Chng 1 NG HC CHT IM 1.1 Nhng khi nim m u 1.1.1 Chuyn ng v h quy chiu Chuyn ng l mt khi nim c bn ca c hc. Chuyn ng ca mt vt l s chuyn di v tr ca vt i vi cc vt khc trong khng gian v thi gian. Mun xc nh v tr ca mt vt trong khng gian, ta phi tm nhng khong cch t vt ti mt h vt khc m ta quy c l ng yn. H cc vt c quy c l ng yn gi l h quy chiu. xc nh thi gian ca vt khi chuyn ng, ta gn vo h quy chiu mt ci ng h. Khi mt vt chuyn ng th nhng khong cch t vt n h quy chiu thay i theo thi gian. Chuyn ng hay ng yn ch c tnh cht tng i, ph thuc vo h quy chiu. Mt vt c th l chuyn ng i vi h quy chiu ny nhng c th l ng yn i vi h quy chiu khc. 1.1.2 Cht im v h cht im Cht im l mt vt c kch thc nh, khng ng k so vi nhng khong cch, nhng kch thc m ta ang kho st. V d: khi xt chuyn ng ca vin n trong khng kh, chuyn ng ca qu t xung quanh mt tri. ta c th coi qu t l cht im. Nh vy, vic xem mt vt c phi l cht im hay khng cn ph thuc vo iu kin bi ton m ta nghin cu. Mt tp hp cht im gi l h cht im. Vt rn l mt h cht im trong khong cch tng h gia cc cht im ca h khng thay i. 1.1.3 Phng trnh chuyn ng ca cht im xc nh chuyn ng ca mt cht im, ngi ta thng gn vo h quy chiu mt h trc ta . Trong h ta Descartes, v tr ca mt cht im M trong khng gian c xc nh bi 3 ta : x, y, z. Ba ta ny cng l ba ta ca bn knh vc t OM = rz M

rOHnh 1-1

(c) y

trn ba trc ta (hnh 1-1).

x

1

Khi M chuyn ng th x, y, z l cc hm ca thi gian: x = f(t) M y = g(t) z = h(t) r = r (t)

(1-1)

Bn knh vc t r ca cht im chuyn ng l hm ca thi gian t: (1-2) (1-1) v (1-2) gi l phng trnh chuyn ng ca cht im M. 1.1.4 Qu o Qu o ca cht im chuyn ng l ng to bi tp hp tt c cc v tr ca n trong khng gian trong sut qu trnh chuyn ng. xc nh qu o ca chuyn ng, ngi ta kh t trong (1-1). 1.1.5 Honh cong Trn qu o (C) (hnh 1-2), ta chn mt im A no c nh lm gc v mt chiu dng, khi v tr ca cht im M ti mi thi im c xc nh bi:AM=S , S gi l honh cong ca M.

ta c: S = S(t)A + M (C)

(1-3)

Hnh 1- 2

1.2 Vn tc Vn tc l mt i lng c trng cho phng chiu v s nhanh, chm ca chuyn ng. 1.2.1 nh ngha vn tc a. Vn tc trung bnh v Xt chuyn ng ca cht im trn mt ng cong (C). Trn (C), ta chn mt gc A v mt chiu dng (hnh 1-2). thi im t, cht im v tr M c xc nh bi AM=S, thi im t'=t+t cht im i c qung ng: MM'=S'-S= S. Khi vn tc trung bnh ca cht im trong khong thi gian t c xc nh:v= S t

(1-4)

b. Vn tc tc thi Khi cho t0 th :

2

lim

t 0

S dS = = v t dt

(1-5)

v: vn tc tc thi (gi tt l vn tc). Vy: Vn tc ca cht im c gi tr bng o hm honh cong ca cht im i vi thi gian. Nu chn A trng vi v tr ban u ca cht im (lc t=0) th AM=S l qung ng cht im i c trong khong thi gian t. Vy: Vn tc ca cht im c gi tr bng o hm qung ng i ca cht im i vi thi gian. v (1-5) l i lng i s, n cho bit: - S nhanh chm ca chuyn ng. - Chiu chuyn ng (v>0: chiu chuyn ng trng vi chiu dng, v 0 ; tng: chuyn ng trn nhanh dn. - < 0 ; gim: chuyn ng trn chm dn. - = 0 ; khng i: chuyn ng trn u. Trng hp = const: chuyn ng trn thay i u. Ta c cc cng thc: = 0 + t (1-28) = 0 t + t2 2

(1-29) (1-30)

2 - 20 = 2

Ngi ta biu din gia tc gc bng vc t gi l vc t gia tc gc. c:

8

- Phng trng vi trc quay ca qu o. - Chiu song song cng chiu vi khi tng (>0) v ngc li. d = - ln: (1-31) dt d = (1-32) Di dng vc t: dtH qu: Mi lin h gia a t v :a t = [ .R

]

(1-33)

v ln:

at = .R

V d: Mt v lng sau khi bt u quay c mt pht th thu c vn tc 700 vng/pht. Bit bn knh ca v lng l 30cm. Tnh: a. Gia tc gc v gia tc tip tuyn ca v lng trong pht y. b. S vng quay ca v lng trong pht y. Gii

a. p dng cng thc(1-28):

= 0 + t.

Khi bt u quay th 0 =0; = 700 vng/pht = 73,30 rad/s 73,30 =1,22(rad / s 2 ) = = suy ra: t 60 Ta c at = .R = 1,22.0,3 = 0,37(m/s2) b. S vng quay ca v lng trong pht y l: n= ; 2 trong suy ra := t2 2

n=

t 2 1,22.60 2 = = 350(vng ) 2 2.2 2.2.3.14

1.4.3 Chuyn ng vi gia tc khng i (chuyn ng nm nghing)

Thc nghim chng t rng trong mt phm vi khng ln lm, mi cht im u ri vi cng mt gia tc g theo phng thng ng, hng xung di v c tr s khng i. Ta kho st chuyn ng ca mt cht im xut pht t mt im O trn mt t vi vc t vn tc ban u l v 0 hp vi mt phng nm ngang mt gc (hnh 1-6) (chuyn ng nm nghing). Gia tc ca chuyn ng: a = g .

9

Ta c: vc t gia tc a ca chuyn ng c cc thnh phn theo hai trc ta Ox v Oy:a x = 0 a a y = gy

(1-34)

v0

S Mg

A

O

x Hnh 1- 6

Tch phn 2 v (1-34) ta c: v x = v 0 cos v (1-35) v y = gt + v 0 sin Tch phn 2 v (1-35) ta c phng trnh chuyn ng ca cht im: x = v 0 tcos M (1-36) 1 y = gt 2 + v 0 tsin 2 v sin . Thay gi tr ca t vo Khi cht im t cao S th vy=0, suy ra t = 0 g(1-36) ta tm c ta nh S ca Parabol, l im cao nht m cht im t c:

v2sin2 xS = 0 2g S 2 2 y = v0sin S 2g Cht im chm t ti A th y =0, suy ra t =

(1-37)

2v 0 sin .Thay gi tr ca t vo g

(1-36) ta tm c ta ca im A, l im xa nht m cht im t c:2 v 0 sin2 x = A A g y = 0 A

(1-38)

V d 1: T mt nh thp cao H=30m, ngi ta nm mt hn xung t vi vn tc v0=10m/s theo phng hp vi mt phng nm ngang mt gc =300. Ly g=10m/s2. Tm:

10

a. Thi gian t lc nm ti lc hn chm t. b. Khong cch t chn thp n ch ri ca hn . c. Dng qu o ca hn . Gii a. Chn h trc ta nh hnh 1-7.O

x

v0

y Hnh 1-7

ta c: v0x = v0co v0y = v0sin Phng trnh chuyn ng ca hn theo phng Oy : 1 1 y = v 0y t + gt 2 = v 0 tsin + gt 2 2 2 Thay cc gi tr vo, ta c: 5t2 + 5t 30 = 0 gii phng trnh trn ta c: t1 = 2s t2 = -3s m2). Coi ma st l khng ng k. Xc nh: a. Gia tc ca hai vt. b. Sc cng ca dy. Gii Tc dng ln m1 v m2 c cc lc: Trng lc P+P2 T2 T1

- Lc cng dy T Cc lc c phng chiu nh hnh 2-10 a. Gi a l gia tc ca h, phng trnh chuyn ng ca h c vit:P1 + T1 + P2 + T2 = (m 1 + m 2 )a

(a)

P1

trong :

T1 = T2

Hnh 2-10

Chiu (a) ln chiu dng chn, ta c:P1 P2 = (m1 + m 2 )a

Suy ra:

a=

(m1 m 2 ) g (m1 + m 2 )

b. Ta vit phng trnh chuyn ng cho vt m1:P1 + T1 = m 1a

(b)

28

Chiu (b) ln chiu dng chn, ta c:P1 T1 = m1a

Suy ra

T1 =

2m1 m 2 g (m1 + m 2 )

BI TP 2.1 Mt xe c khi lng 20 tn, chuyn ng di tc dng ca lc hm c ln 6000N, vn tc ban u ca xe bng 15m/s. Hi: a. Gia tc ca xe. b. Sau bao lu xe dng li. c. on ng xe chy c k t lc hm cho n khi xe dng hn. p s: a/ a = 0,3m/s2 b/ t = 50s c/ s = 375m 2.2 Mt vt trt xung trn mt mt phng nghing hp vi mt phng nm ngang gc =450. Khi trt c qung ng s=36,4cm, vt thu c vn tc v=2m/s. Xc nh h s ma st gia vt v mt phng nghing. p s: k = 0,22 2.3 Mt xe c khi lng 1 tn chuyn ng trn mt on ng nm ngang, h s ma st gia bnh xe v mt ng l 0,1. Tnh lc ko ca ng c trong cc trng hp: a. Xe chuyn ng u. b. Xe chuyn ng nhanh dn u vi gia tc bng 2m/s2. p s: a/ Fk =1000N b/ Fk =3000N 2.4 Mt vt A c t trn mt mt phng nm A ngang. Vt A c ni vi vt B bng mt si dy vt B qua rng rc c nh (Hnh 2-6). Khi lng ca vt A l Hnh 2-6 200g, ca vt B l 300g. Khi lng ca rng rc v dy coi nh khng ng k. a. Tnh lc cng ca dy nu cho h s ma st gia vt A v mt phng nm ngang l k=0,25. b. Nu thay i v tr ca vt A v vt B th lc cng ca dy bng bao nhiu? p s: a/ T =1,5N b/ T =1,5N 2.5 Mt vt c khi lng m=500g c buc vo u si dy di l=100cm. Mt ngi cm u kia ca dy m quay vt trong mt phng thng ng vi tn s n=3 vng/s. Ly g=10m/s2. a. Tnh lc cng ca dy khi vt i qua v tr cao nht v thp nht ca qu o.

29

b. Ngi y quay dy nhanh dn ln cho n khi dy t. Hi dy b t v tr no v tn s vng khi dy b t l bao nhiu? Bit rng dy b t khi lc cng bng 205N. p s: a/ T1 =172N; T2 =182N b/ n =3,18vng/s 2.6 Mt vt c khi lng m=200g c treo u si dy di l=40cm, vt quay trong mt phng nm ngang vi vn tc khng i sao cho si dy vch mt mt nn. Gi s khi dy to vi phng thng ng mt gc =450. Tm vn tc ca vt v lc cng dy. p s: =5,95rad/s; T =2,84N 2.7 Mt ngi dng dy ko mt vt c trng lng P=50N trt u trn mt sn nm ngang. Dy nghing 1 gc =300 so vi phng ngang. H s ma st trt gia vt v mt sn l k=0,3. Xc nh ln ca lc ko. p s: Fk =14,78N

30

Chng 3 NG LC HC VT RN QUAY 3.1 Phng trnh c bn ca vt rn quay 3.1.1 M men lc: a. Tc dng ca lc trong chuyn ng quay: Lc F tc dng ln vt rn ti im M lm cho vt rn quay xung quanh trc .(hnh 3-1).

F2 F

O M Fn

Ft F1

Hnh 3-1

Ta phn tch F ra cc thnh phn nh hnh v:

F = F1 + F2 = F2 + Fn + Ft trong :F2 khng gy ra chuyn ng quay.Fn khng gy ra chuyn ng quay.

Ft gy ra chuyn ng quay.

Vy: Trong chuyn ng quay ca vt rn xung quanh 1 trc, ch nhng thnh phn lc tip tuyn vi qu o ca im t mi c tc dng thc s.b. M men ca lc i vi trc quay: nh ngha: M men ca lc Ft i vi trc quay l mt vc t M xc nh bi:M = r .Ft

[

]

(3-1)

M c phng trng vi trc quay , c chiu thun i vi chiu quay t r sang Ft ,

c tr s: M = r.Ft Nhn xt: - M = r .F (3-2)

[ ] = [r . F ] = [r . F ]t 1

31

- M = 0 khi Ft = 0 hay Ft //. - M l m men ca Ft i vi im O.3.1.2 Thit lp phng trnh c bn ca chuyn ng quay:

Mi l cht im th i bt k ca vt rn nm cch trc quay mt khong ri viOM i = ri , c khi lng l mi v chu tc dng ca F ti ,gi a ti l gia tc tip tuyn ca

Mi (hnh 3-2), ta c:m i a ti = Fti Nhn hu hng 2 v ca biu thc trn vi r i :

[m i ri . a ti ] =

[r . F ] = Mi ti

i

(3-3)

Mi

O ri Mi

ati

Fti

Hnh 3-2

Ta c: Vy (3-3) thnh

[ri .a ti ] = [ri .[ . ri ]] = ( ri . ri ) ri ( ri . ) = ri2 m i ri2 = M i

(3-4) (3-5)

Cng cc phng trnh (3-4) v vi v theo i, ta c:

m ri

2 i i

= Mii

i

M

i

= M

l tng m men cc ngoi lc tc dng ln vt rn i vi trc .

mi

i i

r 2 = I gi l m men qun tnh ca vt rn i vi trc .

Vy:

I = M

(3-6)

(3-6) l phng trnh c bn ca chuyn ng quay ca vt rn xung quanh mt trc. T (3-6) suy ra: = M I

(3-7)

32

Kt lun: Gia tc trong chuyn ng quay ca vt rn xung quanh mt trc t l vi tng hp m men cc ngoi lc v t l nghch vi m men qun tnh ca vt rn i vi trc.3.1.3 Tnh m men qun tnh:

Theo kt qu trn, ta c cng thc tnh m men qun tnh:I=

m ri

2 i i

Nu khi lng ca vt rn phn b mt cch lin tc th ta p dng cng thc:I=

r

2

dm

(3-8)

(tch phn trn ton b vt rn) trong r l khong cch t dm n trc quay .V d 1: Tnh m men qun tnh I ca mt thanh ng cht chiu di l, khi lng M i vi trc 0 i qua trung im G cu thanh v vung gc vi thanh. Gii

Xt mt phn t ca thanh khi lng dm, chiu di dx, cch G mt on x (hnh 3-3).o

G

x

dx

l

Hnh 3-3

M men qun tnh ca dm i vi trc 0 l: dI = x2dm dm dx = V thanh ng cht nn: M l M dm = dx Suy ra: l M dI = x 2 dx (1) thnh: l Suy ra:M I = l Ml l x dx = 122 2 l 2 2

(1)

(2)

V d 2: Tnh m men qun tnh I ca mt a ng cht bn knh R, khi lng M i vi trc 0 i qua tm 0 cu a v vung gc vi a. Gii

33

Chia a thnh nhiu phn t hnh vnh khn c bn knh l x, b rng ca hnh vnh khn l dx (hnh 3-4).o

O

x

dx

Hnh 3-4

din tch ca hnh vnh khn l: dS = d(x2) = 2xdx p dng cng thc (1) : V a ng cht nn: dI = x2dmdm dS 2 xdx 2xdx = = = 2 M R R 2 R2

suy ra:

dm = 2dI =

M xdx R2

(1) thnh Suy ra:

2M 3 x dx R2R

2M MR2 I = dI = 2 x 3dx = R 0 2

(3)

M men qun tnh ca mt s vt rn ng cht c hnh dng i xng: - Vnh trn bn knh R c trc quay i qua tm O v vung gc vi mt phng ca vnh: I = MR2 - Khi cu bn knh R c trc quay i qua tm O:2 I = MR 2 5

- Mt ch nht c chiu di a, chiu rng b c trc quay i qua tm O v vung gc vi mt phng ca mt ch nht:I= 1 M(a 2 +b 2 ) 12

nh l Stene-Huygens:

trn ta tm c m men qun tnh ca cc vt rn i vi trc i xng 0 (i qua khi tm G) ca chng. Trong nhiu trng hp ta phi tm m men qun tnh

34

ca cc vt rn i vi mt trc bt k. Khi ta c th p dng nh l SteneHuygens, c pht biu nh sau:M men qun tnh ca 1vt rn i vi 1trc bt k bng m men qun tnh ca vt i vi trc 0 song song vi trc i qua khi tm G cu vt cng vi tch ca khi lng M ca vt rn vi bnh phng khong cch d gia 2 trc. Xt trng hp thanh ng cht chiu di l, khi lng M, hai trc 0 v cch nhau mt khong d,song song vi nhau v cng vung gc vi thanh (hnh 3-5).dG

0o

x

dx

l

Hnh 3-5

Khi m men qun tnh I ca cc vt rn i vi trc c xc nh bi cng thc (3-9): I = I0 + Md2 (3-9)3.1.4 ng dng Bi ton: Hai vt c khi lng ln lt l m1=2kg, m2=1kg, c ni vi nhau bng 1 si dy vt qua rng rc c khi lng m=1kg (hnh 3-6). Tm: 1. Gia tc ca cc vt. 2. Sc cng T1 v T2 ca cc dy treo. Coi rng rc l 1 a trn, cc dy ni khng gin c khi lng rt nh, ma st khng ng k. Gii Cc lc tc dng vo m1, m2 v rng rc nh hnh (3-6): p dng phng trnh c bn ca c hc cht im cho 2 vt m1 v m2:P1 + T 1 = m 1aP2 + T2 = m 2a

(1) (2)

p dng phng trnh c bn ca vt rn quay cho rng rc: R.(T1' + T2') = I (3) Chiu (1), (2) v (3) ln chiu dng chn, ta c: P1 - T1 = m1a (1') (2') T2 - P2 = m2a (3') (T1' - T2').R =I

35

v

a = R

(4)

mR 2 Trong m men qun tnh ca rng rc l: I = 2

v T1 = T1' , T2 = T2'.

T2 '

R

R

T1 'T1

T2

m2P2

+m1

P1

Hnh 3-6

Gii 4 phng trnh (1),(2),(3) v (4), ta c:(m1 m 2 )g m m1 + m 2 + 2 m m1 (2m 2 + )g 2 T1 = m m1 + m 2 + 2 m m 2 (2m1 + )g 2 T2 = m m1 + m 2 + 2 a=

Thay cc gi tr vo ta c kt qu : a = 2,8m/s2 T1 = 14N T2 = 12,6N3.2 M men ng lng ca mt h cht im 3.2.1 nh ngha

Mt h cht im M1, M2 Mi ln lt c khi lng m1, m2mi chuyn ng vi nhng vn tc v1 , v 2 ...v i ... i vi mt h quy chiu gc O. Ti thi im t, v tr

36

nhng cht im y c xc nh bi cc vc t bn knh r1 , r2 ... ri ... . M men ng lng ca h i vi O c nh ngha:L = L i = ([ri . m i v i ])i i

(3-10)

bng tng m men ng lng ca cc cht im trong h i vi O.3.2.2 Trng hp ring a. H cht im quay xung quanh mt trc c nh M men ng lng ca cht im th i ( m i , ri ) l:Li = Iii

(3-11)

Ii = miri2 l m men qun tnh ca cht im mi i vi trc quay , i l vn tc gc ca cht im mi trong chuyn ng quay xung quanh , khi m men ng lng ca h cho bi:L = Ii ii

(3-12)

b. Vt rn quay xung quanh mt trc c nh Khi 1 = 2 = = i = = n = . Vy:L = ( I i ) = Ii

(3-13)

trong I =

I = (m ri i i

2 i i

) l m men qun tnh ca vt rn i vi trc quay .

3.2.3 nh l v m men ng lng ca mt h cht im

i vi cht im ( m i , ri ) ca h, khi p dng nh l v m men ng lng ta c:dL i = M/ O (Fi ) dt

Cng cc phng trnh trn ta c

i

dL i = M/ O (Fi ) dt i dL i d dL = Li = dt dt i dt

trong : Ta c kt qu sau:

i

d L = M/ O (Fi ) = M dt

(3-14)

nh l: o hm theo thi gian ca m men ng lng ca mt h bng tng m men cc ngoi lc tc dng ln h (i vi mt gc im O bt k).

37

Trng hp ring: h cht im l mt vt rn quay xung quanh trc c nh th m

men ng lng ca h c dng L = I . Khi nh l v m men ng lng c th vit nh sau:dL d(I) = =M dt dt

Ta c:t2

L = L 2 L1 = Mdtt1

t2

(3-15)

Mdtt1

gi l xung lng ca m men lc M trong khong thi gian t = t2 - t1.L = Mt

Nu M = const th:

(3-16)

Ch thch: i vi vt rn quay xung quanh mt trc c nh, m men qun tnh I = const do :d(I) d =I = I = M dt dt

3.3 nh lut bo ton m men ng lng 3.3.1 Thit lp

Gi s c mt h cht im c lp hoc c chu tc dng ca cc ngoi lc nhng tng m men cc ngoi lc y i vi im gc O bng 0, khi ta c: theo nh l v m men ng lng:dL = M = 0 L = const dt

(3-17)

Vy: i vi mt h cht im: a. C lp b. Chu tc dng ca cc ngoi lc sao cho tng m men ca ngoi lc y i vi im gc O bng 0 th: Tng m men ng lng ca h l mt i lng bo ton.3.3.2 Trng hp h quay xung quanh mt trc c nh

p dng nh l v m men ng lng:d (I11 + I 2 2 + .... + I i i ...) = M dt

Khi M = 0 ta c:I11 + I 2 2 + .... + I i i ... = const

(3-18)

38

3.3.3 Mt vi ng dng ca nh lut bo ton m men ng lng

i vi mt h quay xung quanh mt trc vi vn tc gc , nu tng hp cc m men ngoi lc tc dng ln h bng khng th m men ng lng ca h l mt i lng bo ton: I = const Nu v mt l do no m men qun tnh I ca h tng th gim, h quay chm li; ngc li nu I ca h gim th tng, h quay nhanh ln. Ta xt mt v d sau: V d: Gii thch hin tng mt ngi ma xoay trn Ngoi lc tc dng vo ngi l trng lc v phn lc ca t; nu b qua ma st th hai lc u c phng thng ng tc l song song vi trc quay: m men ca cc lc i vi trc quay bng khng. Nu ngi giang tay ra (I tng) th vn tc quay s gim; Nu ngi co tay li hay h hai tay xung (I gim) th vn tc quay s tng.

39

BI TP 3.1 Mt a trn khi lng m1=100 kg quay vi vn tc gc 1=10 vng/pht. Mt ngi c khi lng m2=60 kg ng mp a. Hi vn tc ca a khi ngi i vo ng tm ca a. Coi ngi nh mt cht im. p s: = 22vng/pht

3.2 Hai vt c khi lng ln lt bng m1 v m2 (m1> m2) c ni vi nhau bng mt si dy vt qua mt rng rc (c khi lng l M) (hnh 1). Tm: a. Gia tc ca cc vt. b. Sc cng ca si dy. Coi rng rc l a trn, ma st khng ng k.

p s: a/ a = ( m1 m 2 ) gm1 + m 2 +

M 2

m2

m1

Hnh 1

M M )g m 2 ( 2 m1 + )g ;T = 2 2 b/ T = 2 1 M M m1 + m 2 + m1 + m 2 + 2 2 m1 ( 2 m 2 +

3.3 Mt h gm mt tr c khi lng M=2,54 kg v mt vt nng khi lng m=0,5 kg c ni vi nhau bng mt si dy vt qua mt rng rc (hnh 2). B qua khi lng ca dy, ca rng rc v khung gn vi tr. Tm gia tc ca vt nng v sc cng ca dy. p s: a = 1,16m/s T= 4,42N

m

Hnh 2

3.4 Mt vt A khi lng m trt trn mt phng nghing v lm quay mt bnh xe c bn knh R (hnh 3). M men qun tnh ca bnh xe i vi trc quay bng I. B qua khi lng ca dy. Tm gia tc gc ca bnh xe.

40

RA

Hnh 3 p s = mgR (sin k2 cos )I + mR

3.5 Mt rng rc c hai rnh vi bn knh ln lt l R v r (R>r), mi rnh c mt dy khng dn qun vo, u t do ca cc dy c ni vo mt vt c khi lng ln lt l m1 v m2 (m2>m1) (hnh 4). Tm: a. Gia tc gc ca rng rc. b. Lc cng ca cc dy treo.

m2 m1

Hnh 4 p s: = ( m22 R m1r ) g 2m1r + m 2 R + I

T1 = m1(g + r); T2 = m2(g - R)

3.6 Mt bnh xe ang quay vi vn tc gc 0 = 20 rad/s th b hm, bnh xe quay chm dn u ri dng li sau thi gian t = 20s. a. Gia tc gc ca bnh xe. b. S vng m bnh xe quay c k t lc b hm n lc dng.p s: a/ = 3,14(rad / s 2 ) b/ N = 100 (vng)

41

Chng 4 CNG - NNG LNG 4.1 Cng v cng sut ca lc-cng v cng sut ca vt rn quay 4.1.1 Cng v cng sut ca lc a. Cng nh ngha: Gi thit c mt lc F khng i, im t ca n chuyn di mt on thng MM' = S (hnh 4-1).F

M M'

Hnh 4-1

Theo nh ngha: cng A do lc F sinh ra trong chuyn di MM' l mt i lng c tr s cho bi: A = F.MM'.cos (4-1) = F.S.cos hay:A = F.S

(4-2)

trong l gc gia vc t lc F v vc t dch chuyn MM' = S t Fs = Fcos, ta c: A = Fs.S - A > 0 khi l gc nhn => F sinh ra cng pht ng. - A < 0 khi l gc t => F sinh ra cng cn. - A = 0 khi l gc vung. Trng hp tng qut: im t ca F chuyn di trn mt ng cong t C n D, trong qu trnh F thay i (hnh 4-2).ds

M

D

F

CHnh 4-2

Khi cng trong mt chuyn di v cng nh ds c tnh:dA = Fd s

(4-3)

42

Cng trong chuyn di CD c tnh:A = dA = Fd sCD CD

(4-4)

b. Cng sut Gi thit trong khong thi gian t, mt lc no sinh cng A. Theo nh ngha t s:P= A t

(4-5)A dn v mt gii hn gi l cng sut tc t

gi l cng sut trung bnh ca lc trong khong thi gian t. Khi cho t 0, theo nh ngha t s

thi (gi tt l cng sut) ca lc v c k hiu l: dA (4-6) = P = limt 0

t

dt

Vy: Cng sut c gi tr bng o hm ca cng i vi thi gian. Ta c: hay:dA = Fd s P = FP = F.v

ds dt

(4-7)

Vy: Cng sut bng tch v hng ca lc tc dng vi vc t vn tc ca chuyn di. 4.1.2 Cng v cng sut ca vt rn quay Trng hp vt rn quay xung quanh mt trc c nh (hnh 4-3), cc lc tc dng u l lc tip tuyn.

o

r M

Ft

Hnh 4-3

Ta c: dA = Ftds (gi s Ft hng theo chiu chuyn ng) mt khc: ds = r.d => dA = r.Ft.d (4-8)

Theo nh ngha: r.Ft = M => dA = r.Ft.d = M.d

43

T : hay:

P=

dA d =M dt dt

P = M .

(4-9)

4.2 Nng lng - nh lut bo ton nng lng Nng lng l mt i lng c trng cho mc vn ng ca vt cht. Mt vt mt trng thi xc nh th c mt nng lng xc nh. Khi h chuyn t trng thi 1 (c nng lng l W1) sang trng thi 2 (c nng lng l W2): qu trnh ny h nhn t bn ngoi mt cng A. Thc nghim chng t: (4-10) W2_- W1= A Vy: bin thin nng lng ca mt h trong mt qu trnh no c gi tr bng cng m h nhn c t bn ngoi trong qu trnh . -A > 0: nng lng ca h tng. -A < 0: nng lng ca h gim. (4-11) -A = 0: h c lp, khi W1 = W2 = const Vy: Nng lng ca mt h c lp c bo ton. 4.3 ng nng 4.3.1 ng nng v nh l ng nng ca cht im chuyn ng ng nng l phn c nng tng ng vi s chuyn di ca cc vt. Xt mt cht im c khi lng m chu tc dng ca mt lc F v chuyn di t v tr 1 sang v tr 2 (hnh 4-4).ds

MF

2

1Hnh 4-4

Cng ca lc F trong qu trnh l:A = Fd s(1) (2)

Trong :

F = ma = m(2)

dv dt(2)

dv dS A = m dS = mdv dt dt (1) (1)

44

v2 A = mvdv = md 2 (1) (1) A=

(2)

(2)

mv 2 = d (1) 2

(2)

2 mv 2 mv1 2 2 2

(4-12)

v1 v v2 ln lt l vn tc ca cht im ti im 1 v im 2. Ta c th nh ngha:2 mv1 = W1 l ng nng ca cht im ti v tr 1. 2

mv 2 2 = W2 l ng nng ca cht im ti v tr 2. 2

Tng qut:

W =

mv 2 2

(4-13)

l ng nng ca cht im c khi lng m v vn tc v. (4-12) c vit li: (4-14) W2 - W1 = A nh l: bin thin ng nng ca mt cht im trong mt qung ng no c gi tr bng cng ca ngoi lc tc dng ln cht im sinh ra trong qung ng . 4.3.2 ng nng v nh l ng nng ca vt rn quay Khi vt rn quay xung quanh mt trc , biu thc cng vi phn:dA = FdS = M dt

Ta li c:

M=I

d dt

dA = I

2 d dt = I d = Id 2 dt

Trong khong thi gian hu hn, vn tc bin thin t 1 n 2, ta c:A=2 I 2 I1 2 2 2

(4-15)

Suy ra biu thc ng nng ca vt rn quay l: W =I 2 2

(4-16)

Ch thch: Trng hp vt rn va quay va tnh tin th ng nng ton phn ca vt rn l: W =mv 2 I 2 + 2 2

(4-17)

Trng hp ring: vt rn i xng trn xoay ln khng trt, khi v =R do ng nng ton phn ca vt rn l:

45

W = 4.4 Trng lc th v th nng 4.4.1 Trng lc th

1 I 2 m + 2 v 2 R

(4-18)

a. nh ngha: Mt cht im c gi l chuyn ng trong mt trng lc nu ti mi v tr ca cht im u xut hin mt lc F tc dng ln cht im y (hnh 4-5).dsN

F

MHnh 4-5

Ni chung F = F( r ) = F(x, y, z) , khi cht im chuyn ng t v tr M n v tr N bt k th cng ca lc F bng:A MN = Fd sMN

Nu AMN khng ph thuc vo ng dch chuyn m ch ph thuc vo M v N th ta ni rng F( r ) l mt trng lc th. b. Nhng th d v trng lc th - Trong phm vi khng gian khng ln lm, trng trng u l mt trng lc th. Ta tnh cng ca trng lc P khi cht im dch chuyn trong trng trng u t M n N (hnh 4-6):A MN =MN

Pds

Trong mt dch chuyn nh ds :dA = P.ds = P.ds.cos

Hay

dA = P.ds = P.dz

46

zzM

MPds

zz-dz

zN

N

Hnh 4-6

Suy ra

A MN = - P.dz = PZM -PZ NM

N

AMN = mgZM - mgZN - Trng tnh in Coulumb l mt trng lc th. 4.4.2 Th nng a. nh ngha: Th nng ca cht im trong trng lc th l mt hm Wt ph thuc v tr ca cht im sao cho: AMN = Wt(M) - Wt(N) (4-19) => th nng ca cht im ti mt v tr c nh ngha sai khc nhau mt hng s cng. (4-20) V d: Wt(Z) = mgZ + C b. Tnh cht - Hiu th nng gia hai v tr hon ton xc nh.A MN =MN

Fds

= Wt(M) - Wt(N)

(4-21)

Nu M N (ng cong kn) th:

Fds = 04.4.3 ngha ca th nng

(4-22)

Th nng l dng nng lng c trng cho tng tc. V d: Dng th nng ca cht im trong trng trng ca qu t l nng lng c trng cho tng tc gia qu t vi cht im; ta cng c th ni l th nng tng tc ca qu t vi cht im.

47

4.5 nh lut bo ton c nng trong trng lc th 4.5.1 C nng Khi cht im c khi lng m chuyn ng t v tr M n v tr N trong mt trng lc th th cng ca trng lc cho bi: AMN = Wt(M) - Wt(N) Nhng theo nh l v ng nng ta c: AMN = W(N) - W(M) Suy ra hay Vy: Wt(M) - Wt(N) = W(N) - W(M) (Wt + W)(M) = (Wt + W )(N) W = Wt + W = const (4-23) (4-24)

Tng ng nng v th nng ca cht im c gi l c nng ca cht im. nh lut: Khi cht im chuyn ng trong mt trng lc th (m khng chu tc dng mt lc no khc) th c nng ca cht im l mt i lng bo ton. V d: Cht im m chuyn ng trong trng trng u:W = mgh + mv 2 2 = const

(4-25)

mv2 = const nn trong qu trnh chuyn ng ca cht im, nu H qu: V W = mgh + 2

Wt tng th W gim v ngc li, ch no Wt cc i th W cc tiu v ngc li. Ch : Khi cht im cn chu tc dng ca mt lc F khc (nh lc ma st chng hn) th ni chung c nng ca cht im khng c bo ton. bin thin c nng ca cht im bng cng ca lc F . 4.5.2 S th nng Ta c Wt = Wt(x,y,z). Trng hp th nng ch ph thuc vo mt to (v d x): Wt = Wt(x), khi th ca Wt theo x gi l s th nng. S th nng ca cht im trong trng lc th c th gip ta suy ra mt s kt lun nh tnh v chuyn ng ca cht im .mv 2 + Wt (x) = W = const 2

vi

mv 2 0 => Wt(x) W : trong qu trnh chuyn ng, cht im ch i qua nhng 2

v tr ti th nng ca cht im khng vt qu c nng, hay to x ca cht im ch bin thin trong mt phm vi no . V d 1:Mt t khi lng 1 tn, khi tt my chuyn ng xung dc th c vn tc khng i v = 54km/h. nghing ca dc l 4%. Hi ng c t phi c cng sut bao nhiu n ln c dc trn cng vi vn tc v = 54km/h? Gii

48

Gi F l lc ko ca ng c t khi ln dc. t ln dc vi vn tc khng i v th: F = mg(kcos + sin) Cng sut ca ng c t khi ln dc vi vn tc v c tnh theo cng thc: P = F.v = mg.v(kcos + sin) (*) Theo gi thit, khi tt my xung dc t c vn tc khng i nn phi c: mgsin = fms = kmgcos suy ra h s ma st : k = tg Thay gi tr ca k vo (*), ta c: P = 2mg.vsin = 2.1000.9,8.15.0,04 = 11800 (W) V d 2: T nh thp cao h = 20m, ngi ta nm mt hn khi lng m= 50g theo phng nghing vi mt phng nm ngang, vi vn tc ban u v0 = 18m/s. Khi ri ti mt t, hn c vn tc v = 24m/s. Tnh cng ca lc cn ca khng kh ln hn . Gii Chn mt t lm gc tnh th nng. C nng ca hn lc nm:2 mv0 W0 = mgh + 2

Khi ri ti mt t, c nng ca hn :W = mv 2 2

Do c ma st nn c nng ca hn khng c bo ton. Hiu c nng ca hn bng cng ca lc ma st:Ac = m 2 2 (v -v 0 ) mgh 2 50.103 (242 182 ) 50.103.10.20 = 3,5( J ) = 2

BI TP 4.1 Mt vin n khi lng m=10 kg ang bay vi vn tc v=100m/s th gp mt bn g dy v cm su vo bn g mt on s=4 cm. Tm: a. Lc cn trung bnh ca bn g ln vin n. b. Vn tc vin n sau khi ra khi bn g ch dy d=2 cm p s: a/ Ftb =125N b/ v = 7,1m/s 4.2 T mt nh thp cao h=20 cm ngi ta nm mt hn khi lng 50g theo phng nghing vi mt phng nm ngang, vi vn tc ban u v0=18m/s. Khi ri ti mt t hn c vn tc v=24 m/s. Tnh cng ca lc cn ca khng kh ln hn . p s: Ac =3,7J 4.3 Tnh cng cn thit lm cho mt v lng hnh vnh khn ng knh 1m, khi lng 500 kg ang ng yn quay ti vn tc 120 vng/pht.

49

p s: A =9859J 4.4 Hai vt c khi lng m1=150kg v m2=100kg c ni bng dy vt qua rng rc t nh mt mt phng nghing gc 300 so vi ng nm ngang (hnh 1). Vt m1 trt trn mt phng nghing vi h s ma st l k. Th cho hai vt chuyn ng, m2 i c qung ng h = 0,8m th c vn tc v=0,5m/s. Tnh: a. H s ma st k . b. Lc cng dy (ly g=10m/s2). Gii bi ton trn bng hai phng php: -Phng php ng lc hc. -Phng php nng lng.

m1 m2 Hnh 1

p s: a/ k = 0,16 b/ T = 984N 4.5 Mt vt nng trt trn mt phng nghing ri trt trn mt phng nm ngang. h s ma st l k=0,1 trn c hai on ng (hnh 2). Bit cao h = 1m v a = 5m. a. Tnh on ng x m vt i c trn mt phng nm ngang. b. Tm iu kin v k vt trt ti mt phng nm ngang.

h aHnh 2

p s: a/ x =5m h b/ k < a 4.6 Mt vt c khi lng m = 100kg c dch chuyn trn mt mt sn nm ngang di 10m. H s ma st gia vt v mt sn l k = 0,1. Tnh cng ti thiu m mt ngi cn thc hin trong hai trng hp sau: a. y vt theo phng lm vi ng nm ngang gc 300 v hng xung di. b. Ko vt theo phng lm vi ng nm ngang gc 300 v hng ln trn. p s: a/ A = 1061,7J b/ A = 945,7J

50

Chng 5 TRNG HP DN 5.1 nh lut Newton v lc hp dn v tr 5.1.1 nh lut Newton Hai cht im c khi lng m v mcch nhau mt khong r s ht nhau bng nhng lc c phng l ng thng ni hai cht im , c cng t l thun vi hai khi lng m v m v t l nghch vi bnh phng khong cch r.F = F, = G mm' r2

(5 1)mFr

F'

m'

Hnh 5-1 Vi G l h s t l, ph thuc vo h n v v gi l hng s hp dn v tr. Trong h n v SI, thc nghim cho ta gi tr ca G: G = 6,67.10-11 Nm2/ kg2 Ch thch: a. Cng thc (5-1) ch p dng cho nhng cht im. Mun tnh lc hp dn v tr gia cc vt c kch thc ln ta phi dng phng php tch phn. b. Ngi ta chng minh c v i xng nn (5-1) cng p dng c cho trng hp hai qu cu ng cht. Khi r l khong cch gia hai tm ca hai qu cu . 5.1.2 Vi ng dng a. S thay i ca gia tc trng trng theo cao Lc ht ca qu t ln mt cht im khi lng m (lc trng trng) chnh l lc hp dn v tr (hnh 5-2).mP R

hM

Hnh 5-2

Nu m ngay trn mt t th lc hp dn do qu t tc dng ln m l:

52

P0 = G

Mm R2

(5-2) (5-3)

Mt khc: P0 = mg0 g0 l gia tc trng trng ti mt t. T (52) v (53) suy ra:g0 = G M R2

(5-4)

Trong R = 6370 Km l bn knh qu t. Ti mt cht im cch mt t cao h (hnh 5-2), lc trng trng tc dng ln cht im khi lng m tnh bi:P=G Mm = mg (R + h) 2M (R + h) 22

(5-5)

T suy ra gi ca gia tc trng trng cao h:g=G

(5-6)

(5-5) v (5-6) cho ta: R g = g0 R+h

Nhng:1 h R = 1 + = 2 R+h R h 1 + R h Ta ch xt cc cao h0)

q2

F12

F21

q1F21

(q1.q2>0)F12

q2

q1

(q1.q20 : E r - q0 q

(C)

Hd l = 2

I

dlcos r (C)

127

Suy ra: a. Trng hp I thuc (C ):

(C)

Hd l = 2 d(C)

I

(C)

Hd l = I Hd l = 0

b. Trng hp I khng thuc (C ):(C)

Tng qut: trng hp dng in c hnh dng bt k v (C) c hnh dng bt k th kt qu trn vn ng. Nu H gy bi n dng in th:I=

Ii =1

n

i

nh l: Lu s ca H dc theo mt ng cong kn (C ) bt k (1vng) bng tng i s cng ca cc dng in xuyn qua din tch gii hn bi ng cong :(C)

Hd l =

Ii =1

n

i

(11-21)

Ii l dng nu n nhn chiu dch chuyn lm chiu quay thun xung quanh n, Ii l m nu ngc li. Ch : a. Khi p dng (10-21) ta khng quan tm n nhng dng in khng xuyn qua din tch gii hn bi (C). b. Nu (C) bao quanh dng in nhiu vng th phi ch n du ca I i vi mi vng dch chuyn trn ng cong y. 11.4.3 ng dng a. T trng ti mt im trong cun dy in hnh xuyn Xt mt cun dy in hnh xuyn n vng c dng in cng I chy qua (hnh 11-7). xc nh cng t trng H ti im M cch tm ng dy mt on R: t M ta v mt vng trn (C) bn knh R cng tm vi cun dy .p dng nh l Ampe ta c: Do tnh cht i xng ca cun dy nn cng t trng H ti mi im trn vng trn (C) u c gi tr nh nhau. p dng (11-21) ta c:

(C)

Hd l = Hdl = nI(C)

128

Suy ra: Hay :

H2 R = nI

H=

nI 2 R

(11-22)

(C )o

M R

HI

Hnh 11-7

b. T trng trong ng dy thng di v hn Mt ng dy thng c chiu di l rt ln so vi ng knh ca ng, c xem l ng dy thng di v hn. C th xem ng dy thng l mt phn ca dng in hnh xuyn. Do t trng trong ng dy thng: H = n0I trong : n 0 = (11-23)

n l s vng dy trn mt n v di ca ng dy. 2 R

11.5 Tc dng ca t trng ln dng in 11.5.1 Tc dng ca t trng ln mt phn t dng in-Lc Ampe Theo nh lut Ampe, nu ti M c vc t cm ng t l B th lc t tc dng ln Id l l:dF = Id l , B

[

]

(11-24)

dF gi l lc Ampe.11.5.2 Tc dng tng h gia hai dng in thng song song di v hn Xt 2 dng in thng di v hn t song song ti M v N (hnh 11-8). Dng in I1 gy ra ti M vc t B1 c phng chiu nh hnh v, ln:B1 = 0 I1 2

mt on l ca dng in I2 s chu tc dng ca lc t:

129

F21 = I 2 l , B1

[

]

F21 c xc nh nh hnh v, c ln:

F21 = 0

I1 I 2 l 2

=> I1 ht I2.I1M B2

I2

F21 F12d

N

B1

Hnh 11-8

Xt ngc li, ta c: I2 ht I1 : hai dng in cng chiu th ht nhau. L lun tng t ta thy hai dng in ngc chiu th y nhau. 11.5.3 Tc dng ca t trng u ln mt mch in kn Xt khung dy hnh ch nht ABCD (hnh 11-9).I(C)

o

rM

H dl

P

Hnh 11-9

Vc t cm ng t B vung gc vi AB v CD, B lm vi Pm mt gc . Khung dy ABCD cng v ch quay xung quanh trc . p dng quy tc bn tay tri, ta thy: - Lc t tc dng ln hai cnh AD v BC trit tiu nhau. - Lc t tc dng ln hai cnh AB v CD c phng chiu nh hnh 11-10, c ln bng nhau v bng: FAB= FCD = F = IaB

130

Hai lc FAB v FCD to thnh ngu lc lm cho khung quay xung quanh trc (hnh 11-10) cho n khi = F.d = F.b.sinFCDdB P m ( = 0).

M men ca ngu lc i vi trc c ln:

C D PmB

A BFAB

Hnh 11-10

= PmBsin = Pm , B

(11-25) (11-26)

[

]

11.5.4 Cng ca lc t Xt mch in nh hnh 11-11, thanh MN = l c th trt trn 2 thanh kim loi song song, t trong t trng u. Lc Ampe tc dng ln thanh l: F = IBl Khi on dy dn l dch chuyn mt on nh ds, cng ca lc Ampe l: dA = Fds = Iblds = IbdS = Idm Khi on dy dn l dch chuyn t v tr (1) n v tr (2) th cng ca lc Ampe l:A=m 2 m1

Id m = I mB

(11-27)N N'

B F

+A

M

I M'

Hnh 11-11

m1 v m2 ln lt l t thng qua din tch ca mch in lc thanh l v tr 1 v v tr 2. (11-27) cng ng cho mch in bt k dch chuyn trong t trng bt k.

131

Vy: Cng ca lc t trong s dch chuyn mt mch in bt k trong t trng bng tch gia cng dng in trong mch v bin thin ca t thng qua din tch ca mch . V d 1: Hnh 11-12 v mt ct vung gc ca hai dng in thng song song di v hn ngc chiu nhau. Khong cch gia hai dng in AB = 10cm. Cng ca cc dng in ln lt bng: I1=20A, I2=30A. Xc nh vc t cng t trng tng hp ti cc im M1, M2, M3. Bit M1A = 2cm; AM2 = 4cm; BM3 = 3cm. (Hai dng in t trong khng kh).I1 M1A

I2 M2B

M3

Hnh 11-12

Gii Gi H1 v H 2 l vc t cng t trng ln lt do I1 v I2 gy ra. 1. Vc t cng t trng tng hp ti im M1:H M1 =H1 +H 2H M2I1 H1 H 2 I2B

H2M1

H1 H M3 H2M3

H M1 A

M2

H1

Hnh 11-12a

H1 v H 2 do I1 v I2 gy ra ti M1 cng phng ngc chiu. Do H M1 c xc nh

nh hnh 11-12a, ln:H M1 = H1 -H 2 = = I1 I2 2.AM1 2.BM1

20 30 = 120( A / m) -2 2.2.10 2.12.10-2

2. Vc t cng t trng tng hp ti im M2:H M2 =H1 +H 2 H1 v H 2 do I1 v I2 gy ra ti M2 cng phng cng chiu. Do H M2 c xc nh

nh hnh 11-12a, ln:

132

H M2 = H1 +H 2 = =

I1 I2 + 2.AM 2 2.BM 2

20 30 + = 159, 23( A / m) -2 2.4.10 2.6.10-2

3. Vc t cng t trng tng hp ti im M3:H M3 =H1 +H 2 H1 v H 2 do I1 v I2 gy ra ti M3 cng phng ngc chiu. Do H M3 c xc nh

nh hnh 11-12a, ln:H M3 = H 2 -H1 = = I2 I1 2.BM 3 2.AM 3 30 20 = 135( A / m) -2 2.3.10 2.13.10-2

V d 2: Mt dy dn c un thnh mt hnh thang cn, c dng in cng I =6,28A chy qua (hnh 11-13). T l chiu di hai y bng 2. Tm vc t cm ng t ti im A- l giao im ca ng ko di ca hai cnh bn. Cho bit y b ca hnh thang l = 20cm, khong cch t A ti y b b = 5cm v dng in t trong khng kh.E

I

B

lCD

A

b

Hnh 11-13

Gii V dng in EB v CD c ng ko di i qua A nn vc t cm ng t do 2 dng in ny gy ra ti A bng 0. Vc t cm ng t ti A ch bng tng vc t cm ng t do hai dng in BC v DE gy ra. Gi B1 l vc t cm ng t do dng in DE gy ra ti A, B2 l vc t cm ng t do dng in BC gy ra ti A. Ta c:B =B1 +B 2 B1 vung gc vi mt phng hnh v, c hng i vo v c ln:

B1 =

o I ( cos1 cos2 ) (*) 4 R

133

trong : R = 2b (v BC l ng trung bnh ca tam gic ADE);cos1 = l l +(2b) 22

Gc 1 v 2 b nhau do : cos 2 = -cos 1 . Thay cc gi tr trn vo biu thc (*), ta c:B1 = o I I 2cos1 = o cos1 4 2b 4 b I l = 0 4b l2 +(2b) 2

E

2IB

1 l 1D C

A

B

2 b

Hnh 11-13a

B2 vung gc vi mt phng hnh v, c hng i ra v c ln:

B2 =

o I ( cos'1 cos'2 ) (**) 4 R

trong : R = b, 1 = '1 v 2 = '2 Thay cc gi tr trn vo biu thc (**), ta c:B2 = o I I 2cos1 = 2 o cos1 4 b 4 b I l =2 0 2 4b l +(2b) 2

So snh B1 v B2 ta thy: B1 < B2. Vy B ti A c phng vung gc vi mt phng hnh 11-13a, c hng i ra v c ln:B =B1 - B 2 =

0 I l (***) 4b l2 +(2b) 2

Thay cc gi tr ca I, b,l vo (***) ta tnh c: B 9.10-6 T

134

BI TP 11.1 Hai dy dn thng di song song xuyn qua v vung gc vi mt phng hnh v (hnh 1). M N + I1 I2 Khong cch gia hai dy l 32cm, khong cch t dng in I1 n im M l 8cm, Hnh 1 khong cch t dng in I2 n im N l 8cm. Dng in I2 c chiu nh hnh v v c cng l 5A. a. Hi dng in I1 phi c chiu v cng l bao nhiu cm ng t ti N bng khng? b. Xc nh vc t cm ng t ti im M trong trng hp dng in I1 va tm c trn. p s: a/ I1 = 25A v ngc chiu vi I2 b/ B= 6.10-5TD

11.2 Mt dy dn c gp li thnh hnh tam gic vung cn ADC c AD=AC=10cm (hnh 2). Khung dy c t trong B mt t trng u cm ng t B=0,01T. Cho dng in A I=10A chy trong khung theo chiu CADC. Xc nh lc t Hnh 2 tc dng ln cc cnh ca khung dy. p s: FAD = FCA = 10-2NFDC = 1,41.10 2 N

C

11.3 Mt dy dn c un thnh hnh ch nht c cc cnh a=16cm, b = 30cm, c dng in cng I = 6A chy qua. Xc nh vc t cng t trng ti tm ca khung dy. p s: H = 27A/m 11.4 Mt dy dn c un thnh hnh tam gic u mi cnh a = 50cm. Trong dy dn c dng in cng I = 3,14A chy qua. Xc nh vc t cng t trng ti tm ca khung dy. p s: H = 9A/m 11.5 Mt khung dy trn bn knh R = 5cm, Khung gm 12 vng dy, trong mi vng dy c dng in cng I = 0,5A. Xc nh cm ng t ti tm ca khung dy. p s: B = 7,54.10-5T 11.6 Mt dy dn di, an gia c un li thnh mt hnh vng trn nh hnh 3. Bn knh vng trn dy dn l R = 6cm. TrongHnh 3

I

135

dy dn c dng in cng I = 3,75A chy qua. Xc nh vc t cm ng t ti tm ca vng dy. p s: B = 2,68.10-5T 11.7 Mt khung dy trn bn knh R = 10cm c dng in cng I = 1A chy qua. Xc nh vc t cm ng t ti: a. Mt im trn trc ca vng dy v cch tm O mt on h = 10cm. b. Tm O ca vng dy. p s: a/ BM = 2,3.10-6 T b/ B0 = 6,3.10-6 TB

136

Chng 12 CM NG IN T 12.1 Cc nh lut v hin tng cm ng in t 12.1.1 Th nghim Faraday Th nghim gm nam chm vnh cu, ng dy in c ni vi in k G thnh mt mch in kn. Th nghim c b tr nh hnh 12-1.S N S N

G

G

B

IC B a. Khi a nam chm vo trong lng ng dy Hnh 12-1 B B

IC

b. Khi a nam chm ra khi ng dy

Th nghim chng t: - Nu a thanh nam chm vo trong lng ng dy th kim in k s lch: trong ng dy xut hin dng in. l dng in cm ng. - Nu a thanh nam chm ra dng in cm ng Ic s c chiu ngc li. - Di chuyn thanh nam chm cng nhanh th Ic cng ln. Qua th nghim Faraday rt ra nhng kt lun tng qut sau: a. S bin i m qua mch kn l nguyn nhn sinh ra dng in cm ng trong mch. b. Dng in cm ng ch tn ti trong thi gian m gi qua mch thay i. c. Dng in cm ng t l vi tc bin thin ca m. d. Chiu ca dng in cm ng ph thuc vo m gi qua mch tng hay gim.

137

12.1.2 nh lut Lenx Nghin cu hin tng cm ng in t, Lenx tm ra nh lut tng qut v chiu ca dng in cm ng, gi l nh lut Lenx: Dng in cm ng phi c chiu sao cho t trng do n sinh ra c tc dng chng li nguyn nhn sinh ra n. Ta vn dng nh lut ny xc nh chiu ca dng in cm ng trong hai trng hp hnh 12-1. - Trng hp hnh 12-1a, nguyn nhn gy ra dng in cm ng l do dch chuyn cc bc ca thanh nam chm vo trong lng ng dy, lm cho t thng gi qua ng dy theo chiu t trn xung tng. Theo nh lut Lenx, dng in cm ng Ic phi c chiu sao cho t trng B' do n sinh ra chng li s tng : tc l B' phi ngc chiu vi t trng B ca nam chm. Bit B' , dng quy tc vn inh c ta c th xc nh c chiu ca dng in cm ng Ic nh trn hnh 12-1a. - Trng hp hnh 12-1b, nguyn nhn gy ra dng in cm ng l do dch chuyn cc bc ca thanh nam chm ra xa ng dy, lm cho t thng gi qua ng dy theo chiu t trn xung gim. Theo nh lut Lenx, dng in cm ng Ic phi c chiu sao cho t trng B' do n sinh ra chng li s gim : tc l B' phi cng chiu vi t trng B ca nam chm. Bit B' , dng quy tc vn inh c ta c th xc nh c chiu ca dng in cm ng Ic nh trn hnh 12-1b. 12.1.3 nh lut c bn ca hin tng cm ng in t: S xut hin ca dng in cm ng trong mch chng t trong mch c mt sut in ng. Sut in ng y gi l sut in ng cm ng. Dch chuyn mt vng dy kn (C) trong t trng, gi s trong khong thi gian dt t thng qua (C) bin thin mt lng dm: c dng Ic trong vng dy (C) (hnh 12-2).n

Ic

t+dt

+n

+Hnh 12-2

t

138

Cng ca lc t tc dng ln Ic l: dA = Icdm dA l cng cn. dch chuyn (C) ta tn mt cng dA = -dA = -Icdm. Theo nh lut bo ton v chuyn ha nng lng: dA bin thnh nng lng ca dng in Ic: ecIcdt vi ec l sut in ng cm ng, ta c: ecIcdt = -Icdm Suy ra:ec = d m dt

(12-1)

nh lut: Sut in ng cm ng lun bng v tr s nhng tri du vi tc bin thin ca t thng qua din tch ca mch in. 12.1.4 Nguyn tc to ra dng in xoay chiu Da vo hin tng cm ng in t, khi cho khung dy quay trong mt t trng u th trong khung dy s xut hin sut in ng bin thin theo quy lut hm s sin i vi thi gian: chnh l nguyn tc to ra dng in xoay chiu. 12.2 Hin tng t cm 12.2.1 Th nghim v hin tng t cm Gi s mch in kn, kim in k G nm v tr a no (hnh 12-3).

0

a

I

G

+

-

Hnh 12-3

- Nu ngt mch in, ta thy kim in k G lch v qu v tr s 0 ri mi quay tr li s 0. - Nu ng mch, kim in k lch qu v tr a lc ny ri mi quay li v tr a. Gii thch: - Khi ngt K: m qua mch bin thin Ic cng chiu vi I qua mch, Ic phng qua in k G lm kim ca G lch qu v tr 0 ri mi quay v 0. - Khi ng K: Ic ngc chiu vi I qua ng dy, dng in qua nhnh c in k G nhiu hn do kim in k G vt qu v tr a ri sau mi v v tr a. Hin tng trn gi l hin tng t cm, dng in xut hin trong mch gi l dng t cm.

139

12.2.2 Sut in ng t cm Theo nh lut c bn ca hin tng cm ng in t: tc = Mt khc: m B m B I m I m = L.I L l h s t l ph thuc vo hnh dng, kch thc ca mch in v tnh cht ca mi trng trong t mch in, gi l h s t cm. Vi mt mch in nht nh L l hng s. Suy ra: 12.2.3 H s t cm H s t cmL= tc = LdI dt

d m dt

(12-2)

mI

(12-3)

trong h SI: L c n v l Henry (H) H s t cm ca ng dy in thng di v hn:B = 0 n 0 I = 0 n l I

m = nBS = 0

n 2 .S I l m n 2 .S L= = 0 I l

(12-4)

12.3 Hin tng h cm 12.3.1 Hin tng Hai mch in kn (C1) v (C2) t cnh nhau c I1 v I2 chy qua. Nu I1 v I2 ng thi bin thin th t thng do mi mch sinh ra v gi qua din tch ca mch kia cng thay i c hai mch u xut hin dng in Ic. Hin tng gi l hin tng h cm, cc dng in Ic gi l dng in h cm. 12.3.2 Sut in ng h cm, h s h cm Theo nh lut c bn ca hin tng cm ng in t h = - m12 : l t thng do I1 gi qua mch in (C2): m12 I1 m12 = M12 I1 - m21 : l t thng do I2 gi qua mch in (C1) . m21 I 2 m21 = M 21 I 2d m . Gi: dt

140

M12 v M 21 l h s h cm ca (C1) v (C2); (C2) v (C1). Ngi ta chng minh

c rng: M12 = M21 = M hc2 = hc1 L m12 dI = M 1 dt dt L dI = m21 = M 2 dt dt

(12-5) (12-6)

M cng c th nguyn nh L (n v l H ). 12.4 Nng lng t trng 12.4.1 Nng lng t trng ca ng dy in Xt mch in gm mt b ngun v mt ng dy in c mc nh hnh 12-4. Lc mch in ng, p dng nh lut Ohm cho mch in: + tc = Ri Ldi Ldi = Ri = Ri + dt dt

Nhn hai v biu thc trn cho idt ta c:

idt = Ri 2 dt + LidiL

I

+Hnh 12-4

trong :

idt : nng lng do ngun sinh ra trong khong thi gian dtRi 2 dt : phn in nng chuyn sang nhit trn R

Lidi = dWm : phn in nng tim tng trong cun dy di dng nng lng t trng.

Trong c qu trnh thnh lp dng in, nng lng t trng trong ng dy l:Wm =Wm

dWm = Lidi =0 0

I

LI 2 2

(12-7)

12.4.2 Nng lng t trng

Gi V = Sl l th tch ca ng dy mt nng lng t trng trong ng dy c tnh:

141

m =

Wm 1 Li 1 = . = . V 2 Sl 2

2

0

n 2 SI 2 2 2 2 l = 1 n I = 1 B 0 S.l 2 2 0 l2

(12-8)

Nng lng t trng trong th tch dv:dWm = m dv = 1 1 B2 dv = BHdv 2 0 2

Nng lng t trng trong mt vng khng gian c th tch V bt k:Wm = dWm =V

1 B2 1 0 dv = 2 V BHdv 2V

(12-9)

V d 1: Mt ng dy thng di c t cm L=0,5H, in tr thun R=2. Khi c dng in cng I chy qua ng dy th nng lng t trng tch ly trong ng dy l: W=100J. Tnh cng dng in I v cng sut ta nhit trn ng dy. Gii

p dng cng thc tnh nng lng t trng trong ng dy in:1 W = LI 2 2

Suy ra : Ta c cng thc tnh cng sut:

I=

2W 2.100 = 20( A) = L 0,5

P = R.I2 = 2.202 = 800(W)

V d 2: Mt cun dy phng gm n vng. Hai u cun dy c ni vi mt in k o in lng. in tr ca cun dy v in k l R. Lc u cun dy c t

trong t trng u, vc t cm ng t B vung gc vi mt phng cun dy. Sau di chuyn nhanh cun dy ra khi t trng. Tnh in lng Q di chuyn qua in k. Bit din tch gii hn bi cun dy l S.Gii

Chn vc t php tuyn n ca cun dy cng phng cng chiu vi vc t cm ng t B . Ti thi im t1 t thng qua cun dy l: m1 = nBS Ti thi im t2 (khi a cun dy ra khi t trng) t thng qua cun dy l: m2 = 0 Suy ra: |m| = nBS

Sut in ng xut hin trong cun dy c ln:= m nBS = t t

142

Cng dng in qua in k:I= nBS = R t

in lng Q:Q = It = nBS R

BI TP

12.1 Mt ng dy dn thng di gm N = 500 vng c t trong mt t trng c ng sc t trng song song vi trc ca ng dy. ng knh ca ng dy d=10cm. Tm sut in ng cm ng trung bnh xut hin trong ng dy nu trong thi gian t=0,1s cm ng t thay i t 0 n 2 Tesla. p s: EC=78,5V 12.2 Mt cun dy dn gm N = 100 vng quay trong mt t trng u vi vn tc khng i =5vng/s. Cm ng t B=0,1T. Tit din ngang ca ng dy S=100cm2. Trc quay vung gc vi trc ca ng dy v vung gc vi vi ng sc t trng. Tm sut in ng cm ng xut hin trong cun dy v gi tr cc i ca n. p s: E=NBScos(2n+) Em=NBS =3,14V 12.3 Ti tm ca mt khung dy trn phng dn gm N1 = 50 vng, mi vng c bn knh R=20cm, ngi ta t mt khung dy nh gm N2 = 100 vng, din tch mi vng S=1cm2. Khung dy nh ny quay xung quanh mt ng knh ca khung dy ln vi vn tc khng i =300 vng/s. Tm gi tr cc i ca sut in ng cm ng xut hin trong khung dy nu dng in chy trong khung dy ln c cng I = 10A. p s: Em =4,7.10-3V 12.4 Mt my bay bay vi vn tc v=1500km/h. Khong cch gia hai u cch my bay l=12m. Tm sut in ng cm ng xut hin gia hai u cch my bay bit rng thnh phn thng ng ca cm ng t trng tri t cao ca my bay B=0,5.10-4T p s: Ec =0,23V

143

Chng 13 TRNG IN T 13.1 Lun im th nht ca Mawell 13.1.1 Pht biu lun im Bt k mt t trng no bin i theo thi gian cng sinh ra mt in trng xoy. 13.1. 2 Phng trnh Mawell - Faraday Xt mt vng dy dn kn (C) nm trong mt t trng ang bin i theo thi gian (hnh 13-1).

dSBoHnh 13-1

(C)

Theo nh lut c bn ca hin tng cm ng in t, sc in ng xut hin trong vng dy l:c = d m d = BdS dt dt (S)

Mt khc, theo nh ngha ca sc in ng ta c: c = Ed l(C)

suy ra:

(C)

Ed l =

d BdS dt (S)

(13-1)

l phng trnh Mawell - Faraday di dng tch phn. Vy: Lu s ca vc t cng in trng xoy dc theo mt ng cong kn bt k th bng v gi tr tuyt i nhng tri du vi tc bin thin theo thi gian ca t thng gi qua din tch gii hn bi ng cong . ngha ca phng trnh (13-1) l: n cho php ta tnh c in trng xoy nu bit quy lut bin i ca t trng theo thi gian. Trong gii tch vc t ngi ta chng minh c:

145

(C)

Ed l = rotEdS(S)

Mt khc ta c: d BdS = dt (S) dB dt B ( ) ddt dS (S)

suy ra:

rotE =

(13-2)

Trng hp tng qut: vc t cm ng t c th bin i theo c thi gian v khng gian nhng ch c t trng bin i theo thi gian mi sinh ra in trng xoy, do (13-2) c vit li:rotE = B t

(13-3)

13.2 Lun im th hai ca Mawell 13.2.1 Pht biu lun im Bt k mt in trng no bin i theo thi gian cng sinh ra mt t trng. 13.2.2 Phng trnh Mawell - Ampe a. Gi thuyt ca Mawell v dng in dch: Dng in dch l dng in tng ng vi in trng bin i theo thi gian v phng din sinh ra t trng. Theo Mawell in trng bin i gia hai bn ca t in sinh ra t trng ging nh mt dng in (dng in dch) chy qua ton b khng gian gia hai bn ca t in, c chiu l chiu ca dng in dn trong mch v c cng bng cng dng in dn trong mch . Nu gi Id l cng dng in dch chy gia hai bn t in, S l din tch ca mi bn th mt dng in dch gia hai bn l:Jd = Id I = S S

vi I l cng dng in dn trong mch. Ta c:I= dq dt

suy ra:

Jd =

1 dq d q d = = S dt dt S dt

: l mt in mt trn bn dng ca t in. Ta c: D = , suy ra:

146

Jd =

dD dt

(13-4) (13-5)

Di dng vc t:

Jd =

dD dt

Biu thc (13-5) chng t: vc t mt dng in dch bng tc bin thin theo thi gian ca vc t cm ng in. Trong trng hp tng qut, vc t cm ng in D = D(x, y, z, t) nhng ch c in trng bin i theo thi gian mi sinh ra t trng, do :Jd = D t

(13-6)

M rng gi thuyt trn v dng in dch cho trng hp mt dng in bt k, Mawell i ti gi thuyt tng qut sau: Xt v phng din sinh ra t trng th bt k mt in trng no bin i theo thi gian cng ging nh mt dng in gi l dng in dch c vc t mt dng bng:Jd = D t

trong D l vc t cm ng in ti im ta xt. b. Thit lp phng trnh Mawell -Ampe Theo Mawell t trng do c dng in dn v in trng bin i theo thi gian tc dng in dch sinh ra. V vy Mawell a ra khi nim dng in ton phn bng tng dng in dn v dng in dch. Do ta ni rng t trng do dng in ton phn sinh ra. Mt ca dng in ton phn c tnh theo cng thc:J tp = J + D t

(13-7)

Theo nh l v dng in ton phn:

Hd l = IC

tp

vi:

D dS I tp = JdS = J + t S S

suy ra:

Hd l = J + C S

D dS t

(13-8)

l phng trnh Mawell -Ampe di dng tch phn.

147

Vy: Lu s ca vc t cng t trng dc theo mt ng cong kn bt k th bng cng dng in ton phn chy qua din tch gii hn bi ng cong . Ta cng chng minh c rng:rotH = J + D t

(13-9)

l dng vi phn ca phng trnh Mawell-Ampe, p dng c i vi tng im trong khng gian. ngha ca phng trnh (13-9) l: n cho php ta tnh c t trng H nu bit s phn b dng in dn quy lut bin i ca in trng theo thi gian. 13.3 Trng in t v h thng phng trnh Mawell 13.3.1 Nng lng trng in t in trng v t trng ng thi tn ti trong khng gian to thnh mt trng thng nht gi l trng in t. Trng in t l mt dng vt cht c trng cho tng tc gia cc ht mang in. Mt nng lng t trng:1 w = w e + w m = ( 0 E 2 + 0H 2 ) 2

(13-10)

Nng lng t trng:W=

V

wdV = 2 ( E0 V

1

2

+ 0H 2 )dV (13-11)

13.3.2 Phng trnh Mawell -Faraday - Dng tch phn:

(C)

Ed l = dt BdS (S) B t

d

(13-12)

- Dng vi phn:

rotE =

(13-13)

13.3.3 Phng trnh Mawell -Ampe - Dng tch phn: D dS Hd l = J + t C S

(13-14)

- Dng vi phn:rotH = J + D t

(13-15)

13.3.4 nh l Ostrogradski-Gauss (O-G) i vi in trng:

148

- Dng tch phn:

DdS = qS

(13-16)

- Dng vi phn:divD =

(13-17)

13.3.5 nh l O-G i vi t trng: - Dng tch phn:

BdS = 0S

(13-18)

- Dng vi phn:divB = 0

(13-19)

13.3.6 Cc phng trnh lin h cc i lng c trng cho trng Trong cc phng trnh Mawell cc i lng c trng cho trng u c xc nh ti tng im trong khng gian v ni chung u l cc i lng bin thin theo thi gian:E = E(x, y, z, t) B = B(x, y, z, t) D = D(x, y, z, t) H = H(x, y, z, t)

a. in trng tnhE = E(x, y, z) D = D(x, y, z) B=0 H=0

h phng trnh Mawell thnh:

Ed l = 0C

hay rotE = 0 hay divD =

DdS = qS

D = 0 E

b. T trng khng iE=0 D=0 B = B(x, y, z) H = H(x, y, z)

h phng trnh Mawell thnh:

Hd l = I hay rotH = JC

149

BdS = 0S

hay divB = 0

B = 0 H

c. Sng in tE = E(x, y, z, t) ; B = B(x, y, z, t) ; D = D(x, y, z, t) ; H = H(x, y, z, t) ;=0

J =0

h phng trnh Mawell thnh:rotE = divD = 0

B t

; ; ;

rotH =

D t

divB = 0

D = 0 E

B = 0 H

V d: Chng t rng trong chn khng, vc t cm ng t B tha mn phng trnh sau:E 0 0 2E =0 t 2

Gii Trong gii tch vc t ta chng minh c ng thc:rot rotE = grad divE E

(*)

i vi chn khng: J = 0 , = 0 h phng trnh Mawell thnh:rotH = D t

; divD = 0

Do : ta c:

grad divE = 0rotE = B t

B rot rotE = rot = rotB t t v tri ca (*) c dng: D 2E = 0 0 2 = 0 rotH = 0 t t t t

(

)

(

)

hay l:

E 0 0

2E =0 t 2

150

BI TP 7.2 Chng minh rng in th tnh in tha mn phng trnh sau y: = 0

7.3 Trong mt th tch hu hn c vc t cm ng t B vi cc thnh phn: Bx=0 ; By=0; Bz=B0+ax, trong a l mt hng s v lng ax lun lun nh hn so vi B0. Chng minh rng nu trong th tch khng c in trng v dng in th t trng y khng tha mn phng trnh Mawell. 7.7 Mt t in c in mi vi hng s in mi =6 c mc vo mt hiu in th xoay chiu u = U0cost vi U0= 300V, chu k T=0,01s. Tm gi tr ca mt dng in dch bit rng hai bn ca t in cch nhau 0,4cm. p s: jd =2,5.10-3sin20t(A/m2) 7.11 Cho mt trng in t bin thin trong chn khng vi cc vc t cng trng E(0,0,E) v , trong H(H,0,0) , trong H = H0cos(t-ay) vi a= 0 0 . Chng minh rng gia cc vc t cng trng c mi quan h sau y: 0 E = 0 H

151