Phần 1 tài liệu Phương pháp giải bài tập trắc nghiệm cơ học 12 do Trận Trọng Hưng biên soạn cung cấp cho người đọc các phương pháp giải bài tập trắc nghiệm về: Động lực học vật rắn, dao động cơ. Mời các bạn cùng tham khảo nội dung chi tiết.
P^-| 'S^^ P T R A N T R O N G Hl/NG canoe 12 I NHA XUJIlT BAN T £ N G H9P THANH PHO HO CHf MINH cty P H l / d N G PHAP G I A I BAI TAP T R A C N G H I E M C d H Q C 12 C h i r o n g I R A N T R Q N G HLfNG Bai D O N G LlXC H O C V A T R A N C H U Y E N DONG Q U A Y C U A VAT RAN QUANH MOT TRgC C O DjNH C h j u t r a c h n h i e m xua't b a n TOM TAT LI T H U Y E T N G U Y E N T H I T H A N H HUCfNG B i e n tap : HAI AU TNHH MTV DVVH Khang Vl§t T o a goc - \a ban in : H O N G H A I V i t r i cua vat tai m o i thcJi diem se diTdc xac dinh bang goc cp giffa mot mat phang dong T f i n h bay : Cong ty K H A N G V I E T P gan vcti vat va mat phang co dinh PQ (ca Bia : Cong ty K H A N G V I E T hai mat phang deu chiJa true quay A Z ) Goc (p goi la toa goe cija vat NHA XUAT BAN TONG H P TP HO CHf MINN NHA SACH TONG H P 62 N g u y e n T h j M i n h K h a i , Q l D T : 38225340 - 38296764 - 38247225 - Dcfn vi cua la radian (rad) - N e u chon chieu du'dng la chieu quay cua vat thi (p > T S c d o g o c - Toe goc la dai lu'dng dac trUng cho mu^c nhanh, cham cua chuyen dong quay Fax: 84.8.38222726 E m a i l : tonghop@nxbhcm.com.vn Acp Toe goc trung binh: V^ebsite: w w w n x b h c m c o n i v n / w w w f i d i t o u r c o m Toe dp goc ttfc thdi: co = Tong phdt hanh - d(p — dt D d n vj toe dp goc la rad/s ; • G i a toe g(>c C O N G T Y T N H H MTV DjCH Vg VAN HOA KHANG V I E T Dia chl: 71 Dinh Tien Hoang - P.Da Kao - Q.1 - TP.HCM Dien thoai; 08 39115694 - 39105797 - 39111969 - 39111968 Fax: 08 3911 0880 Email: kliangvietbookstore@yalioo.com.vn - -•.;», 1/^ !r\ J Gia toe goc la dai liTdng dac trUng cho sir bien thien nhanh hay cham cua toe dp goc Aco - Gia toe goc trung binh: Ytb - Gia toe goe tii'e thdi: Y= - D d n vj cua gia toe goe la rad/s^ Website: www.nhasachkhangvlet.vn At dco dt ' j' 4, Cac phrf t = , = , ' (yt) y.t ^y.tana / = N/3 ~ 1,3 ( S ) , is G6c q u a y c u a t h a n h : Acp = coot + - y t ^ ChondapanB L u c d a u (t = 0) t h a n h b a t d a u q u a y (cOo = 0) nen: • 2A(p A(p = - y t ^ y = — ^ v d i A(p:=60" = - r a d Lf THUYET VA BAI TAP TRAC NGHIEM ^^^Aq 1.1 Chpn cau dung Toe goe eua vat rin quay t = ls A tinh bang cong thiJc At B tinh bang ddn vj rad/s _27t j , „ ^ N e n : y = - ^ = — (rad/s^) ^ I C dae tru'ng cho mu'c nhanh, cham cua ehuyen dong quay D du"dng neu vat quay nhanh dan va am neu vat quay cham dan Gia toe tiep tuye'n cua dau A : a, - r.y = 0,3 — = 0,628 (m/s^) •3 ChondapanA ' ^ -.p ^ V ^ ' : ' 1.2 Chon cau sai Chuyen dong quay deu c6 n^a/rS-i'f^' V I du M o t dTa m a i bat dau quay nhanh dan deu tiT nghi K h i dia quay du-dc I goc l,5rad thi mot d i e m tren vanh dTa dat gia toe tiep tuyen la 2m/s^, gia toe ; cua d i e m CO Idn x a p x i bhng ; A a - 2,5m/s^ ; B a = 12,6m/s^ D a = 6,3m/s^ )V'!(3 ff :•:•!; ;',M ""-^ ' ; ->^:.2Acp Y JV • ^7- Vay ^ (ifh B toe goe khong doi ' ' " C toa goc du'de tinh bdi: cp = cpo + cot ' ' " 1.3 Chon cau diing V d i mot ehieu du'dng bat k i thi ' - , • B vat quay theo ehieu du'dng thi co > 0, vat quay theo chieu a m thi co < r? T , a,, =ft> r a„ CO ^ , v a = Ja;, + a, v d i — = — ma co = 2yA(p (vi 03o = 0) _ _ [a, = r y a, y j i • A toa goe cp > Giai ; I A gia toe goc la hhng so" ' D goe quay ti le vdi thdi gian quay , C a = 5,8m/s^ " C vat quay (co y ;^ 0) theo ehieu du'dng thi coy > D vat quay nhanh dan thi y > - • '• ''^ '•••y':i 'CH-^y-^'Aiy: : v ! ^ : \ 1.4 Chon cau diing K h i vat rjfn quay thi eac d i e m khac tren vat ran (khong '< J ft I = 2A(p - > a„ = 2A(p.a, > 1* : ' : u : ' t ' H ' : , : sa?v' a, nam tren true quay) co cung j •,- , J " uU IV D o do: a = ^/(2A(p.a,)^+a^ = a, ^(2Acp)^ + = 27(2.1,5)'+ ^ , m / s ^ ' ^ ^ ' ,,,,, C h o n d a p a n D - ^ ;W;;ô.Hf r f - ã W ' ' : A - V i du 4, M o t banh xe dang quay nhanh dan deu v d i gia toe goc Irad/s^ quanh A toa dp goe va goe quay ' C to'c dp goc va toe dp (dai) B goe quay va gia to'c g6c D gia toe goe va gia toe (dai) 1.5 M o t vat rKn quay nhanh dan deu tif nghi, to'c dp goc ti le vcjii A goc quay B binh phu'dng thdi gian C Ccin bac hai cua thdi gian D can bac hai eua goc quay I 1-6 Chpn cong thiJc diing >;•!; / l u tarn O Sau thdi gian t ke tiT luc banh xe bat dau quay, d i e m M tren vanh banh xe c6 vectd gia toe a hdp vdi ban kinh C M goc a = 30° T h d i gian t xap x i bang A t = l , s dt At ( ; i At • -7 Chpn ehieu diTdng la ehieu quay eua vat ra'n thi B.t=l,3s C t = 2,5s D t = 0,9s i r j A 0) > B y > C co.y > D co.y < 1.8 Chon cau sai noi ve chuyen dpng quay cua vat rin 1.18 Canh quat dai 50em quay deu vdi to'c dp 90 vong/phiit To'c dp dai cua dau A Gia to'c goc dac trU'ng cho siT bien thien cua toe dp goc d moi thdi diem canh quat la B Khi vat ran quay deu thi veetcf van to'c eua mot diem tren vat ran khong A v = 18,84 m/s B v = 6,58 m/s C v = 24,32 m/s D v = 12,65 m/s 1.19 Mot dTa mai co ban kinh 20cm bat dau quay nhanh dan deu ttr nghi Sau C Gia toe hu'dng tam dae trU'ng cho siT thay doi ve hifctng eua vectd van toe thdi gian lOs ke tu* liie bat dau quay toe dp dai ciia mot diem tren vanh dTa la D Gia toe tie'p tuye'n dac trifng cho su" thay doi ve Idn cua veetcf van to'c 20m/s Gia toe gde quay cua dTa bhng , , , i , 1.9 Chon cong thiJc dung A a„ = V ^ r B a„ = C a = r^w'^ +y- A y = 5rad/s' B y = 8rad/s^ C y = 18rad/s' D y = Orad/s' 1.20 Mot dTa tron ban kinh 30cm bat dau quay nhanh dan deu tCi" nghi vdi gia toe D a, = r.co goc 2rad/s^ Sau 10s thi gia toe ht/dng tarn cua mot diem tren vanh dTa lii 1.10 Vat ran dang quay vcti toe dp 10 vong/phiit Toe dp goc cua vat rftn luc la A a„ = 60m/s^ B a„= 120m/s^ C a„= 80m/s^ D a„ = lOOm/s^ 1.21 Dau kirn giay cua dong ho de ban coi nhu" chuyen dpng tron deu tren A Q) = - rad/s B co = - rad/s C to = —rad/s D co = - rad/s du'dng tron ban kinh 3cm Toe dp dai va gia to'c dai ciia dau kim xa'p xi bang 1.11 DTa mai dat toe dp goc 12,56 rad/s sau 5s ke tij" Itic khdi dpng DTa mai da quay dUdc A lOvong B 20v6ng C vong 120 vong/phiit Ke tijT lile ham banh xe se diJng lai sau B 25 vong C vong B v = 0,526cm/s vii a = 0,248em/s- C V = 1,224em/s va a = 2,140cm/s' D v 2,568em/s va a = 1,634cm/s' 1.22 Coi chuyen dpng ciia cac dau kim giay va kim phiit la cac chuyen dpng D vong 1.12 Mot banh xe dang quay vdi toe dp 300 vong/phi'it thl bi ham va sau 6s no A 10 vong A v = 0,314em/s va a = 0,033cm/s' D 50 vong tron deu tren cac du'dng tron vdi ban kinh eua dau kim giay bang — Ian ciia dau kim phut Ti so'gia to'c dai ciia dau kim giay va dau kim phiit la 1.13 Canh quat dang quay vdi toe dp 300 vong/phiit thi li'ie t = canh quat bat dau quay cham dan deu va sau 5s canh quat toe dp 180 vong/phiit Canh quat se duTng lai vao Itic A t = s B 120 C 300 D 360 1.23 Trai Dat coi nhu" hinh eau ban kinh 6400km chuyen dpng tron deu xung B t=10s C t = 12,5s D t = 7,5s quanh true ctia no vdi thdi gian quay moi vong la 24 gid To'c dp dai va gia 1.14 Sau thdi gian 10s, toe dp quay cua banh xe tang deu tir 120 vong/phiit len y A 60 toe dai cua mot diem tren xich dap gan bang ii ,m q 480 vong/phiit Biet ban kinh banh xe la 20cm, thdi gian xe da chay A V = 465in/s ; a = 0,0338m/s^ B v = 465m/s; a = 34m/s^ ' du^c quang du'dng C v = 228m/s ; a - 0,0338m/s' D v = 228m/s; a = 34m/s^ A s = 15,7m Uj"'-! B s = 125,6m C s = 31,4m 1.24 Trai Dat eoi nhu" hinh cau ban kinh 6400km chuyen dpng tron deu xung D s = 62,8m 1.15 Mot banh xe bat dau quay tu" nghi vdi gia toe goc rad/s' Biet ban kinh banh to'c dai cua mot diem tren mat dat d vT dp 60" xa'p xi bang xe la 50cm Sau 4s mot diem tren vanh banh xe dii quay dtfdc mot cung dai g A s = 8m B s = 4m C s = - m ãô ã D 16m n 1.16 Mot dTa mai bat dau quay cham dan deu va dCfng lai sau 10s Trong 10s dTa quay du"dc 25 vong, lay A l,57rad/s^ TC = 3,14 Dp Idn gia toe goe eua dTa la B 3,14rad/s^ C 6,28rad/s^ D 12,56rad/s^ 1.17 PhiTdng trinh chuyen dpng quay eua vat ran cho bdi cp = 2t^ + 3t - 6, t tinh bang giay, cp tinh bang rad Toe dp goc Itic t = 5s la A CO = 15rad/s B co = 42rad/s C co = 23rad/s quanh true ciia no vdi thdi gian quay moi vong la 24 gid To'c dp dai va gia A V = 465 m/s ; a = 0,034 m/s^ B v = 465 m/s ; a = 0,017 m/s' C v = 232,6 m/s; a - 0,034 m/s^ D v = 232,6 m/s ; a = 0,017 m/s^ ••25 Biet ban kinh banh xe dap la 0,5in Liic xe chay deu vdi toe dp 18km/h thi gia to'c dai ciia mot diem tren vanh banh xe la , A 25m/s^ D 30m/s^ B 15m/s^ 1-26 Mot dTa mai chuyen dpng nhanh dan deu ttr nghi vdi gia toe gde la 2rad/s Biet ban kinh dTa la 20cm Dp Idn gia toe eua mot diem tren vanh dTa sau Is ke tir luc dTa ba't dau quay la D co = 8rad/s C 50m/s^ A a 0,4 yfs m/s^ C a = 0,4 ^ Cty TIMHH M T V DVVH Khang Viet B a = 0,5N/5 m/s^ m/s^ D a = 0,8 ^ 1.8 D a p an B Sai, vi vat ran quay deu thi hu'dng cua vectcf van to'c thay doi m/s^ 1.27 M o t cu'ng O A dai I m co the quay d6 dang mat phang nam ngang quanh mot true qua dau O Dau A dat du'dc to'c 2m/s sau k h i quay nen vectd van toe thay doi (chi co Idn ciia van to'c la khong d d i ) 1.9 Dap an C V i a = ^a^ + af = ^J((a^rf C a - 5,5m/s' B a = 8,6m/s^ A a = 12,3m/s^ A sai, D a = 9,2m/s' 1.28 M o t dTa compact (CD) bat dau quay nhanh dan deu tiT nghi xung quanh tam O K h i quay du'dc goc — r a d thi vectd gia to'c dai a tai d i e m M tren vat ran hdp v nhifng neu la chuyen dpng cham dan thi y < 0, liic coy < ; ' l.l2.DapanB D sai, V I neu chieu du'dng ngu'dc vdi chieu quay thi chuyen dong nhanh dan CO y < ) ) 1.4 Dap an B ^ = (vong) 271 , ' ' , » vi- , no = 300 v/ph ' , , ,< = IOT: (rad/s) 120 i 300 60 n, = 120 v/ph A sai, V I cac d i e m khac tren vat r ^ n eo toa goc cp khae C sai, V I toe (dai) v = co.r; r khae thi v khae coo = 27ino = 27t C0| = 2Kni = 27i = 47i (rad/s^) 60 471-1071 « ; Y=- ^ = - (rad/s ) D sai, V I a = -^a^ + a^ , a„ = co^r va a, = yr khae neu r khae nhau, luc a khae • ' - co^ =2y.Acpz^ A c p - - - ' ^ " ^ 2(-7i) 2y 1.5 Dap an D , v d i coo = thi co = yt va co = ^2yAcp , ; V a y toe goc co ti le vdi can bac hai cua goc quay Acp (va ti le vdi thdi gian t) 1.6 D a p an C, co = - ^ = cp'(t) ' ' ' ' ' dt :.: = t ( r a d ) ^ = 25(v6ng) 271 113 Dap an C A p dung: co coo + yt d Trong 5s dau: CO] - coo = yti (1) Tuf luc t = de'n dij'ng (co = 0): - cOo = yh (2) 1.7 Dap an A V i vat quay theo chieu du'dng thi co > cOn = 271 B sai, V I neu chuyen dong cham dan thi y < 12 " C sai, V I CO > va y < (cham dan) thi co.y < (11 D sai, V I CO > va y > (nhanh dan) thi coy > (2) ^ - ^ -COn t, t -COo coi - co,'0 , vdi = 107t (rad / s) 60 ^ 180 ^ , , CO, = 2n = 67t (rad/s) 60 ti =5s PhLfOng p h j p nen: giai b^i t j p trSc nghigm Co hgc = 12 - Trgn Trgng Hung Cty TIMHH MTV D W H Khang Vi|t 1.18 Dap an A = 12,5 (s) 90 CO = 27in = 27: — = 371 (rad/s); 60 671-1071 1.14 D a p an D CO < ' (0 - MQ = yt vdi = ^ Y _^ CO - COQ ^ t ©0 = — = 47t(rad/s) © - (OQ ' 03 = — = 1671 (rad/s) 60 „ v = i ^ ^ - ^ = T = l 8,84 (m/s) R 0,5 LL.t = - ( c o + 0)()).t ' i l Dap an D CO = u ' I'n-'H y = 2 y| =3,14(rad/s^) > o g * i r Goi k i m giay la so" , k i m phut la so 2A(p ^ t V d i Acp = 25.271 = 157 (rad) nen y = - « 0,033 (cm/s^) 30 1.22 Dap an C = COQ + yt - > COQ = - y t • ' ;f!