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1'" =i2 ..Jj./.g J2 1O.Jj'g =.j609,81 = 24,26 mls "'VlC\."l. e..O'.....:.~J a.."",.."...~~!.

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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'

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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):

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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'

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?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 .

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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

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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:

\P=O.l5t' t'=L 0,15

r=I-O,15,' :::> r=I-O.l5 L . 0,15

1'=1-90

x

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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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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 ~

r - 0/4 Ij" : G~,Of-IY)/~\ ?o. G\A ck,r 2.( w'p'~-~~J4"'/~?

O~Q' "'~ 141"\1~~ n o,r

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-l> ---'" . f"p ::: + NG./4 = 1-01\: 11M

t:!!?- = 0,40;.!AI" - '0;0.

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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(.=

" v(

ocito,!p' Vt=2,

IA = o,f tOfu. oj c. //I,C '10;;

rv \. t I ~ ~ Y.L ..&li:." HI:::.:?