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Chng I:ng lc hc chtim

CHCChc nghin cu dng chuyn ng n gin nht ca vt l chuyn ng c.

Chc gm hai phn chnh: ng hc v ng lc hoc.

CHNG I

NG LC HC CHT IMNi dung ca chng I nghin cu cc c trng ca chuyn ng chc (phng

trnh chuyn ng, phng trnh quo, qungng dch chuyn, vn tc, gia tc) vnguyn nhn gy ra sthay i trng thi chuyn ng.

1. NG HC CHT IM

I. Nhng khi nim mu

1. Chuyn ng.

Theo nh ngha, chuyn ng ca mt vt l schuyn di v tr ca vt i vicc vt khc trong khng gian v theo thi gian. xc nh v tr ca mt vt chuyn ng,ta phi xc nh khong cch t vt n mt vt (hoc mt h vt) khc c qui c lng yn.

Nh vy, v tr ca mt vt chuyn ng l v tr tng i ca vt so vi mt vthoc mt h vt c qui c l ng yn. T ngi ta a ra nh ngha v h qui chiu.

Vtc qui c l ng yn dng lm mc xc nh v tr ca cc vt trong khnggian c gi l hqui chiu.

xc nh thi gian chuyn ng ca mt vt, ngi ta gn h qui chiu vi mt

ng h. Khi mt vt chuyn ng th v tr ca n so vi h qui chiu thay i theo thi gian.Vy chuyn ng ca mt vt ch c tnh cht tngi ty theo h qui chiu c

chn, i vi h qui chiu ny n l chuyn ng, nhng i vi h qui chiu khc n c thl ng yn.

2.Chtim, hchtim, vt rn.Bt k vt no trong t nhin cng c kch thc xc nh. Tuy nhin, trong nhiu bi

ton c th b qua kch thc ca vt c kho st. Khi ta c khi nim v cht im:Chtim l mt vt m kch thc ca n c thb qua trong bi ton c xt.

Kch thc ca mt vt c th b qua c khi kch thc rt nh so vi kch

thc ca cc vt khc hay rt nh so vi khong cch t n ti cc vt khc. Vy, cng cthnh ngha:

Mt vt c kch thc nh khngng k so vi nhng khong cch, nhng kchthc m ta ang kho stc gi l chtim.

Nh vy, ty thuc vo iu kin bi ton ta nghin cu m c th xem mt vt l chtim hay khng.

Th d: Khi xt chuyn ng ca vin n trong khng kh, chuyn ng ca qutquay quanh mt tri, ta c th coi vin n, qut l cht im nu b qua chuyn ngquay ca chng.

Tp hp cc cht im c gi l h chtim. Nu khong cch tng i gia cccht im ca h khng thay i, th h cht im c gi l vt rn.

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3.Phng trnh chuyn ng ca chtim xc nh chuyn ng ca mt

cht im, ngi ta thng gn vo h quichiu mt h ta , chng hn h ta Descartes c ba trc ox, oy, oz vung gc

tng i mt h p thnh tam din thunOxyz c gc ta ti O. H qui chiuc gn vi gc O. Nh vy vic xt chtim chuyn ng trong khng gian sc xc nh bng vic xt chuyn ngca cht im trong h ta chn.V tr M ca cht im sc xc nh

bi cc ta ca n. Vi h ta Descartes Oxyz, cc ta ny l x,y,z.

Bn knh vect rMOr

r

= cng c cc tax,y,z trn ba trc ox,oy,oz (hnh 1-1), vc mi lin h:

k)t(zj)t(yi)t(xrr

rr

r

++= .

Khi cht im chuyn ng, v tr Mthay i theo thi gian,cc ta x, y, z ca Ml nhng hm ca thi gian t:

x = x(t)y = y(t) (1-1)z = z(t)

rr

ca cht im chuyn ng cng l mt hm ca thi gian t:Do bn knh vect(1-2))(trr

rr

=

Cc phng trnh (1-1) hay (1-2) xc nh v tr ca cht im ti thi im t vc gi lphng trnh chuyn ngca cht im. V mi thi im t, cht im c mtv tr xc nh, v khi thi gian t thay i, v tr M ca cht im thay i lin tc nn cc hm

x(t), y(t), z(t) hay l nhng hmxc nh, n tr v lin tc ca thi gian t.)(trr

4. Qy o

Quoca chtim chuyn ng l ng cong to bi tp hp tt c cc v trca chtim trong khng gian trong sut qu trnh chuyn ng.

Tm phng trnh Quo cng c ngha l tm mi lin h gia cc ta x,y,z cacht im Mtrn quo ca n. Mun vy ta c th kh thi gian t trong cc phng trnhtham s (1-1) v (1-2).

5. Honh congGi s k hiu quo ca cht im l (C) (Hnh 1-1). Trn ng cong (C) ta

chn im A no lm gc (A ng yn so vi O) v chn mt chiu dng hng theochiu chuyn ng ca cht im. Khi ti mi thi im tv tr M ca cht im trnng cong (C) c xc nh bi tr i s ca cung AM, k hiu l:

AM = s

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Ngi ta gi s l honh congca cht im chuyn ng. Khi cht im chuynng,s l hm ca thi gian t, tc l:

s = s(t) (1-3)Khi dng honh cong, th qung ng cht im i c trong khong thi gian

t=t-to l s=s-s0, trong s0 l khong cch t cht im n gcA ti thi im ban u

(to = 0), s l khong cch t cht im n gcA ti thi im t. Nu ti thi im ban ucht im ngay ti gc A th s0 = 0 v s = s, ng bng qungng m chtim i

c trong khong thi gian chuyn ng t.

II. Vn tcc trng cho chuyn ng v phng, chiu v nhanh chm, ngi ta a ra

i lng gi l vn tc. Ni cch khc: vn tc l mti lngc trng cho trng thichuyn ng ca chtim.

1. Vn tc trung bnh v vn tc tc thiGi s ta xt chuyn ng ca cht im trn ng cong (C) (hnh 1-2). Ti thi

im t, cht im v tr M,ti thi im t=t+tcht im i c mt qung ng s

v v tr M Qung ng i c ca cht im trong khong thi gian t = ttl:

MM = s s = s

T ss/tbiu th qung ng trung bnh m cht im i c trong mt n vthi gian t M n M, v c gi l vn tc trung bnh ca cht im trong khong thi

gian t(hoc trn qung ng tMn M)

t

svtb

= (1-4)

Vn tc trung bnh chc trng cho nhanh chm trung bnh ca chuyn ngtrn qung ng MM. Trn qung ng ny, ni chung nhanh chm ca cht im thayi t im ny n im khc. V th ctrng cho nhanh chm ca chuyn ng ti

tng thi im, ta phi tnh t s s/t trong

nhng khong thi gian t v cng nh, tc l

cho t0.

Theo nh ngha, khi t0, MM, t

ss/ts tin dn ti mt gii hn gi l vn

tc tc thi (gi tt l vn tc ) ca cht im tithi im t:

dt

ds

t

sv

t=

= 0

lim (1-5)

Vy: Vn tc ca chtim chuyn ngbngo hm honh cong ca chtim theo thi gian.

S gia s cng chnh l qung ng m cht im i c trong khong thi gian t= t-to. Do ni chung c th pht biu (1-5) nh sau:

Vn tc ca chtim chuyn ng bngo hm qungngi c ca chtim theo thi gian.

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Biu thc (1-5) biu din vn tc l mt lng i s.

giy

mtn vo ca vn tc trong hn v SI l: (m/s).

2. Vectvn tcc trng y c v phng chiu v

nhanh chm ca chuyn ng ngi ta a ramt vectgi l vectvn tc.

nh ngha:Vect vn tc ti v tr M lvectc phng nm trn tip tuyn vi quo tiM, c chiu theo chiu chuyn ng v c lnc xc nh bi cng thc (1-5).

c th vit c biu thc ca vectvntc, ngi ta nh ngha vectvi phn cung sd

r

lvectnm trn tip tuyn vi qu o ti M, hng theo chiu chuyn ng v c ln

bng tr s tuyt i ca vi phn honh cong ds. Do ta c th vit li (1-5) nh sau:

dt

sdv

r

r

= (1-6)

3.Vectvn tc trong hto DescartesGi s ti thi im t, v tr ca cht im chuyn ng c xc nh bi bn knh

vect rOMr

= (hnh1-4). thi im sau t=t+t, v tr ca n c xc nh bi bnknh vect:

rrOMrr

+=

Khirdr,M'M,0trr

, do

MM ,'MM .sdrdrr

=

Hai vect sdrdrr

, bng nhau, do

ta c th vit li biu thc (1-6) cavn tc nh sau:

dt

rdv

r

r

= (1-7)

Tc l: Vect vn tc bngohm bn knh vect v tr chuyn ngca chtim theo thi gian.

Gi ba thnh phn zyx vvv ,, ca

vc tvn tc vr

theo ba trc ta c di i s ln lt bng o hm ba thnh phn tng ng ca bn knh vc ttheo ba trcta :

dt

dzv

dt

dyv

dt

dxv zyx === ,, (1-8)

ln ca vn tc c tnh theo cng thc:

sdr

M vr

Hnh.1-3nh ngha vectvn tc

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222222

+

+

=++=dt

dz

dt

dy

dt

dxvvvv zyx (1-9)

III. Gia tc

M

M

v r

Hnh 1-5Vn tc ti nhng im khc nhau

vr

c trng cho s bin thin ca

vectvn tc, ngi ta a ra mt i lnggi l vectgia tc. Ni cch khc, gia tcl i lungc trng cho sbin i trngthi chuyn ng ca chtim.

1. nh ngha v biu thc vectgia tc

Khi cht im chuyn ng, vectvn tc ca n thay i c v phng chiu v

ln. Gi s ti thi im tcht im im M, c vn tc l vr

, ti thi im sau t = t+t

cht im v tr M c vn tc vvvrrr

+= (hnh 1 -5). Trong khong thi gian t=t- t,

vectvn tc ca cht im bin thin mt lng: vvvrrr

=

t

v

r

T s xc nh bin thin trung bnh ca vectvn tc trong mt n v thi

gian v c gi l vectgia tc trung bnh ca cht im chuyn ng trong khong thi

gian t:

t

vatb

=

r

r

(1-10)

Nhng ni chung ti nhng thi im khc nhau trong khong thi gian t xt, bin thin vectvn tc trong mt n v thi gian c khc nhau. Do , c trng cho

bin thin ca vectvn tc ti tng thi im, ta phi xc nh t stv

r

trong khong thi

gian v cng nh, ngha l cho t 0, khi t s s tin dn ti gii hn gi l vect

gia tc tc thi (gi tt lgia tc) ca cht im ti thi im tv c k hiu l

t

v

r

ar

.

dt

vd

t

va

t

rr

r

=

= 0

lim (1-11)

Vy: Vectgia tc ca chtim chuyn ng bngo hm vectvn tc theo thigian.

Nu phn tch chuyn ng ca cht im thnh ba thnh phn chuyn ng theo batrc ox, oy, oz ca h ta Descartes, ta c:

2

2

2

2

2

2

,,dt

zd

dt

dva

dt

yd

dt

dva

dt

xd

dt

dva zz

yy

xx ====== (1-12)

v ln ca gia tc sc tnh nh sau:

2

2

22

2

2

2

2222

+

+

=++=

dt

zd

dt

yd

dt

xdaaaa zyx

2. Gia tc tip tuyn v gia tc php tuyn

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Trng hp tng qut, khi cht im chuyn ng trn quo cong, vectvn tcthay i c v phng chiu v ln. c trng ring cho s bin i v ln phngv chiu ca vectvn tc ngi ta phn tch a

r

thnh hai thnh phn: gia tc tip tuyn vgia tc php tuyn.

Xt chuyn ng ca cht im trn quo trn (hnh 1-6). Ti thi im t, cht

im ti v tr M c vn tc ; Ti thi im t cht im v tr M, c vn tcvr

v r

. Ta vvect 'v

r

== AB c gc ti M.

Ta t trn phng MA mt on sao cho . Khi , nh trn hnh v (1-

6), bin thin vectvn tc trong khong thi gian

'vMCr

=MC

tl:

=vr

CBACAB +=

Theo nh ngha (1-11) v gia tc, ta c:

t

CB

t

AC

t

va

ttt lim

lim

lim

000

+==r

r

(1-13)

Theo (1-13), vectgia tc gm hai thnh phn. Sau y ta s ln lt xt cc thnhphn ny.

a. Gia tc tip tuyn.

Ta k hiu thnh phn th nht ca (1-13) l:

tAC

at

t =

0lim

r

Thnh phn ny lun cng phngvitip tuyn ca quo ti thi im t, v vyc gi lgia tc tip tuyn.

tar

ACChiu ca trng chiu vi . V

vy khi tar

vr

th cng chiu vi , khivv >'

, th ngc chiu vitar

vr

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ln bngo hm tr svn tc theo thi gian.b. Gia tc php tuyn

Thnh phn th hai ca gia tc, c k hiu l v theo (1-13), ta c:nar

t

CBa

tn

= 0

limr

v'vrr

CB AC Khi t 0, , dn ti vung gc vi , tc vung gc vi tip tuynca quo ti M. V vy na

r

c gi lgia tc php tuyn.

t

CBa

tn =

0lim ln ca gia tc php tuyn l:

Ta t MOM= CMB = . Trong tam gic cn MCB c:

222

=

CMB MCB =

Khi t 0, MM, 0, MCB 2

. Vy n gii hn, ACCB do

phng ca ACan r

tc l vung gc vi tip tuyn ca quo ti M.

nar

Chiu ca lun hng v tm ca quo, do c gi l gia tc hng

tm.

tCB

at

n lim

0 =na

r

ln ca cho bi:

Ch rng cc gc: BMC = MOM= . Khi t0, MM, vvrr

' , gc rt nh, cth coi gn ng:

s =MMR,

R

svvCB

== '.'.

=

tsv

t

'lim

0 ts

t lim

0

R

1'lim

10v

R ttCB

at

n lim

0 = = . (1-15)

vdtds

ts

t==

lim0

vvt

=

'lim0

v

Thay cc kt qa va tnh c vo (1-15), cui cng ta sc:

Rvan

2

= (1-16)

Cng thc (1-16) chng tan cng ln nu cht im chuyn ng cng nhanh v quo cng cong (R cng nh). Vi cc iu kin ny, phng ca vectvn tc thay i cngnhiu. V th,gia tc php tuyn c trng cho sthay i phng ca vectvn tc.

Tm li vectgia tc php tuyn c trng cho sthay i phng ca vectvn tc,n c:

Phng: trng vi phng php tuyn ca quo ti M;Chiu: lun hng vpha lm ca quo;

Rvan

2=C ln bng:

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c. Kt lunTrong chuyn ng cong ni chung vect gia tc gm hai thnh phn:

gia tc tip tuyn v gia tc php tuyn, tc l:

nt aaarrr

+= (1-17)

Gia tc ti p tuyn c trng cho s bin i v ln ca vect

vn tc.

Gia tc php tuyn c trng cho s bin i v phng ca vectvn tc.

ln

22222

+

=+=R

v

dt

dvaaa nt

Trong trng hp tng qut qu o ca cht im lmt ng cong bt k, ngi ta chng minh c rng timi v tr, vc tgia tc tip tuyn v php tuyn vn cho

bi cc biu thc trn, nhng ch rng trong biu thc anth R l bn knh cong ca quo ti M (tc l bn knhca vng trn mt tip ca quo ti M)

Chng ta xt mt s trng hp c bit:

- Khi an = 0, vectvn tc khng thay i phng, cht im chuyn ng thng(quo chuyn ng l ng thng ).

- Khi at= 0, vectvn tc khng i v tr s v chiu, n chuyn ng congu.

- Khi a = 0 vectvn tc khng i, cht im chuyn ng thngu.

IV. Mt s dng chuyn ng cn gin

1. Chuyn ng thng bin iuTrong trng hp ny a = 0, a = const, nn ta c:n tGia tc

constdt

dvaa t === (1-18)

==

tv

v

adtdv

dt

dva

00

T suy ra:

atvv o += (1-19)

ng i:

( ) +==v

v

o

v

v

s

dtatvvdtds00

0

2

2attvs o +=chn gc ta l v tr ban u ta c: (1-20)

T (1-19) v (1-20), kh thng stta sc20

22 vvas = (1-21)

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Trong chuyn ng thng, nu a=0, vn tc chuyn ng khng thay i, do chuyn ng ny c gi l chuyn ng thngu. Trong chuyn ng thng u:

v = const, s = vtRi t do l chuyn ng ca vt di tc dng ca trng lc vi vn tc ban u v0=

0 v gia tc a = g.

2. Chuyn ng trnTrong chuyn ng, nu bn knh cong ca qu o khng thay i ( R = const),

chuyn ng sc gi l chuyn ng trn.

OMTrong chuyn ng trn, do c s thay i gc quay ca bn knh vect , ngoicc i lngv, a, a , at n, ngi ta cn a ra cc i lng vn tc gc vgia tc gc.

a.Vn tc gcGi s cht im Mchuyn ng trn quo trn tm O, bn knhR. Trong khong

thi gian t = t tcht im i c qung ng s bng cung MMng vi gc quay

= MOMca bn knhR = MO (Hnh 1-8). i lng /t biu th gc quay trung bnh cabn knh trong mt n v thi gian v c gi l vn tc gc trung bnh trong khong thi

gian t:

M'

M

sO

R

Hnh 1-8Lp cng thc vn tc gc

ttb

=

(1-22)

t

Nu cho t0, t s s tin ti gii hn, k

hiu l , biu th vn tc gc ca cht im ti thi imt:

dtd

tt

lim

0==

(1-23)

Vy: Vn tc gc bngo hm gc quay theo thi gianVn tc gc c n v l radian trn giy (rad/s).

Vi chuyn ng trn u (R= const, = const, v = const) ngi ta cn a ra nhngha chu k v tn s. Chu k l thi gian cn thitchtim i c mt vng trn. Dochuyn ng trn u, gc quay trong khong thigian t l:

= .t.

Trong mt chu k t =T, =2.

V ta suy ra: .

2T ==

Vy:

2T =

Tn s(k hiu l f) l svng (schu k)quay c ca chtim trong mtn v thi gian.

Trong khong thi gian mt giy cht im i

c cung trn , mi vng trn c di 2, do theo nh ngha tn s, ta c:

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T

1

2f ==

n v ca chu k l giy (s), ca tn s l 1/s hoc cn gi l Hertz (Hz).

Ngi ta biu din vn tc gc bng vc t , nm trn trc ca vng trn quo, thun

chiu i vi chiu quay ca chuyn ng v c gi tr bng .

r

* Lin hgia cc vect vr

v . Gia bn knh R, cung MMv gc c mi

lin h (xem hnh 1-8): MM = s = R , do :t

Rt

s

.

= Khi t 0, ta c:

Rv = (1-24)

ROMr

= vRr

r

r

,,(hnh 1-9) ta thy ba vectNu t theo th t to thnh mt

tam din thun ba mt vung. Ngoi ra theo cng thc (1-24) ta c th vit:

Rvr

rr

= (1-25)

* Lin hgia a v n

RR

an2)(

=Rv2

= R2a , v=R, ta suy ra:n=

Ran2= (1-26)

b. Gia tc gc

Gi s trong khong thi gian t = t t, vn tc gc ca cht im chuyn ng trn

bin thin mt lng = - . Theo nh ngha, lng /t gi l gia tc gc trung bnh

trong khong thi gian t, n biu th bin thin trung bnh ca vn tc gc trong mtn v thi gian:

ttb =

Nu cho t 0,gia tc gc trung bnh tin ti gii hn gi l gia tc gc ca cht

im ti thi im t, k hiu l . Do :

tt

lim

0 =

Theo nh ngha vo hm, ta c:

2

2

dt

d

dt

d== (1-27)

Vy: Gia tc gc bngo hm vn tc gc theo thi gian v bngo hm bchai ca gc quay theo thi gian.

Gia tc gc c n v bngRadian trn giy bnh phng(rad/s2).

Khi > 0, tng, chuyn ng trn nhanh dn,

Khi < 0, gim, chuyn ng trn chm dn.

Khi = 0, khng i, chuyn ng trn u.

Khi = const, chuyn ng trn bin i u (nhanh dn u hoc chm dnu). Tng t nh chng minh cho trng hp chuyn ng thng bin i u, ta cng

c th chng minh c:t += 0 (1-28)

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20 2

1tt += (1-29)

(1-30)

2 202 =

Vi ch l: ti thi

im ban u t = 0, o o = 0, vntc gc c gi tr .o

Ngi ta biu din gia tcgc bng mt vc tgi l vc tgia tc gc, c:

- Phng nm trn trcca qu do trn

- Cng chiu vi khi

> 0 v ngc chiu vi hi < 0

k

- C gi tr bng Vy ta c th vit h thc

sau:

dt

d = (1-31)

* Lin hgia a vtdtdv

at =Thay v=.R vo ta c:

( )

R

dt

dR

dt

Rdat === (1-32)

taR,,Theo nh ngha ca cc vect , ta thy ba vect theo th t lun to

thnh tam din thun ba mt vung; Kt hp vi (1-32) ta c th vit:

Rat = (1-33)

3. Chuyn ng vi gia tckhngi

Xt chuyn ng ca mtcht im xut pht t mt im Otrn mt t vi vc t vn tc ban

u l 0vr

hp vi phng nm ngang

mt gc (hnh 1-11) B qua milc cn khng kh.

Chn mt phng hnh v l

mt phng thng ng cha , haitrc ta Ox nm ngang v Oy

0vr

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thng ng hng ln trn (hnh 1-11). Quo ca cht im s nm trong mt phng Oxy.

a. Phng trnh chuyn ng

0vr

Ta phn tch vectvn tc thnh 2 thnh phn theo 2 trc Ox, Oy:

vox = v cos, vo oy = vosinCoi chuyn ng gm hai thnh phn: thnh phn theo phng Ox, c vn tc ban u

v = 0; thnh phn Oy c vn tc ban u vox,c gia tc bng khng a oy, gia tc bng ayx =g, giatc ny ngc chiu vi trc Oy. Vy phng trnh chuyn ng ca cht im l:

x = (v cos)t (1)o

20 2

1)sin( gttvy = (2)

b. Phng trnh quoKhtt hai phng trnh (1) v (2) ta c:

xtg

v

gxy +=

220

2

cos2

(3)

Vy qu o ca cht im l mt parabol, b lm hng xung di (Hnh1-11).

c. Thi gian riKhi vin n ri chm t, y = 0, t (2) ta c:

02

sin0 =

tgt

v

Phng trnh ny c 2 nghim:Nghim t =0ng vi thi im xut pht, t1 2ng vi lc chm t. Vy thi gian cn

thit cht im bay trong khng kh l t =t t2 1=t2.

g

vtt

sin2 02 == (4)

d. cao cc iKhi t n im cao nhtp, vn tc ca cht im theo phng Oybng khng:

00 == gtvv yy

Thi gian t cao nht:

g

vt

sin0=

cao ln nht m cht im t c:

( )g

vgttvy

2

sin

2sin

220

2

0max

== (5)

f. Tm bay xa ca chtimKhi cht im chm t, n cch gc O mt khong OR = x. Khi y=0.

g

v

g

vx

2sinsin.cos2 2020 ==T (3) ta c: (6)

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2. NG LC HC CHT IM

ng lc hc nghin cu mi quan h gia s bin i trng thi chuyn ng cacc vt vi tng tc gia cc vt . C s ca ng lc hc gm ba

nh lut Newton v nguyn l tng i Galilo.I. Cc nh lut Newton

Cc nh lut Newton nu ln mi quan h gia chuyn ng ca mt vt vi tc dngt bn ngoi v quan h gia cc tc dng ln nhau gia cc vt.

1.nh lut Newton thnhtChtim c lp: L chtim khng tc dng ln chtim khc v cng khng

chu tc dng no tchtim khc.nh lut Newton th nht pht biu nh sau:Mt chtim c lp nu angng yn, stip tc ng yn, nu ang chuyn ng,

chuyn ng ca n l thng v u.0=v

r

Trong c hai trng h p, cht im ng yn ( ) v chuyn ng thng u

( ) u c vn tc khng i. Khi vn tc ca cht im khng i, ta ni trng thichuyn ng ca n c bo ton.

constv =r

Nh vy theo nh lut Newton I: Mt cht im c l p lun bo ton trng thichuyn ng ca n.

Tnh cht bo ton trng thi chuyn ng c gi l qun tnh. V vy nh lut thnht ca Newton cn c gi l nh lut qun tnh.

C th vn dng nh lut qun tnh gii thch nhiu hin tng thc t.V d, ontu ang ng yn bng chuyn ng t ngt. Khi , hnh khch ang ng yn hoc ngitrn tu s b ng ngi v pha sau do qun tnh. Tng t, khi on tu ang chuyn ngthng u b dng t ngt, hnh khch s b chi ngi v pha trc.

2.nh lut Newton thhainh lut th hai ca Newton xt cht im trng thi khng c lp, ngha l chu

tc dng ca nhng vt khc. Tc dng t vt ny ln vt khc c c trng bi mt i

lng l lc, thng k hiu bng vectFr

.

,...,, 321 FFFrrr

Khi mt vt chu tc dng ng thi ca nhiu lc th ta c th thay tt c

cc lc bng mt lc tng hp: ...321 +++= FFFFrrrr

.

Lc tc dng ln mt vt lm thay i trng thi chuyn ng ca vt. V trng thica mt vt c xc nh bi vn tc v v tr ca n, do khi chu tc dng ca mt lc,vn tc ca vt b bin i, tc l vt thu c gia tc. Lc tc dng cng ln, gia tc m vtthu c s cng ln. Th nghim chng t rng gia tc ca mt vt cn ph thuc vo quntnh ca vt. Qun tnh ca mt vt c c trng bi khi lng ca vt, k hiu l m.

Ba i lng l lc, khi lngv gia tc lin h vi nhau theo mt nh lut thcnghim do Newton nu ra, gi l nh lut Newton th II v c pht biu nh sau:

F

r

Chuyn ng ca mt chtim chu tc dng ca lc l mt chuyn ng cgia tc a

r

,

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Gia tc chuyn ng ca mt chtim t l thun vi lc tc dng v t lnghch vi khi lng ca chtim y, t c th vit:

mF

ka

r

r

= (1-34)

Trong , k l mt h s t l ph thuc vo cch chn n v cc i lng trong

cng thc (1-34). Trong hn v quc tSI, ngi ta chn k = 1, do :

m

Fa

r

r

=

Hoc c th vit:

amFr

r

= (1-35)

R rng cng mt lc tc dng ln vt nu khi lng m ca vt cng ln th gia tcca vt cng nh, ngha l trng thi chuyn ng ca vt cng t thay i. Nh vy khilng m ca vt c trng cho qun tnh ca vt.

Thc nghim chng tnh lut Newton 2 ch nghim ng i vi h qui chiu quntinh (sc nu r di y).

Biu thc (1-34) bao gm cnh lut Newton I v II, c gi l phng trnh cbn ca ng lc hc chtim.

T phng trnh:

amFr

r

=

Vi nh lut Newton I:

constvaF ===rr

r

00 Vi nh lut Newton II:

00 =mFaF

r

r

r

3. Hqui chiu qun tnhnh ngha: H qui chiu trong mt vt c lp nu angng yn sng yn mi

mi cn nu ang chuyn ng schuyn ng thngu c gi l h qui chiu qun tnh.Ni cch khc, h qui chiu trong nh lut qun tnh c nghim ng l h qui

chiu qun tnh.Thc nghim cng chng tnh lut Newton II ch nghim ng i vi h qui chiu

qun tnh.

4. Lc tc dng trong chuyn ng congTrong chuyn ng cong, gia tc ca cht

im gm hai thnh phn gia tc tip tuyn tar

v

gia tc php tuyn nar

. Gia tc tng hp ca cht

im l ar

ntaa arrr

+=

Nhn 2 v ca phng trnh ny vi khilng ca cht im, ta c:

nt amamamrrr

+=

Theo nh lut Newton II:

tFr

tar

Hnh 1-12Lc hng tm v lc ly tm

nar

ar

Fr

nFr

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nntt amFamFamFr

r

r

r

r

r

=== ,,

nt FFFrrr

+=ta c:

tt amFr

r

=Thnh phn c gi l lc ti p tuyn, lc tip tuyn gy ra gia tc tip

tuyn, tc lm thay i ln v chiu ca vn tc; cn thnh phn nn amFr

r

= c gi l

lc php tuyn hay l lc hng tm, lc hng tm gy ra gia tc hng tm, lm thay iphng ca vectvn tc.

Nh vy iu kin cn thitcho chtim chuyn ng cong l phi tc dng lnn mt lc hng tm, c ln:

R

vmmaF nn

2

==

5.nh lut Newton th baTrong t nhin khng bao gi c tc ng mt

pha. Newton chng minh rng khi cht im A tcdng ln cht im B th ngc li cht im B cng tcdng ln cht im A. Newton a ra nh lut NewtonIII pht biu nh sau:

Khi chtim A tc dng ln chtim B mt lc

Fr

th ng thi chtim B cng tc dng ln chtim

A mt lc Fr

Fr

Fr

. Hai lc v ng thi tn ti, cng phng, ngc chiu, cng cngv t ln hai chtim A v B khcnhau (hnh 1-13):

FF =rr

Fr

Fr

l lc phn tc dng, thng gi tt lphn lc. Hai vectlcNgi ta gi v

Fr

c im t khc nhau nn chng khng phi l hai lc cn bng, tc l khng trit tiunhau.

Nu mt h gm hai cht im A v B tng tc nhau th cc lc tng tc gia A v

B (Fr

Fr

v ) khi c gi l ni lc tng tc trong h, tng hp hai vectni lc ny

ca h bng khng: .0' =+ FFrr

Trng hp tng qut, nu h c n cht im, trong h ch c cc ni lc tng tcgia cc cht im ca h (khng tng tc vi cc cht im khc ngoi h) th hc

gi l h c lp (hay cn gi l h kn). Khi nu xt tng i cht im ca h th tng hailc tng tc gia chng bng khng. Do nu xt c h th: Tng hp cc ni lc ca mth c lp lun bng khng.

II. Cc nh l vng lngTnh lut Newton II ta c th suy ra mt s pht biu khc, l cc nh l v

ng lng.

1.nh l 1

Fr

Gi s cht im c khi lng m chu tc dng ca lc , theo nh lut Newton

II, cht im s chuyn ng vi gia tc a

r

sao cho:Famr

r

=

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Fdt

vdm

r

r

=Hay

Gi thit khi lng m khng i, ta c th vit:

( )(1-36)

Ta t: , v gi l vectng lng ca cht im, do c th vit li (1-36) nh sau:

vmKr

r

= Kr

Fdt

Kd rr

= (1-37)

nh l 1: o hm ng lng ca mt chtim theo thi gian bng tng hp ccngoi lc tc dng ln chtim .

2.nh l 2T (1-37) ta suy ra:

dtFKd

rr

= (1-38) bin thin ca vect t thi im tK

r

1 c vectng lng 1Kr

n thi im t2 c

vectng lng 2Kr

c th tnh c nh sau:

===2

1

2

1

. 12

t

t

K

K

dtFKdKKKrrrrr

r

r

(1-39)

Ngi ta gi l xung lng ca lc2

1

.t

t

dtFr

Fr

trong khong thi gian t t1n t2. Biu

thc (1-39) c pht biu thnh nh l 2 nh sau:nh l 2: bin thin ng lng ca mt chtim trong mt khong thi gian

no bng xung lng ca lc tc dng ln chtim trong khong thi gian .

Trng hp ring khi Fr

khng i theo thi gian, (1-39) trthnh:

tFK =rr

(1-40)hay:

Ft

K rr

=

(1-41)

Tc l: bin thin ng lng ca chtim trong mtn v thi gian bng lctc dng ln chtim :

3. ngha ca ng lng v xung lnga. ngha ca ng lngn y ta c hai i lng c trng cho trng thi chuyn ng l vn tc v ng

lng. Vn tc c trng cho chuyn ng v mt ng hc. Cn ng lng c trng chochuyn ng v mt ng lc hc, v ng lng khng ch lin quan n vn tc m cn linquan n khi lng ca cht im.

Hn na ng lng cn c trng cho kh nng truyn chuyn ng ca

chtim.

Fdt

vmdr

r

=

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minh ho, ta ly v d sau. Mt qu cu khi lng m chuyn ng vi vn tc 1v1r

n p thng vo mt qu cu khi lng m ang ng yn. Sau va chm, qu cu m2 2 s

chuyn ng vi vn tc . Thc nghim chng t2vr

2vr

1vr

khng nhng ph thuc vo m

cn ph thuc vo m 111 vmKr

r

=, ngha l ph thuc vo1 (ng lng ca qa cu th nht).

Vn tc cng ln nu cng ln, ch khng phi ch ring do ln.2v

r

11vm

r

1v

r

Vy kh nng truyn chuyn ng ph thuc vo ng lng ca vtb. ngha ca xung lng

Xung lng ca mt lc tc dng trong khong thi gian tc trng cho tc dngca lc trong khong thi gian . Thc vy, cc cng thc (1-39) v (1-40) chng t tcdng ca lc khng nhng ph thuc vo cng ca lc m cn ph thuc vo khongthi gian tc dng. Cng mt lc tc dng, bin thin ng lng t l thun vi khongthi gian tc dng.

III.ng dng phng trnh cbn ca chc kho st chuyn ng ca cc vt

Tnh lut Newton th III ta suy ra rng: tng tc l hin tng ph bin ca tnhin. Do gia vt chuyn ng v vt lin kt vi n lun c cc lc tng tc gi l cclc lin kt. Di y ta s xt mt s loi lc lin kt thng gp.

1. Cc lc lin kta. Lc ma st* Lc ma st trtThc nghim chng t khi mt vt rn m trt trn gi S, n tc dng mt lc nn

ln mt gi S. Theo nh lut Newton III, mt ny li tc dng ln vt m mt phn lc Rr

gm hai thnh phn Nmsf v (hnh 1-14) sao cho:msfNRrrr

+=

- Thnh phn gi l phn lc php tuyn, n hng vung gc vi gi S ti

im tip xc v lun trc i vi p lc

Nr

'Nr

(lc nn vung gc vi mt ti p xc) ca vt mtc dng ln mt gi S sao cho iu kin sau y c tho mn:

'Nr

=- .Nr

- Thnh phn gi l lcma st trt, n

c phng trng vi tip tuyn vi mt gi Sti

im ti p xc, ngc chiu vn tc

msfr

vr

v cn trchuyn ng ca vt. Nu vn tc ca vt khng quln th lc ma st trt c ln t l vi phn lc

php tuyn:fms = kN

Trong , kl h s t l, gi l h sma sttrt, lun c gi tr nh hn n v ( k

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Lp cao su trn t cng Thp trn t cng0,40,6 0,20,4

* Lc ma st ln l lc ma st xut hin mt tip xc gia mt vt ln trn mt ca mt vt khc.

ln ca lc ma st ln cng t l vi ln ca phn lc php tuyn v c tnh theo

cng thc:

Nr

fms = r

N

trong r l bn knh ca vt ln, l h s ma st ln.Thc nghim chng t lc ma st ln nh hn lc ma st trt. V vy trong k thut,

ngi ta thng s dng cc bi chuyn ma st trt thnh ma st ln ca cc vin bi haythanh tr trong cc bi.

* Lc ma st nht l lc ma st xut hin mt hai lp cht lu (cht lng hay cht kh) chuyn ng

i vi nhau. Nu mt vt chuyn ng trong cht lu vi vn tc khng ln lm, th lc mast nht (gia lp cht lu bm dnh vo mt ngoi ca vt vi lp cht lu nm st n) t lv ngc chiu vi vn tc:

vrfmsr

r

=

y r l h sma st nhtca cht lu. Tr s ca r ph thuc vo bn cht vnhit ca cht lu, n nh hn nhiu so vi h s ma st trt v ma st ln. V vy ngita thng dng du nht bi trn mt tip xc gia cc vt chuyn ng gim lc ma st.

Nu vt c dng hnh cu ng knh dth lc ma st nht tnh theo cng thc Stokes:

f = 3dVmstrong, c gi l h snhtca cht lu.2. Lc cngGi s c mt vt no b buc vo mt si dy khng dn, di tc dng ca mt

ngoi lc Fr

vt c mt trng thi ng lc hc no (ng yn hay chuyn ng vi gia tc

xc nh). Si dy s b ko cng. Ti mi im ca dy s xut hin nhng lc Tr

v phn

lc Tr

. Cc lc ny l cc lc tng tc gia hai nhnh hai pha ca si dy v c gi llc cngca si dy. Theo nh lut Newton IIIta c:

TT =rr

ln ca cc lc cng ph thuc vo

trng thi ng lc hc ca si dy.Mun tnh lc cng cu si dy, ta tng

tng ct si dy ti mt im Mbt k thnh haiphn. t vo mi u (b ct) ca si dy cc lc

cng Tr

v Tr

sao cho trng thi ng lc hc cami nhnh dy (v ca c h) vn gi nguyn nh

khng ct dy. Sau p dng phng trnh cbn ca ng lc hc cho mi phn ca h vtchuyn ng (mi phn gn vi mt bn dy).

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3.V dTa hy xc nh gia tc chuyn ng ca h hai vt A vB v sc cng ca si dy

ko hai vt (hnh 1-15). Hai vt ln lt c khi lng m mvA B. VtA trt khng ma sttrn mt phng nghing mt gc so vi phng nm ngang. B qua khi lng ca rng rcv ca si dy. Tc dng ln vtA c:

B

* Sc cng Tr

,* Trng lc ,AP

r

* Phn lc php tuyn ca mt phng nghing.Nr

Trng lc tc dng ln vt A c phn tch thnh hai thnh phn:APr

=APr

21 PPrr

+

Trong vung gc vi mt phng nghing v trit tiu vi2Pr

Nr

:P = Pcos = m2 Agcos, cn P =m1 Agsin song song vi mt phng nghing.

TPrr

,1Vy cc ngoi lc tc dng ln A cn li l: . Hai lc ny cng phng nhng

ngc chiu nhau.Gi s P1>T, vt A b ko xung dc, vt B b ko ln. Chn chiu chuyn ng l

chiu dng, phng trnh chuyn ng ca A l:P -T = m1 Agsin - T= mAa (*)

Tc dng ln vt B c trng lng ca vt B, sc cng ca si dyLy chiu chuyn ng ca h lm chun, ta c phng trnh chuyn ng ca B l:

T - P = m a (**)B BT phng trnh ny ta c:

T = m a+m g =m (a+g).B B B

Thay T t phng trnh ny vo (*) ta c:

gmm

mma

BA

BA

+

=

sin

Sc cng si dy T

( )gmm

mmT

BA

BA sin1++

=

IV. Mmen ng lng

1. Mmen ca mt vc ti vi mtim

aCho vc tA gc ti A v mt im O cnh.Theo nh ngha mmen ca

B = ,a i v i im O l mt vc t

( )aOM rr

/

v:

( ) araOAM aOrrr

r

(1-42)r ==/

( )aOM rr

/Mmen l mt vc t:

- Gc ti Oar

- Phng vung gc vi mt phng cha o v

OAsang- Chiu l chiu thun i vi chiu quay t

B

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( ) ( )raraM aOrrrr

r

r ,sin../ =- C ln

Tnh cht

( )aOM rr

/a. = 0 khi hoc0=ar

ar

c phng i qua O.

b. Mmen ca mt vc ti vi mt im l mt hm tuyn tnh ca vc t:

( ) ( ) ( )bOaObaO MMM rrrrrrr

/// +=+

( ) ( )aOaO MM rrrr

// =

bar

r

,c. Khi hai vc t cng phng, ngc chiu v cng ln th:

( ) ( ) 0// =+ bOaO MM rrrr

2.nh l vmmen ng lng

Fr

Xt cht im M chuyn ng trn quo (C) di tc dng ca ngoi lc , theo(1-37) ta c

( ) Fdt

vmddtKd r

r

r

==

Nhn hu hng hai v ca phng trnh vi

OMr=r

(O l gc ta )

( )Fr

dt

vmdr

r

r

r

r

=

Ch :

( )( ) ( )Kr

dt

dvmr

dt

d

dt

vmdr rrrrrr

==

( )( ) ( )

dt

vmdr

dt

vmdrvm

dt

rdvmr

dt

dr

r

r

rr

r

rr

=+=V

Vy ta c th vit :

( ) FrKrdt

d rrrr= (1-43)

Krr

r

Trong gi l mmen i vi im O

ca vc tng lng , c gi l vc tmmenng lng ca cht im i vi im O, k hiu :

Kr

vmrLrr

r

= Phng trnh (1-43) c th vit li :

( )aOMdt

Ldr

r

r

/=

nh l vmmen ng lng : o hm theothi gian ca mmen ng lngi vi im O cachtim chuyn ng bng tng mmen i vi imO ca cc lc tc dng ln chtim.

Trong trng h p cht im chuyn ng trn

quo trn th :

( ) ImRRmvLvmOML ==== 2r

r

r

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Trong I = mR2c gi l mmen qun tnh ca cht im i vi im O.T hnh v (1-18) ta c :

Vy mmen ng lng ca mt cht im chuyn ng trn bng tch ca mmenqun tnh ca cht im vi vc tvn tc gc ca cht im y.

r

r

IL =

V. Chuyn ng tng i v nguyn l tng i Galilo1. Khng gian v thi gian theo chc cin.Ta xt hai h qui chiu O v O gn vi 2 h trc ta Oxyz v Oxyz. H O ng

yn, h Ox trt dc theo trc Ox sao cho OxOx, OyOy, OzOz (hnh 1-19).

Ta gn vo mi h ta mt ng h ch thi gian. Ta xt mt cht im chuyn ngtrong h O. Ti thi im tn c cc ta x,y,z. Cc ta khng gian v thi gian tngng ca cht im trong h O l x,y,z', t.

Chc cin c xy dng trn c snhng quan im ca chc Newton v

khng gian, thi gian v chuyn ng. Cc quanim ca Newton nh sau:

a. Thi gian chbi cc ng h tronghai h O v O l nhnhau:

t=t (1-44)

Ni cch khc, thi gian c tnh tuyti, khng ph thuc h qui chiu.

b. V tr M ca chtim trong khnggian uc xc nh ty theo h qui chiu, tc lta khng gian ca n ph thuc h quichiu. Trong trng hp c thhnh 1-19, tac:

'OOx = x+ , y =y, z = z. (1-45)Vy: v tr ca khng gian c tnh cht tngi, ph thuc h qui chiu. Do :

chuyn ngc tnh cht tngi, ph thuc h qui chiu.c.Khong cch gia 2 im ca khng gian c tnh cht tuyti, khng ph thuc h

qui chiu.Tht vy, gi s c mt ci thc AB t dc theo trc Ox gn vi hO. Chiu di

ca thc o trong hOl:

l0 = xB-xABChiu di ca thc trong h O l:

l = xB-xA.BTheo (1-45) ta c:

xA = xA+ , xB = x'OO B BB+ ,'OODo : xB-xB A= xBB-xAtc l: l = l0,

chiu di ca thc bng nhau trong hai h qui chiu (khng ph thuc h qui chiu).Ta xt cht im chuyn ng trong hO. Coi rng ti thi im u t0=0 gc O v

O trng nhau, O chuyn ng thng u dc theo trc Ox vi vn tc V.Khi :

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= Vt,'OO

Theo (1-44) v (1-45)x = x+ Vt, y =y, z = z, t = t (1-46)

v ngc li:x= x - Vt, y= y, z= z, t= t (1-47)

Cc cng thc (1-46) v (1-47) c gi lphp bin i Galilo.

2. Tng hp vn tc v gia tc

Ta hy tm mi lin h gia vn tc v giatc ca cng mt cht im i vi hai h quichiu O v O khc nhau.

Gi sOxyzchuyn ng i vi Oxyzsao cho lun lun c:

OxOx, OyOy, OzOz (hnh 1-20).

rMOrOM ==rr

,t theo hnh (1-20)

rOOr +=rr

MOOOOM +=c: hay (1-48)o hm hai v ca (1-48) theo thi gian ta

c:

( )DT

OOd

dt

rd

dt

rd +

=rr

(1-49)

vdt

rd rr

= vdt

rd = r

r

l vn tc ca cht im i vi hO,Ch rng: l vn tc ca cht

im i vi h O,( )

Vdt

OOd r=

l vn tc chuyn ng ca O i

vi O. Nh vy:

Vvvr

rr

+= (1-50) c gia tc, ta ly o hm hai v ca (1-50) theo thi gian:

dt

Vd

dt

vd

dt

vdr

rr

+

=

Aaar

rr

+=Ta c: (1-51)

Trong , l gia tc ca cht im i vi hOar

l gia tc ca cht im i vi hOa r

Ar

l gia tc chuyn ng ca hOi vi hO.Hai cng thc (1-50) v (1-51) l cc cng thc tng hp vn tc v gia tc.

3. Nguyn l tngi GaliloTa hy xt chuyn ng ca cht im trong hai h qui chiu khc nhau O v O nh

nu trn. Ta gi s O l h qun tnh, cc nh lut Newton c tha mn. Nh vyphng trnh cbn ca ng lc hc ca cht im s l:

Famr

r

= (1-52)l gia tc ca cht im i vi h Oa

r

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Fr

l tng hp cc lc tc dng ln cht im xt trong h O.

Ar

Gi l gia tc ca cht im i vi h O,ar

l gia tc chuyn ng ca h O

i vi h O, theo (1-51), ta c:

Aaar

rr

+=

Ar

= 0 do Nu h O chuyn ng thng u i vi h O thaa =rr

Famamr

rr

== Vy

Famr

r

= (1-53)Nh vy nh lut Newton cng c tha mn trong hO, vy hOcng l h qui

chiu qun tnh v ta c th pht biu nh sau:Mi h qui chiu chuyn ng thngu i vi h qui chiu qun tnh cng l h qui

chiu qun tnh.V cc nh lut Newton c nghim ng trong cc h qui chiu qun tnh cho nn

cng c th pht biu:Cc phng trnh ng lc hc c dng nhnhau trong cc h qui chiu qun tnh

khc nhau. l nguyn l tngi Galilo.V cc phng trnh ng lc hc l cs m t v kho st cc hin tng chc

cho nn ta c th pht biu:Cc hin tng (cc nh lut ) chc xy ra ging nhau trong cc h qui chiu qun

tnh khc nhau.Vy cc phung trnh chc bt bin qua php bin i Galilo. c mt h qui chiu qun tnh, ta phi chn mt h qui chiu sao cho khng gian

trong n ng nht v ng hng, cn thi gian trong n l ng nht. iu ny bo mcho nh lut I ca Newton c nghim ng ti bt k thi im no v ti bt k v tr notrong h qui chiu . Trong thc t khng th c mt vt c lp tuyt i v mt khng giantha mn iu kin trn. Do ch c th chn mt h qui chiu qun tnh mt cch gn ng

bng cch gn khi tm ca thi dng h vi gc ca mt h trc ta , cc trc hng ncc v sao ng yn i vi khi tm. V khi lng ca mt tri rt ln nn c th coi khitm ca thi dng h trng vi tm ca mt tri. H qui chiu qun tnh ny c tn l h

Nht tm. Trong mt s trng hp ngi ta gn gc ca h trc ta vi tm ca qutnhng b qua chuyn ng quay quanh mt tri va s quay quanh trc ring ca n. H ny

c gi l ha tm. Tuy chnh xc ca n khng cao nh h Nht tm nhng cng cth coi n l h qui chiu qun tnh trong nhiu bi ton thc t.

4. Lc qun tnh

Ar

By gita gi s h qui chiu O chuyn ng c gia tc i vi h O. Khi nucht im chuyn ng trong h O th theo (1-51):

Aaar

rr

+=

nhn hai v vi m ta c:

Amamamr

rr

+=

V O l h qui chiu qun tnh nn trong h ny nh lut Newton c nghim ngcho nn:

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amFr

r

=

AmamFr

r

r

+=Do :

Hay )AmFamrr

r

+= (1-54)

Nh vy trong h O chuyn ng c gia tc i vi h O, cc nh lut chuyn ng cacht im c dng khng ging nh trong h O. Trong h O, ngoi cc lc tc dng ln cht

im cn phi k thm lc ( )AmFqtrr

= ( )AmFqtrr

=. Lc c gi l lc qun tnh, n lun

cng phng ngc chiu vi gia tc Ar

ca chuyn ng ca h O i vi h O.H quichiu O nh vy c gi l h qui chiu khng qun tnh. Phng trnh ng lc hc cacht im trong h O l:

qtFFamrr

r

+= (1-55)

Nhkhi nim lc Qun tnh ta c th gii thch s tng gim trng lng v khngtrng lng trong con tu v tr v nhiu hin tng khc xy ra trong thc t, nh cc hin

tng do chuyn ng quay ca qut xung quanh trc ca n gy ra (s gim dn ca giatc trng trng v pha xch o, s ldn ca mt bn b ca cc con sng chy theohng bc nam).

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HNG DN HC CHNG I

I. MC CH, YU CU

Sau khi nghin cu chng 1, yu cu sinh vin:

1. Nm c cc khi nim v c trng cbn nh chuyn ng, h quy chiu, vn

tc, gia tc trong chuyn ng thng v chuyn ng cong.

2. Nm c cc khi nim phng trnh chuyn ng, phng trnh quo ca chtim. Phn bit c cc dng chuyn ng v vn dng c cc cng thc cho tng dngchuyn ng.

3. Nm c cc nh lut Newton I,II,III, cc nh l vng lng v mmmen nglng.

4. Hiu c nguyn l tng i Galilo, vn dng c lc qun tnh trong h quichiu c gia tc gii thch cc hin tng thc t.

II. TM TT NI DUNG1. V tr ca mt cht im chuyn ng c xc nh bi ta ca n trong mt h

ta , thng l h ta Descartes Oxyz, c cc trc Ox, Oy, Oz vung gc nhau, gc Otrng vi h qui chiu. Khi cht im chuyn ng, v tr ca n thay i theo thi gian.

Ngha l v tr ca cht im l mt hm ca thi gian:

( )trrrr

= hay x=x(t), y=y(t), z=z(t).

V tr ca cht im cn c xc nh bi honh cong s, n cng l mt hm cathi gian s=s(t). Cc hm ni trn l ccphng trnh chuyn ng ca chtim.

Phng trnh lin h gia cc ta khng gian ca cht im l phng trnh quoca n. Kh thi gian t trong cc phng trnh chuyn ng, ta s thu c phng trnh quo.

dt

sd

dt

rdv

rr

r

==2. Vectvn tc c trng cho nhanh chm, phng chiu ca chuyn

ng, c chiu trng vi chiu chuyn ng.

dt

vda

r

r

=3.Vectgia tc c trng cho s bin i ca vctvn tc theo thi gian. N

gm hai thnh phn: gia tc tip tuyn v gia tc php tuyn.

Gia tc tip tuyn c trng cho s thay i v ln ca vectvn tc, c ln:

dt

dvat =

c phng tip tuyn vi quo, c chiu cng chiu vi vctvn tc nu chuyn ngnhanh dn, ngc chiu vi vctvn tc nu chuyn ng chm dn.

nar

Gia tc php tuyn c trng cho s bin i v phng ca vectvn tc, c

ln

Rvan2

=

c phng vung gc vi tip tuyn ca quo, lun hng v tm ca quo.

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tn aaarrr

+=Nh vy gia tc tng hp bng:

4. Trng hp ring khi R = , quo chuyn ng l thng. Trong chuyn ngthng, a = 0, a = a .n t

Nu a = const, chuyn ng thng bin i u. Nu tt 0= 0, ta c cc biu thc:

atvv +=0

20 2

1attvs +=

20

22 vvas =

a >0 nhanh dn u, a 0 nhanh dn u,

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nh phn lc, lc mast ca mt bn, lc cng ca si dy, lc Hng tm v lc ly tm trongchuyn ng cong

Ta t: vmKr

r

= , v gi l vectng lng ca cht im, do c th vit li biuthc nh lut Newton th hai nh sau:

Kr

Fdt

Kd rr

=

nh l : o hm ng lng ca mt chtim theo thi gian bng tng hp ccngoi lc tc dng ln chtim .

Krr

r

Ta gi gi l mmen i vi im O ca vc tng lng , c gi l vctmmen ng lng ca cht im i vi im O, k hiu :

Kr

vmrLrr

r

=

( )aOMdt

Ldr

r

r

/=Khi ta c :

nh l vmmen ng lng :o hm theo thi gian ca mmen ng lngi viim O ca chtim chuyn ng bng tng mmen i vi im O ca cc lc tc dng lnchtim.

7. Cc nh lut Newton I v II ch nghim ng trong cc h qui chiu qun tnh, l hqui chiu trong nh lut qun tnh c nghim ng.

Nguyn l tng i Galilo pht biu: Mi h qui chiu chuyn ng thngu ivi h qui chiu qun tnh cng l h qui chiu qun tnh, ni cch khc, cc hin tng chc xy ra ging nhau trong cc h qui chiu qun tnh khc nhau, do dng ca cc

phng trnh chc khng i khi chuyn t h qui chiu qun tnh ny sang h qui chiu

qun tnh khc.Chc cin (chc Newton) c xy dng da trn 3 nh lut Newton v nguyn

l tng i Galilo. Theo chc cin, thi gian c tnh tuyt i, khng ph thuc vo hqui chiu. Nh, rt ra mi lin h gia cc ta khng gian v thi gianx,y,z,ttrong hqui chiu qun tnh O v cc ta x,y,z,ttrong h qui chiu qun tnh Ochuyn ngthng u i vi O. T ta rt ra kt qu:

t = t, l =l0

8. Ta cng thu c qui tc cng vn tc:

Vvvr

rr

+=

Aaar

rr

+= v qui tc cng gia tc:

trong v l vn tc v gia tc ca cht im xt trong hO, cn v v avr

ar

r

r

l vn

tc v gia tc cng ca cht im xt trong hO chuyn ng vi vn tc V Arr

so vi O.l gia tc ca hOchuyn ng so vi O.

Nu hOchuyn ng thng u i vi O (khi Ocng l h qui chiu qun tnh)

th Ar

= 0, a , do :r

FamamF ===r

rr

r

Ngha l cc nh lut chc gi nguyn trong cc h qui chiu qun tnh.

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Ar

Aaar

rr

+=Nu hOchuyn ng c gia tc so vi hO th 0, . Trong hO, nhlut Newton II c dng:

qtFamAmamamFr

r

r

rr

r

+===

lc qun tnh cng phng, ngc chiu vi gia tcAmFqtrr

= Ar

ca h qui chiu

Ochuyn ng so vi O.

III. CU HI N TP

1. H qui chiu l g? Ti sao c th ni chuyn ng hay ng yn c tnh cht tngi. Cho v d.

2. Phng trnh chuyn ng l g? Quo chuyn ng l g? Nu cch tm phngtrnh qy o. Phng trnh chuyn ng v phng trnh quo khc nhau nh th no?

3. Phn bit vn tc trung bnh v vn tc tc thi? Nu ngha vt l ca chng.

4.nh ngha v nu ngha vt l ca gia tc? Ti sao phi a thm khi nim gia tc

tip tuyn v gia tc php tuyn? Trong trng hp tng qut vitdtdva =

r

c ng khng?

Ti sao?

5. Tnh ngha gia tc hy suy ra cc dng chuyn ng c th c.

6. Tm cc biu thc vn tc gc, gia tc gc trong chuyn ng trn, phng trnhchuyn ng trong chuyn ng trn u v trn bin i u.

7. Tm mi lin h gia cc i lng a, v, R, , , a , a trong chuyn ng trn.t n8. Ni gia tc trong chuyn ng trn u bng khng c ng khng? Vit biu thc

ca gia tc tip tuyn v gia tc php tuyn trong chuyn ng ny.

9. Chuyn ng thng thay i u l g ? Phn bit cc trng hp:a = 0, a >0, a< 0.10. Thit lp cc cng thc cho to, vn tc ca cht im trong chuyn ng thng

u, chuyn ng thay i u, chuyn ng ri t do.

11. Thit lp cng thc lin h gia gc quay, vn tc gc v gia tc gc trong chuynng trn u.

12. nh ngha h c lp. Pht biu nh lut Newton th nht v nh lut Newton thhai. Hai nh lut ny p dng cho h qui chiu no? Ti sao?

13. Phn bit s khc nhau gia hai h: h khng chu tc dng v h chu tc dngca cc lc cn bng nhau. H no c coi l c lp.

14.Chng minh cc nh l vng lng v xung lng ca lc. Nu ngha ca cci lng ny.

15. Pht biu nh lut Newton th ba. Nu ngha ca n.

16. H qui chiu qun tnh l g? H qui chiu qun tnh trong thc t?

17. Lc qun tnh l gi? Nu vi v d v lc ny. Phn bit lc qun tnh ly tm v lcly tm. Nu v d minh ha v trng thi tng trng lng, gim trng lng v khng trnglng.

18. Chc cin quan nim nh th no v khng gian, thi gian?

19. Trnh by php tng hp vn tc v gia tc trong chc Newton.20. Trnh by php bin i Galilo v nguyn l tng i Galilo.

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IV. BI TP

Th d 1. Mt t chuyn ng nhanh dn u, i qua hai im A, B cch nhau 20mtrong thi gian 2s. Vn tc ca t khi i qua B l 12m/s. Tm:

a. Gia tc ca chuyn ng v vn tc ca t khi i qua im A.b. Qung ng m t i c tim khi hnh n im A.

Bi gii:

atvvt

vva BA

AB =

=a. (1)

tvatAB A+=2

2

1(2)

T (1) v (2) ta c:

( ) 22

/2.2

smt

ABtva B =

=

vA =vB at = 8m/sB

b. V vn tc t lc khi hnh v = 0 nn ta c:0

ma

v

a

vatas

tav

AAA

A

162

1

2

1

2

1 22

2 ==

==

=

Th d 2. Mt vt c nm ln t mt t theo phng thng ng vi vn tc ban uv 2.o = 20 m/s. B qua sc cn ca khng kh, ly gia tc trng trng g = 10 m/s

a.Tnh cao cc i ca vt v thi gian i ln c cao .b.T cao cc i vt ri ti mt t ht bao lu? Tnh vn tc ca vt khi vt chm

t.

Bi giia. Khi vt i ln theo phng thng ng, chu sc ht ca trng trng nn chuyn

ng chm dn u vi gia tc g 10m/s2; vn tc ca n gim dn, khi t ti cao cc ith vn tc bng khng.

vy = voy gt = 01

l thi gian cn thit vt i t mt t ln n cao cc i.vi t1

sg

vt 201 ==T ta suy ra:

mg

vgttvh y 2022

20

21

10max ===Ta suy ra: cao cc i:

2b.T cao cc i vt ri xung vi vn tc tng dn u v=gtv h=gt /2=20m. T ta tnh c thi gian ri t cao cc i ti t t :2

sg

ht 2

10

2.202 max2 ===

Lc chm t n c vn tc

smgtv /202 ==

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Th d 3. Mt vlng ang quay vi vn tc 300vng/pht th b hm li. Sau mt phtvn tc ca v lng cn l 180 vng/pht.

a.Tnh gia tc gc ca vlng lc b hm.

b.Tnh s vng vlng quay c trong mt pht b hm .

Bi gii

2.60

180)s/rad(2.

60

300 = =10 (rad/s), = = 6 (rad/s)1 2

a.Sau khi b hm phanh, vlng quay chm dn u. Gi , 1 2 l vn tc lc hm vsau mt pht. Khi t 12 +=

2212 s/rad209,0-s/rad60

4-

t

-=

==

2-0,21rad/s=

b.Gc quay ca chuyn ng chm dn u trong mt pht :

)rad(480).6060

4-(5,060.10t

2

1t 221 =+=+=

S vng quay c trong thi gian mt pht l:

2402

n =

= vng

Th d 4. Mt t khi lng m = 1000kg chy trn on ng phng. H s ma stgia bnh xe v mt ng bng k = 0,10. Ly gia tc trng trng g = 10m/s2. Hy xc nhlc ko ca ng ct khi:

a. t chy thng nhanh dn u vi gia tc 2m/s2 trn ng phng ngang.b. t chy thng u ln dc trn ng phng nghing c dc 4% (gc nghin

ca mt ng c sin = 0,04).Bi gii:

a. Khi vt chuyn ng trn mt ng phng ngang tc

dng vo vt c PFF msk Nrrrr

,,, , p dng nh lut II

Newton ta c: NPFFam mskrrrr

r

+++= , chiu phng

trnh ln trc ox cng phng chiu chuyn ng ca vtta c:

ma = FkFmsFk= m (a + kg) = 3000N

NPFF mskrrrr

,,, b. Khi vt chuyn ng trn mt ng phng nghing tc dng vo vt c ,

phn tch thnh 2 thnh phn: 1Fr

2Fr

Pr

cng phng vi mt nghing, c phng vung gc

vi mt nghing. p dng nh lut II Newton ta c: NPFFam mskrrrr

r

+++= , chiu phng

trnh xung trc ta cng chiu chuyn ng:( ) NkmgFFFFFFma mskmsk 1375cossin0 11 +=+===

Th d 5. Ngi ta gn vo mp bn (nm ngang) mt rng rc c khi lng khngng k. Hai vt A, B c khi lng bng mA= mB = 1kg c ni vi nhau bng mt sidy vt qua rng rc. H s ma st gia vt B v mt bn k = 0,1. tm:

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a. Gia tc chuien ng ca h.b. Sc cng si dy. Coi ma st rng rc khng ng k.Bi gii:

amFNPPF msBAr

rrrrr

=+++=Lc tng hp tc dng ln h:

Chiu phng trnh trn phng chuyn ng ca

tng vt v chn chiu dng l chiu chuyn ng cavt ta c:( ) 2/5,4 sm

mm

kmmgamaFPF

BA

BAmsA =+

===

b. tnh lc cng si dy, ta vit nh lut II

Newton cho vt A:

TPam AArr

r

+= V tr s:

( ) NagmTTPam AAA 5,5===

BI TP TGIItbytax sin;cos ==1.1- Phng trnh chuyn ng ca cht im c dng:

srad/4,31=Cho bit a = b = 20cm; . Tm:

1. Qy o chuyn ng ca cht im.2. Vn tc v chu k ca chuyn ng.3. Gia tc ca chuyn ng.

12

2

2

2

=+ b

y

a

x

p s: 1. Qy o chuyn ng ca cht im:Nu a = b th , vy quo l ng trn.222 ayx =+

; T = 0,2 (s)2. )/(2 smRv ==

smaaatbdt

dvataos

dt

dva yx

yy

xx /282;sin;

2222 =+===== 3.

1.2- Hai t cng chy trn mt on ng t A n B . Chic t th nht chy na uon ng vi vn tc v v na sau ca on ng vi vn tc v1 2 . Chic t th hai chyna thi gian u vi vn tc v v na thi gian sau vi vn tc v1 2.Tm vn tc trung bnhca mi t trn on ng AB. Cho bit v = 60km/h v v = 40km/h1 2

p s: 1. hkmv /48=

2. hkmv /50=

1.3- Mt ngi cho mt con thuyn qua bsng theo hng vung gc vi bsng vi vntc 7,2km/h. Nc chy mang con thuyn v pha xui dng mt khong 150m. Tm:1. Thi gian cn thit thuyn qua c sng. Cho bit chiu rng ca sng bng 0,5km.2. Vn tc ca dng nc vi bsng.

p s: t = 250(s); v23 = 0,6m/s

1.4- Mt xe la bt u chuyn ng gia hai im (nm trn mt ng thng) cch nhau1,5km. Trong na on ng u xe la chuyn ng nhanh dn u, cn na on ng

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sau xe la chuyn ng chm dn u. Vn tc ln nht ca xe la gia hai im bng50km/h. Bit rng tr s tuyt i ca cc gia tc trn hai on ng bng nhau. Tnh:1. Gia tc ca xe la.2. Thi gian xe la i ht qung ng gia hai im .

p s: a 0,13m/s2; t 213,84 (s)

1.5 Mt vt ang ng yn bt u chuyn ng nhanh dn u, bit rng trong giy th 5n i c mt qung ng 18 m. Hi trong giy th 10, vt i c qung ng bng

bao nhiu ?

p s: s = 38 (m)

1.6 Mt ngi ng sn ga nhn mt on tu ang bt u chuyn bnh , bit rng toath nht chy ngang qua trc mt ngi trong 6s . Coi chuyn ng ca on tu lnhanh dn ln. Hi toa th n i qua trc mt ngi quan st trong bao lu ? p dng vitrng hp n = 7 .

( ) ( )snntt 676)1( == p s:1.7- Th vt ri t do t cao h = 20m. Tnh:1. Qung ng m vt ri c trong 0,1s u v 0,1s cui.2. Thi gian cn thit vt i c 1m u v 1m cui ca cao h. Cho g = 10m/s2.

p s: 1. h1 = 0,05 (m) ; h = 1,95 (m)2. t1 = 0,45 (s) ; t = 0,05 (s)

1.8- Phi nm mt vt theo phng thng ng t cao h = 45m vi vn tc ban u v0bng bao nhiu n ri ti mt t:

1. Trc 1 giy so vi trng hp vt ri t do?2. Sau 1 giy so vi trng hp vt ri t do. Cho g = 10m/s2.

p s: 1. Nm xung vi vn tc v0 = 12,5 (m/s)2. nm ln vi vn tc v0 = 8,75 (m/s)

1.9- Mt hn c nm theo phng nm ngang vi vn tc ban u v0 = 15m/s. Tm giatc php tuyn v gia tc tip tuyn ca hn sau khi nm 1 giy. Cho g = 10m/s2. B quami lc cn.

p s: at = 5,6 m/s2; an = 8,3m/s

2

1.10- Ngi ta nm mt qu bng vi vn tc ban u v0 = 10m/s theo phng hp vi mtphng nm ngang mt gc = 300. Ga s qu bng c nm i t mt t. Hi:

1. cao ln nht m qu bng c tht c.2. Tm bay xa ca qu bng.3. Thi gian t lc nm qu bng ti lc bng chm t. Cho g = 10m/s2. B qua mi lc cn.

9

2sin;

2

sin 2022

0max

vx

g

vy ==p s:

1.11- T cao h = 25m ngi ta nm mt hn ln pha trn vi vn tc ban u v0 =15m/s theo phng hp vi mt phng nm ngang mt gc = 300. Xc nh:1. Thi gian chuyn ng ca hn .2. Vn tc ca hn lc chm t.

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Cho g = 10m/s2. B qua mi lc cn.

p s: 1. t = 3,1 (s), 2. v = 26,85 9m/s)

1.12- T mt nh thp cao h = 30m, ngi ta nm mt hn xung t vi vn tc ban uv0 = 10m/s theo phng hp vi mt phng nm ngang mt gc = 30

0. Tm:1. Thi gian chuyn ng ca hn .

2. Khong cch t chn thp n ch ri ca hn .3. Dng quo ca hn .Cho g = 10m/s2. B qua mi lc cn.

p s: 1. t = 2(s)2. x = 17,3(m)

220

2

cos2.

v

gxtgxhy =3.

1.13- Mt v lng sau khi bt u quay c mt pht th thu c vn tc 700vng/pht.

Tnh gia tc gc ca vlng v s vng m vlng quay c trong pht y nu chuyn ngca vlng l chuyn ng nhanh dn u.

p s: 1. = 1,22 (rad/s2), 2. n = 350 vng

1.14- Mt on tu bt u chy vo mt on ng trn, bn knh 1km, di 600m vi vntc 54km/h. on tu chy ht qung ng trong 30s. Tm vn tc di, gia tc phptuyn, gia tc tip tuyn, gia tc ton phn v gia tc gc ca on tu cui qung ng. Coi chuyn ng ca on tu l chuyn ng nhanh dn u.

p s: v = 259m/s); an = 0,625(m/s2); a = 0,7(m/s2)

1.15- Mt ngi di chuyn mt chic xe vi vn tc khng i. Lc u ngi y ko xe vpha trc, sau ngi y y xe v pha sau. Trong c hai trng hp, cng xe hp vi mtphng nm ngang mt gc . Hi trong trng hp no ngi y phi t ln xe mt lc lnhn? Bit rng trng lng ca xe l P, h s ma st gia bnh xe v mt ng l k.p s: Ko xe v pha trc : f = k( P - F.sin )ms

= k( P + F.sin )y xe v pha sau: fms

1-16 Mt bn g phng A c khi lng 5kg b p gia hai mt phng thng ng song song.Lc p vung gc vi mi mt ca bn g bng 150N. H s ma st ti mt tip xc l 0,20.Ly g = 10m/s2. Hy xc nh lc ko nh nht cn dch chuyn bn g A khi nng n lnhoc h n xung.p s: Ko g ln pha trn: F = mg + 2kNmin

= 2kN mgKo g xung di: Fmin

1.17- Mt tu in, sau khi xut pht, chuyn ng vi gia tc khng i 0,5m/s2. Sau khi btu chuyn ng c 12s, ngi ta tt ng cca tu v tu chuyn ng chm dn ucho ti khi dng hn. Trn ton b qung ng h s ma st bng k = 0,01. Tm:1. Vn tc ln nht ca tu.2. Thi gian ton b k t khi tu xut pht cho ti khi tu dng hn.

3. Gia tc ca tu trong chuyn ng chm dn u.4. Qung ng ton b m tu i c. Cho g = 10m/s2.

p s: 1. v = 6m/s; 2. t = 72s; 3. s = 216m

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1.18- Mt t c khi lng 5 tn ang chy b hm phanh chuyn ng chm dn u. Sau2,5s xe dng li. T lc bt u hm phanh cho n khi dng hn n i c 12m. Tm:1. Vn tc ca t lc bt u hm phanh.2. Lc hm trung bnh. Cho g = 10m/s2.

p s: v = 9,6m/s; F = -19,2.103N

1.19- Mt on tu gm mt u my, mt toa 10 tn, v mt toa 5 tn ni vi nhau theo tht trn bng nhng l xo ging nhau. Bit rng khi chu tc dng mt lc 500N th l xo gin1cm. B qua ma st. Tnh gin ca l xo trong hai trng hp:1. on tu bt u chuyn bnh, lc ko ca u my khng i vsau 10s vn tc ca on tu 1m/s.2. on tu ln dc c nghing 5% vi vn tc khng i( 2). Cho g = 10m/s05,0sin =

p s: 1. x = 3cm; x = 1cm; 2. x = 14,7cm; x = 4,9cm1 2 0 2

= 300g, m1.20- Hai vt nng c khi lng m1 2 = 500g c bucvo hai u si dy vt qua rng rc c khi lng khng ng k.Vt m di vt m mt khong h = 2m (hnh 1-1bt). Xc nh:1 21. Gia tc chuyn ng ca h vt v sc cng si dy.2. Sau bao lu hai vt m1 v m cng cao. Cho g = 10m/s

22 , b

qua khi lng ca dy, si dy khng gin, b quama st trc ca rng rc.

p s: a = 2,5m/s2; T = 3,75N

1.21- Cho h gm ba vt nh hnh v, khi lng ccvt ln lt m = 1kg, m = 2kg, m1 2 3 = 3kg(hnh 1-2bt). Khi lng ca rng rc khng ng k. Ditc dng ca trng lng vt m3, h vt s chuynng. Ma st gia cc vt v mt ngang k = 0,2. Tm:1. Gia tc chuyn ng ca h vt.2. Sc cng ca cc si dy ni gia cc vt. Cho g =10m/s2, b qua khi lng ca dy, si dy khng gin, b qua ma st trc ca rng rc.

p s: a = 4m/s2

Lc cng gia vt m , m l T =6N, gia vt m v m l T1 2 1 2 3 2 = 18N

1.22- Mt toa xe khi lng 20 tn chuyn ng vi vn tc ban u 54km/h. Xc nh lctrung bnh tc dng ln xe, nu toa xe dng li sau thi gian:1. 1pht 40giy.2. 10giy.

p s: 1. F1 = 3.103N; 2. F = 3.104N2

1.23- Mt vin n khi lng 10g chuyn ng vi vn tc v0 = 200m/s p vo mt tm gv xuyn su vo tm g mt on s. Bit thi gian chuyn ng ca vin n trong tm g

t = 4.10-4

s. Xc nh lc cn trung bnh ca tm g ln vin n v xuyn ca vin n.4p s: F = -0,5.10 N; s = 4.10-2m

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1.24- Mt ngi khi lng 50kg ng trong thang my ang i xung nhanh dn u vi giatc 5m/s2. Hi ngi c cm gic nh th no v trng lng biu kin ca ngi trongthang my. Cho g = 10m/s2.

p s: P= P Fqt = 250N

1.25- Mt thang my c treo u mt dy cp ang chuyn ng ln pha trn. Lc u

thang my chuyn ng nhanh dn u sau chuyn ng u v trc khi dng li chuynng chm dn u. Hi trong qu trnh , lc cng ca dy cp thay i nh th no? Cmgic ca ngi trn thang my ra sao?

p s: Nhanh dn u: T = m(g + a)

Chuyn ng u T = mg

Chuyn ng chm dn u: T = m(g a)

1.26- Mt t khi lng 2,5 tn chuyn ng vi vn tc khng i 54km/h qua mt chiccu. Xc nh lc nn ca t ln cu, nu:1. Cu nm ngang.2. Cu vng ln vi bn knh cong l 50m.3. Cu lm xung di vi bn knh cong l 50m (tng ng vi v tr t gia cu). Cho g= 10m/s2.

p s: Cu vng ln: N = 13750N

Cu vng xung: N = 36250N

1.27- Mt phi cng li my bay thc hin vng nho ln vi bn knh 200m trong mt phngthng ng. Khi lng ca phi cng bng 75kg. Xc nh:

1. Lc nn ca phi cng tc dng ln gh ngi ti im cao nht v thp nht ca vng nholn khi vn tc ca my bay trong vng nho ln lun khng i v bng 360km/h.2. Vi vn tc no ca my bay khi thc hin vng nho ln, ngi phi cng bt u b rikhi gh ngi? Cho g = 10m/s2.

3p s: Thp nht: N = 4,5.10 N; Cao nht N = 3.103N

v 44,72m/s

1.28- Mt vt nh khi lng m = 1,0kg c t trn mt a phng ngang v cch trc quayca a mt khong = 0,50m. H s ma st gia vt v mt a bng

k = 0,25. Hy xc nh:a. Gi tr ca lc ma st vt c gi yn trn mt a khi a quay vi vn tc n = 12vng/pht (vg/ph). Ly gia tc trng trng g = 10m/s2.

b. Vi vn tc gc no ca a quay th vt bt u trt trn a?

p s: 1.Fms = aht . m = m (2n2) r 0,79N.

)/(2,2min sradm

kg

r

kg= 2.

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