phân loại và phương pháp giải chi tiết bài tập trắc nghiệm vật lí 12 trọng tâm

• At: T h d i gian quay goe Aep s b Toe do goc tiic thai co Toe dp goc ttre thdi tai thdi diem t la dai lupng dac tntog cho miJc dp quay cham hay quay nhanh cua v^t r ^ quanh mpt true e

ThS TRAN THANH BINH

PHAN LOAI VA PHUOING PHAP

GIAI CHI TIET BAI TAP TRAC NGHI|M

V A T L I 12

B I ^ N SOAN T H E O C H U O N G T R I N H M(5|

DANH C H O HQC S I N H BAN N A N G C A O VA BAN C O BAN

• T6m t i t If thuyet

• Phan logi theo tCfng van de

• C6c phudng phap gi^i bai t§p + b^i t§p

mlu cho tCrng v§'n de

• B^i t$p va If thuyet trie nghiem

• Bai tap luyen tap cuoi m6i chUdng

N H A Y I I A T R A M FiAi w n r o i i o c G I A T P HO CHf MINH

1

I / I

Ldf N O I D A U

C^' h k m giijp cdc em hoc sinh c6 t a i lieu tot k h i hoc m o n Vat l i 12,

chiing t o i bien soan quyen "Phan loai va philcfng phap giai chi tiet b ^ i tap trac nghiem Vat h' 12 trong tam"

Trong quyen sach nay c6 10 chUofng M o i chucfng deu c6 cau triic nhif sau:

PHAN I: TOM TAT LI THUYET TRQNG TAM CUA CHUdNG

( N h ^ m giup cac em hoc sinh nMm vufng l i thuyet de c6 the l a m cdc cau h o i trSc nghiem l i thuyet cua chucfng)

P H A N II: P H A N LOAI TL/NG V A N DE VA PHLTONG P H A P G I A I B A I T A P TLTNG VAN DE

(Moi van de, m o i loai deu c6 cac b ^ i tap mau, cac cau t r i e nghi$m bai tap ciia tiTng van de, tifng loai, bai tap luyen tap cua chucfng)

PHAN III: DAP AN CUA CAC CAU TRAC NGHIEM LI THUYET + BAI TAP TRAC NGHIEM VA BAI TAP LUYEN TAP

(Tat ca cac cau t r i e nghiem va bai tap deu c6 hudng dan giai va dap dn) Chic c h i n r i n g , neu cae em hoc sinh chiu kho l a m cac eau t r i e nghiem va

hki tap trong quyen sach nay v i n i m that vi^ng tCrng v a n de trong quyen

sach nay mpt each day dii t h i cic em se eo ket qua t o t mon Vat l i trong cie

k i t h i Day eung la t a i lieu tham khao cho g i i o vien eung n h u phu huynh hoc sinh trong viee giup cie em hoc to't mon Vat l i 12

Tuy t i c gia c6 r a t nhieu eo g i n g trong qua t r i n h bien soan, nhung chic

r i n g quyen s i c h khong the t r a n h khoi nhiJng thieu sot Mong qui v i gido vien, cac bac phu huynh va cac em hpc sinh c6 nhiirng y kien d6ng g6p de l i n

tdi ban quyen s i c h se hoan ehinh han

Chan thanh cam an

Tdc gia

* K h i vat r a n quay quanh mot true eo' dinh t h i :

- Moi diem tren vat vach mot dUcfng tron n k m

trong mat phang vuong goe vdi true quay, c6

ban kinh bkng khoang caeh tiT diem do den

true quay, eo tarn d tren true quay

- M o i diem tren vat deu e6 eung mot goe quay

* Tpa do goe (p = (Ox, OM )

De don gian ta chi xet vat quay theo mot ehieu va thiJdng ehon chieu quay eiia vat la chieu diTcfng -> k h i do (p > 0

• (po: Toa do goe lue dan tai thc(i diem to (rad)

• (p: Toa dp goc liie sau cf thdi diem t (rad)

• to: Thdi diem lue dau k h i vat eo cpo (s)

• t i : Thdi diem liic sau k h i vat c6 (p (s)

• cotb: Toe dp goe trung binh trong thdi gian At (rad/s)

• Acp: G6c ma mot vat quay dupe trong thdi gian At (rad)

• At: T h d i gian quay goe Aep (s)

b) Toe do goc tiic thai (co)

Toe dp goc ttre thdi tai thdi diem t la dai lupng dac tntog cho miJc dp quay

cham hay quay nhanh cua v^t r ^ quanh mpt true eo' dinh tai thdi diem do

m = l i m Acp dcp

^ ' - 0 At dt = <p'(t)

• Ytb: Gia toe goe trung b i n h (rad/s^)

• Aco: Do bien t h i e n toe do goe (rad/s)

• At: T h 6 i gian ma v a t bien t h i e n Aco (s)

b) Gia toe goc tiic thai (y)

Gia toe goc tiJc t h d i t a i t h d i diem t l a d a i lugng dac t r i m g cho sir bien

t h i e n eiia toe do goe c3r thcfi diem t k h i v a t xin quay quanh mot true

CO khong doi -> T, f k h o n g doi

* Neu chon chieu ducfng cCing chieu quay cua vat r ^ n t h i :

• Y > 0 k h i v a t quay nhanh d a n deu (co t a n g dan deu theo At)

• Y < 0 k h i vat quay cham dan deu (co giam dan deu theo At)

• Tong quat:

• co.Y > 0 -> vat quay nhanh dan deu

• oj.Y < 0 ^ vat quay cham dan deu

5 van t6c Mh gia t6c cua c^c di^m tr§n v§t

CHUYEN DQNG QUAY TRON DEU

CHUYEN DQNG QUAY TRON BIEN DOI DEU

K h i v a t r a n quay t r o n deu t h i vectcf v a n toe m o t diem nao do

t r e n v a t c h i thay doi ve hUcing

ma khong thay doi ve do IcJn

K h i vay r a n quay t r o n deu t h i gia toe toan phan cung ehinh l a gia toe hucJng t a m a„ (gia toe phap tuyen)

ea ve hudng va do IcJn

K h i vat r d n quay tron bien doi deu t h i gia toe toan phan bao gom gia toe tiep tuyen va gia toe phap tuyen (gia toe hirdrng tam)

a = a„ + a^

+ a:

Bg Ian:

V(Ji:

• a„ vuong goe v , dac trUlig cho

sU thay doi ve hucjrng cua v

V 2

a„ = — = CO r

CO phuong eiia v , dac trUng

cho sU thay doi ve do 16n eiia v

a, = r.Y

Van de 2: PHL/ONG TRINH DQNG Ll/C HQC CUA VA T RAN QUA Y

QUANH MQT TRUC CO DINH

1 Momen lUc ddi v6i true quay Momen lue M cua life F doi v d i v a t r a n co true quay co d i n h l a d a i lucfng dac trung cho tdc dung c o d i n h l a d a i lifcfng dae trUng cho tae dung l a m quay v a t r ^ n quanh true co d i n h do va dUOe do bang tich so ciia lUc va each tay don

M = F.d = F.r.sina

Trong do:

• M : Momen cua life (N.m) ,

• F: Luc tAc dung (N) I

• d: Canh tay don (m)

(Khoang each tCr true quay den gid cua life)

2 Momen quSn tinh (I)

- Momen quan t i n h I doi v6i m o t true la d a i lu'Ong dac trutig eho miJe

quan t i n h cua vat rMn trong chuyen dong quay quanh true ay

1 = 1 m, r

- Momen qudn t i n h c6 dp Idn phu thupe vao k h o i iMng vat r ^ n , phu thupc

vao sir phan bo' k h o i li^png eiia vat r ^ n doi vcJi true quay (gan hay xa

• y: Gia toe goe (rad/s^)

Nhqn xet: Ydi cung mot Momen t h i

• Neu I i d n y nho -> kho thay doi toe dp goe

• Neu I nho -> y i d n -> de thay doi to'e dp goe

4 Momen quSn tinh cilia m$t s6 v$t dfing ch^t

a) V a t la t h a n h m a n h dong chat, k h o i lupng m, chieu dai I eo true quay la

dudng trung true cua t h a n h

(A)

I = —ml'

12

J

b) V a t la t h a n h m a n h dong chat, k h o i liTpng m, chieu dai / c6 true quay d i

qua mpt dau t h a n h va vuong gde v d i thanh

1 Momen dOng ladng

Momen dong liTpng eiia v a t r ^ n d o i vdi mpt true quay b^ng tich so eua momen quan t i n h cua vat doi v d i true quay do va toe dp goe cua vat quay quanh true do

L = I.a) ( u) > 0 - » L > 0 ; a ) < 0 - ^ L < 0 )

Trong do:

• L: Momen dong lirpng (kgm^/s)

• I : Momen quan t i n h cua vat r ^ n (kgm^)

• (o: Toe dp goe eiia vat (rad/s)

2 Djnh lu$t b^o to^n momen d6ng lUdng

L = h k n g so hay L , + + = L\ L ' j +

Dieu kien dp dung dinh ludt:

K h i tong dai so cua cac momen ngoai lye dat len mot vat r ^ n (hay h$ vat)

doi vdi mpt true quay b^ng khong (hay cac momen ngoai lire t r i $ t tieu)

Luu y: K h i I doi vdi true quay khong doi -> to = 0 hoac w - h k n g so

(Vat khong quay hoae vat quay deu)

Van de 4: DQNG NANG CUA VAT RAN QUA Y QUANH

MQTTRUC CODfNH

1 D6ng nSng cCia m6t vat rjn quay quanh m6t true c6 dinh

Trong do: • W,i: Dong n^ng eua vat r a n quay quanh mpt true (J)

• I : Momen quan t i n h cua vat (kgm^)

• co: Toe dp gdc cua vat (rad/s)

2 Dinh li dong nSng

A = W, - W d,,,-,„

Trong do: • A: Cong eiia ngoai lue (J)

• ,^ : Dong nftng lue dau (J)

• Wj : Dong nftng lue sau (J)

Luu y: Neu vat chi c6 chuyen dpng quay quanh true t h i : W j =

• Neu vat thuc h i e n dong thcfi hai ehuyen dpng la quay quanh true va t:

3 Djnh li vl true song song

true va t i n h t i e n t h i : W j = +

1(A) = 1 ( 0 ) + inx^

Trong do:

• 1(A): Momen quan t i n h eiia mpt vat doi vdi true quay ( A ) (kgm^)

• IQ: Momen quan tinh eiia true di qua trong tarn G song song vdi (A) (kgm^)

• m: K h o i lupng vat rSn (kg)

• x: Khoang each vuong gdc giSa true ( A ) va true song song qua (G) (m)

Hinh minh hga dinh li:

(A) (G) '

0)

G

TRAC NGHrEM LI THUYET

C a u 1 M p t diem tren vanh dia t r o n each true quay d i qua t a m mpt khoang R

k h i dia quay t r 6 n deu quanh true t h i toe dp d^i va toe dp goc eiia diem do c6

quan he vdi nhau theo bieu thiJe nao t r o n g cac bieu thiifc sau?

A v - - B v = - C 0 3 = - D R = vlco

C a u 2 H a i hpc sinh A va B diing t r e n mpt ehiec du quay t r o n deu quanh true

CO dinh di qua t a m Hoe sinh A d ngoai ria, hoc sinh B d each t a m mpt doan

b ^ng mpt phan t u ban k i n h chiee du quay Gpi TA, TR la ehu k y quay cua hoc

sinh A va hoe sinh B Lue nay ta cd:

A. T A = T B - ' ^ B T A < T B - C T A > TR D T A = 4T„

C a u 3 Klpt dia CD coi n h u chuyen dpng t r d n deu xung quanh true di qua t a m eiia dia Gpi E va F Ian lupt la diem d ngoai r i a va diem d each t a m dia mpt doan b^ng nuTa ban k i n h cua dia Gpi VR, VF, fs, fp Ian lupt la toe dp dai va

t ^ n so ciia cac diem E va F Ket luan nao sau day la diing?

A. V E = V p ; f e = B V E = V p ; = 2iv

C vp = 2VE; ffi = fp- D- V E = 2VF; h =

fp-C a u 4 M p t dia quay t r o n deu quanh true doi xii'ng di qua t a m Gpi A la diem d

ngoai r i a , B la diem each true quay mpt doan b a n g mpt phan ba ban k i n h

dia Gpi (DA, COB, YA, YB Ian lupt la toe dp gdc va gia toe gdc cua ede diem A va

B Chpn cau ket luan diing

A. COA = 2(0B; YA = YB- B (OA = cou; YA = 2YB

C. tOA = COB; YA = D - " A = COR; YA =

YB-C a u 5 M p t dia YB-CD quay t r d n deu quanh true d i qua t a m dia M p t diem bat k y

n k m d mep dia se:

A khong cd gia toe tiep tuyen Ian gia toe phap tuyen

B cd gia tdc phap tuyen nhulig khong cd gia toe tiep tuyen

C CO gia tdc tiep tuyen va ed ea gia toe phap tuyen

D ed ea gia tdc tiep tuyen va gia to'e phap tuyen nhuiig gia toe phap tuyen Idn hon gia toe tiep tuyen

C a u 6 M p t vat rSn quay t r o n deu quanh true eo' dinh di qua vat t h i mpt diem

t r e n vat d each true quay mpt doan r 0 se cd

A toe dp gdc thay doi B ehu ky quay thay doi

C tdc dp dai thay doi

D vecto van tdc dai thay doi nhung to'e dp Aki eiia diem dd, khong ddi

C a u 7 M p t vat r ^ n quay tron deu quanh mpt true cd dinh Gpi N la so dao dpng

ma vat thue hien trong thdi gian At, cpo va cp la tpa dp gdc lue dau va lue sau, co

la tdc dp gdc cua vat r a n Chpn bieu thiire sai trong cac bieu thde sau:

A T : — B T = — C N = ^ ^ ^ D (p = (po - «At

C a u 8 Bieu thiJc nao sau day la dung k h i n 6 i ve dp Idn cua gia toc ph^P tuyen

(gia toc hudng tarn)?

A an = - B an = w.R C an = — D a^ = (w.R)^

C a u 9 Bieu thiJc nao sau day la dung k h i noi ve do Idn ciia gia toc tiep tuyen

cua vat rSn quay bien doi deu quanh mot true co dinh?

A. at = r.y B a t = - C a t = - D at = —

C a u 10 M o t vat r ^ n quay bien doi deu quanh mot true d i qua vat r ^ n Chon

goc t h d i gian to = 0 la luc v a t bat dau quay Goi t i , t2 la cac t h d i diem liic sau

(t2 = 2t,) Neu xet m o t d i e m t r e n v a t r ^ n each true quay T ^0 t h i :

A a,, = 2a,^ B. a,^ = 2 a , C a,_ = ^ D. a, = a,^

C a u 11 Bieu thiJc nao trong eac bieu thiic sau khong the ap dung cho vat r ^ n

quay bien doi deu quanh mpt true co d i n h d i qua vat?

A. 9 - cpo = o)o-t + • B. (0^ - (OQ = 2Y ((P -

cpo)-C. cp = (po + cD.t D © = 0)0 + y.t

C a u 12 Bieu thiJe nao trong eae bieu thiJe dudi day la sai k h i t i n h gia toe toan

phan ciia vat r ^ n quay bien d6'i deu quanh m o t true eo d i n h d i qua vat?

I r j

C a u 13 Phdt bieu nao sau day 1^ dung doi v d i vat r ^ n eo chuyen dpng quay

deu quanh m o t true?

A Toc do goc l a m o t hSng so

B Toe dp goc la m o t h a m bae n h a t doi v d i t h d i gian

C PhUdng t r i n h chuyen dpng la m o t h a m bac h a i doi v d i t h d i gian

D Gia toe goc 1^ m o t hkng so

C a u 14 Chpn phat bieu sai k h i noi ve momen quan t i n h eiia vat r ^ n

A Momen quan t i n h phu thupc vao h i n h dang va k i c h thude ciia vat

B Momen quan t i n h phu thupe vao k h o i liTpng ciia v a t r ^ n

C Momen quan t i n h k h o n g phu thupc vao v i t r i true quay ciia v a t r ^ n

D Momen quan t i n h k h o n g phu thupc v^o toc dp goc ciia vat

C a u 15 M o t chiee du quay ehiu tac dung eiia mot momen lue khong doi Chon

phat bieu sai t r o n g eac phat bieu sau:

A Gia toc goc ciia chiec du quay la m o t h^ng s6'

B K h d i lUdng cua chiec du quay la m o t h k n g so

C Toe dp cua chiec du quay 1^ mot h^ng so

D Momen qudn t i n h l a m o t h k n g so

C a u 16 M p t vat rMn quay bien doi deu quanh mpt true co dinh d i qua vat rin

Chon phat bieu sai t r o n g cac phdt bieu sau:

A Gia toc goc la h k n g so

B Toe dp goc la mpt h a m bac n h a t d o i v d i t h d i gian

C Toe dp gdc 1^ mpt h k n g so

D Trong chuyen dpng quay bien doi deu ciia vat r k n quanh mpt true co' d i n h

di qua no t h i toe dp goc tSng hay giam nhiJng luang eo dp Idn bkng nhau trong nhiJng khoang t h d i gian b k n g nhau

C a u 17 Mpt vat r k n quay v d i toc dp goc khong doi quanh mpt true ed dinh d i

qua vat So' vong quay m a v a t quay dUde trong t h d i gian t ke til luc v a t b a t

dau quay se:

A t i le vdi t l B t i le v d i t C. t i le vdi D t i le v d i -

t

C a u 18 M p t vat r k n quay v d i gia toe goc khong doi quanh mpt true ed' dinh d i

qua vat Goc ma v a t quay dupe sau t h d i gian t , ke tCr luc b k t dSu quay se

A t i le vdi t^ B t i le vdi t C t i le v d i \ D t i le v d i -

t t

C a u 19 M p t vat r ^ n quay nhanh dan deu quanh mpt true eo' dinh xuyen qua

vat M p t diem t r e n v a t rSn khong n k m t r e n true quay va each true quay m p t doan r ^ 0 Chpn phat bieu dung t r o n g eac phat bieu sau:

A Td'c dp gdc khong phu thupc r

B Gia to'c gdc phu thupc r

C Gia toe tiep tuyen k h o n g phu thupc r

D Gia toe phap tuyen khong phu thupc r

C a u 20 M p t vat r k n quay quanh mpt true co' dinh xuyen qua vat Cac diem

t r e n v a t r k n khong thupc true quay se

A vaeh n e n cac dudng t r o n n k m t r o n g m a t p h k n g vuong gdc v d i true quay

B ed eung v a n toe d a i d eung mpt t h d i diem

C quay dUde nhi^ng gdc k h o n g bkng nhau t r o n g eung mpt khoang t h d i gian

D ed gia to'c gdc va to'c dp gde la h k n g so

C a u 21 H a i dia tron dang quay

dong true va eiing chieu v d i toe dp gde coi, 0)2 Momen quan

t i n h cua h a i dIa 1^ I i , I 2 M a

sdt d true quay khong dang

ke Sau do cho h a i dia d i n h vao nhau va quay v d i to'c dp gdc 0) Bieu thde the h i e n m d i quan he giOfa ede dai li/dng coi,

h)o^-C a u 2 2 H a i d i a t r 6 n m 6 n g n ^ m

n g a n g c6 c u n g t r u e q u a y t h S n g

diJng d i qua t a r n cua h a i d i a

M o m e n q u a n t i n h ciia 2 d i a l a I j ,

I2 L u c dau d i a 1 diJng y e n , d i a 2

quay v d i to'c do CO2 B o qua m a

C a u 23 M o t ngir&i diJng t r e n m o t chiee b a n x o a y d a n g quay L u c d a u n g u d i a y

d a n g t a y r a t h i ghe va n g u d i q u a y v d i toe dp goe l a coi B o qua m a s a t d t r u e

quay Sau do ngurdi a y t h u t a y l a i s a t n g i r d i t h i ghe v a n g i r d i q u a y v d i toe dp

goe C02 B i e u thiirc n a o d i i n g t r o n g eac b i e u thuTc sau?

A Iicui - I2CO2 B I1CO2 = l2Wi- C IiCOi + 12(1)2 = 0 D coi >

12-C a u 25 M o t v a n d p n g v i e n triTpt b a n g quay q u a n h m o t t r u e t h ^ n g diJng v d i toe

dp goe (iJi, m o m e n q u a n t i n h I i k h i h a i t a y t h u l a i s a t n g u d i Sau do v a n d p n g

v i e n d a n g t a y r a t h i t h a y toe dp goe lue n a y l a 0)2 = — Bo qua m a s a t giiJa

CO toe dp goe (o M a sdt d t r u e quay k h o n g d d n g k e M o m e n d p n g l u p n g v a

d p n g n a n g quay se b i e n d o i n h u t h e n a o n e u toe dp goe cua d i a t a n g l e n 3 I a n ?

A M o m e n d p n g liTpng v a d p n g n a n g quay deu t a n g 3 I a n

d l u d i a m o t diJng y e n , d i a 2 quay v d i toe dp goe

u)2 B 6 qua m a sat d t r u e quay Sau do h a i d i a

d i n h vao n h a u va quay c u n g toe dp goe (o C h p n

C a u 30 H a i dia t r b n m o n g c6 cung dpng n a n g quay, t i so giOfa m o m e n quan t i n h eiia

dia 1 va dia 2 d i qua tarn cua dia 1 va dia 2 la i - = 4 H m t i so' toe dp goe —

A 1 B 2 C 3 D 4

C a u 31 M o t t h a n h m a n h d o n g c h a t d i e n d i e n d e u , c h i e u d a i t h a n h l a I, k h o i

l i i p n g t h a n h l a m T h a n h ed t h e q u a y x u n g q u a n h m o t t r u e n ^ m n g a n g d i qua

m o t d a u eiia t h a n h v a v u o n g goe v d i t h a n h B d qua m p i m a s a t v a sdc c a n

B i e t m o m e n q u a n t i n h cua t h a n h l a I = v a g i a toe r o i ta do l a g H o i

sat va sure can T h a n h d a n g ddng y e n d v i t r i can bkng H o i can p h a i t r u y e n cho

t h a n h mot toe dp gdc l a bao nhieu de t h a n h quay den v i t r i n k m ngang

C a u 33 M o t t h a n h A B d o n g cha't cd c h i e u dai / cd t h e quay t r o n g m a t phSng

t h i n g d d n g qua m o t t r u e nkm n g a n g d i qua m o t d a u ciia t h a n h v a v u o n g goe

eua t h a n h T h a n h ed t i e t d i | n deu, khd'i lifpng m , gia to'c r p i t u do la g, I = ^

K h i vat r ^ n quay t r o n deu t h i :

• co: K h o n g doi (la h^ng so)

• oj > 0: Neu vat quay theo chieu

• V: Toe do dai(van toe dai) (m/s)

• co: Toe do goe (van toe goc) (rad/s)

• r: B a n k i n h quy dao cua vat (m)

• Sin Gia toe phap tuyen (gia toe

hirdng tarn) (m/s^)

• at: Gia toe tiep tuyen (m/s^)

• a: Gia toe toan phan (m/s^)

• cpo: Toa do goe liic dau (rad)

• 9: Toa do goe luc sau (rad)

B A I T A P M A U

B a i 1 Mot banh xe quay deu quanh mot true eo dinh vdi tan so 1 2 0 0 v5ng/phut

a) T i m toe dp goe cua banh xe

b) Goc ma bdnh xe quay duoc trong t h d i gian 4 s

Hudng ddn gidi Nhdn xet: Do banh xe quay deu quanh mot true co d i n h nen ta can silr dung

eac k i e n thiJe cua chuyen dpng t r o n deu

B a i 2 Roto eiia mot dong ecf quay t r o n deu quanh mot true co dinh Biet r k n g

eiJ m6i phut t h i Roto quay d M c 1 2 0 0 vong T i m :

a) Chu ky quay ciia Roto

b) Toe do dai eua mot diem t r e n v a n h bdnh xe

e) Gia toe huc?ng t a m cua mot diem t r e n v a n h banh xe

Tom tat

• cpo = — rad

3

a

Hiidng dan gidi

i) Toe do g6c ciia banh xe

Ap dung cong thiic:

(p - cpo = co.t

C a u 2. K i m p h i i t ciia 1 chie'e d o n g h 6 g a p 4/3 I a n c h i e u d a i k i m g i d C o i cac

k i m n h u q u a y deu T i so g i a toe h i f d n g t a m giura d a u k i m p h u t v a d a u k i m

* CO,): V a n to'c goc liic b a n dau (rad/s)

* (o: to'c do goc l i i c sau ( r a d / s )

* y: to'c do goc (rad/s^)

* cpo: T o a do goc l i i c d a u ( r a d )

* (p: T o a do goc l i i c sau ( r a d )

* a„: G i a tdc h u d n g t a m ( g i a to'c

p h a p t u y e n ) (m/s^)

B a i 1. M o t b a n h xe q u a y n h a n h d a n deu q u a n h m p t t r u e tiT t r a n g t h a i durng

y e n v a sau 5 s t h i d a t dugc toe do goc l a 10 r a d / s T i m :

a) G i a toe goc ciia b a n h xe

b) Goc q u a y cua b a n h xe t r o n g t h d i g i a n 5 s k e t i r t r a n g t h d i d i l n g y e n

c) So v o n g m a b a n h xe q u a y dugfe t r o n g t h d i g i a n 5 s d t r e n

Hiidng dan gidi

C h p n c h i e u q u a y cua b d n h xe l ^ m c h i e u ducfng, goe

t h d i g i a n ( t = 0) l a liic v a t b ^ t d a u q u a y , a) G i a toe goc eiia b a n h xe

D o luc sau R o t o dCtog l a i - > co = 0

co^ - cof, = 2y (cp - cpo)

B a i 3 M p t d I a m a i c6 b a n k i n h 40 c m hAi d a u q u a y k h o n g to'c dp goc luc d a u

v d i g i a toe goc k h o n g d o i c6 dp I d n l a 27i (rad/s^) T i m :

a) Toe dp goe m a d i a m a i d a t dugtc sau 4 s k e iii l u c t = 0

b) G i a to'c t i e p t u y e n , g i a toe p h d p t u y e n cua m p t d i e m t r e n v d n h d i a t a i _ t h d i d i e m t = 4 s k e t d liic t = 0

c) Gia toe toan ph^n ciia mot diem tren vanh dia tai thdi diem 5 s ke tCf

luc t = 0

Hu6ng ddn gidi

Chon ehieu quay ciia dia mai lam chieu dUcfng va goe thdi gian (t = 0) la luc dia mai bdt dau quay, a) To'c do goe cua dia mai dat duoe sau 4 s

b) Gia to'c tiep tuyen ciia mot diem tren vanh dia tai t = 4 s

at = r.y = R.y = 0,4.27i

w = 87t (rad/s)

a, = 0,87t (mis')

• Gia to'c phap tuyen ciia mot diem

tren vanh dia tai t = 4 s

an = w l r = (olR = (87t)^.0,4 a„ = 252,4 (m/s=^)

c) Gia toe toan phan ciia mot diem tren vanh dia tai thdi diem 5 s

Nlian xet: O eau nay chung ta khong the silr dung lai toe do goe (JJ = 871

(rad/s) nhii cau b ma chung ta phai di tim lai toe do goc liie t = 5 s

• To'c do goe khi t = 5 s

• (0 = (Oo + y.t = 0 + 271.5 = IO71 (rad/s)

• Gia toe tiep tuyen

C a u 9 Mot banh xe quay nhanh dan deu quanh true Liie t = 0 banh xe eo toe

do goc 4 rad/s Sau 2 s, toe dp g6c ciia no tSng len den 8 rad/g Gia toe goe

ciia bdnh xe la

A 1 rad/sl B 2 rad/sl C 3 rad/sl D 4 rad/sl

C a u 10 Tai th6i diem t = 0, mpt bdnh xe dap bdt dau quay quanli I U U L true vdi gia toe goc khong doi Sau 4s no quay dupe mpt goc 20 rad Toe dp goc va gia toe goc ciia banh xe tai thcfi diem t = 5 s la

A 12,5 rad/s; 2.5 rad/s' B 20 rad/s; 2,5 rad/s^

C.IO rad/s; 22 rad/s^ A 10 rad/s; 12 rad/s""

C a u 11 Mpt banh xe eo ban kinh I m quay nhanh dan deu trong 4 s to'c dp goc tang tCf 20 rad/s len 30 rad/s Gia to'c goe eiia banh xe v^ gia to'c hir(Jng tarn ciia 1 diem tren vanh banh xe sau 2s la

A 2 rad/ s'^; 400 mJs\ 2,5 rad/ s^; 625 m / s l

C 4 rad/s^ 300 m / s l D 5 rad/ s^; 196 m/s^

C a u 12 Mpt banh xe C<D difdng kinh 4 m quay v(5i gia toe khong doi la 6 rad/s^

Gia toe tiep tuyen ciia diem P tren vanh banh xe la

A 4 mJs- B 8 m / s l C 12 m / s l D 16 mJs\

C a u 13 Mpt banh xe dang quay vcJi van toe goc la 40 rad/s thi bdt dau quay cham dan deu Sau 8s thi xe dCfng lai So vong ma banh xe quay diJpc la

A 24,5 vong B 25,4 vong C 22,5 vong D 25,2 vong

C a u 14 Mpt banh xe c6 difdng kinh 4m quay v(Ji mpt gia toe goc khong doi bang 2 rad/s^ Luc t = 0, banh xe n&m yen va ban kinh ciia diem P tren vanh lam vdi ducfng thang n&m ngang mpt g6c7r/3 rad Gia to'c toan phan sau 3 s ciia diem P la

A 40,6 mls\ 72 m / s l C 72,11 mJs\ 60,05 m / s l

C a u 15 Mpt banh xe eo du6ng kinh 2m quay vdi gia to'c goe khong doi b&ng 4

rad/s^ Liic t = 0 banh xe nam yen va ban kinh eiia diem P tren vanh banh

xe hop vdi difdng thang nkm ngang mpt goc (po- Goc quay va so vong ma

banh xe quay diJpc sau 4s ke t\i liie t = 0 la

A 30 rad, 4 vong B 20 rad, 2 vong

C 32 rad, 5,1 vong D 40 rad, 6 v6ng

C a u 16 Mpt banh xe quay c6 ban kinh 3m va gia toe goe dp Mn khong doi la

2 rad/s^ T a i thdi diem t gia to'c toan phan tai mpt diem tren vanh banh xe

la 10 m/s^ Gia toe phap tuyen va toe dp goc tai thdi diem do la

C a u 19 1 l o n g cac pliuoiig t i i n h sau, phuong t r i n h n^o m6 t d quy ludt quay

cham dan deu ciia vat r ^ n quanh mot true co' dinh?

A (p = - 4 t ' ( r a d , s) C cp = 7i/6+4t +15t'(rad; s)

B co= 10+ 2t (rad/s; s) D cp = - 4 t + 2t^ (rad s)

Van de 3: MOMENLl/C- MOMEN QUAN TINH CUA VATRAN QUAY

QUANH TRIJC CO DINH

• y: Gia to'e goc (rad/s^)

I I I Momen q u a n tinh c u a mpt so' v^t r ^ n

1. Thanh c6 tiet d i | n nho so vdi ehieu dki I quay quanh true d6'i xiJng 1^

duc(ng trung trUc cua thanh

I = — m.l'

12

Trong do

• m: kho'i lifdng thanh (kg)

• /: chieu dai thanh (m)

2 Vat la thanh manh dong chat kho'i lifcfng m, chieu dai / cd true quay di qua

mpt dau thanh vk vuong goc vdi thanh

1 = - m.l^

3

Trong do

• m: Khoi lucfng thanh (kg)

• /: Chieu dki thanh (m)

3 Vat Ik h i n h tru rSng hoSc vanh tr6n quay quanh true di qua tam

l = m.R'

Trong do

• m: K h o i lucfng tru rong hoac v ^ n h trdn (kg)

• R: Bkn k i n h t r u rong hoac yknh tr6n (m)

4 Vat la hinh tru d&c hay dia t r o n mong quay quanh true di qua tam

• m: Khoi luang qua cau (kg)

• R: Ban k i n h qua cau (m)

BAI TAP MAU

B a i 1 Mot chat diem chiu tdc dung cua mot momen 0,64 N m lam chat diem

ehuyen dong tron vdi gia toe goc khong doi la 2,5 rad/s^

a) Tim momen quan tinh ciia chat diem doi vdi true quay di qua tam eiia diTcfng tron va vuong goc vdi mat phSng chufa dirdng tron

b) Tim kho'i luorng eua chat diem Biet dudng tron cd ban kinh 20 cm

Tom tat

• M = 0,64 N m

• Y = 2,5 rad/s^

a) T i m I = ? b) Tim m = ? Biet R = 20 cm = 0,2 m

HUdng dan gidi

a) Momen quan t i n h eiia chat diem do'i vdi true quay di qua t a m eiia dudng tron va vuong gdc vdi mat phSng chufa dudng tron

M 0,64

M = I.Y ^ ' I = — =

I = 0,256kgm'

2,5 b) Khoi lugng chat diem

I = niR^ ==> m = 0,256

0,2'

<=> m = 6,4 kg

B a i 2 Mot dia mai chiu tac dung ciia mot life F = 6 N tai mot diem tren vanh

diem Biet ban kinh dia mai la 40 em va dia c6 khoi liTOng la 2 kg Tim momen lire va momen qudn tinh cua dia mai trong hai triTdng hgrp sau:

a) ^ b)

Hu&ng dan gidi

a) • Momen lufc tac dung len dia mai

M = F.d = F.R = 6.0,4

M = 2,4 N.m Momen quan ti'nh ciia dia mai

I = - m.R' = i 2 0 , 4 '

2 2

I = 0,16kg.m' b) » Momen life tac dung len dia mai

B a i 3 Mot thanh AB c6 khoi lupng 0,6 kg, chieu dai thanh la 80 cm Biet

thanh manh, dong chat c6 the quay xung quanh true quay la dudng trung

true cua AB H a i dau A va B gan cac qua cau mA - 100 g va niu - 300 g

Biet thanh quay nhanh dan deu k h i chiu tac dung ciia momen liie 60 N m

khong doi T i m :

a) Momen quan t i n h ciia he (gom thanh AB va hai qua cau)

b) T i m gia toe goc cua he (gom thanh AB va hai qua cau)

y 0.12 + 0,02y + 0,02 = 0

BAI TAPTRAC NGHIEM

C a u 20. Mot dia tron mong, phang, dong chat co the quay xung quanh 1 true di qua tam va vuong goc vdi mat ph^ng dia Tac dung vao dia mot liTc co phiicfng tiep tuyen vdi vanh dia co dp Idn la 40 N Biet ban k i n h vanh dia la

40 cm Dia ehuyen dong quanh true vdi y = 2 rad/s^ Bo qua moi siJe can Khoi lirong cua dia la

A 100 kg B 200 k g C 300 kg D 400 k g

C a u 21. Mot rong roc co ban k i n h 10 em va co momen quan t i n h doi vdi true quay la 0,04 kgm^ ban dau rong roc diing yen, tae dung vao rong roe mot lire khong doi F = 6N tiep tuyen vdi vanh ngoai cua no Bo qua moi sufe can Gia toe goc ciia rong roc la

A 5 rad/s' B 10 rad/sl C 15 rad/s^ D 20 rad/sl

Van de 4: BAI TAP VE PHL/ONG TRJNH CHUYEN DQNG CUA VAT

RA N QUA Y QUANH MQ T TRUC

• a = Vaf, + a?

a n = r , a t = r y

BAI TAP MAU

B a i 1 Mot thiing nxidc dugc tha xudng gieng nhcJ

mot soi day dai quan quanh h i n h t r u c6 ban k i n h

R = 10 cm va momen quan t i n h I - 0,02 kgm^

Khoi lu'ong ciia day va momen quan t i n h cua tay

quay khong dang ke H i n h t r u coi n h a tay quay

t y do khong ma sat quanh mot true c6 dinh Kho'i

iMng thung nUdc la 8 kg T i n h :

a) Gia toe eiia thijng niidc

b) Luc cang day treo Cho g = 10 m/s^

Hiidng dan gidi

Nhaii xet: Trong bai tap nay, chung ta can xac

d i n h duoc vat nao chuyen dong quay quanh true va vat nao chuyen dong t i n h tien

Trong bai toan nay, chung ta thay r a n g rong

r p c C O chuyen dong quay va thung n U d c t h i chuyen dong t i n h tien

a) Ap dung dinh luat I I Niuton cho thijng n U d c va phuang t r i n h dong lire hoc

cho chuyen dong quay ciia h i n h t r u

ma he thufc lien he giuTa gia t o e dai

va gia toe goe la a = R.y y = —

R -> T.R - I - T

b) Lyc cang day treo

The a = 8 mJs^ vao (2)

la 0,02.8 (2) ^ T =

1 ^ B a i 2 H a i vat m, = 2 kg, m2 = 2 k g dUcfc lien ket vdi nhau b&ng mot day nhe, khong dan, v^t qua mot rdng

roc CO ban k i n h 8 cm va momen quan

t i n h la 0,04 kgm^ Biet day k h o n g trUOt t r e n rong roc nhulig khong biet giiJa vat m2 va ban c6 ma sat hay khong Luc dau eac vat duoc giii' duTng yen, sau do he vat ducfc tha ra NgUdi

ta thay rang sau khoang t h d i gian la

4 s t h i rong roc quay quanh true cua no ducfc 3 vong Biet gia toe cua cac vat mi va m2 khong doi Bo qua ma sat d true cua rong roc Lay g = 10 m/s^ a) T i n h gia toe goe cua rong roc

b) T i n h gia toe eiia m j va m^

c) T i n h lire cang day d hai ben cua rong roc

d) Co ma sat giiTa vat m2 va mat san hay khong? Neu cd, hay t i n h he so

ma sat giiifa m2 va mat san

c ) T i = ? T 2 = ?

d) Cd ma sat giufa m2 va san khong? Neu cd

t i n h he so ma sat

Hitdng dan gidi

a) Gia tdc gdc cua rong roc

Nhau xet: Trong t h d i gian 4 s rong roc quay 3 vong

N =

2TI

- » cp - (pi, = N.27t = 3.271 = 671 rad

A p dung cong thdc (p - (PO = 0)0.t + 6n = 0.4 +

e) Lire cSng day d hai ben true rong roc

Nhau xet: Chieu (+) cua cac vat nhir h i n h

• Lire cang day T|

* X e t v a t m i : Pi - T, = mi.ai

m,g - T i = m i a i c=> 2.10 - T, = 2.0,1884

d) Nhdn xet:

• T2 = 1 8 , 4 4 5 7 ( N )

• m2a2 = 2.0,188 4 = 0 , 3 7 6 8 ( N )

D o T2 > m2a2 n e n giOfa m2 va m a t s a n c6 lire m a sat - > Liic n ^ y lire m a s^t

giOra m2 v a s a n ngiroc h u d n g v 6 i T2 nhxi h i n h ve b e n d u d i

Day k h o n g trucrt t r e n r o n g roc va r o n g roc c6 t h e

quay quanh m o t true n k m ngang B a n dau h a i vat

dUOc giU dutog y e n r o i sau do buong nhe cho he

chuyen dong Cho g 10 m/s^

a) T i m g i a toe gdc cua r o n g roc v a g i a toe eiia 2

v a t m i , m2^

b) T i m lue cang cua day treo h a i ben r o n g roc

c) T i m q u a n g d u d n g m a m i , m2 d i duoc sau 2 s

k e tii lue t h a cho he c h u y e n d o n g v a goe m a

r o n g roc quay duac sau 2 s t r e n

V a t m i , m2, r o n g roc co e h i e u c h u y e n

d o n g c u n g c h i n h l a c h i e u ducfng d a c h o n

t r e n n h u h i n h v e

(1) (2)

B a i 4 M o t banh xe chiu tdc dung cua m p t momen lire M , c6 dp Idn khong

doi Trong t h d i gian 5 s dau toe dp goc ciia banh xe tSng deu tii 0 den 10

rad/s Sau do momen M , ngifng tac dung, banh xe quay cham dan deu

ditog h d n sau 20 s ke tCf lue M j ngCrng tae dung Biet r k n g trong qua t r i n h

quay t h i banh xe luon ehiu tdc dung ciia momen lye ma sat k h o n g doi va

CO dp Idn la 10 N m

a) T i n h gia toe goc cua bdnh xe trong hai giai doan quay n h a n h d a n deu

va quay eham dan d i u

b) T i n h momen quan t i n h eua banh xe doi vdi true quay va momen life M j

c) T i m so vong ma banh xe quay dupe trong giai doan quay nhanh dan deu

HUdng dan gidi Tom tdt

CO, - (0

72 = 0, 0 - 1 0 Y2 = -0,5 rad/s'

t^ 20 b) Momen quan t i n h cua banh xe doi vdi true quay va momen M i

Vay tii (•) N = f ^ = 3,98(v6ng)

2.71

BAI TAPTRAC NGHIEM

C a u 33 Cho m o t banh xe ehiu tdc dung ciia m o t momen lUc M i khong doi

Tong cua momen M i va momen lire ma sdt e6 gid t r i bkng BON.m Trong 2s

dau, toe dp goc cua banh bien doi deu tii 0 rad/s den 10 rad/s Sau do momen

M l ngi/ng tac dung, bdnh xe quay cham dan deu v^ difng h ^ n l a i 50s Gia suf momen lire ma sat l a k h o n g doi trong suot t h d i gian banh xe quay Dp I d n momen life ma sat la

A 12 N.m B 2 N m , C 3 N.m D 4 N m

C a u 34 Cho m p t banh xe chiu tac dung ciia m o t momen lire M i khong d6i

Tong eua momen M i va momen life ma sat cd gia t r i bkng 18N.m Trong 4 s

dau, toe dp goc cua banh xe bien doi deu ti^ 0 rad/s den 12 rad/s Sau d6 momen M i ngii'ng tac dung, banh xe quay cham dan deu va dCfng h ^ n l a i 24s Gia sijf momen luc ma sat l a khong doi t r o n g suot t h d i gian bdnh xe quay Momen liic M J a

A 20 N.m B 21 N m C 22 N.m D 23 N m

C a u 35 M p t banh xe n h a n dupc mpt gia toe goc 2 rad/s^ quay nhanh d a n deu

ti^ t r a n g t h a i dUng y e n t r o n g 4s dudi tac dung eiia momen life M j v^ momen

liie ma sat Sau do momen lUc M i ngUng tac dung t h i banh xe quay cham

dan deu va difng l a i sau 10 vong quay Toe dp gde sau 4s va t h d i gian tii liic

banh xe b^t dau quay cho den k h i diing l a i la

A 8 rad/s; 19,7 s B 9 rad/s; 12,3 s

C 8 rad/s; 12,3 s D 9 rad/s; 11,3 s

C a u 36 Mpt banh xe c6 k h o i lupng 2 k g va ban k i n h 20 cm quay nhanh d a n

deu tii t r a n g t h a i ddng y e n va sau 4 s d a t toe dp gde l a 40 rad/s Biet r l i n g

trong 4 s nay v a t chiu tac dung cua life F va liTc ma sat theo phupng tiep

tuyen banh xe Sau 4 s t r e n luc F mat d i , banh xe quay cham dan deu diTcfc

10s t h i dCmg l a i Dp Idn cua lUc F tdc dung len bdnh xe \k

A 0,7 N B 1,4 N C 2,8 N

C a u 37 M p t m a y A - t i i t dung de nghien cUu chuyen

dpng cua cac vat C6 k h o i lupng khac nhau n h i i

h i n h Cho m i = 1 k g , m2 = 3 k g rbng rpe quay

quanh m p t true n k m ngang Gia thuyet spi day khong dan va khong trUpt t r e n rong roc Cho kho'i lupng rong rpe la 2 kg, r = 10 em Cho g = 10 m/s^

Gia toe moi vat m i va m2 la

A 1 m / s l B 2 m / s l ' C 3 m / s l

D 3,6 N

D 4 vols

C a u 38 Mot dia dac ban k i n h 0,25 m c6 the quay quanh mot true doi xiJng di

qua tam eua no Mot soi day manh nhe dUdc quan quanh vanh dia ngudi ta

keo dau sgi day bkng mpt life khong doi 12 N Hai giay sau ke tix lue tdc

dung lire l a m dia quay vdi to'c do g6c b^ng 20 rad/s Gia toe goe eua dia va

gia toe eiia dau day la

A 10 rad/s^; 2,5 m / s l C 12 rad/s^ 4 m / s l

B 11 rad/s'^; 3 m/s^ D 10 rad/s^ 3 m / s l

C a u 39 Mot dia dac ban k i n h 0,25 m c6 the quay quanh mot true doi xvJng di

qua tam cua no Mot soi day manh nhe du'cfc quan quanh vanh dia, ngxXdi ta

keo dau soi day bkng mpt liic khong doi ION Bon giay sau ke tii luc tde

dung life t h i dia quay vdi to'c dp g6c hkng 40 rad/s Cho g = 10 m/s^ Goc quay

dupe ciia dia va c h i l u d ^ i doan day difpc keo la

A 40 rad; 5 m B 80 rad; 20 m

C 80 rad; 4 m D 40 rad; 10 m

C a u 40 Mpt vat nang 90 N diipc bupc vao dau 1 spi day nhe quan quanh 1

rong roc dac c6 ban k i n h 0,2 m, khoi liipng 2 kg Rong roe c6 true quay co

dinh nkm ngang va di qua t a m cua no Ngiidi ta tha cho vat rpi t i i dp cao h

xuong dat Cho g = 10 m/s^ Luc c&ng eua day va gia toe cua vat la

• co: Van toe toe (rad/s)

I I D i n h lu$t bao toan m o m e n dpng li^dng

-Cde triidng hpp dac biet:

• Hp gom 1 vat quay

HUc/ng dan gidi

Momen quan t i n h ciia dia mai

gia toe cua bdnh xe khong doi

• Gia toe cua banh xe

2 0 - 0

(1) - C O n

• To'c dp goc eua bdnh xe t a i th&i diem 8 s

(0 = coo + y.t = 0 + 4.8 = 32 rad/s

• Momen dpng iMng eua banh xe t a i thdi diem t = 8 s

L = I.co = 0,4.32 <^ L = 12,8kgmVs

B a i 3 Mpt thanh A B dai 80 cm co khoi liipng khong dang ke CJ mpt dau cua thanh, ngudi ta gMn vat m = 1 kg long vao thanh va co the triiPt tren thanh Lue dau vat d dau thanh va thanh dang quay vdi toe dp goc la 25 rad/s quanh true doi xiifng Sau do vat triipt den v i t r i each true quay 0,2 m T i m toe _ dp goc ciia vat luc nay

Tom tat

• ^ = 80 cm = 0,8 m

• m = 1 k g

HUdng ddn gidi Nhan xet: Do de cho thanh co khoi lupng khong

dang ke va quay quanh true doi xiJng nen

D i a c h i u m o t m o m e n life k h o n g d o i l a 2 N m n e n b ^ t d a u quay n h a n h d a n deu

v d i toe do goc luc d a u cOg = 0 M o m e n d o n g liTpng ciia d i a t a i thcfi d i e m 10s l a

C a u 4 3 M o t d i a dac c6 b a n k i n h R, d i a c6 t h e quay x u n g q u a n h t r u e d o i xiJrng

d i qua t a m v a v u o n g goc v d i m a t p h ^ n g d i a D i a c h i u t a c d u n g ciia 1 m o m e n

life k h o n g d o i M = 5 N m Sau 2 s k e t i f luc b d t d a u quay, toe do goc ciia d i a

ngu'di a y se c h u y e n d o n g n h u t h e nao? V d i v a n toe hkng bao n h i e u ? B i e t

m o m e n q u a n t i n h ciia n g u d i d o i v d i t r u e quay l a I = 2 k g m ^ v a ciia b a n h xe do'i v d i t r u e q u a y l a I = 1 k g m ^

A N g U d i a y c h u y e n d o n g c i i n g c h i e u quay cua b a n h xe v d i v a n toe gdc l a 2 0

0) = 20 r a d / s quay q u a n h t r u e do'i x i i n g Sau do v a t t r u p t d e n v i t r i each t r u e

quay 0,25 m Toe dp gdc cua v a t luc n a y l a

A 80 rad/s B 100 r a d / s C 200 r a d / s D 500 r a d / s

Van de 6: DQNG NANG DINH LI DQNG NANG CUA VATRAN CO

TRI^C QUAY CO DINH

Hitdng dan gidi

D o n g n a n g ciia b a n h xe d o i v6i t r u e quay eua n o

^ _ lo)^ _ 0,8.(807:)''

W , = 25240,576 ( J )

B a i 2. M o t b a n h xe d a n g d t r a n g t h a i diJng y e n c6 m o m e n q u a n t i n h d o i v d i

t r u e quay ciia n o l a 0,2 k g m l H o i p h a i t o n m o t c o n g l a bao n h i e u de b a n h

xe d a t duoe toe do goc l a 4 0 r a d / s B o q u a m o i siJc c a n

quay d e u vdi toe d p goe co = 5 r a d / s q u a n h m o t t r u e v u o n g goc vcJi d i a v a d i

qua t a m ciia d i a D p n g n a n g eiia d I a l a

t r u e quay l a 1,8 k g m ' Sau d o , ngiTdi n a y d o t n g p t t h u t a y l a i dpc t h e o t h a n

BAI TAP LUYEN TAP

B a i 1. M o t dia mong, phSng, dong chat c6 ban k i n h R quay quanh m o t true di

qua tarn va vuong gdc vdi mat ph^ng dia Tac dung vao dia m o t momen luc

20 N m khong doi, dia chuyen dong quanh nhanh dan deu v d i gia toe g6c 10

rad/s^ khong toe do goc ban dau Biet dia c6 khoi lurgng 4 kg Bo qua m o i luc

can T i m ban k i n h dia mong

Dap so: R = 1 m

B a i 2 Mot dia tron dong chat k h o i luong 2 k g b a n k i n h 20 em dang quay deu

quanh true vuong goc vdi m a t dia va d i qua t a m ciia dia v d i toe do goc 40

rad/s Tac dung l e n dia mot momen h a m c6 dp Idn l a 0,8 N m D i a quay

cham dan deu va diing l a i sau t h d i gian t T i m t

Dap so: t = 2 s

B a i 3 T a i t h d i diem t = 0, mot vat rSn bMt dau quay quanh mot true co dinh

xuyen qua vat v d i gia toe goc khong doi Sau 4 s no quay dircfc m o t goe 24

rad T i m toe do goe tiJc t h d i t a i t h d i diem 15 s

Dap so: M = 45 rad/s

B a i 4 Mot vat r a n quay quanh mot true co phuong t r i n h : (p = 20 + 4t^ (rad; s)

T i m toe do goc va goe ma v a t quay dirge sau t h d i gian 10 s ke tCr t h d i diem

t = 0 la bao nhieu?

Dap so: 80 rad; 400 rad

B a i 5. M o t ban t r o n phang n^m ngang ban k i n h 0,4 m co true quay co d i n h d i

qua t a m ban Momen quan t i n h ciia ban doi v d i true quay la 3 kgm^ B a n

quay deu vdi toe do goc 4 rad/s t h i ngu'di ta dat nhe mot vat nho khoi liTOng

400 g vao mep ban va v a t dinh chat vao do Bo qua ma sat d true quay va

sure can moi triidng T i m toe do gdc eua h$ (ban va vat)?

Dap so: 3,916 (rad/s)

B a i 6. V a t r ^ n thur nha't quay quanh mot true ed dinh cd momen quin ti'nh la

Ii = 2 kgm"^ va momen dong \uang la L j Vat r a n thiif hai quay quanh true cd

dinh ed momen quan t i n h l-z - 6 kgm^, momen dong lirgng la L2 Biet dong

nang ciia vat thu" nha't gap 3 Ian dong nang eua vat thuf hai T i m t i so' —

Dap so: ^ = 1

B a i 7. M o t v a t r a n quay deu quanh mot true ed' dinh cd dpng nang Ik 200 J ,

momen quan t i n h l a 4 kgm^ T i m momen dong lirong ciia vat

Dap so: 40 (kgm^/s)

B a i 8 Mot thanh A B dong chat ed chieu dai 1 m cd the quay trong mat phang

t h i n g dufng qua mot true nhm ngang di qua mot dau ciia thanh va vuong gdc vdi

thanh Thanh ed tiet dien deu, khoi lirgng m, gia td'c roi tif do la g = 10 m/s^ Bd

qua moi ma sat va siife can thanh dang diimg d v i t r i can bang H o i can phai

truyen cho thanh mot td'c do gdc bao nhieu de thanh quay den v i t r i nkm ngang

Dap an: 5,47 rad/s

B a i 1

rtr

JJai 9- t a n g to'c tir t r a n g t h a i diJng yen, m o t banh xe tieu to'n m o t cong la

2000 J Biet momen quan t i n h eua banh xe la 0,8 kgm^ Bd qua cac Ivic can

T i m td'c do eua xe dat dUdc

Dap so: 70,7 rad/s

B a i 10. Cho eo he nha h i n h ve:

ni, = 400 g, m^ - 100 g, rong roe cd R = 10 cm

C a u 12 Chpn D

Bieu thurc dung c6 dang a = ^a^, +

hay a = Jioy^rf + (ry)'

Khi vat rdn quay quanh mot true eo dinh xuyen qua vat thi cac diem tren

vat ran khong thupc true quay se vaeh len eae dudmg tron nkm trong mat

ph^ng vuong goe vdi true quay

C a u 21 Chpn B

Dung dinh luat bao toan momen dpng lupng L i + L 2 = Lj + L j

=> I i M i + I2CO2 = IiO) + l^Oi O I i Q i + l2(02 = d i + l2)-<»

- > CO va I t i le nghieh Ngirdi van dpng vien nay thu tay lai la lam I giam

di > I2) -> toe dp goe tang (C02 > coi)

C a u 32 Chon B ^

De thanh quay den v i t r i n^m ngang t h i dpng nSng ban dau phai hhng

the nang cua t h a n h a v i t r i k h i t h a n h n k m ngang

T a c o : i | = m g — = m g ^ o = ^

-C a u 33. Chon C

De thanh quay den v i t r i th^ng duTng cr phia ben true quay t h i dong nang ban

dau phai bkng the nang eua thanh d v i t r i th^ng diing d phia tren true quay

C a u 10

cp - cpo = coo.t + ^ -> 20 = 0.4 + ^ ^ y = 2,5 (rad/s^)

• CO = coo + y.t = 0 + 2,5.5 - 12,5 (rad/s) Chon A

C a u l l y ^ = ^ ^ = 2 , 5 ( r a d / s ^ )

t 4 Toe do goc ciia diem t r e n v a n h bdnh xe sau 2 s

—> Gia toe phap tuyen luc t = 3 s

y = 8 (rad/s') Chon A

C a u 27

M = I y ^ y = M = ^ = 7^5 (rad/s^)

1 4

• (0 = (Oo + yt -> t = 10 (s) Chon A

= | m R ' + mx^ = 0,502 (kgm') Chon A

C a u 30

8 ) 1

= I,„ + m x ' = | m R ' + m x ' = 1.1.0,2' + 1.0,04'

= 0,0216 (kgm') Chon D

• To'c do goc sau 4 s

(Oi = (Uo, + yi.ti = 0 + 2.4 =: 8 (rad/s)

• Gia toe g6c liic sau

Ta c6: 4 - co^, = 2y, ((p - cpo) ^ y^ = = = -0,509 (rad/s')

C a u 37

Do P, = m , g = 30N1

P, = m,g = lONJ ^ > ^ ' Chieu chuyen dong ciia cac vat nhir h i n h

• C h i l u dai doan day dUdc keo

- Gia toe cua dau day

CO = coo + y.t = 0 + 4.10 = 40 (rad/s)

L = Ico = 0,5.40 = 20 (kgm'/s) Chon C

C a u 4 2 Toe do goe cua T r d i D a t trong sir quay quanh true cua no

Do T r a i Dat quay 1 vong m a t 24 gid

CO - 1.27irad 24.3600s 43200 (rad/s) I = - m R ' = -.6.10'\6400000' = 9,8304.10" kgm^

5 5

• Momen dong Itrong ciia T r a i D a t trong sir quay quanh true cua n6

L = Ico = 7,14.10^^ (kgm^/s) Chon C

C a u 43

Y = (0-0)o 3 0 - 0 2x = = 1 5 (rad/s ) , M = I y ^ I = M = A = i (kgm^)

C a u 4 5

• I = Ithanh + I i + I2 = ^rd} + m^Rf + m , R ^ = 0.53 (kgm^)

• L = Ico = 0,53.40 = 21,2 (kgm^/s) ^ Chon B

C a u 4 6

• I = I , 4 n + Ivat = I M R 2 + m R 2 = 416 (kgm^)

• L = Ico = 416.20 = 8320 (kgm^/s) Chon D

( A )

M

O

• G i a toe goc ciia v a t r ^ n

y = (p - cpo = o)o.t + - — 2 4 = 0,4 + ^ <^ y = 3 (rad/s^)

2

• Toe do g6c tufc t h d i t a i t h d i d i e m 15 s

03 = 0)0 + y.t = 0 + 3.15 = 45 ( r a d / s )

B a i 4 Ta CO cp = 20 + 4t'^ (rad/s)

(p = (Po + COflt +

(Po = 20 rad

1 = 4 - ^ 7 = 8 rad/s' [2

• K h i t = 10 s t h i cp = 20 + 4.t^ = 20 + 4.10^ = 420 (rad) nen goc quay

(p - (Po = 420 - 20 = 400 rad

• Toe do goc k h i t = 10 s

CO = (Oo + y.t = 0 + 8.10 = 80 (rad/s)

B a i 5 A p dung dinh luat bao toan momen dpng liTgfng

• De t h a n h quay den v i t r l n ^ m ngang t h i

dpng nSng ban dau p h a i bkng the nSng

CHl/OfNG I I : DAO DONG CGf

P H A N I: T O M T A T LI T H U Y E T

yan de 1: DAO DQNG DIEU HOA

1 Dao d6ng

Dao dong la chuyen dong c6 gidi han trong khong gian, lap di lap lai nhieu

Ian quanh mot v i t r i can hkng V i t r i can b^ng thudng Ik v i t r i k h i vat

dufng yen

2 Dao d6ng tu^n hôn

a) Dinh nghla: Dao dong tuan hoan la dao dong ma trang thai chuyen dong

cua vat diTOc lap lai nhu cu sau nhulig khoang thdi gian bSng nhaụ

b) Chu ky cua dao dgng tudn hoan (T): Khoang thdi gian ng^n nhat sau

do trang thai dao dong lap lai nhu cụ

CO

c) Tan so cua dao dgng tudn hoan (f): So Ian dao dong (so trang thai dao

dong) duoc lap lai nhiT cu trong mgt dcrn vi thdi gian

VẠ/ V g ' Vgsina V Ạ/

Nghi^m ciia phucrng t r i n h (*) la x = Ạcosl cọt + cp )

Trong do:

* A: Bien do dao dong (la do Idn cua l i do dao dong ciTc dai) (m; cm)

* (cot + cp): Pha dao dong a thdi diem t (rad)

* cp: Pha ban dau (rad)

* co: Tan so goc ciia dao dong (rad/s)

Vay: Dao dgng dieu hoa la dao dgng diCgc mo td b&ng dinh ludt ham so

cosin (hoac sin) x = A cos(a)t + cp), trong do A, co, cp la cdc h&ng sọ

4 Li d6, vSn t6c gia t6c trong dao đng didu h6a

a) Li do: x = Ạcos( cọt + cp)

* |x| max = A: K h i vat d vi t r i bien

* |xj min = 0: K h i vSt qua vi t r i can bang

b) Van toe: v = x' (t) = - co Ạsin( cọt + cp ) = oj Ạcos( cọt + cp + ^ )

2

V max = : Klai vat qua vi t r i can bang (x = 0)

v| min = 0: K h i vgt a v i t r i bien (/x/mux = A)

c) Gia toe: a = x"(t) = - cộẠcos(oj.t + cp) = - cộx = cộẠcos((ọt + cp + 7t)

* a max = A : K h i vat d v i t r i bien

* ja| min = 0: K h i vat qua vi t r i can b^ng

V

d) Moi lien he vi pha cua x, v, ạ

* Van toe vuong pha vdi l i do (van toe nhanh pha hcfn l i do mpt goc —)

• Gia toe vuong pha vdi van toe va ngUgfc pha vdi l i do (gia toe cd do Idn t i 1$ vdi do Idn eiia

li do)

• X , V , a bien thien dieu h6a vdi chu ky T tan

so f, tan só goc co

Van de 2: CON LAC DON - CON LAC VAT U

Nghiem ciia phtfcmg t r i n h (*): s = Acos(cot + cp) ( p h i r o n g t r i n h t h e o c u n g l e c h ) ,

3. Ong dung: Dung c o n l i e v a t l i do g i a toe t r o n g t r U c m g g. Biet g i a t r i c u a g

CO the suy r a sir p h a n bo k h o i liTcfng k h o a n g v a t d dirdi m a t d a t t r o n g v u n g

do ( g i u p cho viee t i m mo d a u , n g u o n m/de diidi d a t )

Dao dpng c u a c o n l i e dpn, va c o n l i e 16 xo dadi t a c dung c u a lire the (trong

l a c va lire d a n hoi) va khong c6 ma sat n e n c o nSng c i i a no dupe bao toan

Do do, CP iiSng eua v a t dao dpng duoe bao toan

Wt = - m c o ' A W ( c o t + cp) (**)

* Do t h i Wt vJng v d i trirdng hpip (p = 0

3 Bilu thL/c dong nSng

* T a i t h 6 i diem t bat k i v a t nang m c6 v a n toe l a :

K e t l u a n : C P nSng eua con l i e 16 xo bao toan va c P nSng t i le v d i b i n h

phuong bien dp A ciia dao dpng

Do t h i Wt, Wd ve t r o n g cung m o t he true toa dp

• Dong nang vat naug W , i v a l l u ; n a n g i o \ cung bie'n t h i e n tuan ho^n

vdi chu k y —, t a n so 2f, t a n so goc 2 co Do'i vdi ca n&ng con Idc W la mpt

2 h&ng so khong thay doi

Van de 4 : DAO DQNG TA T DAN VA DAO DQNG DUY TRJ

1 Dao d(5ng t^t dSn

a) Dink nghia: Dao dong c6 bien dp giam dan theo t h d i gian

hi Nguyen nhan gay ra dao dgng tdt dan: Do ma sat va lire can cua moi trUdng

c) Ifng dung: Cai giam rung

2 Dao d6ng duy tri

a) Dink nghia: Dao dong c6 ma sat va lire can ciia moi trircfng nhiTng duorc

chong l a i sir t ^ t dan sao cho bien do va t a n so cua dao dong du'oc duy

t r i khong doi

b) Cdch thiCc hien: Bang each bu nSng lu'Ong diing b^ng nang lircfng bi tieu

hao do ma sat sau moi chu ky

c) ling dung: Dong ho qua l^c, dua vong

Van de 5: DAO DQNG Cl/dNG BLfC - CQNG HifdNG

1 Dao d6ng cuflng bQc

Tac dung len vat mpt ngoai lire F bien doi dieu h6a theo t h d i gian c6 dang

F - Fo.coscu.t

Chuyen dong eiia vat gom h a i giai doan:

- Giai doan chuyen Hep: Dao dong cua he chira on d i n h , bien do cii t a n g

dan, cue dai sau Idn hon circ dai trUdc

- Giai doQn on dinh: Gia t r i bien dp khong doi va keo dai cho den k h i

ngoai lire dieu h6a t h o i tac dung

Dao dpng cua vat trong giai doan on d i n h diroc goi la dao dpng cirSng biJc

* D S c d i e m c i i a dao d p n g cu'Sng hvlc

- Dao dong cUdng biJc la dieu hoa

- Tan so gde cua dao dong cudng hiic hhig t a n so goc cua ngoai liTc

- Bien do eua dao dong eiTorng biJc t i le thuan vdi bien dp Fo cua ngoai lire

va phu thupe vao t a n so goc co ciia ngoai lire

2 Cflng hadng: H i e n tirong bien do cua dao dpng cUdng biJc t a n g den gia t r i

cire dai

Dieu hien: Tan so goc co = t a n so goc rieng cOo (hay tan so f = tan so rieng fo)

Neu ma sat giam t h i gia t r i circ dai cua bien dp tang H i e n tiTPng cpng

hucfng ro n e t hpn

^ " p h a n bi$t dao dflng cuang bQc v6i dao dflng duy tri

• G i o n g n h a u :

_ Deu xay ra dirdi tac dung ciia ngoai lire

_ Ca hai deu co tan so goc gan dung bkng tan so goc rieng cua he dao dpng

• K h a c n h a u :

_ Trong giai doan o n d i n h t h i t a n

so goc dao dpng ciTomg biJe luon

b^ng t a n so' goc ngoai lye (hoac

f = fo)

- Dao dpng ciTSng bdc xay r a trong he diTcii tac dung eua ngoai lire doc lap vcii he

- T a n so goe ngoai lire luon dieu chinh de b^ng t a n so' goc dao

dpng tir do cua he (hoac f = fo)

- Dao dpng duy t r i l a dao dpng

rieng ciia he diTpc bu t h e m nang liTOng do m o t lire dirpc

dieu k h i e n bdi ehinh dao dpng

ay qua mot eP cau nao do

4 Dng dung hi6n tUdng cflng hudng

Che tao t a n so k e , l e n day dan

cpng hiTcing co h a i nhiT co the dan

dpng cirdng biire

Trong mot so trircfng hop, hien tirpng

t d i ket qua l a m gay, v5 cac vat bi dao

Van de 6: TONG H0P DAO DQNG

Tong hap hai dao dgng bang phitamg phdp Fre-nen

Mot v a t dong t h d i t h a m gia h a i dao dpng dieu h o a cung t a n so co cac phiTPng t r i n h Ian liTPt l a :

X i = Aicos(cot + (pi), X2 = AaCOsCcot + cpa)

Dao dpng tong hop eua hai dao dpng t r e n

X = xi + x^ = A eos( cot + ip)

A , e o s c p i + A j C o s c p j

P H A N I I T R A C N G H I E M L I T H U Y E T

Cau 1. Tai mot noi xac dinh, mot con lac dcfn dao dong dieu hoa vdi chu ki T,

khi chieu dai con lAc t&ng 4 Ian thi t i n so con Mc

A Khong doi B TSng 4 Ian C Giam 2 \in D TSng 16 Ian

Cau 2. Mot chat diem dao dong dieu hoa tren true Ox vdi chu k i T Vi t r i can

b^ng ciia chat diem trung vdi goc tpa dp, khoang thdi gian ng^n nhat de no

di tii vi tri c6 l i do x = - A/2 den vi tri c6 l i do x = A/2 la

A T/3 B T/6 C T/2 D T/4 (p

C a u 3. Mot vat nho dao dong dieu hoa tren true Ox theo phiiong trinh

x = Acos(a)t + cp) Van toe ciia vSt c6 bieu thiifc la

A V =Asin(a)t + cp) B v = -coAsin(cot + cp)

C V = - 2a)Acos(cot + cp) D v = a)Acos((ot + (p)

Cau 4. Trong dao dong dieu hoa, van toe tufc thdi ciia vat dao dong tai mpt thdi

diem t luon

A CCing pha vdi l i do dao dong

B Sdm pha TC/6 S O vdi l i do dao dong

C Ngucfc pha vdi l i do dao dong

D Lech pha n/2 so vdi l i do dao dong

Cau 5 Mot vat dao dong dieu hoa vdi bien do A, tan so gdc co Chon goc thdi

gian la luc vat di qua vi t r i can hkng theo chieu du'cfng Phuong trinh dao

dong ciia vat la

A X = AcosCcot + n/2) B x = Acos(cot - 7t/2)

C X = Acoswt D X = Aeos(wt + n/4)

Cau 6. Van toe va gia toe ciia mpt v^t dao dong luon bien thien dieu hoa cCing

tan so va

A Cijng pha vdi nhau B NgUcfc pha vdi nhau

C Lech pha vdi nhau n/2 D Lech pha vdi nhau n/4

C a u 7. Mot con l^c Id xo gom mot 16 xo c6 khdi liipng khong dang ke, do cdng

k, mot dau co dinh va mot dau g^n vdi mot vien bi nho khoi liipng m Con

lie nay dang dao dong dieu h6a ed ecf n&ng

A. Ti le nghieh vdi kho'i luorng m cua vien bi

B Ti le thuan vdi binh phuong bien do dao dong

C .Ti le thuan vdi binh phuong chu ki dao dong nhutig t i le nghieh vdi tan so

dao dong

D Ti le nghieh vdi do edng k eua 16 xo nhiing t i 1? thuan vdi binh phMng

bien dp dao dong

Cau 8. Mot con lac 16 xo gom mot 16 xo c6 kho'i lifpng khong ddng ke, mot dau

CO dinh va mot dau g^n vdi mot vien bi nho Con lac nay dang dao d6ng dieu

h6a theo phiicfng nkm ngang Lire dan hoi eua 16 xo tde dung len vien bi luon

A

B

C

A Hudng ve vi tri can hkng cua vien bi

B Cung hading vdi v cua vien bi khi con l^c 16 xo chuyen dong cham dan

C Luon C l i n g hudng vdi v ciia vien bi

D NgiTpc hudng vdi v cua vien bi khi con l^c 16 xo chuyen dong nhanh dan

C^u 9. Mot con lAc 16 xo gom mpt vat ed kho'i liTpng m va 16 xo ed dp ciifng k,

dao dong dieu h6a Neu tSng dp edng k len 4 Ian va giam kho'i lupng m di 4

Ian thi tin so dao dong cua vat se

A Tang 4 l l n B Giam 4 Ian

C Tang 16 Ian D Khong ddi

Cau 10. Mpt vat nho dao dong dieu h6a c6 bien dp A , chu k i dao dong T, d thdi

diem ban dau to = 0 vat dang qua vi t r i can hhng Quang diidng m^ vat di

dupe tCr thdi diem ban dau den thdi diem t = T/2 la

A 'A/4 • B. A/2 C 2A. D A

Cau 11. Chu ki cua con l^c 16 xo

A T = 2 ^

k

B T = 2nJ— C T = 27T ,

(m k m D T = 27t

CSu 12. Mpt vat dao dong dieu h6a vdi tan so gdc co O l i dp x vat c6 van toe

V. Bien dp dao dong ciia vat dupc tinh bdi cong thdc

Cau 13 Mpt con lac 16 xo treo th^ng ddng, gom mpt h6n bi nang co kho'i liipng

m treo vao 16 xo c6 dp edng k Khi vat can bang, 16 xo dan ra mpt doan A/

Kich thich cho eon lac dao dong, eon l^c dao dong vdi chu ki la

VA7

A, T = 271 A( B T = 27i _g_

AC C T = 1 A( D T =

Cau 14. Xet dao dong dieu h6a cua mpt con Idc 16 xo Gpi O la vi t r i can hhng

M, N la 2 vi tri bien P la trung diem OM, Q la trung diem ON Trong 1 chu

ki, con Me se chuyen dong nhanh dan trong khoang

A. Tir M den O B TCr P den O, tir O den P

C TCr M den O, tCr N den O D TCr O den M, tCr 0 den N

Cau 15. Chu ki cua eon Me vat l i diTpc xac dinh hkng cong thuTc

1

A T =

2n C.T = 2n

mgd

I mgd

C a u 17 Chon cau sai

Tir cong thiJc T = 2 TI

mgd ciia con lac vat l i cac dai luong trong cong

thufc dMc hieu nhu sau

A I : Momen quan t i n h cua vat r ^ n doi vdi true quay

B m : Khoi lUcfng vat rMi;

C d: Khoang each giufa h a i diem bat k y tren vat

D T: Chu k y ciia con Idc v a t l i

C a u 18 K h i dira mot con Idc 16 xo len cao theo phucfng t h i n g diing (coi chieu

dai cua con lie khong doi) t h i t a n so' dao dong dieu h6a ciia no se

A giam v i gia toe trong triTdng giam theo do cao

B khong doi v i chu k i dao dong dieu h6a ciia no khong phu thuQC vko gia toe

trong trirdng

C tSng v i chu k i dao dong dieu h6a cua no giam

D tang vi tan so' dao dong dieu hoa cua no t i le nghich v<Ji gia toe trong trifdng

C a u 19 Mot vat dao dong dieu h6a c6 chu k i la T Neu chon go'c thdi gian t = 0

luc vat qua v i t r i can bkng, t h i trong mpt chu k i dau tien, v a n toe ciia vat

bkng khong may Ian

A 2 Ian B 4 Ian C 6 Ian ' D 8 Ian

C a u 20 Ccf n&ng cua mpt vat dao dong dieu hoa

A tang gap bon Ian k h i bien dp dao dong cua vat tang gap doi

B bien thien tuan hoan theo th6i gian vdi chu k i b^ng chu k i dao dong ciia vat

C bien thien tuin hokn theo thdi gian vdi chu k i bang mpt nila chu k i dao

dong ciia vat

D bkng dong nang ciia vat k h i vat cf v i t r i bien

C a u 21 Mpt vat dao dong dieu h6a doc theo true Ox, quanh v i t r l can hkng 0

vdi bien dp A va chu k i T Trong khoang th6i gian T/4, quang diTdng nho

nha't ma vat c6 the d i difctc la

C a u 22 Doi vdi eon l i e 16 xo dao dong dieu h6a t h i the nang cua 16 xo v ^ dpng

nang cua hon bi dao dong dieu h6a bien doi theo th6i gian

A tuan hoan vdi chu k y —

C a u 23 M p t vat nho c6 kho'i liTpng m dao dpng dieu hoa t r e n true O x theo

phuong t r i n h x = Aeos(cot +(p) Dong nang ciia vat t a i thdi diem t la

C f i ^ 24 Phat bieu nao sau day sai k h i n6i ve dao dong ca hoc?

A- Dao dong tit dan c6 bien dp giam dan theo thdi gian

B T a n so' cua ngoai lire cifdng biJc hkng t a n so' dao dpng rieng cua h f dao

kglH, dpng t h i xay ra cong hufdng

^ H ^ C Dao dpng cirdng buTc xay ra trong hp dudi tdc dung cua ngoai luc dpc lap

^ Q Dao dpng t a t d i n eo bien dp khong doi theo thdi gian

Cau 25 Mpt vat tham gia dong thdi hai dao dpng dieu hda cung phuong, cimg

t i n so: x i = Aieos(cot + cpi) va X i = A2cos(cot + cp2) Bien dp dao dpng tdng hop la

A A - yJAl + AI + 2 A i A 2 cos((p2 - cPi) •

B A = V A ? + A ^ -2AiA2Cos((p2 -cpi) •

C A = A i + A2 + 2 AiA2Cos(cp2 - cpi)

D A = A i + A2 - 2 AiA2Cos(cp2- cpi)

C a u 26 Hai dao dpng dieu hda eo cung t i n so Goi Acp la dp l§ch pha giOra hai

dao dpng Trong dieu kipn nao t h i h a i dao dpng ciing pha ?

kn

D. Acp = (2k + 1 ) -

C Acp = k27i

C a u 27 H a i dao dpng dieu hoa c(3 ciing t a n so Goi Acp la dp lech pha giiJa h a i

dao dpng Trong dieu k i ^ n nao t h i h a i dao dpng ngupc pha ?

Trong dieu k i $ n n^o t h i h a i dao dpng vuong gdc pha

C Acp = k27i D Acp = ( 2 k + 1 )

-2

C S u 29 T a i mpt ncfi x a c dinh, chu k i dao dpng dieu h d a ciia con l i e dPn

A T i le thuan vdi chieu dai con l i e

B T i Ip thuan v d i gia toe trong trudng

C T i le thuan v d i c S n bae h a i cua chieu dai con l i e

D T i le thuan vdi t i n so' dao dpng dieu h d a cua eon l i e

C a u 30 Chu k i dao dpng dieu h d a ciia mpt con l i e dcfn ed chieu dai day treo / tai noi cd gia to'c trong trUdng l a

A T = 2 , j l B T = 2 , ^ C T = i | ^ ^ = i f

C a u 31 d nai c6 gia t6'c t r o n g trU6ng g, con l i e don c6 day tree dai / dao dong

dieu hoa vdi t a n so goc l a

Vg ^

C a u 32 T a i mot nai t r e n mSt dat, chu k i dao dong dieu h6a cua con l^c dcfn

A Tang k h i kho'i \Mng vat nang cua con lac tang

B Khong doi k h i kho'i lu'cfng vat nang ciia con Idc thay doi

C Khong doi k h i chieu dai day treo ciia con ISc tang

D Tang k h i chieu dai day treo ciia con l^c giam

C a u 33 M o t con Mc dcfn ducfc treo d t r a n mot thang may K h i thang may diJng

yen t h i con l^c dao dong dieu hoa v6i chu k i T K h i thang mdy di xuong t h ^ n g

duTng, cham dan deu vdi gia toe c6 do Idn bang mot niira gia toe trong trUdng

t a i ncfi dat thang may t h i con Mc dao dong dieu hoa vdi chu k i T ' b^ng

A T j - B C T/2 D 2T

C a u 34 N h a n d i n h nao sau day diing k h i ndi ve dao dong cd hoc t ^ t dan?

A Dao dong t ^ t dan l a dao dong c6 bien dp giam dan theo t h d i gian

B Luc ma sdt ckng Idn t h i dao dong t a t dan cang cham

C Trong dao dong t ^ t dan, cd nang t a n g dan theo t h d i gian

D Dao dong t ^ t dan c6 cof nang khong doi

C a u 35 Mot con Idc ddn gom sdi day c6 k h o i Idcfng khong dang ke, khong dan,

C O chieu dai / va vien b i nho cd kho'i Iddng m Kich thich cho eon lac dao

dong dieu hoa d noi cd gia toe t r o n g t r u d n g g vdi bien dp goc Q Q Chon md'c

the nang t a i v i t r i can bang cua vien b i t h i the nSng ciia con l i e nay d l i dp

goc a cd bieu thufc l a

A mgl (1+eosa) B mg^ (1 - eosa)

C mgl (3cos - 2eosao) D mgl (3cosa +2cosaQ)

C a u 36 Phat bieu nao sau day sai k h i ndi ve dao dong ciia con ISc ddn (bd qua

life can eiia moi trUdng)?

A Vdi dao dong nho t h i dao dong cua con lie l a dao dong dieu hoa

B K h i vat nang d v i t r i bien, cd nang cua con l i e b k n g the nSng eua nd

C Chuyen dong cua con Ide tii v i t r i can bkng ve v i t r i bien la cham dan

D K h i v a t nang d i qua v i t r i can bang, t h i t r o n g life tac dung len nd can

hhng vdi lye cang cQa day

* A: Bien dp dao dong cua vat ( L i dp ciTc dai) (m; cm )

* CO : Tan so gde (rad/s)

* u) t + ip : Pha dao dong d thdi diem t (rad)

* cp : Pha ban dau t = 0 (rad)

* L = 2A: Chieu dai quy dao (m; cm )

2 Chu ky T: T h d i gian de he thuc h i e n mot dao dong toan phan

271 ^ At

oj N

Trong do: ^

* T: Chu ky (s)

* A t : T h d i gian thiTc h i e n N dao dong (s)

* N : So dao dong ma vat thue h i e n trong t h d i gian t

3 TSn s6 f: So dao dong toan phan ma vat thiic hien trong mot ddn v i thdi gian

= 2nf

Trong do:

* f : Tan so (Hz)

4 Bilu thQc van t6c gia t6c cua v$t dao dong di^u hda

a) Bieu thiCc van toe

-BAI TAP MAU

B a i 1. M o t v a t dao d o n g d i e u h o a c6 p h i r o n g t r i n h :

X = 6cos(27it) ( c m ) a) T i m do I d n ciia v a n t o e cife d a i v a g i a toe eye d a i

b) T i m g i a toe cija v a t d t h d i d i e m t - 2s

Tom tat

X = 4cos(7it + — ) ( c m )

6 a) V i e t b i e u thiJc v a n A = 4 ( c m )

Hu&ng dan gidi

a) D e cho: B i e u thiJc h do: x = 4eos(nt + — ) (cm)

B a i 3. M o t v a t n S n g dao d o n g dieu hoa v 6 i chieu d a i quy dao l a 8 em V a t thire

h i e n diroc 10 dao d o n g m a t 3,14s T i m v a n toe cua v a t k h i qua v i t r i +2 c m

; N

Tom tat

L = 8 c m , N = 10 dao d o n g

c) Cac t h d i d i e m v a t q u a v i t r i 4 e m t r o n g m o t c h u k y d a u t i e n

Tom tat

x = 8eos 47:t + ^

6 ( c m ) a) T i m cac t k l i i qua x = 4 em

b) T i m t k h i X = - 4 c m theo

chieu dircmg I a n 2 c) T i m cac t k h i X = 4 c m

Do v a t c h u y e n ( l o n g t l i e o chieu duang —> v > 0

V = —coA.sin(a)t + 9 ) — 4 7 t 8 si n 47rt +

-6) > 0

<-> sin 47lt + < 0 ( 3 ) (do CO, A > 0)

Doi chieu (1), ( 2 ) v6i (3)

Ket hdp (1), (2)

(1)

(2)

t = ^ ( s ) ; = | , s ,

B a i 5. M o t chat diem M chuyen dong t r o n deu t r e n dudng t r o n t a m 0 c6 ban

k i n h 10 cm v d i toe do d a i la 20 cm/s H i n h chieu cua chat diem M l e n mot

dudng k i n h ciia dudng t r o n n a m trong m a t ph^ng quy dao ciia chat diem

la mot dao dong dieu hoa c6 bien do A , chu k y T T i m A, T

Tom tdt

• R = 10 cm

• V = 20 cm/s

a) T i m 'A, T

Iliidng dan gidi

• Bien do A ciia chat diem dao dong dieu hoa b^ng ban k i n h R cua chat diem chuyen dong

t r o n d e u - > A = R = 10 c m Xet chuyen dong t r o n deu

v = « R ^ c o = ^ = ^ = 2 ( r a d / s ) ; T = ^ = M i i i _ T = 3,14 (s)

Vay chu k y cua chat diem dao dpng dieu hoa cung se c6 gia t r i la

T = 3 , 1 4 s

BAI TAP TRAC N G H I E M

C a u 1- M o t vat dao dong dieu hoa c6 phifong t r i n h : x = 4.cos(2Ttt + - ) ( c m ; s )

Van toe ciia vat t a i t h d i diem t = Is la

^ V = 0 B V = -STI cmys

C a u 2 M o t va t dao dong dieu hoa c6 phucrng t r i n h : x = 2cos(47it)(cm;s) Pha

ban dau ciia v a t va gia to'c d t h d i diem t = 0,5(s) ciia vat la

C 0; 3,271^ cm/s. Q 7i/2 rad; - 327i^ c m / s l

C a u 3 M o t va t dao dong dieu hoa c6 phuang t r i n h : x = 5cos(7xt + cp)(cm;s) K h i

pha dao dong la — t h i v a n toe va gia toe ciia vat Ian luot la

C a u 5. M o t vat dao dong dieu hoa theo phuang t r i n h x - 6cos (4 7:t) (cm; s) Tpa

do ciia vat t a i t h d i diem t = 2s l a

A. X = 3 cm B. X 6 cm C x = - 3 cni D x = - 6 cm

C a u 6. M o t vat dao dong dieu hoa theo phaong t r i n h x - 4cos(47:t) (cm;s) V a n

toe ciia vat t a i t h d i diem t = 1,5s la

C a u 8 M o t va t dao dong dieu hoa c6 chu k y T = 2(s) T i m v a n toe ciia va t t a i

vi t r i +3cm va pha dao dong luc do l a —rad

3

A 3nV3 cm/s B - 37:73 cm/s |

C 0,3^/3 m/s D 7:73 cm/s

C S u 9. T i m toe do ciTc dai cua v a t dao dong dieu hoa c6 tin so f = 2 H z Biet

vat CO l i dp 4 cm k h i d v i t r i bien dUdng

A I67: m/s B 0,167: cm/s C I67: cm/s D 160 cm/s

C a u 1 0 K l i i mot v a t dao d o n g d i e u h 6 a qua v i t r i c a n b k n g t h i toc dp cua v&t l a

67t(cin/s) v a do l<Sn g i a toc cifc d a i 1^ 127:^(cm/s^) T a n so goc \k b i e n do l a

L i f u y : * Quang duang vat di trong 1 chu ky luon la 4A

* Quang dubng vat di trong 1/2 chu ky luon Id 2A

* Quang diCang vat di trong 1/4 chu ky la A neu vat xudt phdt ta VTCB hoac vi tri bien

G o i : S i la quang dirdng vat d i dirac trong t h d i gian N —

S2 la quang dudmg dai n h a t hoac ngdn nha t vat di dugc tron g t h d i

BAI TAP MAU

B a i 1 Mot chat diem dao dong dieu hoa vdi chieu d<\ quy dao la 10 cm T i m

thdi gian ng^n nhat de vat:

a) D i t d vi t r i can bang den v i t r i 2,5 cm

B a i 2 Mot v$t dao dong dieu hoa c6 phddng t r i n h :

X = 6cos Tit + (cm; s) T i m quang dddng ma vat di ddOc

a) trong mot chu ky.' b) trong — chu ky ke t d v i t r i can bang

4

can hhng

Iliidng dan gidi

a) Quang dddng ma vat ddoc di dddc trong 1 chu ky luon la

S = 4A = 4.6 ^ S = 24 cm

b) Quang dddng ma vat di ddoc trong - chu

ky ke t d vi t r i can bang luon la

S = A = 6 c m

c),ti = 0 den ta = 2,5 s c) Quang dudng ma vat di diSac tir t i = 0 den

Bai 3 Mot chat diem dao dpng dieu hoa vdi bien dp A, chu ky dao dpn la T,

tan so goc co Tim quang duorng dki nhat va quang difdng ng^n nhat ma

chat diem c6 the di dupe trong

Hitofng dan gidi

a) S m a x = ? S „ i n - ? trong 1 chu ky

4

• Goc ma chat diem quay dUPc trong —

4 chu ky

Bai 4, Mot chat diem dao dpng di^u hoa co phuong trinh:

X = 4cosl07rt (cm; s) Tim van toe trung binh va toe dp trung binh cua chat diem trong

a) thdi gian mot chu ky b) thdi gian nijTa chu ky tif bien am den bien ducfng ^