Vật lý - Cong Thuc Vat Ly

8/14/2019 Vt l - Cong Thuc Vat Ly

1/16

Chng 1: C HC VT RN

1/ VT RN QUAY Ui lng Cng thc

a. Tc gc constt

==

+>0 vt quay theo chiu dng

+ 0 vt quay nhanh dn u

+ . < 0 vt quay chm dn ub. Vn tc gc 0 t = + ; ( 0 : vn tc gc lc t = 0)c. Ta gc 20 0

1t t

2

= + + ; ( 0: ta gc lc t = 0)

d. Gc quay 20 01

t t2

= = + ; ( gc quay tnh t thi im t = 0 )

2 22 1 0 2 1 2 11

(t t ) (t t )2

= = + (gc quay t t1 n t2.)

e. S vng quay N2

=

f. Lu Khi vt rn quay B th 1 im trn vt rn c. trn B. tc di : v .r= (r: khcch t M n trc quay.)

v = v0 + att

gia tc hng tm 2na .r= gia tc tip tuyn at= .R

gia tc t/phn2 2

n ta a a= + 3/ KHI TM CA VT RN. Khi tm ca cc vt rn ng cht, c dng hnh hc l cc tm i xng ca n. Cng thc tnh ta khi tm ca vt rn:

XG=

i

ii

m

xm

; YG=

i

ii

m

ym(vi i= 1;2,3,..)


2/16

4/ MMEN LC PHNG TRNH NG LC HC VT RN QUAY.a. Mmen lc: Gi d l khong cch t trc quay n gi ca lc (tay n). FM F.d=r , nu Fc xu hng lm vt quay theo chiu dng. FM F.d= r , nu Fc xu hng lm vt quay ngc chiu dng. MF = 0 , nu F c gi i qua trc quay.hoc c phng s.song vi trc quay

b. Phng trnh ng lc hc vt rn quay : FM I= r

(vi l gia tc gc ca vt rn. ; I l mmen qun tnh ca vt rn i vi trc quay.)*. Mmen qun tnh i vi trc quay i qua khi tm G ca vt rn.

Vnh trn bn knh R (hnh tr trn rng):I = mR2 Thc dp: 21

I m12

= l

a trn c, mng (hnh tr trn c): 21

I mR2

= Khi cu c: 22

I mR5

=

*. Mmen qun tnh ca vt rn i vi trc quay : 2GI I md = +( vid l khong cch gia khi tm G v trc quay )

5/ NH LUT BO TON MMEN NG LNG.a. Mmen ng lng L I.= , (n v kg.m2/s )b. nh lut bo tonMmen ng lng

FM 0 L const= =

I khng i vt quay u ( 0) hoc ng yn ( = 0) Nu h ch c 1 vt : 1 1 2 2I . I . = Nu h gm 2 vt : 1 1 2 2 1 1 2 2I . I . I ' . ' I ' . ' + = + = hs

c. Lu F 2 1 FM 0 L L M . t = 6/ NG NNG CA VT RN

a. ng nng quay: W = 21

I2

(J)

+ (rad/s): tc gc ca vt rn+ I: momen qun tnh ca vt rn .

b. nh l ng nng2 2

ngl d 2 1

1 1A W I I M.2 2= = = + M: momen ngoi lc t/dng ln vt rn+ : Gc quay ca vt rn

7/ TN S GC V CHU K DAO NG B CA CON LC VT L:

=I

mg.d T= 2

mg.d

I(vid l k/cch gia khi tm G v trc quay )


3/16

chng 2: DAO NG C HC

1/ CON LC L XO Tan so goc:=K/ m Chu k : T = 2/ Tan so: f= 1/T =N/t = 2f

Li o:x(t)= A.cos(t+ )+ A= |xmax|:Bien o

(cm)+ (t + ): Pha dao

ong+ :Pha ban au

Van toc: v(t)=A.sin(t + ) |Vmax|= A(cm/s) Gia toc: a(t)= 2.x(t) |amax|= 2.A(cm/s2)

. Lu y:+ Chieu dai quy ao = 2.A.

+ Lc hoi phuc: (hp lc gay dh): F= m| a| = K| x|Fmax = KA ( vi K = m2 (N/m))

+ He thc gia x,v, , A : A2

= x2

+2

2

V

V = 22

xA The nang an

hoiong nang C nang DH

Wt = Kx2 (J )+ Goc the nang la

VTCB+ K(N/m) o cng

lo xo+ x= A.Cos(t + )

(m) Wt= KA2.Cos2 (t+ )

W = mv2 (J )+ m (kg) Khoi lng

con lac+ v= A.Sin(t + )

(m/s)+ K = m.2

W= KA2

.Sin2

(t+ )

W = Wt + W (J ) W= Wtmax = KA2 W= Wmax= m2A2

+ A (m) bien o daoong+ (rad/s) tan so

goc

* Con lc l xo treo thng ng bin dng ca l xo Chiu di l xo khi con lc dao ng Khi vt v tr cn bng.P = F0d mg = k.l0 l0 = mg/k Khi vt v tr c li x

l = l0 + x

Khi vt v tr c li x l = l0 + l = l0 + l0 + x lmax= l0 + l0 + A lmin= l0 + l0 A

(vi l0 l chiu di t nhin ca l xo)o ln lc an hoi F= K.|| viK(N/m) va= 0 obien dang o ln lc an hoi khi vat li o x

F= K.|0 + x | (neu truc ox hngxuong)

F= K.|0 x | (neu truc ox hnglen)

Gia tr cc ai Fmax = K.(0 + A)


4/16

Gia tr cc tieu Fmin = 0 Khi A0 Fmin = K(0 A) Khi A < 0

Ch : Nu trc ox thng ng hng ln th: bin dng: l= l0 x Chiu di l xo: l = l0 + l = l0 + l0 x Kssong = K1 + K2 Kn tip = K1 .K2/ K1 + K2 L xo c chiu di , sut n hi E, tit din S th cng ca n: K= ES/

2/ CON LC N*Chu k, li , vn tc khi dao ng iu ha (gc lch 0 100): Tn s gc: g =

l

Chu k: T 2g

= l

Li cong: St = S0cos(t + )( S0 = l0: bin ; 0: gc lch cc i)

vn tc:vt = S0sin(t + )( vi maxv = S0 = l0= 0 gl)

Li gc:t = 0cos(t + )

*. C nng dao ng iu ha: W = W + Wt = 2 2 20 01 1

m S mg2 2

= l

ng nng: W 21

mv

2

=

Th nng trng trng: 2 2 2 2t1 1 g 1

W m S m ( ) mg2 2 2

= = = l ll

*. Lc cng dy treo.2mv

T mg.cos= +l

= mg(3.cos 2.cos0)

gi tr cc i: Tmax = mg(3 2.cos0); ( khi vt qua v tr cn bng = 0.) gi tr cc tiu: Tmin = mg.cos0; ( khi vt ti v tr bin = 0)

Vn tc ca vt ( ti im c cao h) : 0v 2g (cos cos )= l

Khi gc lch 0 100 v = 2 20S S

*. S thay i chu k theo cao h: Th = T ( 1 + Rh

) , ( vi R= 6400 Km )

*. S thay i chu k theo nhit : T2 = T1 ( 1 + ).2

1 0t , ( vi t0= t2 t1)

*. S thay i chu k con lc n khi c thm ngoi lc f tc dng.

T= T'g

g , (vim

fgg+=

' gia tc trng lc hiu dng)

Khi f hng xung g = g + f/m . Khi f hng ln g = g - f/m . Khi f c phng ngang g = 22 )/( mfg +

3/. TNG HP DAO NG IU HA*. Bin dao ng tng hp A= .cos .A2.AAA 212221 ++

( vi = 1 2 lch pha ) = 0 ( cng pha ) A= A1 + A2. = ( ngc pha) A= |A1 - A2|.

*. Pha ban u ca dao ng tng hp

tan =2211

2211

cosAcosA

sinAsinA

+

+

= tan

= , nu mu s c gi tr dng = , nu mu s c gi tr m


5/16

Chng 3: SNG C HC

1/ PHNG TRNH TRUYN SNGa. Bc sng: = v.T = v/f (Vi v l tc truyn sng , tc truyn pha dao ng).b. Phng trnh truyn sng:

u = A cos( t -

x.2) =

t xA cos 2 ( )

T 3. lch pha gia hai im A, B trn cng phng truyn sng

2 d

=

, (vi d l khong cch gia hai im.A, B )

2/ SNG DNGHai u dy l 2 nt Mt u dy l nt, u kia l bng

/2A

B/2

uB phn x = - uB ti Chiu di si dy khi c sngdng

k2

=l ( k: Z s b

sng ) S im nt trn dy: Nnt = k+ 1 S im bng trn dy: Nbng= k.

/4A B

uB phn x = uB ti

Chiu di si dy khi c sng dngk

2 4

= +l ( k: Z s b sng )

S im nt trn dy: Nnt = k + 1 S im bng trn dy: Nbng = k.+ 1

3/ GIAO THOA SNG2 ngun kt hp A, B cng pha 2 ngun kt hp A, B ngc pha

lch pha ca 2 sng thnh phn ti cng 1im

2 12 d d

=

S dy cc i trn on ni 2 ngun

AB AB

< k ZC: hiu in th u(t) sm pha so vi cng dng in i(t). ZL < ZC: hiu in th u(t) tr pha so vi cng dng in i(t). ZL = ZC: hiu in th u(t) cng pha vi cng dng in i(t). on mch ch c in tr thun: = 0 uR(t) cng pha CD i(t). on mch ch c cun cm thun: = /2 uL(t) sm pha CD i(t) gc /2.

on mch ch c t in: = - /2 uC(t) tr pha CD i(t) gc /2.

g. nh lut m: U = I.Z hay U0 = I0.Z on mch ch c in tr thun: UR= I.R on mch ch c cun cm thun: UL= I.ZL on mch ch c t in: UC = I.ZCh. Lu : Cng dng in tc thi: iR(t) = iL(t)= iC(t)= i(t) Ti thi im t uAB(t)= uR(t)+ uL(t)+ uC(t), cn UAB = 2 2R L CU (U U )+

Cun dy c in tr r 0 th coi cun dy trn tng ng.

Hin tng cng hng in: ZL = ZC hay1

LC =

2/ CNG SUT CA DNG IN XOAY CHIUa. Cng sut ca dng in xoay chiu :: P = UIcosb.Cng sut ca mch in xoay chiu R, L, C:


9/16

2

1

2

1

2

1

N

N

U

U

E

E==

H s cng sut: cos =R

Z=U

UR P = I2R= 2

CL

2

2

)Z(ZR

.RU

+

Nu iu chnh L,C,f, mch tiu th cng sut cc i th ta lun c:

+ ZL = ZC hay1

LC =

+ Tng tr Z= R , hay hiu in th hai u mch U= UR

+ Cng sut cc i ca mch PMAX=td

2

R

U

Nu iu chnh Rt mch tiu th cng sut cc i th ta lun c:+ R= | ZL - ZC|

+ Tng tr Z= R 2 , hay hiu in th hai u mch U= UR 2

+ Cng sut cc i ca mch PMAX=td

2

2.R

U=

CL

2

ZZ2.

U

Nu mch in c in tr R v cun dy c in tr hot ng r th khi iu chnh R cngsut tiu th trn R cc i, ta lun c + R= 2CL2 )Z(Zr +

+ Cng sut cc i trn R khi PRmax = td

2

2.R

U

= ( )rR2.U 2

+ 3/ CC LOI MY INI. May phat ien xoay chieu:a. Bieu tho S cam ng: e(t)=E0cos(t + e)

E0= N.B.S.= N.0: S cc ai(V) 0= B.S: T thong cc ai qua1vong (Wb)

b.Chu y: Goi n (vong/s) la toco quay roto va p la so cap cct. May co Roto la phan cam th: f= n.p May co Roto la phan ng th

: f= nc. My pht in xoay chiu 3 pha

Cach mac hnh sao: Ud= 3 Up; Id=

Ip Cach mac hnh : Ud= Up; Id= 3 .Ip Neu 3tai tieu thu oi xngnhau th:

(P= 3UpIp,cos va Itrung hoa = 0 )

2. My bin tha. cc cng thc bin i (Bo qua

ien trca

c cuon dy Bo qua moi hao ph ien

nang th: P1= P2+ U1I1,cos1 = U2I2,cos2+ Neu d.ien va HTcung

pha: U1I1= U2I2b. Truyn taair in nng. Cong suat hao ph tren daytai ien:

P=22

22

.

..

CosU

RPRI = ( P,U: la

cong suat va HT tramphat)


10/16

Chng 6: TNH CHT SNG CA NH SNG

1/ TN SC NH SNG QUA LNG KNH* Lng knh c gc chit quang A 100

Gc lch ca tng tia n sc : D= (n -1)A Gc lch gia tia n sc tm v n sc :

D= (nt n)A Khong cch gia tia n sc tm v trnmn t song song vi phn gic ca gc A:

T= (nt n)A ( khong cch)

* Lng knh c gc chit quang A> 100

Gc lch ca tng tia n sc : D= i1 i2 A khi tia ti vung gc vi mt bn ca lngknh ( i1= 0 ) th gc lch gia tia n sc tmv n sc :

D= Dt D= i2t i2( Vi sini2t= nt sinA; sini2= n sinA )

2/ GIAO THOA ANH SANG BANG KHE YOUNGa. V TR VAN SANG VAN TOI KHOANG VAN:

Khoang van V tr van sang V tr van toi

i =a

D ; vi =

f

c

a: Khoang cachgia 2 kheD: Khoang cachkhe- man

a

Dkx

S= = k.i ; vi kZ

k = 0 x = 0, vansang TT tai O k =l;2 van sangbac 1, bac 2.

xt = (k +2

1)i vi k

Z k = 0; -1van toith nhatk = 1; -2 van toth 2

Bc song anh sang khi truyen trong moi trng co chiet suatn la

n = n

( la bc song anh sang trong khong kh, n =v

c (c

=3.108 m/s) )

b. SO VAN SANG - SO VAN TOI:*.So van em c tren caman quan sat: Goi la be rong trng gi/thoatren man

So khoang van GT tren man la:/ 2i.

ati2

= k + b ( vi 0 b M Phan ng toa NL:E Neu M0 < M Phanng thu NL:Ec. Ch :

nh luat BTNL co the viet: E + WA + WB = WC + WD Nang lng toa ra khi tao thanh 1hat nhan AZ X t cacnuclon : E= Wlk(X) Nang lng phan ng hat nhan tnh theo o hut khoim

E = ( m(C) + m(D) - m(A) - m(B) ).c2

Nang lng phan ng hat nhan tnh theo nang lng lien ket E = ( Wlk (C) + Wlk (D) - Wlk (A) - Wlk (B) ) Nang lng toa ra khi tao thanh n (mol) hat nhan: W = n.NA.E Nang lng toi thieu (hoc tan so nho nhat cua photon)

can cung cap phan nghat nhan xy ra: Wmin = hfmin= E = ( M0 M ).c2

4/ S PHAN HACH PHAN NG NHIET HACHa. S phn hch ht nhn; 23592 U + 10 n X + Y + k. 10 n + E ieu kien e phan ng day chuyen xay ra: He so nhanntron S1 Nha may ien nguyen t S= 1 phan ng day chuyenkiem soat c


16/16

+ Nang lng lo phan ng cung cap cho nha may hoatong trong th/gian t

W=H

tP, ;( trong oP,Hlan lt la cong suat va hieu

suat cua nha may)+ Khoi lng U235 can cung cap cho nha may hoat ong

trong thi gian t

m= E.H.NtP,

A 235 (g) ; n v t (s), P (W),E (J)b. Phn ng nhit hch: D + D 32 He + 10 n + E

T + D + 10 n + E Khoi lng than a (xang) tng ng can phai ot e conang lng W

m=q

.n.N

q

W A= E (kg) ; vi q (J/kg) lnang suat toa nhiet

cua than (xang)

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