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3

Chng 1NG HC CHT IM

TM TT L THUYT

1.

Phng trnh chuyn ng ca cht im r r t hoc ; ;x x t y y t z z t

2. Vn tc- Vc t vn tc ca cht im trong to Descartes

dr dx dy dz v i j k

dt dt dt dt

- Vn tc trong to cong

0lim

t

s dsv

t dt

3. Gia tc- Vc t gia tc trong to Descartes:

2 2 2 2

2 2 2 2

d r d x d y d z a i j k

dt dt dt dt

- Vc t gia tc trong to congt n

dva a a

dt

trong tdv

adt

l gia tc tip tuyn,2

n

va

R l gia tc php

tuyn vi R l bn knh cong ca qu o.4.

Chuyn ng trn

- Vn tc gc: ddt

- Gia tc gc:2

2

d d

dt dt

5.

Tng hp vn tc v gia tc trong chuyn ng tnh tin - Tng hp vn tc: 13 12 23v v v

- Tng hp gia tc:13 12 23

a a a

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4

BI TP V DV d 1Mt cht im chuyn ng c m t bi cc phng trnh sau:

2

2 (1)

4 4 (2)

x t

y t

x, y tnh bng mt, t tnh bng giy.a.

Tm qu o chuyn ng ca cht im.b.

Tm vn tc ca cht im khi 2t s.Li gii:a.

T (1) rt ra / 2t x , thay vo (2) c:2

4y x - Qu o ca cht im l mt nhnh parabol ( 0x ).b.

Vntc ti 2t s2

8

x t

y t

v x

v y t

2 2 24 64x yv v v t

Thay 2 st ta c: 260 16,12v m/s.

V d 2T im O trn mt t mt cht im c nm ln vi vn tc

ban u 0v hp vi phng ngang mt gc .

a.

Tm phng trnh chuyn ng, phng trnh qu o.b.

Tm bn knh cong ca qu o ti O v ti im cao nht caqu o.

Li gii

a.

Phng trnh chuyn ng0

2

0

os t (1)

y= v sin (2)2

x v c

gtt

- Phng trnh quo

2 2

02 os

gy x tg x

v c

(3)

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5

b.

Bn knh cong ca qu o:

- Ti mi im gia tc ton phn a g .

- Ta c2 2

n

n

v va R

R a

- Ti O: cosna g 22

0

osn

vv

R a gc

- Ti im cao nht A: na g , 0 cosxv v v (do 0yv )2 22 2

0 osA

n n

v cv vR

a a g

BI TP P DNG

1.1

Ti cng mt thi im hai tu thu A v B trn mt kinhtuyn. A pha Bc so vi B v cch B mt on d0. A chy v

pha ng vi vn tc khng i Av ; B chy v pha Bc vi vn

tc khng i Bv . cong ca mt t khng ng k.

a.

Xc nh khong cch cc tiu gia A v B.b.

B phi chy theo hng no, sau thi gian bao lu th s gp

A.

xv

A

N

HO

y

x

na g

na ta

a g

0v

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6

1.2 Cho phng trnh chuyn ng ca mt cht im:2

0

2

0

20

x x at

y y bt

z z ct

vi a, b, c l cc hng s.

a.

Tm phng trnh qu o v dng qu o ca cht im.b.

Tnh qung ng cht im i c ti thi imt.c.

Xc nh vn tc v gia tc ca cht im thi im t.

1.3

Mt t ang chy trn ng thng vi vn tc

1 50v km/h. Mt ngi ng A

cch ng y mt khong100AH h m th nhn thy t

va n B cch mnh mt khong500AB m th ngi y bt u

chy ra ng n t.a.

Ngi y phi chy theo hng no c th gp c t,

nu ngi chy vi vn tc 2 20 / 3v km/h.

b.

Mun gp c t ngi y phi chy ra ng vi vntc nh nht bng bao nhiu? Theo hng no?

1.4

Mt ngi cho thuyn qua sng tim A. Vn tc ca nc i vi b l

2v , vn tc ca thuyn i vi nc l

1v . Nu hng mi thuyn ti B th sau

thi gian 1 10t pht thuyn ti C vpha h lu cch B mt on 120BC s m. Nu hng mithuyn v pha thng lu lch mt gc so vi AB th sau thigian

2 12,5t pht thuyn ti B.

a.

Tnh vn tc1v , 2v v chiu rng ca con sng.

b.

Xc nh gc lch .

1.5

Mt ngi ng im A trn rung cch ng (d) mtkhong 1AH a km cn i ti im B trn ng, cch H mt

B H

A

1v

A

B C

D

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7

khong 3BH b km. Vn tc itrn rung l

1 3v km/h, trn ngci l

2 6v km/h. Hi phi i theo

qu o no th thi gian ti B l ngnnht, tnh thi gian y.

1.6 Mt my bay bay t v tr A ti v tr B, AB nm theo hngTy ng v cch nhau mt khong 300s km. Xc nh thigian bay nu:

a.

Khng c gi.b.

C gi thi theo hng Nam Bc.

c.

C gi thi theo hng Ty ng.Cho bit vn tcca gi

1 20v m/s v vn tc ca my bay ivi khng kh l

2 600v km/h.

1.7

Mt vt chuyn ng nhanh dn u trn hai on ng lintip bng nhau v bng 100m ln lt trong 5s v 3,5s. Tnh gia tcca vt.

1.8

Mt ngi ng sn ga nhn thy mt on tu ang btu chuyn bnh, bit toa th nht chy ngang qua trc mt ngi trong 6 giy. Coi chuyn ng ca on tu l nhanh dn u.Hi toa th n qua trc mt ngi quan st trong bao lu? p dngvi 7n .

1.9 Mt on tu chuyn ng thng chm dn u tin vo snga. Mt ngi quan st ng cnh ng tu thy toa th nht chy

ngang qua trc mt mnh trong 4 giy, toa th hai qua trong 5giy. Khi tu dng li, u toa th nht cch ngi y 75m. Hyxc nh gia tc ca on tu.

1.10

Mt cht im chuyn ng thng khng vn tc ban u.Lc u cht im chuyn ng nhanh dn u vi 0,5a m/s2,sau chuyn ng u ri cui cng chuyn ng chm dn uvi giatc c ln nh lc u v dng li. Thi gian tng cng

ca chuyn ng l 25s, vn tc trung bnh trong khong thi gian l 2 m/s. Tnh khong thi gian cht im chuyn ng u.

A

H B(d)

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8

1.11

T cao 80h m phi nm mt vt theo phng thngng vi vn tc ban u 0v bng bao nhiu n ri ti mt t.

a.

Trc 1 giy so vi trng hp ri t do.

b.

Sau 1 giy so vi trng hp ri t do. Ly 10g m/s2

.1.12

Vt A c th ri t do t cao ( )H h so vi mt t.

Cng lc vt B c nm thng ng t mt t vi vn tc 0v

hng ti A.

a.

Tnh 0v hai vt gp nhau cao h.

b.

Xc nh khong cch x gia hai vt trc lc gp nhau theo

thi gian.c.

Nu khng c vt th nht th vt th hai ln ti cao lnnht l bao nhiu? p dng bng s 20H m, 10h m

10g m/s2.

1.13 Mt tu ho chuyn ng chm dn u trn qungng 800s m c dng cung trn bn knh 800R m. Vn tc u qung ng l 0 54v km/h v cui qung ng l

18v km/h. Tnh:a.

Xc nh gia tc ca tu ti im u v im cui qungng.

b.

Tnh thi gian cn thit tu i ht qung ng .

1.14 T ba im A, B, C trn mt vng trnngi ta th ng thi ba vt. Vt th nht

ri theo phng thng ng AM. Vt thhai chuyn ng trn mt phng nghingBM, vt th ba theo mt nghing CM. Hivt no s ri ti im M trc nht. B quama st v sc cn khng kh.

1.15 T my bay ang bay ngang vi vn tc 720 km/h ngi tath mt vt. Tm bn knh cong ca qu o sau khi vt chuyn

ng c 5s. B qua sc cn ca khng kh.

AB

C

M

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9

1.16

T mt nh thp cao 25H m ngi ta nm mt hn theo phng ngang vi vn tc ban u 0 15v m. Tm:

a.

Qu o ca hn .

b.

Thi gian chuyn ng ca hn .c. Khong cch t chn thp n im hn chm t.

d.

Vn tc, gia tc tip tuyn, gia tc php tuyn ca hn khi chm t.

e.

Bn knh cong ca qu o ti im bt u nm v imchm t. B qua sc cn khng kh ly 210m sg .

1.17

Mt vin n c bn ln t mt t vi vn tc ban u

0 200v m/s hp vi phng ngang mt gc030 . Tm:

a.

cao ln nht v tm xa m vin n t c.b.

Gia tc tip tuyn v gia tc php tuyn ca vin n sau lcbn 1 giy.

c. Vi gc bn bng bao nhiu :

-

Tm xa ca n l cc i.

-

cao cc i v tm xa ca n bng nhau.1.18 Hai vin n c bn ln ln lt bi mt sng i bc vivn tc 0 250v m/s. Vin n th nht bn di gc

0

1 60 ,

vin th hai bn di gc 02 45 (trong cng mt mt phngthng ng). B qua sc cn khng kh, ly 10g m/s2. Hy xcnh khong thi gian gia hai ln bn cho hai vin n gpnhau.

HNG DN - LI GII - P S

1.1

Bi ton hai tu thu chy trn cng kinh tuyn

a. Khong cch cc tiu

2

min 0 2 2

A

A B

vd d

v v

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10

b.

Tu B phi chy theo hng chch v pha ng mt gc

so vi ng kinh tuyn vi sin A

B

v

v .

- Tu B ui kp tu A sau thi gian 02 2B A

dtv v

1.2 Cht im chuyn ng trong h to Descartesa.

Kh t trong phng trnh chuyn ng2 0 0 0x x y y z zt

a b c

- Thu c phng trnh qu o 0 0 0x x y y z za a a

- Dng qu o l ng thng.

b. Ti 0 00, v 0t qung ng cht im i c t

2 2 2 2 2 2 2

0 0 0s x x y y z z a b c t

c.

Vn tc ca cht im ti thi im t

2 2 22 2 22

dx dy dz v a b c t

dt dt dt

- Gia tc ca cht im ti thi im t2 2 2

2 2 22 2 2

2 2 22

d x d y d z a a b c

dt dt dt

- Ta thy gia tc ca cht im l mt hng s, nh vy cth thy chuyn ng ca cht im l mt chuyn ngthng nhanh dn u.

1.3 Ngi i b n ta.

Xe (1), ngi (2), t(3)- Ta c:

23 21 13v v v - Theo hnh v:

B H

A

13v C

13v

21v 23v

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11

13 23

sin sin

v v

13

23

sin = sinv

v

viAH

sin =AB

h

01

2

3sin = . 60

2

v h

v hoc 0120

b.

Xc nh hng chy vi vn tc cc tiu

- T h thc phn trn ta c:1

2

.

.sin

v hv

. V

1, , constv h

- 02min axsin 1 90mv . Ngi phi chy

theo hng c 090 vi vn tc 2min 1 10h

v v km/h.

1.4 Ngi cho thuyn qua sng

a.

Tnh1v , 2v ,

- Ln 1: Hng mi thuyn t AB- Vn tc ca thuyn so vi b

13 12 23 1 2v v v v v v

- Do ABC AMN nn ta c

1

2 1

BC AB ACt

v v v (1)

2

1

BC 1200,2

600v

t m/s

- T (1)1 1AB v t (2)

- Ln 2: Hng mi thuyn ti D- Vn tc ca thuyn so vi b

1 2'v v v

A

B C

1v

2v

v M N

A

BD

P Q

1v 2v

'v

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12

- Do ADB APQ nn ta c

2

1 2

AB AD BDt

v v v

2 22 1 2 2.AB v t v v t (3)

- T (2) v (3) c 2 21 1 2 2 11

m/s 1,2 km/h3

v t v v t v

- Ta c1 1 200l AB v t m

b. Xc nh gc

02

1

sin 0,6 arcsin 0,6 37v

v

1.5 Xc nh qu o c thi gian i ngn nht- Qu o c thi gian ti Bngn nht gm 2 on ngAC v CB vi x HC . Muntm

mint th phi chn x thchhp.

- Thi gian i2 2 2

1 2

2 1 3

6

a x b x x xt

v v

- Ly o hm ca t theo x v cho o hm ny trit tiu2

02

1 2 1 1. 0 x 0,58 km=

6 31

dt x xHC

dx x

- Thi gian i 0 0,79 ht (ng vi 0 0,58 kmx )- Vy thi gian ngn nht l 0,79h v phi i theo qu oACB c 0,58HC km).

1.6 My bay bay t A ti B

a.

Khng c gi:2

0,5 h =30s

tv

pht

H(d)

A

BCx

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13

b.

Gi thi theo hng Nam Bc mybay phi chch v pha Ty mt gc so vi phng AB vi:

1

2

3

sin 25

v

v

- Khi 2 22 1 165,46v v v m/s

- Thi gian bay:

1813,127 s 30,2AB

tv

pht

c. Gi thi theo hng Ty ng

1 2

560

3v v v m/s

- Thi gian bayB

1607,143 s 28,78v

At

pht

1.7 Cht im chuyn ng nhanh dn u

- p dng 201

2

s v t at ( 0v l vn tc ban u)

- i ht qung ng u ht1 5t s

1 05 12,5 100s v a m (1)

- i ht ton b 2 qung ng ht 2 5 3,5 8,5t s

2 0

72,258,5 200

2s v a m (2)

- T (1) v (2) ta c 2a m/s2.

1.8 on tu ri ga chuyn ng nhanh dn u- Gi l chiu di mi toa, nt l thi gian n toa tu i

qua trc mt ngi quan st.

- Chiu di l toa l 21

2at (vi 0 0v )

2

1

2t

a

- Chiu di ca (n - 1) toa u l:

T 'v

1v

2v

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14

2 2

1 1

2 111

2 n n

nn at t

a

- Chiu di ca n toa u l: 2 21 2

2 n n

nn at t

a

- Toa th n qua trc mt ngi quan st trong thi gian l

1 12

1 1n n nt t t n n t n na

- p dng vi 7n ta c:

7 6 7 6 1,177t s

1.9

on tu vo ga chuyn ng chmdn u- Chn 0t l lc toa th nht ngang qua ngi quan st,l chiu di mi toa tu. Ta c h

2

0 1 1

2

0 1 2 1 2

2 20

1

2

12

2

2t

v t at

v t t a t t

v v as

0

0

2

0

4 8

2 9 82

150

v a

av

v a

- Gii h ta c 0,25a m/s2.

1.10 Cht im chuyn ng thng- Gi qung ng, thi gian i ht chng, vn tc cui cami chng ln lt l: 1 1 1 2 2 2 3 3 3, , ; , , ; , ,s t v s t v s t v

- Ta c:1 2 3 25 st t t t

1 2 3 50 mtbs s s s v t

- Chng 1: 1 0,5a m/s2, 21 1 1

1

2s a t ,

1 1 1v a t

- Chng 2:2 0a , 2 2 2 1 2s v t v t , ( 2 1v v )

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15

- Chng 3 : 3 0,5a m/s2, 2 23 03 3 3 3 1 3 1 3

1 1

2 2s v t a t v t a t

- Khi dng li: 3 03 3 3 1 1 3 3 10v v a t v a t t t

- Do 2

2 1 13 1 1 1

2a ts a t s

2

1 11 2 1 1 22 2 50

2

a ts s s a t t

1 1 1 2 50a t t t

11 1 1 11

5 s25 2

t =20 s

ta t t t

- Gi tr1 20t s loi (v 12 25 st ). Vy thi gian vt

chuyn ng u 2 125 2 25 2.5 15t t s.

1.11 Nm mt vt theo phng thng ng- Chn chiu (+) hng xung, gc thi gian l lc vt cth ri t do.

- Trng hp ri t do: 21 42h gt t s

a.

Phi nm vt xung di

2

0 0

1 351 1 11,67

2 3h v t g t v m/s

b.

Phi nm vt ln trn

2

0 0

11 1 9

2h v t g t v m/s

1.12 Hai vt A, B chuyn ng thng ng- Chn trc to thng ng, chiu (+) hng ln, gc to ti mt t, gc thi gian l lc th vt.- Qung ng chuyn ng ca vt ri t do v ca vtc nm ln t mt t c dng:

21

2s gt ;

2

0

1

' 2s v t gt

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16

a.

Hai vt gp nhau cao h khi s H v s h - Ta c:

21

2H gt ; 20

1

2h v t gt

0

2

2 2

h H gH gv h H

H H

b.

Khong cchxgia hai vt trc lc gp nhau

2 2 2

2 2

x H h s s

h H gH H hH h H gH

H H

c. Vt 2 nm ln s chuyn ng chm dn n khi 0tv thvt t cao cc i:

0

2

ax 0

0

2

t

m

v v gt

gth v t

22

0ax

2 4m

h Hvh

g H

- p dng 0 15 mv s ; 30 15x t ; ax 11,25 mmh

1.13 Tu ho chuyn ng trn cung trn- Chn chiu (+) l chiu chuyn ng.

2 220 0,125

2tA tB

v va a m s

s

a. Ti A ta c:

A B

Av

Bv

tBa

tAa nAa

nBa

Ba

Aa

s

O

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17

2 2

A tA nAa a a vi2

20 0,28nv

a m sR

22 20,125 0,28 0,31m sAa

- Ti B ta c2 2

B tB nBa a a vi2

20,03m snBv

aR

22 20,125 0,03 0,13m sBa

b.

Thi gian tu i ht qung ng

0 5 15 80 s

2 0,125

v vt

s

1.14

Chuyn ng ca ba vt ti im M- Gi R l bn knh vng trn- Vt 1: 1 2s AM R , 1a g . Vy thi gian chuyn ngt A n M l:

11

2 4S Rt

g g

- Vt 2: chuyn ng t do trn mt phng nghing BM.

2 2 ccosAMB osRs BM AM AMB

2 cosa g AMB

22

2

2 4s Rt

a g

- Vt 3: chuyn ng t do trn mt phng nghing CM.3 2 coscos Rs CM AM AMC AMC

3 cosa g AMC

33

3

2 4s Rt

a g

- D thy1 2 3

t t t ngha l 3 vt s ti M cng mt lc.

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18

1.15 Vt chuyn ng nm ngang- Chn h trc to xOy.- Chuyn ng ca vt c dngchuyn ng ca vt c nm

ngang, vi:0 720km h 200m sv

a g 2 2 2 2 2

0t x yv v v v g t

- p dng cng thc:2

n

va

R (1)

- Trong : 0os = g xnt t

v va gc g

v v (2)

- T (1) v (2)

3 22 2 230

0 0

tv g tv

Rgv gv

- Vi 0 200v m/s, 9,8g m/s2, 5t s c 4500 mR

1.16

Vt c nm ngang t nh thp- Chn mt phng hnh v l mt phng thng ng cha

0v , h trc xOy, 0t l lc nm vt.

a. Ti t, cht im M(x, y) v c a g

0x

y

aa

a g

;0 0

0

x x

y y y

v v vv

v v a t gt

- Phng trnh chuyn ng theo Ox v Oy0

21

2

x v t

y gt

- Kh t trong phng trnhtrn ta c phng trnhqu o:

O x

y

0v

M

v

xv a g

N

yv

O x

y

0v

xv M

v na a

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19

2

2

0

1

2

gy x

v

- Qu o c dng 1 nhnh parabol ON ( 0x ; y H )

b.

Hn chm t khi2H

25 m t = 5 s 2,24g

y H s

c. Tm xa: 0 15 5 33,54s v t s

d.

Vn tc

2 222

0 26,93Ndx dy

v v gt dt dt

m/s

0, sin 0,83 56yN xN N

v gtv v

v v

- Gia tc ton phn2 2

2 22

2 20 10

d x d ya g g

dt dt

m/s2

- Ti N:2sin sin 8,3m sta a g

2cos cos 5,6m sna a g

e.

Bn knh cong ca qu o2

22,5n

vR

a m

- Ti im bt u nm:2

00 n 0; a 22,5

vv v g R

g m

- Ti im hn chm t:

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20

2

n, a os =5,6 m snv v ac

2

129,5 mn

n

vR

a

1.17

Vin n chuyn ng nm xina.

Phng trnh chuyn ng, phng trnh qu o

0

2

0

cos

1sin

2

x v t

y v t gt

; 2 2

02 os

gy x tg x

v c

- Ti A vt t cao cc i 0sin

2OA

vbt

a g

2 2

0ax

sin( ) 500 m

2m OA

vy y t

g

Khi chm t ti N:2

0 sin20 3464 mNv

y xg

b.

Sau lc bn t = 1 s, n M(x, y) vi:

0

2

0

os t=100 3 m

1sin 95 m

2

x v c

y v t gt

- Vn tc n ti M:

0yv

0xv

0v

A

NH

xv

Mv

O

y

x

M

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21

0

0 0

=100 3 m s

sin 90m s

x x

y y

v v

v v gt v gt

- t 09

, 27,510 3

yM x

x

v

v v tg v

- Gia tc ton phn

2 2

2 22 2

2 20 10m s

d x d ya g g

dt dt

- Gia tc tip tuyn v php tuyn2

.sin 10.0,461 4,61m sta a 2os 10.0,887 8,87 m sna ac

- Tm xa2

0 sin2vxg

; cao cc i

2 2

0max

sin

2

vh

g

- tm xa cc i:

axmx 0

axsin 2 1 45

m

- cao cc i v tm xa bng nhau: 2 2 20 0

ax

sin sin 24

2m

v vh x tg

g g

076

1.18

Hai vin n c bn ln ln lt bi mt sng i bc- Hon ton tng t nh bi 1.18, ta c phng trnhchuyn ng ca hai vin n l:

1 0 1 1

2

11 0 1 1

os .

sin .2

x v c t

gty v t

(1)

2 0 2 2

2

22 0 2 2

os .

sin .2

x v c t

gty v t

(2)

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22

- Hai vin n gp nhau khi 1 2

1 2

x x

y y

hay:

0 1 1 0 2 2

2 2

1 20 1 1 0 2 2

os . os . (3)

sin . sin (4)2 2

v c t v c t

gt gtv t v t

- T (3) 12 12

os

cos

ct t

thay vo (4) ta c

1t , 2t

- Khong thi gian gia 2 ln bn 1 20

2 11 2

sin210,7 s

os os

vt t t

g c c

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23

Chng 2NG LC HC CHT IM

TM TT L THUYT

1)

nh lut Niu tn IIF

am

hay ma F

vi F l tng hp lc tc dng ln cht im; m khi lng

ca cht im; a vc t gia tc ca cht im.

2)

Trng lc tc dng ln vt c khi lng m

P mg 3) Lc hng tm

2

n

vF m

R , viR l bn knh cong ca qu o

4) nh l v ng lng

nh l 1:

dK

Fdt ; K mv l vc t ng lng ca chtim.

nh l 2:1

2

2 1

t

t

K K K Fdt

5) Biu thc lc ma st trt (kh)

msF kN

vi k l h s ma st, N l phn lc php tuyn6) nh l v m men ng lng

d LM

dt

trong rL K l m men ng lng ca cht imrM F l m men ca lc tc dng ln cht im

i vi gc O.

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24

Trng hp cht im chuyn ng trn, c dng

d IM

dt

Vi 2I mr l m menqun tnh ca cht im i vi gc O7) nh lut Niutn p dng cho cht im trong h qui chiu

phi qun tnhTrong h quy chiu O' chuyn ng tnh tin so vi h qui

chiu qun tnh O vi gia tc A

qtma F F

Vi a l gia tc cht im trong h O

; F ngoi lc tc dngln cht im; qtF mA l lc qun tnh t ln cht im.

BI TP V D

V d 1Ngi ta gn vo mp bn mt rng rc c khi lng khng

ng k. Hai vt A v B c khi lng ln lt 200Am g v

300Bm g c ni vi nhau bng mt si dy vt qua rng rc.H s ma st gia vt A v mt bn 0,25k . Ly 10g m/s2.a.

Xc nh gia tc chuyn ng ca h vt.b.

Tnh lc cng ca dy v lc nn ln trc ca rng rc. B quakhi lng dy v ma st rng rc.

c. Nu thay i v tr vt A v B cho nhau th lc cng ca dy sbng bao nhiu. Xem h s ma st

gia vt v bn vn nh c.Li giia.

Xc nh gia tc ca h- Theo nh lut II Newton ta c

A A ms A AP N T F m a (1)

B B B BP T m a (2)

- Chiu (1) v (2) tng ng lnphng chuyn ng ca A v B,

A

BP

BT

AT

AP

N

msF

B

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25

chn chiu dng l chiu chuyn ng, ta c:

A ms A AT F m a (3)

B B B BP T m a (4)

- y ta ch , v dy khng gin nn A Ba a a , rng rckhng khi lng nnA BT T T . Mt khc ms AF kN kP

(doAP N ). Kt hp vi (3) v (4) ta c gia tc ca h:

210 0,3 0,25.0,2 5 m/s0,2 0,3

B B A B A

A B A B A B

P Fms P kP m kma g

m m m m m m

b. Tnh lc cng ca dy, lc nn ln trc rng rc- xc nh lc cng T, thay a vo (4) ta c:

( ) 0,3(10 5) 1,5B B B BT T P m a m g a N- Rng rc chu hai lc cng T v phn lc ca trc rng rc,nh vy ta thy lc nn ln trc rng rc chnh l hp lc cahai lc cng T, do ta c lc nn ln trc rng rc:

2 1,5 2Q T Nc.

Thay i v tr vt A v B, ta cng tnh tng t nh trn v stm c nhng biu thc v gia tc ca h v lc cng T cady tng t nh phn trn, ch khc l thay

Am bng Bm v

thay Bm bng Am , d dng thy rng lc cng T ca dy vn

khng thay i.

V d 2Mt vt t trn nh dc di 165m, gc nghing ca dc l ,h s ma st gia vt v mt dc 0,2k . Ly 9,8g m/s2.a.

Vi gi tr no ca vt s nm yn m khng trt.b.

Cho gc 030 ; hy tm thi gian vt trt xung ht ondc v vn tc ca vt chn dc.

Li gii

a.

Xc nh gi tr ca vt nm cn bng

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26

N

P

msnF

x

y

O

- Chn h trc xOy nh hnh v. Khi vt nm cn bng ta c:

0msnP N F (1)- Chiu (1) ln cc trc Ox, Oy:

sin 0msnP F (2)cos 0N P (3)

- T (2) ta c sinmsnF P . vtnm yn khng b trt trn dc th lcma st ngh lc ma st trt, nn:

cosmsnF kN kP 0sin cos 0,2 11P kP tg k

- Vy vt nm yn trn mt dc th 011 .b.

Tnh thi gian trt v vn tc chn dc khi 030 - Vi 030 , vt s trt trn mt dc, ta c:

msP N F ma (4)- Chiu (4) ln Ox, Oy

sin msP F ma (5)

cos 0N P (6)- T (5) v (6) ta tm c:

0 0 2(sin cos )

9,8 sin30 0,2. os30 3,2 m/s

a g k

c

- Thi gian vt trt ht on dc 165s m (ch rng vntc ban u bng 0):

21 2 10,16 s2

ss at ta

- Vn tc ca vt chn dc:3,2.10,16 32,5v at m/s

V d 3Mt cht im c khi lng m c nm ln t mt im O

trn mt t, vi vn tc ban u v0theo hng nghing mt gc vi mt phng ngang. Xc nh m men ng lng ca cht im

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27

i vi im O, ti thi im cht im t cao cc i. (pdng vi 0

0100 ; =30 ; v 25m sm g ).

Li gii

- Chn h trc to xOy, gcthi gian l lc nm.- Phng trnh chuyn ng:

2

0

cos

1sin

2

ox v t

y v t gt

- Ti cao cc i A ta c 0yv

00 1

sinsin 0y

vv v gt t t

g

- Thay vo phng trnh chuyn ng ta c2

0max 1

sin( )

2

vh y t

g

- Mmen ng lng ca cht im ti A i vi O:

A AL OA mv r mv sin( , )A AL mv r r v

- Mt khc ta c: maxsin , Ar r v h , A x oxv v v nn:2 2 3 2

0 0ax 0

sin sin coscos

2 2m x

v mvL h mv mv

g g

- Thay s ta c 28,18L kgm2/s

V d 4Mt vt khi lng m ng yn nh mt ci nm nh ma st.

Tm thi gian vt trt ht nm khi cho nm chuyn ng nhanhdn u sangtri vi gia tc l 0a . H s ma

st gia mt nm v vt m l k, chiu di mtnm l , gc nghing l v gia tc

0 cota g g .

r

oxv

oyv

O

y

xx

AAv

maxh

m

0a

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28

Li gii- Xt trong h quy chiu gn vi nm l h quy chiu chuynng tnh tin vi gia tc 0a , vt m s

chu thm lc qun tnh 0qtF ma .- p dng nh lut Newton cho vt mtrong h quy chiu chuyn ng c giatc gn vi nm, ta c phng trnhchuyn ng:

ms qt ma P N F F (1)

a l gia tc ca vt m trong h quy chiu gn vi nm- Chiu (1) ln hng chuyn ng ca vt:

sin osms qt ma P F F c (2)

- Chiu (1) ln hng vung gc vi hng chuyn ng cavt v chn chiu dng hng ln:

0 cos sinqtP N F (3)

- T (2) v (3) suy ra gia tc:

0sin cos cos sina g k a k - tm thi gian vt trt ht nm, p dng phng trnh:

21

2l a t

0

2 2

sin cos cos sin

l lt

a g k a k

Ch : vi 0 cota g g phn lc php tuyn 0N .

BI TP P DNG

2.1

Mt ngi di chuyn mt chic xe vi vn tc khng i.Lc u ngi y ko xe v pha trc, sau ngi y y xe v

pha sau. Trong c hai trng hp, cng xe hp vi mt phngngang mt gc . Hi trong trng hp no, ngi phi t ln

xe mt lc ln hn. Bit trng lng ca xe l P, h s ma st giabnh xe v mt ng l k.

0a

P

N

qtF

msF

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29

2.2 Mt dy xch c chiu di 1 m c t trn mt bn saocho mt phn ca n bung thng xung t c chiu di l l. Cho

bit h s ma st gia xch v bn l 1/ 3k . Tm chiu di l dy xch bt u trt trn mt bn.

2.3

Mt xe vn ti chy trn ng nm ngang vi vn tc khngi. Sau xe ln dc, nghing vi mt nm ngang mt gc

015 . Mun xe vn chuyn ng u vi vn tc nh c th lcko ca ng c phi ln gp bao nhiu ln so vi khichy trnng nm ngang. Bit h s ma st trong hai trng hp u l

0,05k .

2.4

Mt vt khi lng m c ko ivi vn tc khng i bi mt si dytrn mt mt phng nghing c gcnghing vi mt phng ngang. H sma st gia vt v mt phng nghing

bng k. Xc nh gc gia si dy v mt phng nghing chosc cng nh nht. Tnh gi tr sc cng .

2.5

Hai vt c khi lng 1 300 gm ; 2 480 gm c bucvo hai u mt si dy vt quamt rng rc c khi lng khngng k (hnh v ). Lc u, gi vt

1m di vt 2m mt khong

h = 2 m v trn vt m2c t vt 3 200 gm , sau th cho h vtchuyn ng. Xc nh:

a.

Gia tc ca cc vt v sc cng ca dy.

b. Sau bao lu vt1m v 2m cao nh

nhau.

c.

Lc tc dng ca vt 3m ln vt 2m khi h

chuyn ng. B qua khi lng dy v mast rng rc.

2.6 C hai vt khi lng 1m , 2m lin kt vi nhau bng mt sidy vt qua rng rc nh ca mt phn nghing hp vi mt

m

hm1

m2

m3

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30

ngang mt gc . Vt1m nm trn phng nghing. H s ma st

gia m1v mt nghing l k. Gi thit lc u hai vt ng yn.

a. Vi iu kin no ca t s cc khi lng (

2 1/m m ) cho

vt 2m : i xung; i ln; ng yn.b. Xc nh gia tc ca h vt trong hai

trng hp u. B qua khi lngrng rc v dy, ma st rng rckhng c.

2.7

Trn mt ci bn c khi lng M, t mt h gm ba vt ckhi lng: m, 2m, 3m c lin kt vi nhau bng cc si dy. H

s ma st gia vt 2m v bn l0,1k . Hi h s ma st gia bn v

mt sn phi c gi tr nh nht bngbao nhiu bn ng yn khi h vtchuyn ng. B qua khi lng dyv rng rc, ma st cc rng rc lkhng ng k.

2.8

Trn mt nghing hp vi mt ngang mt gc 030 c thai vt tip gip nhau khi lng ln lt l

1 1m kg, 2 2m kg.H s ma st gia cc vt v mt nghing lnlt

1 0,25k ; 2 0,1k .

a. Xc nh lc tng tc gia hai vtkhi chuyn ng.

b.

Gc nghing phi c gi tr nhnht bng bao nhiu cho cc vt c th trt xung.

2.9 Mt chic xe c khi lng 20M kgc th chuyn ng khng ma st trn mt

phng ngang. Trn xe t mt hn khilng 5m kg. H s ma st gia v xe l

0,2k . Ngi ta t ln xe mt lc F c phng nm ngang v

hng dc theo xe. Hi:

m1

m2

m2

m1

m

MF

2m

m

3m

M

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31

a.

Mun hn khng trt trn xe khi xe chuyn ng th lcF ch c th c gi tr ln nht bng bao nhiu?

b.

Nu lc 60F N hn v xe s chuyn ng th no?Xc nh gia tc ca hn v xe i vi mt t.

2.10

Mt vin n khi lng 10g chuyn ng vi vn tc

0 200 m sv xuyn thng vo mt tm g v chui su vo mton . Bit thi gian chuyn ng ca n trong tm g l

44.10 st . Xc nh lc cn trung bnh ca g v xuynca vin n.

2.11 Mt phn t kh c khi lng 234,65.10 gm chuyn

ng vi vn tc 160 m sv ti va chm n hi vo thnh bnhvi gc nghing 060 so vi php tuyn ca thnh bnh. Tnhxung lng ca lc va chm ca phn t kh ln thnh bnh.

2.12 Mt vt c khi lng 1m kg chuyn ng thng trn mtsn ngang theo phng x, vi vn tc ban u 0 10 m sv v chu

lc cn cF rv , vi 1 kg sr l h s cn v v l vn tc

chuyn ng ca vt.a.

Chng minh rng vn tc ca vt gim dn theo hm s bcnht ca qung ng i.

b.

Tnh qung ng i c ti lc dng.

2.13

Mt vin n c khi lng 10m gbay theo phng ngangtrong khng kh vi vn tc ban u 0 500 m sv . Cho bit lc

cn ca khng kh t l v ngc chiu vi vn tc v ca vin n:cF rv vi

33,5.10 kg sr l h s cn ca khng kh. B quanh hng ca trng lc. Hy xc nh:

a.

Khong thi gian vn tc vin n bng mt na vntc ban u 0v .

b.

Qung ng vin n bay c theo phng ngang trongkhong thi gian trn.

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32

2.14 Mt vt khi lng m chuyn ng t di ln theo phngthng ng Ox vi vn tc ban u 0v . Bit lc cn ca khng kh

t l vi bnh phng vn tc: 2F rv (r l h s t l). Tnh

cao cc i vt ln c v thi gian vt ln n cao cc i.2.15 Mt cht im khi lng m c nm ln t mt im Otrn mt t vi vn tc ban u v0theo hng nghing mt gc vi mt phng ngang. B qua sc cn khng kh. Hy xc nh timt thi im t v i vi im O.

a.

M men ngoi lc tc dng ln cht im.b.

M men ng lng ca cht im.

2.16

Trn trn mt thang my ang i ln vi gia tc2

0 1,2 m sa c gn mt lc k. u di lc k c treo mt rng

rc, ngi ta vt qua rng rc mt si dy v hai u dy treo haivt khi lng ln lt l

1m 200g , 2m 300g . B qua khilng v ma st rng rc, dy khng gin v c khi lngkhng ng kng k. Xc nh:

a.

Gia tc ca vt1

mso vi t v vi thang my.

b.

S ch trn lc k.

2.17 Cho hvtnhhnh vbn. Cn phi dch chuyn mt chicxe theo phng ngang vi gia tc nh nht bng bao nhiu chocc vt m1 v m2 khng chuyn ng ivi xe. Cho khi lng cc vt 1 300 gm ,

2 500m g; h s ma st gia vt 1 2,m m

v xe l 0,2k . B qua khi lng rngrc v dy ni, ma st rng rc khngng k.

2.18 Hi tu ho phi c vn tc bng bao nhiu khi chy qua mton ng vng c bn knh 98R m si dy treo qu cu

buc vo trn toa tu lch so vi phng thng ng mt gc045 . Xc nh sc cng ca dy, bit khi lng qu cu l

500m g.

m2

m1

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33

2.19

Mt vt nh khi lng 1m kg c t trn mt a phngngang v cch trc quay ca a mt khong 0,5r m. H s mast gia vt v a bng 0,25k . Hi:

a.

Lc ma st phi c ln bng bao nhiu vt gi trna, nu a quay vi vn tc 12n vng/pht.b.

Vi vn tc gc no ca a th vt bt u trt khi a.

2.20 Mt my bay thc hin mt vng nho ln c bn knh 400mtrong mt phng thng ng vi vn tc 540km/h.

a. Xc nh lc nn ca phi cng ln gh my bay im caonht v thp nht ca vng nho ln, nu khi lngca phi

cng bng 60kg.b.

Mun cho ngi li trng thi khng trng lng ti imcao nht ca vng nho ln th vn tc ca my bay phi

bng bao nhiu ?

2.21 Mt qu cu khi lng 500m g c treo vo u mt sidy di 50 cm. Qu cu quay trong mt phng nm ngang vivn tc khng i sao cho si dy vch mt mt nn. Cho bit gc

to bi si dy v phng thng ng l 030 . Xc nh lccng dy, vn tc di v vn tc gc ca qu cu.

HNG DN - LI GII - P S

2.1

Trng hp y xe v pha sauphi dng lc ln hn.

- Trng hp ko xe v pha trc,

mun xe chuyn ng u phi c:1 msF F , msF kN v

2 sinN P F P F .

Hay: cos sinF k P F

cos sin

kPF

k

- Trng hp y xe v pha sau, xechu trng lc P, lc y F , phn lc

F2F

1F

N

PmsF

,F

,

2F

,

1F

N

P

,

msF

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34

php tuyn 'N v lc ma st ,msF bng cch phn tch tnh

tng t:

os ksin

kPF

c

- So snhFvF'thy ,F F , suy ra trng hp y xe v phasau phi dng lc ln hn.

2.2 ' 0, 25m - Gi P1 l trng lng ca phn bung thng; P2 l trnglng ca phn dy xch nm trn mt bn. Mun dy bt utrt phi c 1 msP F ; trong 2msF kP (ch : trng lng

ca cc phn dy xch t l vi chiu di), t suy ra:

' 0,25 m1

k

k

2.3 p s 2

1

6,1F

F ln.

- Gi F1, F2l lc ko ca ng c trn ng ngang v ng

dc, mun xe chuyn ng u th tng hp lc tc dng lnt gm: lc ko ca ng c, trng lcP, phn lc php

tuyn N ca mt ng v lc ma st ca mt ng msF phi

bng khng.

0msF P N F - Chiu phng trnh ln hng chuyn ng c biu thc:

F1, F2v suy ra t s 21

F

F

2.4 p s:

min2

sin cos;

1

mg ktg k T

k

- Lc tng hp tc dng ln vt:

0msF T P N F (1)

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35

- Chiu (1) ln hng chuyn ng v ln phng vung gcvi hng chuyn ng, ta c:

os - Psin 0msTc F (2)

sin os 0T N Pc (3)os -TsinN Pc

cos sinmsF kN k P T - Thay vo (2) rt ra:

sin cos

cos sin

mg kT

k

- Theo bt ng thc Bunhiacpxki-Csi:

2 2 2 2

cos sin 1 cos sink k const

- Vy: 2

cos sink ln nht khi:1 os

sin

ctg k

k

v 2max

cos sin 1k k suy ra

min

2

sin cos

1

mg kT

k

2.5 p s:

a.

23,8 , T = 4,08 Na m s

b.

0,73t s

c. 32 1,2F N

Hng dn:a.

Vit phng trnh ca inh lut II Niutn cho tng vt:

1 1 1P T m a

2 3 2 23m m a P T

- Chiu ln phng trnh chuyn ng ca h:

1 1m a P T (1)

2 3 2 3m m a P P T (2)(T T v a: gia tc ca h )

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36

T (1) v (2) gii ra: a v T.

c. Tnh 32 F . Xt ring vt 3m , ta c phng trnh:

3 2 3 23m a P F

( 23F l lc do 2m tc dng ln 3m ) hay:

3 3 23 23 3m a P F F m g a v 32 23F F

2.6

p s:

a.

2m i xung2

1

sin osm

kcm

2m i ln 21

sin osm

kcm

2m ng yn2

1

sin os sin osm

kc kcm

b.

Trng hp 1 2 1

1 2

sin os

m

m m kca g

m

Trng hp 2 1 2

1 2

sin os

m

m kc ma g

m

2.7 p s: min0,6

5,4

mk

M m

- Khi bn ng yn, h vt chuyn ng, ta tm c gia tcca h 3 vt: a = 0,3g (g l gia tc trng trng) v lc cngca dy ni: gia vt 3m v 2m l 1 2,1mgT ; gia vt 2m vm l

2 1,3mgT . Sau xt ring bn, xc nh tng hp ccngoi lc tc dng ln bn v xt thnh phn hnh chiu theo

phng ngang ca tng hp lc. Mun bn ng yn phi ciu kin thnh phn hnh chiu ngang pnF ca lc tng hp:

pn ms MF F

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37

Trong ms MF l lc ma st trt gia bn v mt sn, t

s suy ra h s ma st kmin.

2.8

p s:

a.

1 2 1 212 21

1 2

cos0,9

k k m mF F

m m

N

b.

1 1 2 2min

1 2

k m k mtg

m m

hay 0 'min 8 30

Hng dn:a.

Xc nh lc tng tc gia hai vt

- V h s mast 1 2k k , nn hai vt s cng trt xung trnmt nghing; p dng nh lut II Niutn cho h 2 vt tm cgia tc a ca h. Sau xt ring tng vt

1m (hoc 2m ), s tnh

c lc tng tc21F (hoc 12F ).

b.

Xc nh gcmin

- Mun h hai vt trt xung, cc thnh phn lc 1tP , 2tP ca

trng lc 1P, 2P theo hng song song mt nghing tho mniu kin: 1 2 1 2t t ms msP P F F (1)

1 ms2, FmsF l lc ma st tc dng ln 1m , 2m ca mt phng

nghing.

1 1 2 2

1 1 1 2 2 2

sin ; sin

cos ; cos

t t

ms ms

P m g P m g

F k m g F k m g

- Thay vo (1):1 1 2 2

1 2

k m k mtg

m m

suy ra

mintg v min

2.9

p s:

a. max 49F N

b.

2 2

1 21,96 ; 2,51

da xea m s a m s

Hng dn:

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38

a.

Xc nh lc cc imaxF

- V hn khng trt trn xe, nn h ( + xe) cng chuyn

ng vi gia tcF

a

m M

(1). V lc do xe tc dng ln

khi chuyn ng (xt ring hn ): daF ma (2) chnh l lc

ma st ngh, mun khng trt trn xe: msma F (lc ma

st trt). msF kmg (l lc ma st ngh cc i).

- T (1) v (2) suy ra: F kg m M thay s c 49F N.

Vy max 49F N.

b.

Xc nh gia tc ca hn - Vi 60 N 49 NF hn s trt trn xe. Gi 1a , 2a lgia tc ca hn v xe.- Vi hn 1 1 msm a F (1)

- Vi xe 2 msMa F F (2)

- T (1) 21 1,96m smsFa kg

m

- T (2) 22 2,51 m smsF Fa

M

2.10

p s: 500tbF N, 4 cm.Hng dn:

- p dng nh l ng lng: .t bmv F t suy ra:

500 Ntb mvFt ; xem chuyn ng ca n trong g l

chm dn u: 2 2v al (gia tc 0tbF

am

).

2.11 p s: 24. 7, 44.10 NsF t Hng dn:

msF m

MF '

msF

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39

- Gi 'v l vn tc ca phn t kh bt ra sau khi va chm vothnh bnh; va chm l n hi v v , p dng nh l nglng:

mv mv F t (1)(F lc so thnh bnh tc dng ln phn t khtrong thi gian va chm t ).- Chiu phng trnh (1) ln phng phptuyn ca thnh bnh, chiu (+) hng ra, tac:

cos cos .

2 cos .

mv mv F t

mv F t

- Lc do phnt kh tc dng ln thnh bnh 'F F suy raxung lng ca lc do phn t kh va chm ln thnh bnh:

-242 cos 7,44.10F t mv Ns

2.12

p s:

a. 0

rv v x

m

b. 0 10 m

ms v

r

Hng dn:a.

Theo nh lut II Newton

c

dvma F m rv

dt

;mdv rvdt dx vdt (phng trnh chuyn ng cavt l phng x)r

mdv rdx dv dxm

(1)

- Ly tch phn 2 v ca (1):

0

0

0

v x

v

r rdv dx v v x

m m (2)

b.

Cho 0v suy ra qung ng i n lc dng, t (2) c:

m

v

'v

'v v v

O

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40

0

0

0

10 m

rv x

m

ms x v

r

2.13 p s:

a.

ln 2 1,98 sm

r

b. /0 1 714 mrt m

mvx e

r

Hng dn:a.

Theo nh lut II Newton ta c:dv dv r

m rv dt dt v m

(1)

- Tch phn hai v phng trnh (1) ta c:

0

0

00

ln

v t rt

m

v

dv r v r dt t v v e

v m v m

(2)

- Thay 02

vv vo (2) tm c ln 2 1,98 sm

r

b. Gi x l qung ng i c theo phng ngang

- T (2) ta c: 0 0r r

t tm m

dxv v e dx v e dt

dt

(3)

- Tch phn hai v phng trnh (3) ta c:

00

0 0

1

x t r r

t tm mmvdx v e dt x er

(4)

- Thay 1,98t s vo (4) ta c 714 mx

2.14 p s: 0arctg vm r

rg mg

;

2

0ln 12

rvmh

r mg

Hng dn:

- Theo nh lut II Niutn:

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41

2

2

dv dv r m mg rv dt

mgdt mv

r

- tmg

r ta c 2

dv rdt

v m (1)

- Ly tch phn t 0v n v ng vi thi gian 0t n t tac:

0varctg r

v tg t m

(2)

- Khi vt ln ti cao cc i vn tc 0v .0 0v varctg 0 arctg

r mt

m r

- Thaymg

r ta c thi gian chuyn ng ca vt:

0arctg vm r

rg mg

- cao cc i:2

0 0

0

ln cos arctg ln 12

tv rvm m

h vdt r r mg

2.15 p s:

a. 0 osM mg v c t

b. 20

1os

2L mg v c t

Hng dn:- Xem bi mu (V d 2); chn h to xOy; im O l gc;trc Ox nm ngang, Oy thng ng; gc thi gian l lc btu nm. Ta c phng trnh chuyn ng:

0osx v c t (1)

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42

2

0 sin2

gty v t (2)

- Ngoi ra ta c:

0 cosxv v (3)0 sinyv v gt (4)

a.

M men ca trng lc P mg i vi im gc O:

M r P ( r l bn knh vc t xc nh v tr chtim i vi im gc O).- T suy ra: 0 osM Px mg v c t

b.

M men ng lng ca cht im m vi im gc O:L r mv

- T suy ra ln ca mmen ng lng

y xL m xv yv - Thay (1), (2), (3), (4) vo ta tnh c:

201

os

2

L mg v c t

2.16 p s:

a.

2 2

1 3,4 m/s ; ' 2, 3 m/sa a

b.

5,28F NHng dn:

- Chn h quy chiu O gn vi t (hquy chiu qun tnh), h quy chiu O'gn vi thang my (h quy chiu phiqun tnh); trong h Occ vt chuthm lc qun tnh:

1 1 0 2 2 0;qt qt F m a F m a

- Gi ,1a ;,

2a l gia tc cc vt m1, m2 trong h O. Ta c' '

1 2a a a .

1qtF 2qtF

'T

F

'T T T

1P2P

0a

'

1a '

2a

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43

- Phng trnh ng lc hc ca cc vt m1, m2 trong h O(chiu theo hng chuyn ng ca mi vt):

1 1 1 0m a T m g m a (1)

2 2 2 0m a m g m a T

(2)- T (1) v (2) gii ra:

2 1 0

1 2

2,2m m g a

am m

m/s2.

- Suy ra gia tc ca vt1m vi mt t:

2 1 2 01 0 1

1 2

23,4

m m g m aa a a

m m

m/s2.

- S ch lc k:2F T , t (1): 1 0( ) 2,64 NT m g a a v

2 5,28T T F T N

2.17 p s: Dch xe sang phi vi gia tc 20 10,78 m sa

Hng dn:- C th xem bi mu (vd 4). Khi xe ng yn,h 2 vt c xu hng vt

2m i xung, mun h 2vt khng chuyn ngi vi xe, phi dch xesang phi vi gia tc 0a .

Xt trong h quy chiu

gn vi xe l h quychiu khng qun tnh (h O), cn h quy chiu gn vi mtt l h tuyt i (h O). Trong h O, cc vt m 1, m2chu

thm lc qun tnh: 1 1 0 2 2 0;qt qt F m a F m a . Mun hai vt

m1, m2ng yn trn xe phi c iu kin lin h gia cc lc:

2 1 2 1ms ms qt P F F F (1)

V 2 2 2 0ms qt F kF km a (lc ma st gia 2m v xe);

1 1msF kN hay 2 1msF km g (lc ma st trt gia 1m v xe ).

1qt

F

m2

m1

2N

2P

2qtF

2msF

1N

1P

1msF

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44

- Vy (1) c th vit thnh: 2 1 2 0 1 0m g km g km a m a

2 10 1 2 2 1 01 2

m kma m km m km g a g

m km

- Suy ra gia tc nh nht: 22 1

0 min

1 2

10,87m sm kma gm km

2.18 p s: 31 m/s 111,6 km/hv Rgtg , 6,93T NHng dn:

- Xt trong h quy chiu gn vi toa tu, qu cu chu thm lc

qun tnh li tm: qt ht F ma ; hta l gia tc hng tm khi tu

i qua ng vng, c 2 /hta v R 2

qt

vF m

R

- T rt ra:2v Rgtg

30,99 31 m/sv Rgtg

- Lc cng ca dy theo hnh v: T f , khi vt trng thi

cn bng tng i: 6,93 Ncos cos

P mgT

2.19

p s:

a.

0,79 NmsF

b. 2,2 rads

Hng dn:a.

Xc nh lc ma st- H quy chiu O' gn via quay l h quy chiu phi qun tnh, h quy chiu O gn vimt t l h qun tnh.

- Trong h O, vt m chu cc lc: P, phn lc php tuyn N ,

lc ma st msF v lc qun tnh li tm qt ht F ma ; hta l gia

P

qtF

T

f

P

N

qtF m

ms

F

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45

tc hng tm: 2hta r ( vn tc gc quay ca a;

2 r ). Phng trnh ng lc hc ca vt m trong h O':

ms qt ma P N F F (1)

- Chiu (1) xung phng ngang v phng ng:ms ht F ma ma (2)

0P N (3)- V m ng yn trong h O nn 0a , 0ms ht F ma .

22 2ms ht F ma m r m n r

- Thay s vo ta c: 0,79 NmsF

b.

Vt mun trt trn a, phi c iu kin:

qt ms truotF F

2m r kmg ( msF kN kP )kg

r

- Vt s bt u ri khi a khi 2,2 rad/skg

r .

2.20 p s:a.

2.787 N; 3.963 N

b. 62,61m sv

Hngdn:

a.

Ti im cao nht phi cng chu P;

phn lc gh 1

N .

- Lc hng tm 1htF P N , chiu

ln chiu ca lc htF 2

1ht

vF P N m

R (1)

- Lc nn:2

1 1 2,787 Nv

N N m mg

R

P

P

1N

2N

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46

- Ti im thp nht:2

2ht

vF P N m

R

-Lc nn:3

2 2 3,963 Nv

N N m mg

R

b. Ti im cao nht, phi cng trng thi khng trng lng

th1 0N .

- T (1) suy ra: 62,61m sv Rg

2.21 p s

5,66 N; 1,19 m/s; 4,76 rad/sT v

Hng dn:- Qu cu chu tc dng ca trng lc

P mg v lc cng T , v quay trntrong mt phng nm ngang vi vn

tc khng i nn: htF P T

- Lc hng tm:2

ht

vF m

R

( sinR l bn knh qu o trn).

- Lc cng: 5,66 Nos os

P mgT

c c

2

1,19 m/shtF mv

tg v Rgtg P mgR

- Vn tc gc: 4,76 rad/sv

R

P

T

htF

O R

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47

Chng 3NG LC HC VT RN

TM TT L THUYT

1.

Khi tm ca h cht im- Khi tm G ca h cht im Mi

n

i i

i 1

m M G 0

- To khi tm G trong h to Descartes

n

G i i

i 1

1r m r

m

; vin

i

i 1

m m

- Vn tc khi tm

n nG i

G i i i

i 1 i 1

dr 1 dr 1v m m v

dt m dt m

- Phng trnh chuyn ng ca khi tmn

i G

i 1

F ma

vi2

GG 2

d ra

dt

2.

ng lng- ng lng ca mt h cht im

n n

i i i

i 1 i 1

p p m v

- nh lut bo ton ng lng: Tng ng lng ca mt h

c lp c bo ton: p const

- Bo ton ng lng theo phng: x xF 0 p const - Cng thc Xincpxki cho vn tc tn la

0mv u lnm

3. Chuyn ng ca vt rn- Mmen lc (tip tuyn)

tM r F ; t t tM r.F .sin(r,F ) r.F

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48

- Phng trnh cbn ca chuyn ng quay ca vt rn quanhmt trc c nh

M I.

4.

Mmen qun tnh- Mmen qun tnh ca mtsvtng cht, tit dinuivitrccan

- Thanh st: 21

I m12

- a trn (hoc tr c): 21

I mR2

- Vnh trn (hoc tr rng): 2I mR - Qu cu c: 2

2I mR

5

- nh l Huyghen - Stener: 20I I md

5.

Mmen ng lng- Mmen ng lng ca mt h cht im

n n

i i i

i 1 i 1

L L r mv

- Mmen ng lng ca vt rn quay quanh trc c nh

L I

- nh l v mmen ng lng:dL

Mdt

- nh lut bo ton mmen ng lngdL

M 0 0 L constdt

BI TPV D

V d1

Mt hnh tr c khi lng1m 3 kg c th quay xung quanh

mt trc nm ngang trng vi trc ca n. Trn hnh tr c cunmt si dy mnh, mm, khng gin, khi lng khng ng k.

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49

u t do ca si dy c treo mt vt c khi lng

2m 500 g. Th cho vt m2ri thng ng. Hy tmgia tc ca m2v sc cng ca dy.Bqua sccnkhng kh, bqua ma st trcquay cahnh tr,ly g 10 m/s2.

Li gii

- Chuynngcahbao gmchuyn ng tnh tin cam2v chuyn ngquay cam1. Ta c:

2 2 2P T m a (1)

/O 1M (T ) I (2)

- Chiu phng trnh (1), (2) ln cc chiuchuynngtng ng, ta c:

2 2 2P T m a (3)M I (4)

- y ta c:

1M TR ;a

R

;1 2T T T ;

2

1

1I m R

2

(5)

- Kthp(5) viphng trnh (3), (4) ta cgia tcchuynngcam2v sccng ca dy:

22

1 2

2m 2.0,5a g 10 2,5m s

m 2m 3 2.0,5/

1

1 3.2,5T m a 3,75N

2 2

V d 2

Mtnging thngng giaghGiucpxki sao cho gicatrnglctc dng ln ngitrng vitrcquay cagh. Haitay cangi ang dang ra v cmhai qutging nhau, miqu c khi lng m 2kg , hai qu t cch thn ngi v

khong cch giahai qu t l 1 160cm . Cho hngiv ghquay u vivntcgc

13,14rad s/ .

m2

m1

m2

m1

2P

2T

1T

O

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50

Khi hang quay th ngi ny huhai tay xung sao chokhong cch giahai qu t ch cn l

2 60cm . Hy xc nhvntcgc quay

2 ca h.

Cho bit mmen qun tnh cangiv gh(khng kcc qut) ivi trcquay l 20I 2,5kgm . Bqua sccnkhng kh,

coi kch thccacc qutl rt nh.

Li gii

- Mmen ngoilctc dngln ngiv ghGiucpxki trongtrng hp ny trit tiu, do mmen ng lng ca hcbo ton. Ta c:

1 2L L 1 1 2 2I I 1

2 1

2

I

I

- Khi ngi dang tay v khi ngihtay xung, mmen quntnh cahi vitrcquay cxc nh:

2

11 0I I 2m

2

;2

22 0I I 2m

2

- Vyta c: 2 21

0

2 12 2

20

1,6I 2m 2,5 2.2.

2 23,14 5,5rad s

0,6I 2m 2,5 2.2.

2 2

/

BI TPP DNG3.1.

Mtcon ln hnh tr bn knh R ct gia hai ng bng song song. Ccng bng chuyn ngv cng mt phavi vn tc ln lt l

1v v 2v . Bit conln ln khng trt. Xc nh vn tc gcquay quanh trcv vntcchuynngtnhtincacon ln.

2v

1v

R

O

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51

3.2.

Mtt ang chuyn ngvivntc 0v 72 km/h th hm

phanh v dng li sau mt khong thi gian t 5 s, coi chuynngcat ktkhi hm phanh l chmdn u.

a.

Hibnh t quay thm cbao nhiu vng ktlc hmphanh. Bitbnh t c bn knh R 35 cm.b.

Tnh gia tcgc cabnh xe.

3.3.

Trn mt a trn mng, phng, ng cht bn knh R ckhot mt l trn nh bn knh r, tm Oca l khot cch tmOca a trn mt khong R / 2 . Xc nh v tr khi tm ca a.

3.4. Mt hnh tr rng c khi lng m 50 kg, bn knh

R 50 cm ang quay quanh trc ca n vi vn tc gc600 vng/pht th bmtlchm tiptuynvimttrv vunggc vitrcquay. Sau 1 pht th trdngli.

a.

Tnh cng calchm.b.

Xc nh mmen lchm v lncalchm tiptuyn.

3.5. Trn mt hnh tr rng khi lng m 1kg ngita cun mt si dy mm, nh, khi lng khng ngk. u t do ca si dy c gn vo mt gi c nh.Th cho hnh tr ri thng di tc dng ca trng lc.Tm gia tc ri ca hnh tr v lc cng ca si dy. Bqua sccn khng kh, ly g 10 m/s2.

3.6. Hai vt c khi lng 1m 300g v 2m 100g c ni vinhau bng mt si dy mm, mnh, khng gin, khi lng khng

ng kv c vt qua mt rng rc c nh.Rng rc l mt hnh tr c c khi lngm 200g . Th nh cho h chuyn ng. Bqua mi ma st, dy khng trt trn rng rc.Ly g 10 m/s2. Hy xc nh gia tc chuynng ca h v lc cng ca dy treo hainhnh.

m

m2

m

m1

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52

3.7. Cho hai vt khi lng

1 2m m 1 kg c ni vi nhaubng mt si dy mm, khng gin,khi lng khng ng k v vt quamt rng rc. Rng rc l mt hnhtr c c khi lng m 1 kg. Mast gia mt bn nm ngang v m1c h s k 0,2 . Tnh gia tcchuyn ng ca h v lc cng ca dy hai nhnh.Bqua sccn khng kh, bqua ma st rng rc, ly g 10 m/s2.

3.8. Mt cun ch c khi lng m c t trn mt bn nm

ngang. Vnh ngoi cun ch c bn knh R, phn cun ch c bnknh r. Mmen qun tnh ca cun ch i vi trc ca n l I0.

Ngi ta cm u bung ca si ch ko cun ch mt lc F hp vi phng ngang mt gc .

a.

Tm iu kin v gc cun chchuyn ng v pha trc (pha hngtheo lc ko).

b.

Cho h s ma st gia cun ch v mtbn l k. Hy xc nh iu kin v

ln ca lc F cun ch khng trt.

3.9.

Cho mt rng rc c nh l mt hnh tr c c khi lng

1m 200g , trn hnh tr c cun mt si dy mm, khng gin,khi lng khng ng k. u t do ca si dy cni vi mtvt c khi lng 2m 500g , vt c t

trn mt mt phng nghing gc 045 .Ma st gia m2v mt phng nghing c hs k 0,1 . Thnhm2cho h chuyn ng,ly g 10 m/s2.

a.

Tnh gia tc chuyn ng ca vt m2.b.

Tm qung ng m2 i c sau 2

giy k t khibtu chuyn ng.

1

m2

r

R

F

mm1

m2

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53

3.10.Cho hai hnh tr c, ng cht ging ht nhau,khi lng mi hnh tr l m 2kg . Trn hai hnh trngi ta cun cun mt cch i xng hai si dy nh.Hnh tr pha trn c trc quay c nh. Khi th cho hchuyn ng, hnh tr pha di lun nm ngang. Hytnh gia tc chuyn ng ca h v lc cng ca mi dytreo. Ly g 10 m/s2.

3.11.Tnh mmen nglngcaTri t ivitrcquay can. Coi Tri t l mt hnh cu c ng cht c bn knhR 6400 km v khi lng 24M 6.10 kg, chu k quay quanh trccaTri t l T 24 h.

3.12.Mtatrn ngcht c khi lngm v bn knh R c thquay xung quanh mt trcnmngang vung gcviav cch tm amtkhong R / 2 . Banu giavtr sao cho tm acao nht, sau thnhcho aquay khng vntcu. Hyxc nhvntcgc v mmen ng lngca

ai vitrcquay khi ai qua vtr m tma thp nht.

3.13.Hy tnh ngnng ton phnca cc vtsau:a.

Mtacng cht khi lng m 2 kg ln khng trttrn mtphng nmngang vi vntc v 4 m/s.

b.

Mtqu cucng cht khi lng m 250 g, bn knhR 6cm ln khng trttrn mt bn nmngang vivntc

gc 20 rad/s, trong qu trnh chuyn ng trcquay can c phng khng i(song phng)

3.14.Mt thanh cng c chiu di 50cm v khi lng

1m 200g c th quay t do xung quanh mt trc nm ngang iqua u trn ca thanhv vng gc vithanh. Khi thanh ang nmng yn v tr cn bng bn, mt vin n c khi lng

2m 50g bay theo phng ngang vung gc vi trc quay ca

1

2

OR

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54

thanh vi vn tc v 100 m/s xuyn v mc vo u di cathanh. Hy tm vn tc gc ca thanh ngay sau va chm.

3.15.Mthcht imc tngnglngl p v mmen ng

lng L i vi mt cht im O. Hy xc nh mmen nglngcahiviimO. Khi no th mmen nglngcahcht imkhng phthucvo imO.

3.16.

Chng minh rng mmen ng lng L ca mt h chtimivi imO gnlinvihquy chiuK c thcho bi:

0 0L L r p

trong 0L l mmen ng lngcahcht imivikhi

tm ca n, 0r l bn knh vc t ca khi tm i vi im O

trong hK, p l tng nglngcah.

3.17.Mt chic bt ch c chiu di 20 cm c gi thngng, sau bung nhn xung mtbn nmngang, coirng trong qu trnh u bt ch khng b trt trn bn. Hy

xc nh vn tc gc ca bt ch ti thi im bt ch hp viphng thng ngmtgc . p dngtithi imbt ch nmngang. Ly g 10 m/s2.

3.18.

Tnh camtmtphngnghing c cao h 50 cm ccc vtkhc nhau cthln khng trtkhng vntcu. Lyg 10 m/s2. Tnh vn tc di ca cc vt ti chn mt phngnghing, nu:

a.

Vtcthl mttrc.b.

Vtcthl mttrrng.

3.19.Mt a trn ng cht c khi lng 1m 100 kg v bnknh R 1,5 m ang quay uvivntcgc / 3 rad/s quanhtrc can ctthng ng, trn ac mtngikhilng

2m 50 kg ng timp a.

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55

a.

Xc nhvn tcgc caakhi nginy i vo v ngtitm a(coi nginy nhmtcht im)

b.

Tnh cng thchinbi nginy khi di chuyntmp avo tm a.

3.20.

Mtqucu c, ng chtbn knh rbtu ln khng trt tnhmtbn cubn knh R. Hy xc nh v tr qu cu rimt bn cu v vn tc gc ca qu cu tithiim. Coi rng ( R r ).

3.21.Mt khu pho c khi lng 41m 10 kg c th chuyn

ngkhng ma st trn ng nmngang, khi chuynngkhupho mang mt qu n c khi lng2m 100kg , vin n c

th c bn ra khi nng vi vn tc u 500 m/s so vi nngpho. Hy xc nhvntccakhupho ngay sau khi bn trongcc trnghpsau:

a. Khi bn khupho ng yn, quncbntheo phngngang.

b.

Khi bn khu pho ng yn, qunc bn hng lntrn hpviphng ngang mtgc 060 .

c.

Khi bn khu pho chuyn ng vi vn tc 0v 18 km/h,quncbntheo phng ngang vpha trc.

3.22.Cho mt chic nm c dng mt tam gic vung cn, khilng 1m 5 kg ang nm yn trn mt sn nm ngang. Th ri

vo mt nm mtvtnhkhilng 2m 0,5 kgt cao h 1 m (so vi im chm mt nm).Sau va chm, vt m2 b bt ra theo phng nmngang. Hy xc nh vn tc chuyn ng canm ngay sau va chm. B qua mi ma st, lyg 10 m/s2, coi va chml n hi.

3.23.Mt ho tin lc u ng yn c khi lng tng cng l

0m 270 kg, sau phtkh un ra pha sau vivntckhng

h

m1

m2

R

m,r

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i u 300 m/s so vihotin, lngkh phtra trong migiy lm 90 g.a.

Sau bao lu hotintvntc1v 40 m/s.

b.

Khi khilngtngcngcahotincn lil 2m 90 kgth vntccahotinl bao nhiu?

HNG DN - LI GII - P S

3.1. Con ln v bng chuyn- Giv v ln ltl vntcchuynngtnh tinca conln v vntcquay cacon ln quanh trccan.

- Vn tc di ca mt im nm mp con ln i vi trcquay (htognvicon ln) l R v chnh bng vntccaimtipxc giacon ln v bng chuyn.- Do con ln khng trt, nn trong htognvimt tta c stng hpvntc:

1 t1 1 t1v v v v v v v R (1)

2 t2 2 t2v v v v v v v R (2)

- T(1) v (2) ta tnh c:

1 2

1 2

1v v v

21

v v2R

3.2. Xe t hm phanh chuyn ngchmdn uv dng li- Gia l gia tcchuynngcat khi hm phanh, c:

t 0v v 0 20a 4t 5

m/s2

- Qung ng xe i thm ccho n khi dng li:2 2

0

1 1s v t a t 20.5 4.5 50

2 2 m

a.

Svng m bnh t quay thm cs 50

n 22,742 R 2 .0,35

vng

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b.

Gia tcgc cabnh xe

ta a 4 11,43R R 0,35

rad/s2.

3.3.

Khi tm caaphngbkhot- Khi cha bkhot, khi tm ca atitm O l imtca

trnglc P ccoi l hplcca 1P (ttiO1) l trnglc

tc dngvo phn ckhot bi v 2P l trnglctc dngvo phna bkhot.

- Trng lc 2P c imt tikhi tm G, do tnh i xngnn G nmtrn ng OO

1, khc pha viO

1so viO.

- Theo quy tchplcsong song ta c:

2 2

1 1 1

OG P m

OO P m

- Do aphng ngcht nn khi lngtlvidintch:2 2

2 2

2 2 2 2

1 1

m S r r

m S R r R r

- Vy nn ta c:2 2

12 2 2 2

r RrOG OO

R r 2(R r )

3.4. Hnh trrngchulchm tiptuyn- Chuynngca trl chuyn ngquay chmdn u. Tac vntcgc ban u 0 600 vng/pht 20 rad/s

- Mmen qun tnh ca hnh tri vi trcquay:2 2I mR 50.0,5 12,5 kgm2.- Gia tcgc cachuynng

t 0 0 20

t 60 3

rad/s2.

a.

Cng calchm- Cng calchm bngbinthin ngnng cavt

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58

2

dt d0 0

1A W W W 0 I

2

2 31 12, 5.(20 ) 24, 67.10

2

J

b.

Mmen lchm v lnlchm- Theo phng trnh c bn ca vt rn quay quanh trc cnh, ta c:

M I 12,5.( ) 13,093

Nm

- ln lchm tiptuyn

t t M 13,09M RF F 26,18R 0,5 N

(Trong cc p strn, du - thhintnh cn).

3.5. Trrngc cun dy cthri- Chuynng ca tr gm chuyn ngquay quanh trc vchuynngtnh tincakhi tm. Trc quay tcthil tipimC casidy v hnh tr.

- Ta c:

/ C / C

P T ma

M (P) M (T) I

(1)

- Chiu(1) ln chiuchuyn ngtng ng, cP T ma (2)

/ C CM (P) I (3)

- Mt khc ta c 2 2C 0I I mR 2mR ; ta a

R R ;

/ CM (P) PR mgR . Kt hpvi(2), (3) ta c:

g 10a 5

2 2 m/s2.

T P ma m(g a) 1(10 5) 5 N

m

R

P

T

C

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3.6. Chuyn ng ca h gm 3 chuynng: chuyn ng tnh tin ca m1, m2 vchuyn ngquay carng rc.

- Ta cphng trnh:

1 1 1 1

2 2 2 2

/O 1 /O 2 O

P T m a

P T m a

M (T ) M (T ) I

(1)

- Chiu(1) ln trctota c

1 1 1 1P T m a (2)

2 2 2 2P T m a (3)

' '/ O 1 / O 2 OM (T ) M (T ) I (4)

- Ta c2mR

I2

; '1 1T T ;'

2 2T T ; 1 2a a a ;ta a

R R ;

' '

/O 1 1M (T ) RT ;' '

/ O 2 2M (T ) RT . Kt hp vi (2), (3), (4) ta

c:

1 2

1 2

m m 300 100a g 10 4

m m 0,5m 300 100 0,5.200

m/s2

1 1 1 1T P m a m (g a) 0,3(10 4) 1,8 N

2 2 2 2T P m a m (g a) 0,1(10 4) 1,4 N

3.7. Chuynng cahgm3 chuyn ng: chuyn ng tnhtin ca m1, m2 v chuyn ngquay carng rc., Ta c:

1 1 1 ms1 1 1

2 2 2 2' '

/O 1 / O 2 O

P N T F m aP T m a

M (T ) M (T ) I

- Chiuhtrn ln chiuchuynng camivt, ta c:

1 ms1 1 1T F m a (1)

2 2 2 2P T m a (2)' '

/ O 2 /O 1 OM (T ) M (T ) I (3)

m2

m

m1

O

1P 2P

1T

'

1

T '

2T

2T

x

O

m

m1

m2

1T

2P

1P

1N

2T

'

2T

'

1T ms1F

O

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60

- y ta c1 2a a a ,

'

1 1T T ;'

2 2T T ;ta a

R R ;

2

O

mRI

2

; ' '/O 1 1M (T ) RT ;' '

/O 2 2M (T ) RT ; 1 1P N . Kt hp

vi(1), (2), (3) ta c:

2 1

1 2

m km 1 0,2.1a 10 10 3,2

m m 0,5m 1 1 0,5.1

m/s2

1 1 1 1T kP m a m (kg a) 1(0,2.10 3,2) 5,2 N

2 2 2 2T P m a m (g a) 1(10 3,2) 6,8 N

3.8.

Cunchtrn mtbn nmngang- Khi cun ch ln khng trt trn bn, n c trc quay tcthi l c thng song song vi trc ca n v i qua tipimC cacunchvimtbn.- Mmen qun tnh cacunchi vitrcquay:

2

C 0I I mR

a.

Tm iukinvgc ko

- Phn tch lc F thnh hai lc:t nF F F , tng ng gy ra cc mmen

quay / O tM (F ) , /O nM (F )

- cunchchuynngvpha trc:

/ O t / O nM (F ) M (F )

2 2

Fcos (R r cos ) Fsin .r sin

FR cos Fr(cos sin ) 0rcos

R

- Vy cunchchuyn ngvpha trcthr

cosR

.

b. Cun ch chu tc dng ca cc lc msnP, N,F ,F . Ta c

phng trnh chuyn ng ca cun ch (va chuyn ng

tnh tin, vachuyn ngquay):

r

R

F

O

C

tF

nF

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61

msn

/ C C

P N F F ma

M (F) I

(1)

- Chiu (1) ln phng ngang vphng thng ng, ln chiu

chuynngtng ng, ta c:msnFcos F ma (2)

N Fsin P 0 (3)

/C t /C n CM (F ) M (F ) I (4)- y ta c a R (5)- T(4) ta c:

2

0 2

0

F(R cos r)

F(R cos r) (I mR ) I mR

(6)

- Thay (6) vo (2), kthpvi(5) ta c:

msn 2

0

F(R cos r)Fcos F mR

I mR

msn 2

0

mRF(R cos r)F Fcos

I mR

(7)

- cunchkhng btrt, ta phic iukin:msnF kN (8)

- Kthp(3), (7), (8) ta c:2

0

2

0

kmg(I mR )F

rI (cos k sin ) mR k sin

R

3.9.

Hon ton tng t, chuyn ngca h gm chuyn ng tnh tin cam2v chuyn ngquay cam1, ta c:

2 ms 2'

/ O O

P N T F m a

M (T ) I

(1)

- Chiu (1) ln chiu chuyn ngtng ng, ta c:

2 ms 2P sin F T m a (2)

m1

m2

N

2P

msF

T 'T

O

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62

2N P cos 0 (3)'

/O OM (T ) I (4)

- Ta c T T ' , a / R ,

2

0

mR

I 2 ,

'

/OM (T ) RT ' (5).a.

Gia tcchuyn ngcam2- Kthp(2), (3), (4), (5) ta c:

2

1 2

m (sin k cos )a g

0,5m m

0 0500(sin 45 0,1cos45 )10 5,3

0,5.200 500

m/s2

b.

Qung ng m2i csau 2 giy

- Ta c 2 21 1

s at 5,3.2 10,62 2

m

3.10.Do hai sidy i xng v ging ht nhau cho nn ta gischc 1 sidy.

- Hnh tr (1) ch c chuyn ng quay, hnh tr (2) va

chuynngtnhtin, vachuyn ngquay. Do chuynngc tnh tng inn c thcoi trcquay tcthi cahnh tr(2) l trccan.- Ta c:

2 2 2

/O1 1 0 1

/O2 2 0 2

P T ma

M (T ) I

M (T ) I

(1)

- Chiu(1) ln chiuchuyn ngtng ng:

2 2 2P T ma (2)

/O1 1 0 1M (T ) I (3)

/O2 2 0 2M (T ) I (4)

- y ta c2

0

mRI

2

, 1 2T T 2T ; 2 1 2a ( )R (5)

- T(3) v (4) vitcthta c 1 2 (6).

1

2

2P

2T 1T

O1

O2

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63

- T(2), (3), (4), (5), (6) ta c:4 4

a g 10 85 5

m/s2.

1

1 1 1T T mg 2.10 22 10 10 N

3.11.Mmen nglngcaTri t- Tri t l mtvtrnquay quanh trc, do mmen nglngivitrcquay cxc nh:

0L I - lnca mmen nglng:

2

02 2L I MR 5 T

24 6 2 332 26.10 (6,4.10 ) 7,15.105 24.60.60

kgm2/s

3.12.Mmen nglngcaaphng- Mmen qun tnh caai vitrcquay:

2

2 2 20 R 1 1 3I I m mR mR mR

2 2 4 4

- Gi l vn tc gc tc thi ti v tr thp nht, theo boton cnng ta c (khitm hthp mtonbng R):

2

2

1 2mgR 2mgR 8gmgR I

32 I 3R mR

4

- Mmen nglngtivtr thp nhtcxc nh:L I

23 8g 3L I mR mR gR 4 3R 2

3.13.

Tnh ngnng ton phn cacc vt- ngnng ton phncamtvtcxc nh:

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64

2 2

d dtt dq

1 1W W W mv I

2 2

a.

acng cht ln khng trt2

2 2 2 2

d1 1 1 1 1 vW mv I mv mR 2 2 2 2 2 R

2 23 3mv 2.4 244 4

J

b.

Qucu cng cht ln khng trt

2 2 2 2 2 2

d

1 1 1 1 2W mv I mR mR

2 2 2 2 5

2 2 2 27 7mR 0, 25.0,06 .20 0,25210 10

J

3.14.

Vin nxuyn vo thanh treo thng ng- Mmen qun tnh ca thanh i vitrcquay:

2

2 2 2

1 01 1 1 1 1

1 1I I m d m m m

12 2 3

- Ngay trcv sau va chmmmen ngoilctc dngln hivitrcquay bng0, do mmen nglngcahboton:- Ngay trcva chm: 1 2 2L L L m v - Ngay sau va chm, gi l vntcgc quay ca thanh:

' 2 2

1 2 1 2

1L I (I I ) m m

3

- Mmen nglngboton: 'L L

2 2 22 1 2

1 2

1 3m vm v m m

3 (m 3m )

3.0,05.100120

(0,2 0,05)0,5

rad/s

3.15.

Xt mtimO (cnhso viO).- Mmen nglngcahcht imi vi O:

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65

' ' ' '

i i i i

dL L r m r

dt

- M ta c 'i i i 0r O'O r r r vi 0r OO'

- Nn ta c:'

i 0 i i 0 i 0 i i

i i i 0 i i 0 i i

d dL (r r ) m (r r ) (r r ) m r

dt dtd d

r m r r m r L r m vdt dt

'

0L L r p - mmen ng lng ca h cht im khng ph thuc

vo imO, tcl 'L L vimi 0r th ta phic p 0 .

3.16.Hon ton tng tnhbi 3.19, ta c:- Mmen nglngi viimO cxc nh:

'

0L L r p

- Chnim O' G trng vikhitm. Khi ' G 0L L L l mmen ng lng ca h i vi khi tm ca n;

0r OO' OG ln bn knh vc tcakhi tm ktgcO.Vyr rng ta c:

0 0L L r p

3.17.Chic bt ch trn mtbn nmngang- Trong qu trnh , chuyn ngcabt ch l chuynngquay quanh trc c nh, trc quay nm ngang, i qua im

tipxc cabt ch vimtbn v vung gc vimtphngquoca bt ch.- Mmen qun tnh ca bt i vitrcquay:

2

2 2 2

0

1 1I I md m m m

12 2 3

- Gi l vn tc gc ca bt ch thi im bt hp viphng thng ng mtgc . Theo boton cnng, c:

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66

21mg (1 cos ) I

2 2

2

mg (1 cos ) mg (1 cos )

1I m3

3g(1 cos ) g6 sin

2

- p dng 01 90 0

1

106 sin 45 12,23

0,2 rad/s

3.18.

Vttrttrn mtphng nghing- p dngnhlutboton cnng, ta c:

2 21 1mgh mv I

2 2

a.

Vtl mt trc2

2 2

2

1 1 1 vmgh mv mR

2 2 2 R 3mgh mv

4

4 4v gh 10.0,5 2,56

3 3 m/s

b.

Vt l mttrrng

2

2 2

2

1 1 v

mgh mv mR 2 2 Rmgh mv

v gh 10.0,5 2,24 m/s

3.19.Ngi di chuyn trn a quay nm ngang, mmen nglngcahcboton.

a. Xc nhvntcgc quay

- Khi ngingoi mp a, mmen qun tnh ca hivitrcquay cxc nh:

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67

2 2 21 2D N 1 2

1 m 2mI I I m R m R R

2 2

- Khi ngii vo tm a, mmen qun tnh ca h:2

1 D N 1

1I I I m R 2

- Theo boton mmen nglng, ta c:1L L

1 1

2 21 21 1

I I1 m 2m

m R R2 2

1 21

1

m 2m 100 2.50 2

m 100 3 3

rad/s

b.

Cng thchinkhi ngidi chuyn- Cng thchinbng binthin ng nng cah:

2 2

d1 d 1 1

1 1A W W I I

2 2

2 2 2 21 1 21

m m 2mA R R

4 4

- Thay sta c A 123,37 J

3.20.

Qucu cln trn mtbn cu- Chngctnh thnng timtbn nm ngang. Xt tivtr

M caqucutrn mtbn cuc AOM .- Theo nhlutboton cnng:

2 21 1mg(R r)(1 cos ) mv I

2 2

- Ta c ngnng chuyn ngcaqucu:2

2 2 2 2 21 1 1 1 2 v 7mv I mv mr mv2 2 2 2 5 r 10

- V vyta c:27mg(R r)(1 cos ) mv

10 (1)

- Xt lctc dngln qucu theo phng hngtm, c:

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68

2vPcos N m

R r

(2)

- Rt mv2t(1) thay vo (2) ta c:

1N mg 17cos 107 (3)

- Qu cu cn ln trn mt cu th N 0 , vy thiim qucu bturikhimt cu th ta c N 0 .

1 10

N mg 17cos 10 0 cos7 17

(4)

- Thay (4) vo (1) ta xc nhcvntcdi caqucu:

10g(R r)v17

(5)

- Vyvntcgc caqucu tithiimqucubtu rikhimtcu:

2

v 10g(R r)

r 17r

3.21.

Khupho trn ngnmngang- H kn theo phng ngang, do ng lng ca h cbo ton theo phng ngang. Chn chiu dng l phngngang hng theo chiubn.

a.

Qunbntheo phng ngang khi pho ng yn- Giv l vntccakhupho ngay sau khi bn

1 20 m v m u

2

1

m 100v u 500 5m 10000 m/s

b.

Quncbnxin ln khi pho ng yn

1 20 m v m ucos

02

1

m 100v u cos 500cos 60 2,5

m 10000 m/s

c.

Quncbnngang khi pho chuynng- Vntccaqunso vitkhi bnra l

P

N

O

A

M

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69

0 0v ' u v v ' u v 500 5 505 m/s- Tng tta c:

1 2 0 1 2(m m )v m v m v'

1 2 0 2

1

(m m )v m v 'vm

(10000 100)5 100.5050

10000

m/s

- Trong cc p strn, du- thhin khupho bgitlingcvi hng bn.

3.22.ng lngcahcboton theo phng ngang.- Giv0l vntcca m2ngay sau khi btra khimtnm, vl vntcchuyn ngcanm sau va chm.- Theo boton nglng, ta c:

11 2 0 0

2

mm v m v 0 v v

m (1)

- Mtkhc, do va chmn hinn nng lngcahcbo ton:

2 2

1 2 0 2

1 1m v m v m gh

2 2

2

2 11 2 2

2

1 1 mm v m v m gh

2 2 m

2

1 1 2

mv 2gh

m (m m )

0,52.10.1 0,426

5(5 0,5)

m/s

3.23.

p dngcng thcXincpxki.a.

Git l thigian ktlcbt u chuyn ngnkhi tvntcv1.

- Khilngtithiim:

1 0m m t m

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- T ta c:

0 0 11

0 0

m m vv u ln exp

m t m m t m u

1

0

1

vexp 1m u

tvm

expu

40exp 1

270 300374,48

400,09 exp300

s

b.

Vntccahotin

02

2

m 270v u ln 300ln 329,58

m 90

m/s

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71

Chng 4NNG LNG

TM TT L THUYT

1. Cng ca lc F trong chuyn di CD bt k

S

CD CD

A FdS F dS

dSl vc t chuyn di nguyn t, SF l hnh chiu ca F

trn phng dS.

- Trng hp lc F khng i, chuyn di thngosA FSc

l gc hp bi lc F v phng chuyn di S.

2. Cng sut ca lc (hay ca my)dA

P Fvdt

v l vc t vn tc ca im t lc.

3.

ng nng ca cht im2

2d

mvW

nh l ng nng 2 1d d dW W W A

4.

Th nng ca cht im trong trng trng u

tW mgh h l cao ca cht im (so vi mt t)- nh l th nng: 2 1 1 2t t t t t W W W W W A A l cng ca lc trng trng

5. nh lut bo ton c nng trong trng trng2

2

mvW mgh const

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72

BI TP V D

V d 1

Mt xe t c khi lng 310m kg bt u chy trn ng

nm ngang. ng c sinh ra lc ln nht bng 10

3

N. Tnh thi gianti thiu xe t c vn tc 3u m/s trong trng hp:a.

Cng sut cc i ca ng c t bng 4P kW.b.

Cng sut cc i y l 1P kW. B qua mi ma st.

Li gii

a. Theo gi thit 3max 10F N, suy ra gia tc cc i:

2ax

ax 1 m sm

m

m

F

a - Cng sut ng c:P Fv m v at ; Khi 3 m sv u th

3t s. Lc y cng sut ng c 3. 3.10 w = 3P F u kW,cha vt cng sut cc i l iu kin kh d.

b.

Trng hp sau, v cng sut cc ica ng c 310P W nn vn tc

ch bng: 1 1 m/sPvF

sau thi

gian 11 1v

ta

s. Sau giai on

chuyn ng nhanh dn u ny,

vn tc tip tc tng thP

Fv

phi gim, chuyn ng l

nhanh dn khng u, vn tc tng chm hn giai on trc.Cng A ca ng c chuyn thnh tng ng nng ca xetrong thi gian t2 a vn tc t 1 1 m/sv ln

2 3 m/sv u .

2 22 2 11

2A P t m v v (1)

2 4 st - Vy thi gian tng cng xe t t vn tc u l:

v(m/s)

O 1 5

1

3

t(s)

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1 2 5 st t t Ch :T (1) tm thy s ph thuc ca vn tc vo thi gian (

giai on 2)

2 2 2

2 1

2t

Pv v t vm

hay tv t

Ta c th ( )v t trong trng hp 1P kW.

V d 2Mt b ng ct trt khng vn tc ban u t cao 2 mh

theo mt phng nghing gc 045 so vi phng nm ngang, vachm vi sn ri trt trn mt sn nm ngang. N dng li im

cch chn mt nghing bao nhiu?. Bit h s ma st gia b ct vimt nghing hoc vi sn l 0,5k , ly 210 m/sg .

Li gii:- Xt giai on trt trn mt nghing chiu di:

2 2 2sin

hs h

m.

- Gi v l vn tc chn mt nghing th nh lut bo ton

nng lng cho ta:21 cos

2mgh mv kmgs

m coss h

- Suy ra 2 1 4,5 m/sv gh k (1)

- Trong va chm chn mt nghing, thnh phn thng ng

ca ng lng yP b trit tiu bi phn lc Q ca sn, sinh raxung lc Q t trong thi gian va chm t :

. sinyQ t P mv (2)

Thnh phn nm ngang: cosxP mv ca ng lng khngb trit tiu nhng cng gim trong thi gian va chm, v b ctsinh p lc bng Q ln sn nn c lc ma st msF kQ xut

hin; xung lc msF t trong thi gian t lm gim xP

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74

osmsF t mvc mu ( u l vn tc ca b ct sau vachm).

coskQ t mv mu (3)

- T (1) v (3) ta c cos sin 1,6 m/su v k - Xt giai on chuyn ng trn sn, ng nng 2

1

2mu

chuyn thnh cng calc ma st trn on ng x.

21

2mu kmgx

2

0,25 m2

u

x kg

BI TP P DNG

4.1. Tnh cng cn thit mt l xo gin ra 0 20x cm, bit rnglc ko t l vi gin ca l xo v h s n hi ca l xo l

3000k N/m.

4.2.

ng nng ca mt ht chuyn ng trn ng trn bnknh R ph thuc vo qung ng i c s theo quy lut:2

dW as , a l mt hng s. Tnh lc tc dng ln ht, coi lc l

hm s ca s.

4.3. Mt vt khi lng m c nm ln dc theo mt mt phngnghing gc so vi mt phng ngang. Vn tc ban u l 0v ; h

s ma st bng k . Tnh qung ng i c ca vt n khi dng

li v cng ca lc ma st trn qung ng y.4.4.

Mt vt c khi lng 3 kgm trt t nh mt mt phngnghing cao 0,5m, chiu di mt phng nghing l 1m. Khi tichn mt nghing, vn tc ca vt l 2,45 m/sv .

a.

Tnh cng ca lc ma st khi vt trt trn mt nghing.Bit vn tc ban u ca vt bng khng, ly 210 m/s .g

b.

Xc nh h s ma st gia vt v mt nghing.

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4.5.

Mt vin n khi lng 10 gm ang bay vi vn tc

0 100 m/sv th gp mt bn g dy v cm su vo bn g mt

on 4 cms . Hy tm:

a.

Lc cn trung bnh ca g ln vin nb.

Vn tc vin n sau khi ra khi bn g, nu bn g chdy' 2 cms .

4.6.

Mt l khi lng khng ng k c cng 300 N/mk ,mt u l xo buc vo vt, khi lng

12 kgm nm trn mt phng nmngang. H s ma st gia vt v mt

phng l 0,4 . Lc u l xo cha bbin dng. Sau t vo u t do ca

l xo mt lc F nghing gc 030 so vi phng nm ngang th vt dch chuyn rt chm mt onng 0,4 ms . Hy tnh cng ca lcFtrong dch chuyn trn.

4.7.

Mt vt khi lng 1 kgm c t

trn mt tm g phng, c hai li c ttrn mt sn nm ngang. Vt c ni viim O bng mt si dy nh, n hi, cchiu di t nhin 0 40 cm v dy c

phng thng ng. H s ma st gia vt v tm g l 0,2 .T t ko tm g theo phng nm ngang cho n khi vt bt utrt trn tm g, khi y dy lch khi phng thng ng mt gc

0

30 . Tnh cng ca lc ma st tc dng ln vt k t lc kotm g n lc vt bt u trt trn tm g, cho 210 m/sg .

4.8. Mt vt nh trt khng ma st trn mt ngoi ca mt bncu c bn knh 1,5 mR . Bit vt trt t nh xung. Hi

a.

Vt s ri khi mt bn cu caono? (so vi y bn cu). Nu vntc ban u

0

v ca vt bng khng.

b. Cn phi truyn cho vt vn tc ban

m

F

m

O

F

R

0v

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u 0v theo phng ngang nh nht bng bao nhiu vt

c th ri khi mt bn cu m khng trt. Cho210 m/sg .

4.9.

Mt vin bi st treo vo dy di1 m c ko cho dy nm ngang ri th

cho ri xung khng vn tc ban u. Khigc gia dy v ng thng ng l

030 th vin bi va chm n hi vo mttm st c t thng ng. Hi vin bi nyti cao h bng bao nhiu sau va chm?

(so vi v tr cn bng).4.10.Mt vin bi ri t cao h xung mt mt phng nghinggc so vi mt phng nm ngang.

a.

Tnh t s khong cch giacc im va cm ca vin bivi mt phng nghing. Coi vachm l n hi tuyt i.

b.

Tnh khong thi gian kt tlc vin bi bt u ri t imM n lc n bt u chm t. Bit 5h cm; 0A B 2 m;

210 m/sg ; 030 .

4.11.Mt con lc th n khi lng M,vin n khi lng m bay theo phngnm ngang, xuyn vo bao ct (con lc) v

b mc li trong bao ct ng thi bao ctc nng ln cao h .

a.

Hy lp biu thc tnh vn tc vinn ngay trc va chm vi bao ct.

b.

Tnh t s phn trm ng nng ca vin n bin thnhnhit khi va chm. Cho 20 gm ; 1,2 kgM .

c.

Khi lng vin n 20m g. Hi khi lng bao ct ti al bao nhiu bao ct chuyn ng c. Bit bao ct (c

M

A0A1

A2

h

hm M

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c vin n bn trong) chuyn ng c khi ng nng baoct 1% ng nng vin n trc va chm.

4.12.Mt qu cu c ng cht c bn knh R v khi lng

1m kg, ln khng trt vi vn tc

1 36v km/s n va chm vo

thnh tng ri bt ra vi vn tc2 28,8v km/h. Tnh nhit lng

to ra trong va chm .

4.13.

Mt thanh ng cht c chiu di lv khilng M, c th quay xung quanh mt trc nmngang i qua u trn ca thanh. Mt vin n ckhi lng m bay theo phng ngang ti xuyn

vo u di ca thanh v b mc li trong thanh.Bit sau va chm thanh b lch i gc so viphng thng ng, coi m M . Tm vn tcvin n trc lc va chm.

4.14.Mt u my xe la khi lng m m my chy t nh ga

sao cho vn tc ca n cho bi quy lut v a s , vi a l hng s,s l qung ng i c. Tnh cng tng cng ca tt c cc lc

tc dng ln u my thc hin trong tgiy u k t lc m my.4.15.

Mt khu pho khi lng 450 kgM nh n theo phngnm ngang. n pho c khi lng 5 kgm , vn tc u nng

450 m/sv . Khi bn b pho git v pha sau mton 45 cmS . Tm lc hm trung bnh tc dng ln pho.

4.16.

Mt qu cu khi lng 0,1 kgm c gn u mt

thanh nh, khi lng khng ng k, di 1,27 ml . H quaytrong mt phng thng ng xung quanh u kiaca thanh. Ti im cao nht qu cu c vn tc

0 4,13 m/sv .

a. Tm s ph thuc ca th nng v ng

nng ca qu cu theo gc hp bithanh v phng thng ng. Gc tnh

th nng ti v tr thp nht.

O

m

A

O

Av

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b.

Xc nh lc tc dng Fca qu cu ln thanh theo gc .Tnh Fti cc im thp nht v cao nht?

4.17.Mt vt chuyn ng khi lng1m ti va chm vo mt vt

th hai ang ng yn, khi lng 2m . Coi va chm l n hixuyn tm. Hy xc nh s phn trm ng nng ban u ca vtth nht truyn cho vt th hai sau va chm? p dng cho cctrng hp:

1 2m m ; 1 29m m .

HNG DN - LI GII - P S

4.1. Tnh cng ca lc ko- Theo nh ngha ta c biu thc tnh cng:

0 0

2

0

0 0

1

2

x x

A Fdx kxdx kx

- Thay s ta c 21

3000.0, 2 602

A J

4.2.

Ta c 2 212

dW mv as

2 22 (1)

2 2

mv as

dv as

dt m

- Lc hng tm:2

ht

vF m

R

2 42

2

4 ht

a sF

R (3)

- Lc tip tuyn: tdv

F mdt

2 2 2 4tF a s (4)

- Lc tc dng ln ht chuyn ng2

2 2 2 1ht ts

F F F asR

4.3.

Tnh qung ng v cng ca lc ma st- ng nng ban u ca vt bng tng cng ca lc ma st vth nng vt khi dng li:

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201

sin cos2

msmv mgh A mgs k mg s (1)

- Qung ng vt i c cho n khi dng li:

2

0

2 sin osvs

g kc

- Cng ca lc ma st:

2

0

2ms

kmvA

k tg

4.4.

Vt trt t nh mt phng nghing

a.

Cng ca lc ma st- Theo nh lut bo ton v chuyn ho nng lng

21

2 msmgh mv A

21 62

msA m gh v

J

b.

H s ma st gia vt v mt nghing

- Ta c osms msA F s kmgsc , mt khc2 2

cos s h

s

2 2

2 2 0,23msms

AA kmg s h k

mg s h

4.5. Vin n xuyn vo tm ga.

Xc nh lc cn trung bnh ca tm g ln vin n- Ton b ng nng vin n dng thng cng ca lc cn

22 0

0

1

12502 2c c

mv

mv F s F N S b.

Vn tc ca vin n khi xuyn qua tm g

2 2 2

0 0

2 '1 1' 70

2 2

cc

F smv F s mv v v

m m/s.

4.6. Tnh cng ca lc ko F- Cng ca lc F gm 2 phn: Cng

1A lm gin l xo, to

th nng cho l xo v cng v cng 2A lm dch chuyn vt.

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- V vt dch chuyn rt chm nn:

K msF F N

- Ta c p lc: sinN mg F

- Mt khc cosKF F - Vy cos sinF mg F

45sin cos

mgF

k

N

- gin ca l xo: 0,15F

k m

2

1

1 A 3,42 k J

2 cos 15,6KA F s F s J- Vy cng ca lc F:

1 2 19A A A J

4.7. Tnh cng ca lc ma st- Cng ca lc ma st lm vt dchchuyn so vi sn nh cng bng tng

th nng n hi ca dy lin kt gia vtvi im treo O.- Lc vt bt u trt 030 , vt chu

cc lc sau: Trng lc P, phn lc Q ,

lc cng T , lc ma st ngh cc i msF .

- Vt ln tm g p lc: cosN Q P T

cosmsF N P T (1)- i vi vt, lc ma st msF ng vai tr lc ko, cn thnh

phn sinT ng vai tr lc cn. Lc vt bt u trt trntm g th sinmsF T (2)- T (1) v (2) ta c:

2,97 3

cos sin

PT

N (3)

- gin ca dy l:

P

msF

KF

F

N

nF

msF m

O

T Q

P

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2 2

0 0htv mv

F P Q m Q mgR R

- iu kin vt ri khi mt cu m khng trt xung:

0Q 0minv 3,87 m/sRg 4.9.

T nh lut bo ton c nng suy ravn tc vinbi ngay trc lc va chm:

201

1 os2

mg mg c mv

0 2 cosv g (1)

0v hp vi php tuyn ca tm st gc ,do va chm n hi nn vn tc vin bingay sau va chm c ln bng 0v :

0 2 osv v g c (2)- Vc t cng to vi php tuyn ca tm st gc (gc phnx bng gc ti).- Sau va chm vin bi chuyn ng theo qu o trn i ln,

vi vn tc nv vung gc dy:

os2 2 os . os2nv vc g c c

- Cn thnh phn sin2tv v dc theo si dy c tc dng

ko gin dy v phn ng nng tng ng: 2 / 2tmv ca vin

bi bin thnh ni nng lm dy nng ln.- p dng nh lut bo ton c nng:

2

22

1

2

3cos .cos 2 21,7

2 8

n

n

mv mgh

vh cm

g

-Nhn xt: Do va chm n hi nn khng c s mt mt ngnng ca vin bi khi khng va chm vi tm st, nhng ngay

sau va chm vin bi li b mt mt phn ng nng do dy bgin t ngt.

nv

v 0v

tv

O

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4.10.Thi gian ri on 0MA h l:

0

20,1 s

ht

g

- Vn tc bi ngay trc lc va chm ti A0:

0 2 1 m/sv gh (1)- Chn h quy chiu xOy( 0O A ); 0t l thi imngay sau va chm A0 v vntc ban u v ( 0v v , v hng

ln lch so vi y gc ). Ta cphng trnh chuyn ng:

2

0

2

0

sinsin 2

2

osy t os 3

2

x t v t g t

gcv c t t

0

0

sin sin

os os

x

y

v v g t

v v c gc t

- Sau thi gian1t vin bi li va chm vi mt phng nghing

A1. Ti A1: 0y , t (3) 012

0,2v

tg

1 0 11 ; 2 8 sinAx A A h (4)- Ngay trc va chm ti A1, vn tc 1v ca vin bi c ccthnh phn:

1 0 1 0

1 0 1 0

sin sin 3 sin

os - gcos os

x

y

v v g t v

v v c t v c

(5)

-Ngay sau va chm ti A1, vn tc 1v ca vin bi c cc thnh

phn:

M

0

A1A2

h v

y

x

O

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1 1 0

1 1 0

3 sin

os

x x

y y

v v v

v v v c

(6)

-Nu chn gc to A1 v 0t l ngay sau lc va chm

A1, ta c phng trnh chuyn ng

2

0

2

0

sin3 sin

2

osos

2

gx t v t t

gcy t v c t t

(7)

- Sau thi gian2t vin bi li va chm vi mt nghing A2.

Ti A2: 0y , t (7) 02 12 0,2vt tg s

2 1 216 sinAx A A h (8)

- Lp lun tng t, ta thy vin bi li va chmvi mt nghing

A3: 03 12

0,2v

t tg

s

2 3 24 sinA A h

- Vy 2 31 2

0 1 0 1

2; 3;...A AA A

A A A A

hay 0 1 1 2 2 3: : : ... 1: 2 :3...A A A A A A

- Thay s 0 1 20A A cm, 1 2 40A A cm, 2 3 60A A cm,

3 4 80A A cm, ...- Theo gi thit: 2 mAB m 0 1 1 2 2 3 3 4 2 mA A A A A A A A - Nh vy sau 5 ln va chm vin bi chm vo im B tc l

bt u chm t.- Tng thi gian k t lc bt u ri M n lc chm t B l 0 1 4... 0,9T t t t s

4.11.Con lc th n

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a.

Gi v v V ln lt l vn tc vin n v vn tc bao ctngay sau khi n xuyn vo.

- Theo nh lut bo ton ng lng

mv M m V (1)- Theo nh lut bo ton c nng

21

2M m V M m gh (2)

- T (1) v (2) 2M m

v ghm

(3)

b. Phn ng nng vin n bin thnh nhit khi va chm l:

2 21 12 2

W mv M m V

- T s phn trm ng nng bin thnh nhit2

2

11

2

W M m V Q

m vmv

1 100%m M

Mm M m (4)

- Thay s 98,4%Q

c. iu kin baoct chuyn ng c:

2

2 21 10,01

2 2 100

m vM m V mv M m

V

(5)

- T (1)v (5) 2 298 99 0M Mm m , thay 0,02 kgm tac:

2 1,96 0,0396 0M M (6)- Gii (6) c 1,98 kgM

- Vy: ax 1,98 kgmM .

4.12.Nhit lng to ra trong va chm ng bng gim ngnng ca qu cu. M ng nng qu cu:

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2 2 21 1 7

2 2 10dW mv I mv

- Vi 22 v

; =

5 R

I mR

- Vy 2 21 27

25,210

Q m v v J

4.13. Xt h n v thanh; theo nh lut bo ton m men nglng trc va chm v ngay sau va chm:

2 22

12 4

3

3

M Mmv m

mv

m M

- p dng nh lut bo ton c nng cho h t thi im ngaysau va chm n lc t gc :

2 2 22 21

2 2 12 4

m M Mmgh Mgh

1 os ; h 1 os2

h c c

2 23

1 os2 3 2

m v Mc m g

m M

2 22

2 3 2sin

3 2

m M mM v g

m

- V m M 2

sin3 2

Mv g

m

4.14.

Vn tc v a S gia tc 21

2

dva

dt Lc tc dng

ln u my21

2F m ma (1)

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- Phng trnh chuyn ng

2 2 2 21 1 1 dS=

2 4 2t

S t a t a tdt (2)

- Cng

4 4 2

0

1 1

4 8

t

A FdS FdS ma tdt ma t (3)

4.15.

T nh lut bo ton ng lng suy ra vn tc git li capho

2

2 mV vM

(1)

- ng nng git li ca pho bng ng cng ca lc hm tcdng ln pho

21 .2

dW MV A F S (2)

- T (1) v (2):2 2

125002

m vF

MS N

4.16.Qu cu gn vo thanh cnga.

Th nng v ng nng qu cu c tnh theo cng thc:

1 costW mg (1)

201

1 cos2

dW mv mg (2)

b.

Lc tc dng Fca qu cu ln thanh2

1

mvos +htF P F mgc (3)

- T (2) 2 20 2 1 osmv mv mg c - Thay vo (3) ta c:

2

0v2 3 os +F m g gc

(4)

- Ti im cao nht 0,36 NF , ti im thp nht 6,2 NF

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4.17. nh lut bo ton ng lng , ,1 1 1 1 2 2m v m v m v 2

2 22 , , , ,2 21 1 2 1 2

1 1

2m mv v v v v

m m

(1)

- nh lut bo ton nng lng2 22 , ,

1