18장 1/18

Chapter 18Chapter 18

강체의평면동역학 : 일과에너지

( Planar Kinetics of a Rigid Body : Work and Energy )

18 1운동에너지 (Kinetic Energy)

18장 2/18

18.1 운동에너지 (Kinetic Energy)

강체내의미소질량 dm (속도 v)이가지는운동에너지는1 2

따라서강체의총운동에너지는

v== vvdmdT ,21 2

따라서강체의총운동에너지는

dmvdTT == ∫∫ 21 2

P

rω=×⊥

×+=

rωrωrωvv

이므로

PP

PPP

rωv

v

+++×⋅+=

×⋅×+×⋅+⋅=⋅=

)()()(2

)()()(222

2

jijikrωv

rωrωrωvvvvv

PyPxPyPxP vxωvyωxωyωvvxωyωyxω

+−=+−⋅+=×⋅+−=+×=×

)()()( )(

jijirωvjijikrω

18장 3/18

222

2

)()(2)()()(221

rωvv

dmvdTT

+×+××+×+

== ∫∫rωvrωrωrωvvvvv

v 와 ω는강체의전영역에걸쳐동일한값을가지므로(즉 dm에무관하므로)

22 )()(2

)()(2)()()(2

rωvxωvyωv

rωvv

PyPxP

PPPPP

++−+=

+×⋅+=×⋅×+×⋅+⋅=⋅= rωvrωrωrωvvvvv

( ) ( ) ( )21

21 222 ++−= ∫∫∫∫ PyPxP dmrdmxvdmyvvdmT ωωω

vP와 ω는강체의전영역에걸쳐동일한값을가지므로(즉, dm에무관하므로)

1)-(18 21

21

2222 ++−=

∫∫∫

∫∫∫∫

PPyPxP Imxvmyvvm ωωω

) ( 0

,

이므로잡으면좌표원점으로를대체하면로를

jirrjir

==

+=≡+= ∫∫∫yxGP

dmydmxdmmyx

G

. 2)-(18

2)-(18 21

21 22

있다수얻을를식출발해도부터로ρωvv ×+=

+=

G

GG IvmT ω

)(ρG

질점의경우회전의미없으므로 만고려하면되지만강체의경우회전에의한항이추가된다. 21 2

GmvT =

a. 병진의경우 ( Translation )

18장 4/18

( )= 0 이고강체내의모든점의속도가동일하므로

3)(181 2T

ω

3)-(18 2

2GmvT =

b 고정축에대한회전의경우 ( Rotation About a Fixed Axis )b. 고정축에대한회전의경우 ( Rotation About a Fixed Axis )고정축이점 O를지날때

11

,

4)-(18 21

21 22

ω

ω

GGGOG

GG

rv

ImvT

=∴×+=

+=

rωvv

( ) 5)-(18 21

21 222 ωω OGG ImrIT =+=

c.f. G축에대한회전일경우 vG=0이므로 만남는다. 그러나 o에대한회전이므로 가된다

O점에대한회전의경우, G에대한회전운동에너지 + o축과 G축사이의상대거리에의해생기는 (병진)속도에기인한병진운동에너지

: 21 2ωGI :

21 2ωoI

c 일반적인평면운동의경우 (General Plane Motion )

18장 5/18

c. 일반적인평면운동의경우 (General Plane Motion )

6)-(18 21

21 22 ωGG ImvT +=

22 GG

: 21 2

Gmv 병진운동에너지(translational kinetic energy)2 G

1 2I

( gy)

:강체내의모든점의속도가 와동일하다고볼수

있을때의운동에너지.Gv

:21 2ωGI 회전운동에너지(rotational kinetic energy)

강체내의모든점이질량중심 G에대해상대운동(회전운동)하기때문에추가되는에너지.)

일때, 예를들어서회전축이 G를지날때이에너지가강체의운동에너지 T 가된다.

0v =G

에너지: 스칼라양 전체운동에너지 = 운동하는모든부분의운동에너지의합

G에대한회전운동에너지 + (병진)속도에기인한병진운동에너지

Ex 18.1) The system of three elements shown consists of a 6-kg block B, a 10-kg disk D and a 12-kg cylinder C. If no slipping occurs, determine the total kinetic energy of the system at the instant shown.

sradmsmrv DDDDB /8)1.0(/8.0; === ωωω

smvrvsradrv

GCICGG

CCCICEE

/4.0)4.0)(1.0(;/4)2.0(8.0;

/

/

======

ωωωω

JvmT BBB 92.121 2 ==Block

111 ⎞⎛Disk JrmIT DDDDDD 60.1

21

21

21 222 =⎟

⎠⎞

⎜⎝⎛== ωω

Cylinder JrmmvImvT CCCGCGGC 44111111 22222 =⎟⎞

⎜⎛+=+= ωωCylinder JrmmvImvT CCCGCGGC 44.1

22222⎟⎠

⎜⎝

++ ωω

TTTT C++=the total kinetic energy of the system

J

TTTT CDB

96.444.160.192.1

=++=

++

18 2힘의일 (The Work of a Force)

18장 7/18

18.2 힘의일 (The Work of a Force)변하는힘이한일 (Work of a Variable Force )

S : 최종이동거리

dr : 힘 F의작용점의변위

∫∫∫ =⋅+=⋅=StntF dsFdsFdU θθθ cos)sin(cos uuurF ∫∫∫ S

여기서힘의크기 F와방향 θ는경로 s의함수이다.

일정한 힘이한일 (Work of a Constant Force )

18장 8/18

일정한 힘이한일 (Work of a Constant Force )

병진운동하는강체에작용한크기와방향이일정한힘 F = Fc가한일

sFdsFdsFU cSccFc)cos(coscos θθθ === ∫∫

( Fc의크기 Fc와방향 θ가일정 )

무게가한일 (Work of a Weight )

yWWdydydxWdUy

SW Δ−=−=+⋅−=⋅= ∫∫∫Δ

0)( jijrF

( g )

⎩⎨⎧

=<Δ

>Δ

0) y ( " " :

0) y ( : ,

아래로양

때일어날위로변위가의음즉

GWU

스프링힘이한일 (Work of a Spring Force )18장 9/18

|||kk

dsksdsksdUs

sSs

22 |11

)()( 2

1

>⎟⎞

⎜⎛

−=⋅−=⋅= ∫∫∫ iirF

|| s| sksks 1221

22 | ,

22 >⎟

⎠⎞

⎜⎝⎛ −−=

여기서 si는각위치에서의스프링의변형량

( unstretched position에서각위치까지의거리)( unstretched position 에서각위치까지의거리)

The work is always(인장, 압축) negative Displacement압축 인장

Spring force

일하지않는힘 (Forces That Do No Work)이되는경우0=⋅∫ rF d

① F = 0 즉, 힘이작용하지않는경우② dr = 0 즉, dr/dt = 0 (변위나속도가 0인곳에힘이작용한경우)③ F ⊥ dr 즉, 힘과작용점의변위가수직을이루는경우

∫

W(경우③)와 N(경우②혹은③)이한일은 0 이며

마찰력 이한일은 ⎨⎧ ) (0 : 경우경우구름의 ②

마찰력 Fr이한일은

⎩⎨⎧

) ( :)(0:

거리미끄럼는경우미끄럼의

경우경우구름의

ssFr

②

힘의작용점 IC dr/dt=0 dr = 0, (그 순간힘의방향으로 IC의변위 =0)

18 3우력이한일 (The Work of a Couple)

18장10/18

18.3 우력이한일 (The Work of a Couple)

우력: 크기가같고방향이반대이며작용선이평행한두힘우력

a. 병진하는동안우력이한일

) ( 0)( 2121 . 이므로이다 sssFsF dddd ==⋅−+⋅

우력

b. 회전하는동안우력이한일

sFsF dddUM

⎞⎛⎞⎛

⋅−+⋅= )( 21

θM dMddFr

drFdrF

⋅≡==

⎟⎠⎞

⎜⎝⎛−−+⎟

⎠⎞

⎜⎝⎛=

θθ

θθ2

)(2

그림에서 (미소각변위는벡터)

유한회전의경우

θM dMddFr θθ

, )( kθkM θddFrMM −==−=

∫=2θ θdMU 0∫F d유한회전의경우 ∫=

1θθdMUM

우력은물체가회전할때만일을하게된다.

0=⋅∫ rF d(병진경우)(회전경우)

Ex 18.2) The bar shown has a mass of 10-kg and is subjected to a couple moment of M = 50 N.m and a force of P = 80 N, which is always applied perpendicular to the end of the bar. Also, the spring has an unstretched length of 0.5 m and remains in the vertical position due to the roller guide at B. determine the total work done by all the forces acting on the bar when it has rotated downward from θ = 0 to θ° = 90°.

FBDFBD

JmghUW 2.147)5.1(1.98 ===

JdMU 578)2/(502

∫ θθ

JdMU M 5.78)2/(501

=== ∫ πθθ

JmNU P 37732

)80( =⎟⎠⎞

⎜⎝⎛ ⋅=π

JUs 0.75)5.075.0)(30(21)5.075.2)(30(

21 22 −=⎥⎦

⎤⎢⎣⎡ −−−−=Work by spring force

JU 5280.3770.755.782.147 =+−+=Total work

18 4일과운동에너지의원리 (Principle of Work and Energy)

18장 12/18

18.4 일과운동에너지의원리 (Principle of Work and Energy)

질점동역학에서와같이

1221 TTU −=∑ −

강체가위치 1 에서위치 2 로이동하는동안에강체에작용

한힘(외력) &모멘트가한일강체의운동에너지의변화=

한힘(외력) & 모멘트가한일

내력이한일 =0

(강체이므로내력들사이의상대운동(변위) =0)

또는내력은반대방향 pair로작용, if body moves, work of one force cancels that of its counterparts

Ex. 18-3) The 30-kg disk shown is pin supported at its center. Determine the number of revolutions it must take to attain an angular velocity of 20 rad/s starting from rest. It is acted upon by a constant force F = 10 N which is applied to a cord wrapped around itsacted upon by a constant force F = 10 N, which is applied to a cord wrapped around its periphery, and a constant couple moment M = 5 N.m

T 0속도, 힘, 거리 Principle of Work and Energy

( ) JIT

T

O 12020)2.0)(30(21

21

210

22222

1

=⎥⎦⎤

⎢⎣⎡==

=

ω

UM = Mθ

UFC= Fs as the cord moves downward s = θr = θ(0.2 m)

Principle of Work and Energy

{ } { } { }{ } { } { }TFsMT

TUT

21

2211

=++

=+ ∑ −

θ{ } { } { }

revrevradrad 73.211.171.17

120)2.0()10()5(0

=⎟⎟⎞

⎜⎜⎛

==

=++

θ

θθ

revrad

radrad 73.22

1.171.17 ⎟⎟⎠

⎜⎜⎝ π

θ

Ex 18-4) The 700-kg pipe is equally suspended from the two tines of the fork lift. It is undergoing a swinging motion such that when θ = 30° it is momentarily at rest. Determine the normal and frictional forces acting on each tine which are needed toDetermine the normal and frictional forces acting on each tine which are needed to support the pipe at the instant θ = 0°. Measurement of the pipe and the suspender are shown in.

IG = mr2

22

222 2

1)(21 ω+= GG IvmT

01 =T

( ) [ ] 22

222222 )4.0(700)15.0(700

21

21

21 ωω +=+== mrIIT GOor

[ ] [ ]22

22

222

875.63

)15.0)(700(21)4.0()700(

21

ω

ωω

=

+= 22875.63

222ω=

2

Work (Free-Body Diagram)• The normal and frictional forces on the tines do no work since they do

t th i inot move as the pipe swings.

Δy = 0.4 m – 0.4 cos 30° m = 0.05359 m

{ } { } { }

{ }TUT 2211 =+ ∑ −

Principle of work and energy

{ } { } { }srad /40.2

875.63)05359.0)(81.9(7000

2

22

==+

ωω

)(700;)(←∑+

FF

E.O.M.

{ω

)4.0()40.2(700)81.9(700;)(

)(700;)(2

2

=−=↑+

==←

∑∑

NamF

aFamF

rTnGn

tGTtGt

321

ααα ])4.0(700)15.0(700[][0; 222 +=+==+∑ mrIIM GOO

• Since (a ) = 0 4α a 0)(0 ==α• Since (aG)t = 0.4α

kNNF

a

T

T

tG

48.80

0)(,0

==

==α

• There are two tines used to support the load, therefore

kNkNN

F

t

t

24.42

48.80

==′

=′

2

Ex 18-5) The wheel weighs 20 kg and has a radius of gyration kG = 0.18 m about its mass center G. if it is subjected to a clockwise couple moment of 22 N.m and rolls from rest without slipping, determine its angular velocity after its center Grolls from rest without slipping, determine its angular velocity after its center G moves 0.15 m. The spring has a stiffness k = 160 N/m and is initially unstretched when the couple moment is applied.

01=T 01T

22

222

1121)(

21 ω+= GG IvmT

(v ) = 0 24ω 222

222 ])18.0)(20[(

21))(20(

21 ω+= Gv

(vG)2 0.24ω2 222 9.0 ω=T

FBD

Us = -½ ks2.

• the wheel rotates θ = sG/rG/IC = 0.15/0.24 = 0.625 rad

{ } { } { }TUT 2211 =+ ∑ −

G G/IC

• Hence the spring stretches sA = θrA/IC = 0.625 (0.48) = 0.3 m

{ } { }

{ }

TksMT

121

22

22

1 =⎭⎬⎫

⎩⎨⎧ −+ θ

{ }

srad /70.2

))(9.0()3.0)(160(21)625.0(220

2

22

2

=

=−+

ω

ω

Ex 18-6) The 10-kg rod is constrained so that its ends move along the grooved slots. The rod is initially at rest when θ = 0°. If the slider block at B is acted upon by a horizontal force P = 50 N, determine the angular velocity of the rod at the instant θ = 45° Neglect friction and the mass ofdetermine the angular velocity of the rod at the instant θ 45 . Neglect friction and the mass of blocks A and B.

22

222 2

1)(21 ω+= GG IvmT

22

22

222

)(2670)(5

])8.0)(10(121[

21))(10(

21

22

ω

ω

+=

+= G

v

v

22 )(267.0)(5 ω+= Gv

22/2

40)45tan4.0()( ωω == o

ICGG rv

24.0 ω=

22

22

222 067.1267.08.0 ωωω =+=T

Δy = (0.4 – 0.4 cos 45°) m;

{ } { } { }TUT ∑{ } { } { }{ } { } { }

{ } { } { }TPsyWTTUT

0671)45sin80(50)45cos4040(1980 221

2211

++

=+Δ+

=+ ∑ −

ωoo{ } { } { }srad /11.6

067.1)45sin8.0(50)45cos4.04.0(1.980

2

2

==+−+

ωω

예제 18-1 ( Problem 18-22 )18장 18/29

The 20-kg disk is originally at rest, and the spring holds it in equilibrium. A couple moment of M = 30N·m is then applied to the disk as shown. Determine its angular velocity at the instant its mass center G has moved 0.8 m down along the inclined plane. The disk rolls without slipping.

(sol.)자유물체도 :

sF

8.920×o30

m 2.0G

정적 평형상태 :

rFN

A

N1.98 0)2.0(30sin)81.9(20)2.0( ; 0

=

=+−=∑s

sA

FFM

주어진 Info.거리, 모멘트또는힘, 속도

평형상태에서의스프링의변형 m654.0150

1.981 ===

kFs s

미끄럼없는구름을시작하여 disk 가구른각도 rad4208.0===

rsθ

일과에너지원리!

2.0r

0& 0 == ∑∑ FM ANote. 강체의평형조건

18장 19/29

일-에너지 방정식에서

)( 0 121221 있었으므로상태에정지초기에==−=∑ − TTTTU Q

구름마찰력 Fr과반력 N은일을하지않으므로

{ }21

21

2

1

2

1221 )4(30)30sin)(8.9(20)(21θ ++−+−=⋅+⋅== ∫∫∑ − sssskddTU MrF

MUspU WU

222

111

,21

21

21

2

ωω

⎞⎛

⎟⎠⎞

⎜⎝⎛ ==+= rvmrIvmI GGGG

222 )2.0)(20(21)2.0)(20(

21

21 ωω +⎟

⎠⎞

⎜⎝⎛=

rad/s11=∴ ω ANS.rad/s11 =∴ ω ANS.

주의 disk 위의점이면서경사면과접촉하는점 A의속도가 0 이므로운동에너지d s 위의점이면서경사면과접촉하는점 의속 가 0이 운동에너지

T2는 IAω2으로표현될수도있다. 평행축의원리에따라 IA = IG + mr2임을

염두에두면위의결과와같음을알수있다

21

염두에두면위의결과와같음을알수있다.

( ) 21

21

21

21 22222

2 GGGA vmImrIIT +=+== ωωω

예제 18-218장 20/29주어진 Info.

거리, 초속도, 힘 F=mg일과에너지원리!

A man having a weight of 150 lb crouches down on the end of a diving board as shown. In this position the radius of gyration about his center of gravity is kG = 1.2 ft. While holding this position at θ = 0º, he rotates about his toes at A until he loses contact with the board when θ = 90º. If he remains rigid, determine

(sol.)다이빙대에서 90º 회전하는동안의일-에너지방정식

approximately how many revolutions he then makes before striking the water after falling 30 ft.

W) 0 ( 121221 ==−=∑ − TTTTU Q

( )222 5111)51(150 ωω ===+= rrvkmIvmI

무게가한일 =

+N

W

( )

( )22 21

51 , , 22

)5.1(150

ω

ωω

+==

===+=

rmIII

.rrvkmIvmI

GAA

GGGGG

발에작용한힘(N)이한일 = 0

222222 )5.12.1(32.2150

21)(

21 ωω +=+= rkm G

ft/s 675.7)117.5)(5.1( rad/s117.5

===∴

Gvω

공기저항을무시하면선수가다이빙대를떠난후 G점에작용한모멘트가 0이므로각가속도

18장 21/29

공기저항을무시하면선수가다이빙대를떠난후 G점에작용한모멘트가 0이므로각가속도α = 0이다. ( )초기속도 v0 =7.675 ft/s 를가지고자유낙하하는물체의질량중심의이동거리

∑ = αGG IM∴

121

2

200 ++= gttvssmg = maG

aG = g

s 147.1

)2.32(21675.7030 2

=∴

++=

t

ttaG g

일정한각속도 ω = 5.117 을가지고회전하는물체의

회전각도 )147.1(117.500 +=+= ωθθ t

revolution 934.0)rad(2

)rev(1)(rad870.5 =⎟⎟⎠

⎞⎜⎜⎝

⎛×=

πANS.

18 5에너지보존 (Conservation of Energy)

18장 22/29

18.5 에너지보존 (Conservation of Energy)보존력(Conservative force): 운동경로에무관하고운동의시작점과끝점에만의존하는일

ex) 중력, 스프링힘강체에보존력만작용하는경우즉보존계에서는에너지보존의원리가성립한다.

중력에의한위치에너지 : Gg yWV = 탄성력에의한위치에너지 : 2

21 skVe = 2

Gyg WyWdyyG

=−= ∫0

)(V0

+: when body is above the datum (yG>0)-: below (yG<0)

Always positive

총위치(potential)에너지 : eg VVV +=Gyg WydyWyG

=−−= ∫−0

))(()(V jj

18장 23/29일-에너지방정식에서

1221 TTU −=∑ −

( ) 일한력이∑∑ ( )( ) )) (ex.(

)(

21

21

2121

일한이비보존력

일한보존력이

마찰력noncons

cons

U

UU VV

∑∑∑

−

−−

+

= −≡

따라서 ( ) ( )( ) ( )1212

212121

VVTT

UUUconsnoncons

−+−=

−= ∑∑∑ −−−

( ) ( )( ) ( )

( ) 변화의에너지역학적

1122

1212

VTVTVTVVTT

+=

+−+=+

−=−=+==⇒ ∫∫∫∫∫ FdrFdrFdrFdrFdrUdatumdatumdatum

VV 21

22

21

보존계에선비보존력이한일이 0이므로

⎟⎠⎞⎜

⎝⎛ ==

+⇒

∫

∫∫∫∫∫−

UFdr

FdrFdrFdrFdrFdrUdatum

datum

한일보존력이현재위치

V

VV 21211121

Q

보존계에선비보존력이한일이 0이므로

2211 VTVTVT +=+===+ 일정에너지역학적

Ex 18-7) The 10-kg rod AB is confined so that its ends move in the horizontal and vertical slots. The spring has a stiffness of k = 800 N/m and is unstretched when θ = 0°. Determine the angular velocity of AB when θ = 0°, if the rod is released from rest when θ = 30°.

ksWyV

)30sin40)(800(1)30sin20(198

21

2

2111

+−=

+−=

oo

J19.6

)30sin4.0)(800(2

)30sin2.0(1.98

=

+=

02 =V2

0T 22 1)(1 ω+= IvmT

(vG)2 =(rG/IC)ω2 = (0.2)ω2when θ = 0°, pt A IC, so vB

01 =T

22

222

222

])4.0)(10(121[

21))(10(

21

2)(

2

ω

ω

+=

+=

G

GG

v

IvmT vA

222 267.0 ω=T

(vG)2 =(rG/IC)ω2 = (0.2)ω2

{ } { } { } { }{ } { } { } { }

VTVT

026701960 22211 +=+

Note.θ = 30° 상태에서 IC 를찾지말고θ = 0° 상태에서 IC 를찾아야됨.

{ } { } { } { }srad /82.4

0267.019.60

2

22

=+=+

ωω

Ex 18-8) The disk has a mass of 15 kg and a radius of gyration of kG = 0.18 m, and it is attached to a spring which has a stiffness k = 30 N/m and an unstretched length of 0.3 m. If the disk is released from rest in the position shown and rolls without slipping, determine its p pp g,angular velocity at the instant G moves 0.9 m to the left.

s1 = [(0.92 + 1.22)½ - 0.3] = 1.2 m

s2 = (1.2 – 0.3) = 0.9 m

JksV 621)21)(30(11 22 ===

JksV

JksV

15.12)9.0)(30(21

21

6.21)2.1)(30(22

2222

11

===

===

01 =T

22 1)(1 IT

22

222

22

222

])18.0)(15[(21))(15(

21

2)(

2

ω

ω

+=

+=

G

GG

v

IvmT(vG)2 = 0.225ω2

222 6227.0 ω=T

{ } { } { } { }{ } { } { } { }

VTVT

1512622706210 22211

+=+

+=+

ω{ } { } { } { }srad /90.3

15.126227.06.210

2

2

=+=+

ωω

Ex 18-9) The 10-kg homogeneous disk is attached to a uniform 5-kg rod AB. If the assembly is released from rest when θ = 60°, determine the angular velocity of the rod when θ = 0°. Assume that the disk rolls without slipping. Neglect friction along the guide and the mass of th ll t Bthe collar at B.

JyWV R 74.12)60sin3.0(05.4911 === o

02 =V 02V

(ω ) = 0 and (v ) = 0 ( as the rod is fully extended)(ωD)2 0 and (vA)2 0.

(vG)2 =(rG/IC)(ωR)2 or (vG)2 = (0.3)(ωR)2

( as the rod is fully extended)

222

22

22

22

222

00)]()60)(5(1[1])(30)[5(1

)(21)(

21)(

21)(

21

DAADRGGR IvmIvmT

ωω

ωω

+++=

+++=

22

22

)(3.0

00)]()6.0)(5(12

[2

])(3.0)[5(2

R

RR

ω

ωω

=

+++=

{ } { } { } { }{ } { } { } { }

VTVT

R 0)(3.074.120 22

2211

+=+

+=+

ω{ } { } { } { }sradR

R

/52.6)()(

2

2

=ω

예제 18-318장 27/29

The uniform 80 lb garage door is guided at its ends on the track. Determine the required initial stretch in the spring when the door is open, θ = 0º, so that when it falls freely it comes to rest when it just reaches the fully closed position, θ = 90º. Assume the door can be treated as a thin plate, and there is a spring and pulley system

(sol.)on each of the two sides of the door.

Ss

S

Asθ = 90º θ = 0º

스프링이변형되지| Δ sS | = 4 ft

A 스프링이변형되지

않는위치 (unstr-etched position)

풀리의운동학에서 일정=+ SA ss 2

SA ssΔΔ−=Δ

282

여기서 는스프링끝점 S 의초기위치(θ = 0º )로부터문이닫혔을때(θ = 90º )까지의변위이다.

SsΔ−= 28

SsΔft4 −=Δ∴ SsS

⎪⎧ = (open)0:1 oθ상태

18장 28/29

에너지보존의원리에서⎪⎩

⎪⎨⎧

=

=+=+

(closed) 09 : (open) 0:

2

12211 oθ

θ

상태

상태VTVT

0 VVTT === 이므로 2121 0 VVTT === 이므로

스프링이 2개이므로

)4(12)(120 22 Δ kWk

)4)(9(212)4(80)9(

212

)4(212)(

2120

22

22

⎟⎠⎞

⎜⎝⎛ ++−=⎟

⎠⎞

⎜⎝⎛

+×+Δ=×+

ss

skyWks G

문의중력

ANS.ft 44.2 22

=∴⎠⎝⎠⎝

s위치에너지

c.f.) 스프링이한일 always (-)) y ( )

스프링의위치E always (+)

모든동역학문제

Newton’s law F=ma :힘과가속도만주어진경우Newton s law F ma : 힘과가속도만주어진경우일-에너지 : 거리, 속도, 힘주어진경우운동량-충격량: 속도, 시간, 힘

Homework

Ch. 18 연습문제5 7 14 21 25 30 31 33 40 43 46 48 52 55 585, 7, 14, 21, 25, 30, 31, 33, 40, 43, 46, 48, 52, 55, 58