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Rješeni zadaci iz kinematike
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lJ:::. 2,\.,':' ::JI- ~x :::=. -10.. v: A Uvrstitijcdnadzbu 2.) u L): ytl ... 4
,,~ ,- ?.c... '!~ 1= m" 17,321111 );,,::::. ~ '-4'~ ,~
'Uk. = tl tc.1J>;! '2. t.o-c J:-";I' b '" ~ c:,l., 'B"; u.y,'! IS:!, ~ .:rl.Avri < - - 0..",, I' ~ -",~" ~f;2
1'" =i2 ..Jj./.g J2 1O.Jj'g =.j609,81 = 24,26 mls "'VlC\."l. e..O'.....:.~J a.."",.."...~~!.
V,. (I) = va sin a g' tc 6.,.) ~.,..d.c... ~t&.o. u..~",,~..l.' C1'1f..."u... !; "" ./0 ~L.l rr ~) ~~""- It ..."u.~'-c.I ..... "'" bjt.-\:~
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3.8 Cestica A izbacena je pod kutom a =40 pocetnom brzinom "" = 20 m/s. Na polovici dometa cestice A postavljena je cestica B. Cesticu je B potrebno izbaciti vertikalno uvis tako da se sudari sa cesticom A kada ona bude u svom tjemenu. Potrebno je odrediti: a) kada treba izbaciti cesticu B vertikalno u vis minimalnom brLinom da se sudari sa ccsticom A \l tocki T. Ie
b) kojaje minimalnu brzina cestice B pri tome potrebna?
.,.-.----"f1____ _ \\)u
x
B d!2
Kosi hitac:
d .sin2a:o d 20' sin80 - dome!: 20,OSm
g 2 9,812
1 .. I BT H .sin' a 20' .sin' 40 - rnakSI111a nn visma eta: = 8,424111
2g 29.81
vrijerne leta: d ?
x=vucosa'A :0 d=vocosa'A :0 IA =0---= - =2.62Is vocosu 20 cos 40 '
Vertikalni hilae:
- brzina: v "OB - gtH;
. prijedelli put: y = vo ". 2
Za y=H :0 1'=0:
() v,n-g. t :0 'o=voo g
H = v,,, ;;:> v",," =.j2. 9.81 8,424 = 12.86 mls
---------I" = vo" = 12,86 =:. vrijeme potrebno da cestica B dade do tocke Tg 9,81 = 1,3
=:. cestice treba izbaciti istodobnoI"
f'otokopmlOllIco"UlNUlUPY",M. t.upca !4Jomocopy,SlJ, V, o.jf'
Xl>: :\loCD::,J.,t =':'..)~ Vx" ~o~. " ;'0, Co~60 ~ ~j:; -t't
'1 t ;:'OlS;f)"-O
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L __
sc giba krivocrtno po putanji I' = I + 2q>, (I' jc u merrima, a q> u radijanima). U
trcnutku v, = 4 m/s dokje cijclo vrijcme v, = 3 m/s = kons!.
odrcditi.
ceslice, komponenle ubrzanja tc njihove iznose,
putunje s ucrtanim vektorima.
a- -------\ ~---/'v~ ~ -.. a _..-. .. ... J?' ~
v, =i' 2'=4 I) ;:::. =2radls . / r
Vv =r=(1+2q2 3 2) =:} ~yo~X1 :z
3 1'= 1+2rp=21l1
5m/s
Iz jcdnadzbc 2):
. .. 3 I d v. =(1+2rp) .rp= 3 =:} q>+2rpq>=
1 .. 0 .. 16 dI'qi+ 20 + 2rpqi '" 0 +2-q>= rp=-- ra s' 4 3
1z jcdnadzbc I):
1'=2 If, .. .. 32 1r=2rp= m/s
3
Komponente vektora ubrzanja:
.. . , 32 2 2' 6 .'. I' (I =r-rrp' =--- .. =-18 66ms' , 3 '
.. 2" 3 (16) 2 8 " ' a" =rq>+ 1'11'=2' -3 +24 = ill'S'
----
-----
0'
f ?-t
y.m
IS ',m
,"otOkOpJrlJOOlca"VJNO{JJPY",M. Gupca 14.dlnocopy.sb(pYlp,hr,skype:dmocopy,0351f)25:nt,09~!92415b0 35 ,"otOKOpIrQOnlca"VlNUtOPY",M. trupca 14,dlnocopy.sb~v,p,''r,skype:dlnocoPy,035Jb25211,09~19241560 3b
?4, \:-oW'?-VI~ - 1oNOVLjclJ Grupa:__ ime i pn:zime: potpis , datum:_. _20 3.13 Cestica se giba krivacrtno po pul!lIlji:r =I+ 2",. (r u motrima, 'I' u rndijonirn.j.
U nekorn je trel1utku gdje je Y, ~ 4 mi. dok je cijela veijeme y. =3mls =!cons! . Potrebno je adrediti: (-~.-~ oj polataj estice. komponenre ubrzanja Ie njihave imose te 3.14 tica se gibe. krivoctlllo po putnnji pmb) skicu dijel. putan]. s uertanim vel ~ 2f
Vf ", rf ~ ~j"r,2. ~~~r2. ~ ::.- 01 odZ, 45 2 ::;. {/ ( tA" .. "1 "" 02.0.r o ('-21 ~/o.'''-3:' (j r'-~-__r'jq/~",)t.v.I'V\.
37
1: ~
r,q Jilt, . IUiOJ ;e ,"lIS I ;Rj'eJl: L\ b) polumjet zakrivljenosti PUlllnj. u tom palota]u Ie
c) nOrma1nu i tangencijalnu kompanentu ubrzanj .
V~::'OOMMI~ ~ ~> CI",'>-===,.,;..j~ lOOO'" x-' A. ( 2.00, 'CO1 o,lf,~"Glt
II = .20002~ X
j I ~ :J!2J1QQ_ = -0/; ~'l.
201XJj .2.x 100::0'~'~--X- ::: o,ooS~ 4 - ~,
1 r l.lie=[h(~ltr LHo,S)J :- 2:rs MM
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a) komponentll brzije v, i ukupni iznos brzine cestiee,
Zax=lm:b) ubrzanjc ccstice te
c) pollinucr zarivljenosti putanjc ccsticc ako komponcnta brzinc v, u tom trcnutku izoosi
a,+ =0 => a, v, 0,25 rnls i u tomje trenutku konstantna.
1 ( ) I ,a =-2- -1 m/s', 4 2
L_3.=-112x 2 x
Polumjer zakrivljenosti putanje: d
xy+2x-4=0 p ~ +(y')'f
y" I)
4 4v=--2 -2=2m x 1Zax=l => y 2:
, 4 4
v,. +2v, + =0 => v,.+4v, 0 Y =--=-4f
v =-4v =-4. 1 -!tnls 8 _! 8 y r 4
-I'
Iznos vektora brzine:
P 8 8,762m v ~v~ +v~ ="/0,25' +(_1)' =~m/s
Komponcnte nbrzanja:
v,. konst. => a, =0 . 4 .
DeriviranjcOl jednadzbc I) dobije sc:
+xv+2.i: 0 \..!dt
Jl E ,J) Q .. "" >
.~
.
~
2A',jATAIc.
x o;ti
'l" Zed, 16 C
v i Ibt'':f i v,): X ~ 1, rt1i~
/Z2.fZY'.I (It V'
(I, ,- x' -% 0 iflb'+ ttl 1~6 t? a~ ~ ~ tt r01\'
;).1(%, 3i(l,,'O( " (\., (),r.~ l a{,;; ;. t \::- vllS' J1'
: J.{' 11
o} .il
ct, ';; 1-IT) ~ 2\
b) Vy,J,. r'hS-/ ,,~ (~< II'C,~~ t
-',c) (),' () enS
ai'
~stica se giba po spiralnoj cije su jednadZbe gibanja r 1+I' i -~ I, Potrebno je u trenutku kada je komponenta ubrzanja po iznosu jednaka cirkularnoj odrediti: a) polozaj, b) brt:inu i ubrzanje. Popratiti skicom.
v, = f 2t
vv=rq, (1+1')-1=1+1'
a, r - /'q/ 2 (I [' 1-1' a. =rip+2f 0+22'11 41
a,=a. => I-I' 4t => f'+41-1=0
42J5 II., 2.1 2
4-2J5 --4.236s nemoguceI, 2
Polozaj:
r=I+/' =1+ = 1,056111
rp I = 0,236 rad
Brzina:
v,=2t=2 0,472 m/s
Vv =1+ t' =1+ 0,236' = 1,056 rnls
I,jg
::f:
o
K..
.. '
~lJ>taCijSkO jc gibanjc stapa OA oko ncpomicnc osi 0 ~drcdcno jc jcdnadzbom I \~;~51" 11', rad, dok jc pomicanje kljizaca B duz stapa OA zadano jednadzbOlllv '{}15 t -0 1 56 rti~S
.ce ,. c
- v. ,r~~ c (),62! rfllJ r ~ 1-0.1 51', r, Ill, I, s. Odredi!; brzillu i ubrzanje klizaca u trcllutku kada :llap zatvara s{ O.'~t ---...--........Vf~ (.J 0,36 vodoravnolJl razinolll kut II' =30 . Sklcirati pUlallju cestice te wktorc brzillc i ubrzanja.
( ,O,'!/. , (),11166'M
f 0,3 A i: 0 .(\~ V ,.""
,,","--_"""'''''''Y'M"V''Y'~'''~_''"'''
O/~ e
("-::. t_~ $
0(. r_ " -0,:'- ( Komponcntc brzinc u polamim koordinalallla: ,tJ,'~- O,4JE,6 .0,0,\, ~,ili'l~
V vrer + vq.efl
Vr ::;:;;j+Q.,r -0, t,~1.\
v. = r1>qf' li-i -\ (.( .... -')., IJ,H .tl;'~i: + (I-oJ'n ) - Il,:' ';'Ol Ct ,. Q,6Qi 'mI.; 1.
Komponente ubrzanja u polarnim koordinatama:
a== a,.e,. + a/it;!
(1, =
rip +2r1>Gv
Derivacije:
'II = 0,151' 1'=1-0,151'
;. =-{l,311>=0,31
j; -{l,3
')
.p =0,3
so
o,f -0,425' mis1
'"
~---~~...............,
Trenutak i polozaj klizaca: y ,.If , ~ 11'=30"=-=0,151' 1= --=1,868s
6 60,15
r=I-O,15I' 1-0,151,868' 0,47601
Brzina klizaca:
\', = j. =-0,3 1 = -OJ 1,868 = -0.56 o1!s
,.,. =1'0' = (I - 0.15 (1- (l,15 .1.868'). 0,267 lI1/S r=l-rp
jC 'CO -0,56 e, 0,267 eo
\.= = J(-0,56)' +0.267' =0.62m/s
Ubrzanje klizaca:
= -0.3-0,476 I)' =-0.3-0,476.(0,3.1.868)' =-0,45m/s'", a,. r.p+ 2,..0' =0,4760,3 +2 (-0.3/) (OJ d
0,476 OJ + 1.8(8). (OJ 1,868)= -0,485 mis'
Ci =-0,45 e, -OA85 e"
a J(-0,45)' + (-0,485)' =0,662 mis'
Putanja:
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r=I-O,15,' :::> r=I-O.l5 L . 0,15
1'=1-90
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j4.3.7 Tere! A, !roji je pri~vriCcn os .""",tcz1jivom utotu omolAnom o!ro Orupa:~ ime i prozimei L., polpis , dalum: 1,L, O,L20 10,diska II ( " 20 em ), gib. I. PfOlDa dolj. po zakonu x. 4+ 21' (I u
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11.-+ Slap 0:\(' 1'01 ira uk" kOllslanlnolll kUIIlOItl bl-linllll1 m~ III l'H,l's. a pro lJ \"tll'l'ujc J..rllZ klh:ac f3 koji sc moze y.ihalt samtJ po vcniknl11l1m pmvclI
0, D. Odrcditi hl'zinll i uboanj" kliza':a B. ilkn jc zaoano: 0:\ '" 20 cm.
0,0 .J-O till .~" .J-S".
Popratili skicmn. NapiJlllel1a: Slika .ic II mjcriIu!
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datum.i~,,~ lmci _____~, datum /':1 /I. OK.
loeo CD u kojoj je krutni Z1ijeb Totira u pokazanomG Valjili" A mole se gibati po krutnom flijebtl polurnjcrar, voden kulisom smJeru kutnom brzinom w ~ 12 rndIs i kutnim OB, koj. Totira oka 0 kutnom brzinom '" = IOrndis u:z kutno ubrmnje ubrzanjem a = 6 rndIs' . Na meid OB
a =40 rndIs' Odrediti meladom plana blZin. i ubrzanja, apsolutnu brzinu iljebu ploee
i ubrzanje valjilie.. u trenutku kadaje II' =30 ako je za4ano: r =0,4 rn. Luk flijeb. tang;ra dunnu
OA =0,56 m, CD. Potrebno je odrediti:
a) blZinu valjele.. A. b) apsalutno ubrzanje valjclea A Ie?v e) kutno ubrzanje rueicc OB.
o,,~ O(~ AC ,(.01/"-' Ie,.- t"'{J Vr=W'At."'Al VI'
W OA= VA V? t Vr~l Ic,,~olr""ls
J.oA .LAC 1a,A o.Ar/v-",
VI' v,,~
1'v \feeL o2\~" v,,!;;: ~JCIA = "'1' t 0., et 'I ClCor Vrcl..,:::: 3
j0,,1 t a'l a.;tq""tcl'cltCl lO,A lP,'t l!o~ .l.dA IloA Ie"" ~ 10.,/~'- "" \
o v' M:!- S~' {'l 'POI,,,~ ~ i0; r =o;T :.1'1,1", s
V' 0.1= C\~ta~-t,,;-\o.riaCo{" -L:c lLl. =Sb""/~ 'I. lIoA lOA IIAC JAC IO,~ lO,A .lvr
A ~,b '! 101\ 0P'll "
'" oZ, 0-,\ = 40''1tb' 22,411)/:;;; 1. a~ ~ ~ : ,1'i;,q"1l~OJ If
G\ ~ - V.'L If'C1(b( '" 2. ( p---=~.''" 2 '10, k,-~ YJM;:' t A.t O,'f ~CM'"
Ac= 6'O,IPO,1 ~C) 0\"- v,'L ~
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lmeiprezime: HQ.vo.lG L.ug,II" ,polpis ,datum~.
~ StapOvi AB i Be .globno su ve>.ani u tocki B, I.e se gibaju ~ konstantnim apsolutnim kutnim brzinama. Odrediti, u prika>.anom
polo!.aju, brdnu i ubrLanj. tack. C kao i trenutni pOl brzina ~p, BC
akojezadano: AB=BC=lm, w, =3radls , w,=4rad1s.
v~ ~ Vf +V,et Ac.J..lC lU
At. = ~1"J.ro~1C.At.= :". 1,%2.~S,i~6f1?Jj
t\(.~ l/PZM~to..~ W2:tJ(.=
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IA = o,f tOfu. oj c. //I,C '10;;
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