fcc + y)i + (x + ky)j.

OLUTION

( a )

CU 6UAB = ^kxdx = k

Fx = F , F is normal to BC, UBC = 0

A

- a - .

U CA = =

2

^mcv< = (* - 0 >not conservative -4

(b) From Problem 13.77, ^ k = 1 = ldy dx

Conservative, UABCA = 0A

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PROBLEM 13.81

F .t ,, .k tx* / Certain springs are characterized by increasing stiffness with increasing

\ X ' deformation according to the relationF = k{x + k2x3, where F is the

force exerted by the spring, /c, and k2 are positive constants, andxis the

y ^ 7 .k ,x -k ,^ deflection of thespring measured from its undeformed position.

Determine (a) the potential energy Ve as a function of x, (b) the

maximum velocity of a particle of mass m attached to the spring and

released from rest withx - . Neglect friction.

OLUTION

Ve = k\X2 + k2xA -4e 2 1 4 2

Conservation of energy: 7] = 0, T2 = i m v 2

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Certain springs are characterized by decreasing stiffness with increasing

deformation according to the relationF = kxx - k2x3 , where F is the

F .k \- k x* frce exerted by the spring, kx and k2 are positive constants, andxis thedeflection of the spring measured from its undeformed position.

Determine (a) the potential energy Ve as a function of x, {b) the

maximum velocity of a particle of mass m attached to the spring and

released from rest withx - x0. Neglect friction.

PROBLEM 13.82

OLUTION

) Ul_>2 I ?Fdx = - ^ 2( V + k2x~ )dx

= ~ \ ( X2 ~ X\ ) +1 { X2 -^l4)

Ul^ 2 = Ve l- V e2:

Ve =kxx2 - k2x4 4

) Conservation of energy: T\ = 0, T2 -mv

V\e - - kxxl - ~ k 2x l, V2e - 0

-m v 2 = h x I - k7xt2 2 4

,2*.Requires x0 < 1 *

v = x 20 +\ 2 m j

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Knowing that the velocity of an experimental space probe fired from the

earth has a magnitude = 32.5 Mm/h at pointA,determine the velocity

of the probe as it passes through pointB.

PROBLEM 13.83

SOLUTION

rA = hA + R = A3 Mm+ 6.37Mm

rA = 10.67Mm

rB = hB + R = 72.7Mm +6.37Mm

rB = 19.07 A/m

t A, vA =32.5 Mm/h = 9028 m/s

Ta = m(9028 m/s)2 = 40.752 x 106 m

Va =GMm -gR'm

t B

rA = 10.67 Mm = 10.67 x 106 m

R = 6370 km = 6.37 x 10h m

vB=-

(9.81 m /s ') 6.37 x 106 m)= - 4 ----- 3------------------ - m = -37.306 x 10hm

(10.67 x 10h m

rj* 1 2 GMm -gR'm7* = vb = --------- = ----------

2 rB rB

rB = 19.07 Mm = 19.07 x 106 m

(9.81 m/s )(6.37 x 10h m) mA-------- ------------- = -20.874 x 10 m

(19.07 x l 0 h m

Ta + VA = T + VB\ 40.752 x 10hm - 37.306 x 106m = - m v \ - 20.874 x 106m

I *

v\ =2[40.752 x 106 - 37.306 x 106 + 20.874 x 106]

v j = 48.64 x 106 n r/s 2

vB = 6.9742 x 10 m/s = 25.107 Mm/h vB = 25.1 Mm/h

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A lunar excursion module (LEM) was used in the Apollo moon-landing

missions to save fuel by making it unnecessary to launch the entire

Apollo spacecraft from the moons surface on its return trip to earth.

Check the effectiveness of this approach by computing the energy per

kilogram required for a spacecraft to escape the moons gravitational

field if the spacecraft starts from (a) the moons surface, (b) a circular

orbit 80 km above the moons surface. Neglect the effect of the earths

gravitational field. (The radius of the moon is 1740 km and its mass is

0.0123 times the mass of the earth.)

PROBLEM 13.84

OLUTION

Note: GMmoon = 0.0123GA/earth

By Equation 12.30 G M ^ = 0.0123gR2E

At co distance from moon: r2 = oo, assume v2 = 0

e2 = r2 + v2 = o - GMm m'cm = 0 - 0 = 000

(a) On surface of moon: RM =1740 km = 1.74 x 106 m

v, = 0, 7j = 0 Re = 6370 km = 6.37 x 106 m

K| = ~GMmm Ei = Tl + Vx= 0 - 0 0123gRzm'RM R

, = -(0.0123)1(9.81 m/s2)( 6. 37 xl 06 m)1 "'lem

1(l .740 x 106 m)1

'lem

Where mkm = mass of the lem

, = ( 2.814 x 106 m2/s2 j m,

AE = E2 - | = 0 + ^2.814 x 106 m2/s2 j ml

Energy per kilogram:

lem

A

m.=2810 kj/kg A

lem

Cb) rx = Rm +80 km

r, = (1740 km + 80 km) = 1820 km = 1.82 x 106 m

Newtons second law:

F = km

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y _ GMm1fllem g = J1+ V = 1 i em _ l em

rx 1 1 1 2 rx rx

E = 1 G M a = 1 (0.0123)g/?|/w lem

1 2r, 2 rx

! (0.0123)j9.81 m/s2'{(6.37 x 103 m)mlem

1 " 2 1.82 x 106 m

Ey = ( -1 .345 x l0 6rn2/s2W m

A = E2 - j = 0 + (l.345 x 106 m2/s2)m leffi

AEEnergy per kilogram: = 1345 kJ/kg A

lem


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A=arro km A spacecraft is describing a c ircular orbit at an altitude of 1500 km above

the surface of the earth. As it passes through point A, its speed is reduced

by 40 percent and it enters an elliptic crash trajectory with the apogee at

point A. Neglecting air resistance, determine the speed of the spacecraft

when it reaches the earths surface at point B.

PROBLEM 13.85

1500 km

OLUTION

rcular orbit velocity

v- G M . . . ,- j = GM = gR

Vr =

GM gR2 (9.81 m/s: )(6.370 x 10ft m)

r (6.370 x l0 6 m + 1.500 x l0 6 m)

v l = 50.579 x 106 n r /s 2

vc = 7112 m/s

elocity reduced to 60% of vc = 4267 m/s

onservation of energy:

ta + va = tb + vb

1 / 7 GM 1 / 2 GM /ft~ / f i vA

-------------= ~ / t v B

------------2 rA 2 rB

,2 9.8l(6.370 x 106)2 v2 9.8l(6.370 x 106)4.267 x 103 -------------------- - r L = ^ --------- - J -

v ' (7.870 xl O6 ) 2 6.370 xl O6 )

vB= 6.48 km/sA

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A satellite describes an elliptic orbit of minimum altitude 606 km above

the surface of the earth. The semimajor and semiminor axes are

17 440 km and 13 950 km, respectively. Knowing that the speed of the

satellite at point C is 4.78 km/s, determine (a) the speed at point A, the

perigee, (b) the speed at pointB,the apogee.

PROBLEM 13.86

OLUTION

c Distance OA =6370 km + 606 km = 6976 km

Distance OOe =17,440 km - 6976 km = 10,464 km

3350rl = (13,950)2 + (10,464)2

\ rc =17,438,400 mAt C: r + F= 1m(4780f -

2 rc

Constant = (T +V) = -11.402 x 106 m2/s2m

% 0

, , . . , T+ V 1 2 gR2(a) At point/I, = vA

m 2 6.976 x 10 m

, i , (9.81m/s2)(6.370xl06m)2(-11.402 x 106) = v \ - 4------------ i i ------- r-----' 2 (6.976 x 106 m)

vA =9.56 km/s 4

(b) At points, rB = 2 ( l7.440 x 103km) - j 6.976 x 103km)

= 27.904 x 103 km

T + V I s gR 2

m 2 U 27.904 x 106 m

/ 1 , (9.8l)(6.370 x 106)2( 11.402 xlO6) == v j , ' \1 1 2 (27.904 xl O6)

vB = 2.39 km/s 4

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While describing a circular orbit 200 mi above the earth a space vehicle

launches a 6000-lb communications satellite. Determine (a) the

additional energy required to place the satellite in a geosynchronous orbit

at an altitude of 22,000 mi above the surface of the earth, (b) the energy

required to place the satellite in the same orbit by launching it from the

surface of the earth, excluding the energy needed to overcome air

resistance. (A geosynchronous orbit is a circular orbit in which the

satellite appears stationary with respect to the ground).

PROBLEM 13.87

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(b) Launch from earth

At launch pad EF = - ^ m- = - ^ m = -WRE

Re Re

Ee = -6000(3960 x 5280) = -1.25453 x 1011

& e = Eg* ~ Ee = -9.5681 x 109 + 125.453 x 109

AE = 115.9 x 109 lb-ft f \fp ~ rA

9 = 2 GMrp ~ rA

V rP y \ r*p )

r

VA = 2 GM-TP_ 1

rA r. +\ P

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Substitute vA from (2) into (1)

E0 _ i

W 2 g2G M- -

rn + rA\ P A

GM GM

grA g

rp

/ \1 1

rAu +rJ 1

GM

rp + rArAS

-G M

g {rA +rB)

GM

rAg

- i rP + ta)

On earth:

GM - gR.\ => = -----5 L _W rA + rp

\2

& - . - ,.65598 , , 0 ft-,MbW (50,000 mi x 5280 ft/mi)

EF = TF + Ve , Ve = 0, Te = 0, VE =WGM

g&E

For propulsion:

Em.=-EEL= -MEm. = _R _ - 20.9088 x 106 ft - ib/ibw gRE gRE

Ep _ E0 Ee

W W W

= -1.65598 X 106 - (-20.9088 x 106)

- E = 19.25 x 106 ft lb/lb A W

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A satellite of mass mdescribes a circular orbit o f radius rabout the earth .

Express as a function of r (a)the potential energy of the satellite, (b) its

kinetic energy, (c) its total energy. Denote the radius of the earth by R

and the acceleration of gravity at the surface of the earth by g,

and assume that the potential energy of the satellite is zero on its

launching pad.

PROBLEM 13.89

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Observations show that a celestial body traveling at 1900 Mm/h appears

to be describing about point B a circle of radius equal to 60 light years.

PointBis suspected of being a very dense concentration of mass called a

black hole. Determine the ratio M B/M s of the mass atBto the mass of

the sun. (The mass of the sun is 330 000 times the mass of the earth, and

a light year is the distance traveled by light in one year at a speed of

300 Mm/s.)

PROBLEM 13.90

SOLUTION

v - 1.9 Gm/h = 0.528 x 106 m/s

1 light year is the distance traveled by light in one year.

r O H 4 , r / Speed of light = 0.3Gm/s

r S

M

r - (60 yr)0.3 GM

1.00 xlO9 Gm

x (365 days/yr)(24 h/day)(3600 s/h)

r =0.56765 x 1018 m

f n o , . GMBm mv* J m \ Newton s second law F =-----f =

' r r

m b =

r r

r_S_

G

1000 mGMemh = g^artii = (9.81 m/s2 )(6370 km )2 j ^

- 398.059 x l0 12(m3/s2)

Msun = 330,000M e : GMsun =330 000 GMearth

GMsun = (330 000)^398.059 x 1012)

= 0.131360 x lO 21 m3/s2

0.13136 xlO 21G =

^sun

rv rv MA/f = = -^sunB G 0.13136 xlO 21

M (0.56765 x 1018)(0.528x 106)2M b _ = \------------------- A----- ------- = 1.205 x 10 A

\ A r \ w 1 0 / 1Msun 0.131360x10

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PROBLEM 13.91

(a) Show that, by setting r = R + y in the right-hand member of

Eq. (13.17') and expanding that member in a power series iny /R , the

expression in Eq. (13.16) for the potential energy Vg due to gravity is a

first-order approximation for the expression given in Eq. (13.17').

(b) Using the same expansion, derive a second-order approximation

forVg.

LUTION

V. = -WR 1+ -* I R

add the constant WR,which is equivalent to changing the datum from r = g o to r = R:

Vg =WRR { R j + "

First order approximation:

[Equation 13.16]

Second order approximation: Vg = WRR VR)

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PROBLEM 13.92

How much energy per pound should be imparted to a satellite in order to

place it in a circular orbit at an altitude of (a)400 mi, (b) 4000 mi?

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Two identical 2-kg collars, A and B, are attached to a spring of constant

100 N/m and can slide on a horizontal rod which is free to rotate about a

vertical shaft. CollarBis initially prevented from sliding by a stop as the

rod rotates at a constant rate 0O= 5 rad/s and the spring is in

compression with rA = 1m and rB = 2.5 m. After the stop is removed

both collars move out along the rod. At the instant when rB = 3 m ,

determine (a) rA, (b) 0, (c) the total kinetic energy. Neglect friction and

the mass o f the rod.

PROBLEM 13.93

OLUTION

itial state A *:----=

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Two identical 2-kg collars,A andB,are attached to a spring o f constant

100 N/m and can slide on a horizontal rod which is free to rotate about a

vertical shaft. CollarBis initially prevented from sliding by a stop as the

rod rotates at a constant rate 0Q = 5 rad/s and the spring is in

compression with rA = 1m and rB = 2.5 m. After the stop is removed

both collars move out along the rod. At the instant when the spring is in

compression and the total kinetic energy is 185 J, determine (a) rA,

(b) rB, (c) 0. Neglect friction and the mass o f the rod.

PROBLEM 13.94

SOLUTION

Initial state Y ( ( mr 2

100x0 = 2(1)(5)2

x0 = 0.5 m

(Unstretched length = 2 m )

Conservation o f energy

T>- j ( 2)[5 (l)]! + ^ ( 2)[5(2.5)]2

= 181.25 N m

V0 = ^(100 )(0.5)2 = 12.5 N-m

T + V =185 + ^(lOO)xj2 = 181.25 + 12.5

50xj2 = 8.75, x, = 0.41833 m

For compression:

(rB - r A) - 2 = -0.41833

Conservation o f angular momentum

(A) K = r\d = (l) 2 (5) = 5

CB) hB = rB0 = (2.5)2(5) = 31.25

r | =(2.5)2rj , rB =2.5rA

rB ~r A ~2 = 2.5 rA - r A - 2 =1.5rA - 2

(a).'. 1.5^ - 2 = -0.41833, rA - 1.054 m -4

(b) rB = 2.5rA =2.64 m rB = 2.64 m ^

(c) rAd= 5, 6= 4.50 rad/s 6= 4.50 rad/sA

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A 5-lb co llar^ is attached to a spring o f constant 50 lb/ft and undeformed

length 0.5 ft. The spring is attached to point Oo f the frameDCOB. The

system is set in motion with r = 0.75 ft, ve =1 .5 ft/s, andvr = 0.

Neglecting the mass of the rod and the effect of friction, determine the

radial and transverse components of the velocity of the collar when

r =0.4ft.

PROBLEM 13.95

OLUTION

H 0.75 f-t H a O' - o Conservation of angular momentum about O

mv0(O.75)= mv^(0.40)

_ 0/7 5^ = j g75/ j 5\ _ 2.8125 ft/s

0 0.40 V ;


T + V = r + V

r = I ( v ; + v J ) = A ( o + i.5=) =

V0 = 2.81 ft/s 4

5.265

g

V = 2-lc(r - rB) = (50)(0.75 - 0.5) = 1.5625

v'e = 2.8125 ft/s

r = ^ m ( v + v;2) = + (2 .8125)2) = ~(2.5 v;2 + 19.775)

V = i , k(r ' - rB f = |(5 01 b/f t)(0 .4f t - 0.5 ft)2 = 0.25 lb-ft

T + V = T' + V ^ 2 1 + 1.5625 = 2'5Vr + 19-775 + 0.25g g

5.625 + 1.5625# = 2.5v;2 + 19.775 + 0.25#

2.5v;2 =-14.15 + 1.3125(32.2)

v!2 = 11.245

vl = 3.35 ft/s 4

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A 4-lb collarAand a 1.5-lb collarBcan slide without friction on a frame,

Y consisting of the horizontal rod OEand the vertical rod CD, which is free

to rotate about CD. The two collars are connected by a cord running over

' . a pulley that is attached to the frame at O. At the instant shown, the

/ velocity v A of collarA has a magnitude of 6 fit/s and a stop prevents

collarBfrom moving. If the stop is suddenly removed, determine (a)the

velocity of collarA when it is 8 in. from O, (b)the velocity of collarA

when collar B comes to rest. (Assume that collarB does not hit O, that

collar A does not come off rod OE, and that the mass of the frame is

negligible.)

PROBLEM 13.96

SOLUTION

O)

H

Conservation o f angular momentum about O

( 4 ^

00

-----m v, =

V12 J A l l 2 jmMr

4(6) = 8(v^)r

(v'a) t = 3 ft/s

Conservation of energy

va=0.75 ft/s 71 = -V

Vj =0 (choose datum forBto be its initial position)

M 2-M l+M l =M l+ (3)2- K)2+ 9

T2= aM2+\mBM2 VB=Mr (kinematics)

T2 =g

M l +9I f 1.5 / , \2 2.75 / t\2 18

M r = M r + J

v2=mBg' '

72

= 1.51 | =0.5 lb ft

T\+v\=T2+V2: + 0=HIl(v'A)l+ + 0.5g g g

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2 . 7 5 = 54 - 0.5g = 54 - 0.5(32.2) = 37.9

{VAfR = 13.782 =>(*a)r = 3.712 ft/s

^ = y / M i + M r = >/13-782 + (3)2 = 4-773ft/s

0 = tan-1-^7!2 = 51.06

)

Conservation of angular momentum - Assume x is in feet

) * . .( ) =

x =

( 4 I+ x12

' a _ r

P " 3 .

= j ( 6 ) = + X3

F* = 4.77 ft/s, 0


A t vA 6ft/s, vB = 0

i f l2 1g

t ^ 2 1 ( 4 )/>\2 72 T\ = -im An = - - (6) = , Vj = 0

At CD

F2 - msgx = 1.5x

v y

= 51.1

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Ty + VX = T2 + V2> = - K ) 2 + 1.5xS

72 = 2(v'4)2 + 1.5gx

From conservation of angular momentum


72 = 2 (v ')2 + 1.5g - 1 _ I

v, 3

72 = 2(v )2 + - ^ ( 3 - v' )3v'

216v;, = 6(v',)3 + 1 .5g (3 -v ^)

6(v'/))2 + 144.9 - 264.3v^ = 0

6(v'/()3 - 264.3v^ + 144.9 = 0

Solving using mathcad

v'A = 0.552 ft/s M

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PROBLEM 13.97

A 0.7-kg ball that can slide on a horizontalffictionless surface is attached

to a fixed point Oby means of an elastic cord o f constant k = 150 N/m

and undeformed length 600 mm. The ball is placed at point A, 800 mm

from O, and given an initial velocity v0 perpendicular to OA. Determine

(a) the smallest allowable value o f the initial speed v0 if the cord is not to

become slack, (b) the closest distance dthat the ball will come to point O

if it is given half the initial speed found in part a.

OLUTION

A The cord will not go slack if v2 is perpendicular to the undeformed cord

length, L0, at

\

\Conservation o f angular momentum

\ 0. 8

s. \

0.8v, = 0.6v2 v 7 = v, = 1.333v0

i 0.6 0

O'

^ > Conservation of energy

Point v, = v0 7j = -1mvl = 0.35Vq

Vj ^ \ k { L - L 0f = i ( l5 0 N /m)(0.8m - 0.6 m)2

V, =3 J

Point T2 = ^ m v l = 0.35v2

AL = 0 V = 0 Tx+Vx=T2 +V2\ 0.35vg + 3 = 0.35v^ + 0

From conservation o f angular momentum v2 - 1.3158vfl

0.35vq (1.3158)2 - 1 = 3

vn =(3J)

(0.35 kg)(0.7313)= 11.72 m2/s2

vn = 3.42 m/s 4

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The ball travels in a straight line after the cord goes slack.


(0.8 )(l .71) = dv

1.368



d =

v, = 1.71 m/s

Point

T{= I/nv,2 = |(0.7 kg )(l.71 m /s)2 = 1.0234J

Vl = i k(L - Lq)2 = |(1 5 0 N/m)(0.8 m - 0.6 m)2 = 3 J

Point(D T, = -mv,2 = 0.3 5 v2

3 2 3

r 3 = o

Ti +Vl =T 3 + V3: 1.0234 + 3 = 0.35v2 + 0

v = 3.39 m/s

From conservation of momentum

J 1.368 1.368 ^d = -------= --------- = 404 mm

v 3.39

d =404 mm A

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rJOOmm

600mm

/

A 0.7-kg ball that can slide on a horizontalfrictionless surface is attached

to a fixed point Oby means of an elastic cord of constant k = 150 N/m

and undeformed length 600 mm. The ball is placed at point A, 800 mm

from O, and given an initial velocity v 0 perpendicular to OA, allowing

the ball to come within a distance d = 270 mm of point Oafter the cord

has become slack. Determine (a) the initial speed v0 o f the ball, (b) its

maximum speed.

PROBLEM 13.98

(a) Conservation of angular momentum: About O

0.8v0 = 0.27v

v = 2.963v0


Point v, = v0 7] = = 0.35vo2

Vt = jk (L i - L0f = i ( l50N/m )(0 .8m - 0.6 m)2

V = 3 J

Point v2 = v T2 = |m v 2 = 0.35v2

V2 = 0 (cord is slack)

Ti +V{ =T 2 + V2: 0.35vp + 3 = 0.35v2 + 0

From conservation of angular momentum, v = 3.125v0

0.35v02[(3.125)2 - 1

M

= 3

(0.35kg)(8.7656)

Vq = 0.9779m2/s2

v0 = 0.989 m/sA

(b) Maximum velocity occurs when the ball is at its minimum distance

from O, (when d =0.27 m)

vm = 3.125v0 = (3.125)(0.9889) = 3.09 m/s

vm =3.09 m/sA

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Using the principles of conservation of energy and conservation of

angular momentum, solve part aof Sample Prob. 13.9.

PROBLEM 13.99

OLUTION

R =6370 km

r0 = 500 km + 6370 km

r0 =6870 km

= 6.87 x 106 m

v0 = 36,900 km/h

= 36.9 x 106 m

3 . f> X 1 0 3 s

= 10.25 x 103m/s


r0mv0 = r\ A, *j> = rmin> r\ = max

VA' =

r \rn f 6.870 x l0 6V 3\u

v0 = --------------- 10.25 x 1031, rl , I n ) { }

Va- =70.418 x 109

(1)


Point A

Va = ~

v0 = 10.25 x 103m/s

1 1 2Ta = -m v 02 = ~m (l0 .25 x 103V

Ta = (m)(52.53 x 106)(j)

GM - gR2 =(9.81 m/s2')(6.37 x 106 m )

GM =398 x 1012m3/s2

r = 6.87 x 106 m

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VA(398 x 10 m /s

= - i - , = -57.93 x 106 m (j)(6.87 x 10 m)

Point A'

-mvA'

GMm 398 x 1012 m

ta + va = ta. + va.

(J)

52.53 x 106 /ft - 57.93 x 106 / = - r fv 2A. - 398x10 A .

Substituting for vA>from (1)

(7 0. 418xl0 9 ) 398 x 1012-5.402 x 106 = i y -L - -

m f

-5.402 x 106 =_ (2.4793 xlO 21) 398 x 1012

(5.402 x 106)r,2 - (398 x 1012)q + 2.4793 x 1021 = 0

r, = 66.7 x 106 m, 6.87 x 106 m

''max = 66,700 km 4

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PROBLEM 13.100

Transfer As a first approximation to the analysis of a space flight from the earth to

Mars, it is assumed that the orbits of the earth and Mars are circular and

coplanar. The mean distances from the sun to the earth and to Mars are

149.6 x 106 km and 227.8 x 106 km, respectively. To place the spacecraft

into an elliptical transfer orbit at point A, its speed is increased over a

short interval of time to vA which is faster than the earths orbital speed.

When the spacecraft reaches point B on the elliptical transfer orbit, its

speed vB is increased to the orbital speed of Mars. Knowing that the

mass of the sun is 332.8 x 103times the mass o f the earth, determine the

increase in velocity required (a) atA,(b) atB.

SOLUTION

M = mass of the sun

GM =332.8(l0 )3(9.81m/s2)(6.37 x 106 m)2 - 1.3247(lO)20 m3/s2

I GlbfCircular orbits Earth vE =--j-------- = 29,758 m/s

1149.6(10)

GM1227.8(l0)9

Mars vM = I ^ - 24,115 m/s


Elliptical orbit vA(149.6) = vB(227.8)


1 2 GM J_ 2 GM

2 VA 149.6(10)9 2 VB 227.8(10)9

v a = vb ;~ 2 7 - [ = 1.52273vEA B (149.6) *

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SOL CAP 13 L7

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