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2.19 Covjek se nalazi na obali jezera " polol'.aju A j :!.eli u najkracem vremenu stiei u polotaj B n. dnIgoj strani j""era. Za ovo posUlje dvijc moguenosti: L plivati iz polomja A u poloZaj B iii

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te.

tan fJ

= {I,

3,036 7,402 ~

; rfl'll.

nll~;.

'/,ia

~

Kut izmedu vektora ubrzanja i tangente na putanju:

lI1~'t1l'1)(

\";)

I(~

.Ib~ ~11,i.< m

fJ

22,3

0

<\ " ii ( .:\,

{'I

1,1.,6:1. ~.

'v

~ ,0<)1

',{)16 (:C'I

t(hll

te,'

s l'\(fJt

~~t!.

Vt1't::: '\I'i. ;:

1,,<') L

It)i~

'Iv" ,u~

~::.

~f,n

1,

'"

;;(,

J;7'i't~Nlti

3f j

3.4 Odrediti pocetnu brzinu

VII

i kut a, pod kojim treba izbuciti tijelo zuncmarivih dimenzija

iz tocke A kako bi palo u tocku B zadane kosine, ako je poznato .1'" : xn = 1: 2 .

, 32· g!1 VI~ =s!~.~. ~j

1)

U tocki B:

tan p "

'~" "

)'

1

x"

2

--'L = -

=:>

fJ = 26,6°

X,Ol

x"

Xu:,

=64· cosfJ =57,2 m

YB =-64·cosfJ=-28,6m

, ,

Xn

Xa

vO·/[I·cosa

Yo

"0

1n = vo.cosa --

y,,"""

. 1 , "a ·sma--·g·'i; 2

Yn = x" .tana-l. g' , 0

a~.

=-g

vo·cosa

x;

2)

Uvrstiti I) u 2):

fa,dl + C ,.

v~

2

G,

Vo

1'0,

= . cos a = konst.

)'

t 1',
y

f""dl+C=

1

= Xu' tana-i' g a

gdl +C

X=

.1'"

f

1'0 'CGsa

"0.1' -

dt+c

g·1 =

"0 ·sina - g'l

v" 'I'CGSa

a

-2K6=

. ' 1 , ·sma-g 1 Jdt + C ='·u"·sma-i·g,,,

U tocki C; V;;::;: VOr

1'}

Ye

=lit! 'cosa

Rjescnje kvadratne jednadibe:

O=vfl·sina-g·'c ~ I,

.

I

.

-'e ·,sm a -;. g .!( -

1'(1

tan a",

g

. 1'0' I -sma'-;,s1l1a-:;-' g. g­

a=16m

57,2±~57,2' +4·51,12·28,6 2·51,12

=:>

1 2

x~

--.g.-,-,

vo·cos'a

jcdnadzba putanjc kosog hitca u tocki B

MtokOplraOI1lCQ"UINVU)PY",M, lIupca 14,dmocopy.sO(fpVlp.llr,skype:dlnocopy,035Jb151H,09i:if9241 SoD

o

~o..",o

je:. .;f",it,"'"

)( oo.."\:. ::1

I .. Po+",-\:'v-.o

30

~"-d.","""-cl.'L\.:>o..'-U-,,,

fC-"""""-"-""\o..", \.:"'" cd..-ed.i..+,

j"­

~st~c.e tan a,

a,

56,2°

b) lana,•

v.bQo..·~e

Br-'2.~V>v..

6~f0v..o.~C>...

a, " -20.5"

102,24

x Iz jednadzbe 1):

v0,

C:<1.stlCe

~

m/$

rjesenje za plItanjll s tjemenom U C

lA

tr,,-;~~~u. t

v...~~'2_D...~~

'j

~o(1.t

-.:> -I::

~

o,s"\:.'L+~

:::l

a,s,

0(2.

~

(~)'\ I 0,2. x'L

\..( '" O,S' + ~ ..J 0,0'"

mls

Vo,

\;;:---1 !

v, '" \'2.,S"x

+k

1_-:' _ _ _.

o

,l',m

x

VOl

_~Yu~;l..o..:161n

C --,

,

Vx

x~

V,::>_

':1 '"

O{l.m/s

o..x = x _v)( = YB

0. •

vn/s""

(j

Clj~ ~ = "',-\ ~

it.V -

:::) v=lvx"-+~ t-o,,,-<:>

0\5· 'It

=> o.~\Q.x r

""1

A

2. \

... 0..::\ '"

v . coe"s'

G..x"x ... a.~ v~ - ~ .. ~,,;ti"

.

\....rJ

o

:2.

0(2..'2.

" . .1.'- '" -,r~ ,Ott +1:; . 00"",5

1

('-=(
"

:L

t'l.= (O(L-+t"-) •Z ,".>r.. '1 (~

'1.\

't

-'I

-I:~) ~

-1:"-~ 0,0'1.+'

0,

- O,o'/. ... Cl

,o?

!

t ":""/'1. +1:. ) . 1. "C'~(o\()",

-t'l.

" a,o \i

[t= .'j] 012

'N>

/s

ZA I)A1"IW

Vo pod kutom a =60· i pogoda cilj nn visini h 20 m , a vcktor brzinc ccsticc II tom Ircllutku zatvara knt od 30° s vodoravnom razinom . Potrebno je odrediti: a) pocetnn brzinu cestice v.' b) udaljenost /, e) vrijcmc leta ccsticc do cilja?

3.6 Cestica je izbacena pocetnom brzinom

};

r~

..I

\j:::

4

~~

/"+>f ::

0\"::·

QI~·

"I.

'1-=

~'"

~~ • 0.1· "'~

f

Vy ~ /)

;),y-:'

'"

"'t.

e.

Q~. V~ ~

1 ,,",'.

Of:':':

"

Vi'

0"

Vo

y ,1" '1<\'

-

')l..:'

~,r

.::;

"..f 1,

-?

I'Y'{!

~

t)?~.

/1

.?

J ednadiba putanje kosog hitea:

tl'lt '

}t)) in;:

~

()

<

x' 'g

x·tana

('(:t'l

2·v~·cos'a

Za Yc =h=20m, a

~",l ~v I' (,o$4~(l

'.1,

60°:

;~tyj(t~~\l

f·g

20=/·lan600- 2.v;-eos'60°

t

1.

• 0,)0 1

J,.

"tal"!.: '+t 1 . VI })f

1)

i.l~ (ii!O~~t 1..) . i;I

'IL' 1<

".

Dcrivacija putanjc po koordinati x:

.,,,7

::·o,a"

t· 0,7.

y' 5

~~,-,-<-

~f ,.;w..

;

DI,-1 ~ OIOf! ,t2S~'" 11,~·

V"

-1

mS

V~"cos2a

-

a,ool6 ···f

-II""""

lana-~

'Ie

V:

y'(t) = tan a

V;

" i,tlJ.m

'lIt

o

.1

D,U).. tn5

Y'.l~O,J"vs '2,5~

i

l~

~

·(;1 •

1

..\::

,

1,

/J",/.J ( :

.St---r ~

x,m

JB@Mwr:~hA9,W¥:S~¥ii+%Mhl ,

i

OlIO:;

-".,

1,-' ,

o~

'-"" I

.

;<)

• I

~Mlf~

Ian 60°

~=lan30°

v; -cos'! a

t· g

v~ . COSO 60

Ian 30°

)

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~ Cub'"", -!(J,.q:)\Jic~

~) ~) c..)

-3

2)

=>

7'(vQ",\.ol~

""r

1'''-+0.')'(

"1,~

-:;:. 2..k l..

=> 1=

x~

~x

-

of.;lIt ';JllL4'x~t.,

=

tl tc.1J>;!

'2. t.o-c J:-";I' b '" ~

.:rl.Avri

<"st~\"..I" hl.("',~6o"",

=.j60·9,81 = 24,26 mls

'" ~ ~""'ls1. 0.", .....

Komponcntc vektora brzine u tocki C:

cosa

v" ·cos60° = Vc ·cos30° =>

V,.

(I) = va ·sin a ­

0",.

konst

.Jj vo =3'

.1-·:

6.,.)

~.,..d.c...

,sin 30°

tc

-",~"

~f;2 "'Vl

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6:;(;"'~

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&.' . . . 0·....... (4.i:\",:J,'

ho..oIo..... 0:"0. J.,IoO-Je. ~ tv.......... ?·e~u....v.... ~M':'u...:> C1'1f..."u... !; .,U"'·~~)...,..Q u..vl" ~"U~af~O~ &'1,£·..... .)"(".<

"""

u..~",,~..l.'

~

bjt.-\:~

z.;.:;-A'~~

..ro~ 2 . 0 ......J; .,tIc...

v,, r. r!.c/""'b

""Vi ~f

-("fl.~'O d"",­

a.."",.."...~~!. ..!.(

+("<,,,kP1' J' "'­

oO;':~'LO ...... A

7"': -J..o1.oVc. j"O~
,t

T

..(.::: 11'0·

g 1,428 s

e..O'.....:.~J<.

-~ l

:-~. ;:t...t;" ~Q ~t&.o.

11

"'->< - - 0..",,£ I' ~

tL~~~{,~

:,.... ,&0.

t..f.<,/r­

C\."l.

WJ,Cct. "'" J::t....

~

,~

I

~yt

'

• '-4'~

I

<>,'-1.0B:I .~/.t

"" ./0 ~L.l rr ~) ~~""- It ..."u.~'-c.I ..... "'"

, = _24--,,_26_,_s_in_6_0_o-_14..:..,0_0_6_,s_i_n3_0_° 9,81

-oj

·2426 = 14,006 mls

g' tc

'sina-g·t, =1', ·sin30°

c

3

T'1..

'!~<. ~j -=.

c:,l., 'B"; u.y,'! tJL

­

1~

·1O.Jj'g

,

><~Z-

);,,::::. ~

1z jedlludzbc 2) slijedi:

V,,'

,,~

?.c...

~

,I.

~)

m" 17,321111 'Uk.

(I)

.v..(..,jtl.'-l.o ........... ",),.•:1"-7c ""'"

-10.. v: A

:::=.

4 )

IS:!, ~

V,

+IS,

~ ~

')tot.,

::JI-

-/..G,...\.A

V,

~

':fJ!'u

';:;'2~J" ("<_,,,I) ?o~~IJ'~ jt' oJ~d-~J.· """"-)t .... ;~

ytl ... 4 ~LG.t..

J2

,~ xJ'-U.j

'J>t/'T..

!

31

l"i"'-Q~

,

lJ:::. 2,\.,':'

1'" =i2 . .Jj./.g

~

a1i!te..t":

UJ"''''~.Jv"v-

C"f--")

2._/-'

Uvrstitijcdnadzbu 2.) u L):

20

JO

~~~~~ (~L.:CA. !".. .IIt)'j '" ::.o.,lQI,J.....'... r:...-!. t(..lt.~ ...\J ....'" .J:,,"'1',.:c........:c.~'C J,rr..'::'~..., h It..~~ lQc.......')ol..o-tv..';'" (,Iti...... ,ux \' -U'"I

1.A~/'1.Co.)·':;" "it.. <'...... 4i't. .t:!l ...... ?.. """t.o;~' V':"'L'::"'"4 ctv -: ~:1'

t"'j"tt

.u~

~;~"-

t<

"" • >-0",).1

•

~"!.t'*(t:..4®;

-

ho~

l.v.+"t.:

cl":::l....s()"'t.~

~ ~ ~ol. ~. . 2..r:l.

!1,

"­

A 20 1

.;l~

fl_:..

U.

eo

"-"1°'0""

).~

v~ ....~ 't..1'~o. ~

\,

~

- 'O)t.~l

':l -

':l =¥

~

t..J<"

J . ",:",,(, .

1.

t


~ ~i.

-i.1~'7~U..: .~"€.sTJC-"4

- H)'t,'b"L c4 \V.;"':..: .:i 8 "" 1S01'. _

... ",,)t'-'->.(

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'i~ <.l.~.t \l

"'7


11,9;;; ~/!

-;:::."""::"",

l.t"';t.<.~

1
+.,.-{,

l-t \Sa'?;. ­

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r:t\D}\TO~

lesh(o(rJ ir loCki bf )in(j (es1fU' B pri /(lme pO/ffbflQ ~

b)

=

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:

fYJS' !

In ftC,)

(;

a) kada treba izbaciti cesticu B vertikalno u vis minimalnom brLinom da se sudari sa ccsticom A \l tocki T. Ie

:~J:J {

li{i~]

b) kojaje minimalnu brzina cestice B pri tome potrebna?

70" LI,bid

h

• 12 ,l~'5' "I(j '?IO.~ ~

2,[,'l'! rn

.,.-.----"f1____ _

:.

,/ ;

\\)u

x

/),N{

B d!2 ~)t

Kosi hitac: - dome!:

111ft:

{

/

t)

I

"\

~

1~

20' ·sin80° 9,81·2

. sin' a

BT = H

2·g

20,OSm

20' .sin' 40° 8,424111 2·9.81

• vrij erne leta:

!':)cd~)b;,~.J

bv(;h

.sin2a:o d g 2

· 1nn visma .. Ieta: - rna kSI111a

,:

('C~!)

d

d ? x=vu·cosa·'A :0 d=vo·cosa·'A :0 IA =0---= =2.62Is vo·cosu 20· cos 40° '

'~

Vertikalni hilae:

- brzina: v

. prijedelli put:

,..

II' . '!~\'

"OB - g·tH;

•

Za y=H ()

:0

';.2

2

1'=0:

v,n-g. t•

H = v,,,

1t1l.A1'l\(j;

y = vo• ".

:0

'o=voo

g

;;:>

v",,"

=.j2. 9.81· 8,424 = 12.86 mls

f'otokopmlOllIco"UlNUlUPY",M. t.upca !4Jomocopy,SlJ
3.9

I" = vo" = 12,86 g 9,81 = 1,3

I"

=:. vrijeme potrebno da cestica B dade do tocke T

, datum:/..2., ro,~.

, polpis

1m. i prezime:

J'

Cesticaje izl>aama ~om brzinom v, = 30 mls pod kutom a = 60· prema vodomvnoj razinj, Nakon leta cd 2 sekunde potrebnoje odrediti:

a) polof.aj I':stice (vektor po!obj.), .'

b) brzinu ~ce (komponente i vektor brzine). 0) normalnu i tangencijalnu komponentu ubrzanja te d) polUlrijer zskrivljenosti putanje na tom mjestu.

y.m

~>(t) ___ _

XR

--------... "'-1 B , , V••,,"

=:. cestice treba izbaciti istodobno

o

"

,i..l'D :,

x.m

Vo~ :OUM!S

J.,.~60o

~~OMhiJ

i=2.~

X". ~1>, V, o.jf' Xl>:

:\loCD::,J.,t =':'..)<0.:\60"2 \l

"

:= ~o

Vx" ~o~.

-0, " ;'0,

~j:;

-t't

'i '"

fVxlr~

o

..f..

"'"

4=32,~~M ~ Dlt>~

'1-0,5'

'1 t ;:'OlS;f)"-O

0~t61,;f/

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A£ ,2.:j

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::- ~JO::'~ t>'\{~1

=~

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3.10

Jednad!be gibanjal!estice su:x = 2/-2 i y' = --4t' +4 pri ~emu sux iy u metrimn.. mjem. I u sekundama. Treba odrediti jedna&bu putanje te polumjer zakrivljenosti, U to~ki putanje gdje je oroinatay najmanja !reba odrediti brzinu i ubu.anje.

~ t.:= X~2

x" 2.t - 2.

V'l

~

y -"Y1. ::

::- -,",(n

-, ?,t>. D, '.) ~

,inGO'

;;0· 0/66

t t;

r--!i ~

;'i/-J'(

tL·

h

2

"-

4 ~ +~

4

'in

~')

b

-,

·~o

1,

V' -~-::.. ! ,~ MS"'

Y'1. = 21­

(x - 0--) t

., {

'1-'" r::. \L~

, 'IS/Ii- 49,01.



1.= 2.f'I'

~

f

V'~-

" <{'j • ~t (fl

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0)$

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2

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='7

yt =-'ttl. f4

'1(1

33

Vy =

v'" {h,Z') "'JI~

" 6,'.\6

X"

=1

vy,,'¥t

yt= -H?+ 4 I {~ , ;:

i

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c

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to') "."~'"-""~

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IW'!J1tv Z~f Je V,' II 0101: Jr ('fit/a ¥lyffll(' V~.1,.",~*,,,1. cdlec!.t~i: .:,) poio:ov or.};'ce. t~~"li'CJ"eQie r)b''''''lI Q Ie li.nnE.' 10) ol:rro ,:llle l(> fll)/afire ~ vcr ~IJ/m vel:l-o(l'mC\

\.. "

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.

v, =i'

2'¢=4

I)

a~ ~ -.. -------\ a _.~---¥/'v .-. . . . J?' ~

;:::. ¢=2radls

~y ~ .

Vv

=r·¢=(1+2q»·2

3

2)

=:}

o 3 1'= 1+2rp=21l1

5m/s

Iz jcdnadzbc 2): v.

. =(1+2rp) .rp= 3

=:}

..

q>+2rpq>=

3

I

d

1 .. 0 +2·-·q>= 4

qi+ 20° + 2rpqi '" 0

1z jcdnadzbc I):

1'=2¢

..

..

If,

r=2rp=

32

3

m/s

1

Komponente vektora ubrzanja:

(I

,

.. ., =r-r·rp'

32 2 .. 2' =-18 666ms' .'. I' =--3 '

.. 2"1'11'=2' 3 (16) " ' a" =rq>+ -3 +2·4· 2 =8 ill'S'

.. 16 ra dI' rp=-s' 3

/

1

r

:z

X

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Grupa:_ _• ime i pn:zime:

3b

, datum:_. _20

•potpis

f

Cestica se giba krivacrtno po pul!lIlji:r = I + 2",. (r u motrima, 'I' u rndijonirn.j.

3.13

U nekorn je trel1utku gdje je

Y, ~

4 mi. dok je cijela veijeme

y.

= 3mls = !cons! . Potrebno je adrediti:

(-~.-~ 3.14 tica se gibe. krivoctlllo po putnnji pm­ oj DB Sli:i. Imos se brzine ~c:s~ce mije.nje

P" zakonu mi., gd)e)e y om­

IUIta u melrima, a i ubrzanje Zemijine sile

leU (9,81 mls'). Potrebao je odrediti imos

brzine Ie normalnu iroroponenm ubl2llllja a.

(po imosu i smjeru) u treoutku Imo na sliei,

oj polataj ¢estice. komponenre ubrzanja Ie njihave imose te b) skicu dijel. putan]. s uertanim vel
C

1+1 f

r~

Vr '

Yf ~ 3M/::' ~k.;,,,~ -f~'?

r?

Vf

1,

7

0, -""';>

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0'

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IS

(j

JA.

4

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,

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mY "

V\.
-~t'6

+2.,2, 4

O'f=::', _I';

2.

\

W'.iI.4~E

2 ~ ::.- 01 odZ, 45 ::;. {/ ( tA"

r·f ~ ~j"r,2. ~~~r·2.

",

0.r o ('-21 0.

:2f

y.m

Ivl =.J2iY,

gdjejex= 15m.

"'" ? I

\)" f =""

?-t

[;'1)11'1' {«( -::.

1..

C),t\'=

-,

) jJ

~

~

=~

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viC,

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b'l

i.CRA).~'.:-. ~-c "',016><:

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'"milimelrima, I u sekundama). Potrebno je odredjti: a) komponenle, iznos i smjer ubrunj. ~ce ZIl polotaj A(200, 100),

b) polumjet zakrivljenosti PUlllnj. u tom palota]u Ie

c) nOrma1nu i tangencijalnu kompanentu ubrzanj•.

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V~::'OOMMI~ ~

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1 m:

a) komponentll brzije v, i ukupni iznos brzine cestiee,

b) ubrzanjc ccstice te

c) pollinucr zarivljenosti putanjc ccsticc ako komponcnta brzinc

v,

xji+

+ 2}( 0

+.rj'+

o

0 => xji+

}(

Zax=lm:

v,

u tom trcnutku izoosi

=0 => a,

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0,25 rnls i u tomje trenutku konstantna.

1 (-1 ) a =-2·-·

,

4

I

,

2 m/s'

L_3.=-11·2x 2 x Polumjer zakrivljenosti putanje:

d

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~ +(y')' f

p

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4

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=0 => v,.+4v,

v,. +2v, +

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v=--2 • x

r

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0

Y

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f

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=--=-4

_!

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Iznos vektora brzine:

v

~v~ +v~

="/0,25' +(_1)'

Komponcnte nbrzanja:

v,..

4

konst. => a,. =0

DeriviranjcOl jednadzbc I) dobije sc:

+xv+2.i:

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v, = f 2t

vv=rq, (1+1')-1=1+1'

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: J.{' 11

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a) b)

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i·

rt1i~

(It

x'

i Ibt'':f

v

~

'.I (I, ,-

cije su jednadZbe gibanja r 1+ I' i komponenta ubrzanja po iznosu

-~ I, Potrebno je u trenutku kada je

x o;ti

2.1

2

,,~

t

-',

enS

4-2J5 --4.236s 2

I,

ai'

nemoguce

Polozaj:

r=I+/' =1+

rp

I

= 1,056111

= 0,236 rad

Brzina:

v,=2t=2 Vv

I,jg

0,472 m/s

=1+ t' =1+ 0,236' = 1,056 rnls

en

E

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r-­

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~lJ>taCijSkO jc gibanjc stapa OA \~;~51" 11', rad, dok jc pomicanje rfllJ

---...--

0,36

r ~ 1- 0.1 51', r,

........

Ill,

oko ncpomicnc osi 0

0

I, s. Odredi!; brzillu i ubrzanje klizaca u trcllutku kada :llap zatvara s

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,tJ,'~-

O,4JE,6 .0,0,\,

~,ili'l~

V vre r + vq.efl Q.,r·

-0,

t,~1.\

Vr ::;:;;j+

qf' li-i -\ (.( . . -')., IJ,H .tl;'~i:

+

(I-oJ'n ) - Il,:'

';'Ol

v. = r1>

Ct ,. Q,6Qi 'mI.; 1.

o,f

Komponente ubrzanja u polarnim koordinatama:

-0,425' mis1

a == a,.e,. + a/it;!

'"

(1,

G ~---~~

jc jcdnadzbom

zadano jednadzbOlll

vodoravnolJl razinolll kut II' = 30° . Sklcirati pUlallju cestice te wktorc brzillc i ubrzanja.

0,3

i:

~drcdcno

kljizaca B duz stapa OA

=

v rip + 2r1>

...............,

Derivacije:

')

so

'II = 0,151'

1'=1-0,151'

1>=0,31

;. =-{l,31

.p =0,3

j;

-{l,3

Trenutak i polozaj klizaca:

y 1= ~ --=1,868s 6·0,15

If , 11'=30"=-=0,151' 6

r=I-O,15I'

1-0,15·1,868'

,.

0,47601

Brzina klizaca: \', =

j.

= -0,3 1 = -OJ ·1,868 = -0.56 o1!s

,.,. = 1'0' = (I - 0.15 jC

'CO

-0,56 e,

0,267 =

\.=

(1- (l,15 .1.868').

x 0,267

lI1/S

eo

J(-0,56)' +0.267' = 0.62m/s

Ubrzanje klizaca:

",

I)' =-0.3-0,476.(0,3.1.868)' =-0,45m/s'

= -0.3-0,476

a,. r.p+ 2,..0' = 0,476·0,3 + 2· (-0.3/)· (OJ d

0,476· OJ +

1.8(8). (OJ ·1,868)= -0,485 mis'

Ci =-0,45 e, -OA85

a

e"

J(-0,45)' + (-0,485)' = 0,662 mis'

Putanja: \P=O.l5t'

r=I-O,15,' 1'=1-90

t'=L 0,15

:::>

r=I-O.l5· L .

0,15

r=l-rp

rOtOKOplraomCO"VINUlUP'{",M. Lt/pea

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r ,W!;~ ~

t' .'

V{ c \ ("~.

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r 3!)0

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n<1 dl( r'

p.dCf".i>

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3.22

(r1

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je r u rnetrirna. a 'I' u radijanimo. Odrediti: 0) polofaj /:estice (polarne koordinate), b) polumjer zakrlvljenosti putanje, c) i.mos brz.ine:, te d) norm.lnu i tanS''l1cijalnu komponentu ubrzanja. e) Sldcimti dio putanje i uerlati komponente brzine i ubrznnja :za dani poloiaj Cos';.e,

-:;

~:.~

;'

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b) Kolikije pulumjet7.akrivljcnosti u tom trenutku'! c) Skicitati i2l'lleunate vektore n. dijelu putar\ie.

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Ime i prezitne

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4.4.32

Za mehnnizam. nl! ~lici pOZllato je da 11 pri­

kazallolll polozajn brzina slapa B iZllosi

.ill '" -12. 7/ (dakle prellw dolje 12.7 culls).

POIrebuo je odrediti brziull i locke A 1

brzillll stapa AB.

U ~apnom mehanizmu klir.al); A giba se po vertikalnoj vodHici

s brzinom VA = 0,6 mls i ubrzanjem GA::;; 2,5 mls1 prema slid. Klir.a~i ~ i D mogu kJiziti po vodoravnoj vodilici.

4.4.28

Pomatoje AC=CB=O,2m CO=O,2..j3m. Za po)ohj mehanizma prema slici potr
a) brzinu i ubrzanje klizaea 0,

b) kulnU brzinu i kutno ubrzanje !tapova CO i AB.

i,kljnClvo analiticki.

Zadntak Pv

"b" VA nUl

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v,

Vt , Vl..iVU,

v.

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4.4.34

Za mehaniz3m. 11<1 slici pOZllalo je da i.-ullla brzin8 iznosi (I) =4 rad/s '" kOIlSf.

Polrebllo je odrediti brzilltl i locke C i kutnu

brzillll slapa AC.

w= 2..rod/3

~...

2..V'adjs1..

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2.,

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lmci

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/':1 /I. OK. loeo CD u kojoj je krutni Z1ijeb Totira u pokazanom smJeru kutnom brzinom w ~ 12 rndIs i kutnim ubrzanjem a = 6 rndIs' . Na meid OB

iljebu ploee

Luk flijeb. tang;ra dunnu

Valjili" A mole se gibati po krutnom flijebtl polurnjcrar, voden kulisom OB, koj. Totira oka 0 kutnom brzinom '" = IOrndis u:z kutno ubrmnje a 40 rndIs' • Odrediti meladom plana blZin. i ubrzanja, apsolutnu brzinu

i ubrzanje valjilie.. u trenutku kadaje II' = 30· • ako je za4ano: r = 0,4 rn.

=

OA =0,56 m,

CD. Potrebno je odrediti: a) blZinu valjele.. A. b) apsalutno ubrzanje valjclea A Ie e) kutno ubrzanje rueicc OB. o,,~ O(~ AC

?v

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G'" ~.~

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