phy10t4ucm&ulg

7/24/2019 PHY10T4UCM&ULG

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s or x

UNIFORM CIRCULAR MOTION (UCM)

ANGULAR QUANTITIES

r

r

or

From 1 to 2 there is a change in

linear displacement (arc s), and alsoa change in angle from 0 to . As thearc length increases the angle in thecircle also increases

s s =r

x =r

1. ANGULAR DISPLACEMENT ( or)

= x / r


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=

v=

r


ANGULAR QUANTITIES

2. ANGULAR VELOCITY ()

x r

This is the change indisplacement with respect to an

interal of time

tt = v / r

AVE=

t

INS=d

dt


3/37


ANGULAR QUANTITIES

=

a=r

. ANGULAR ACCELERATION( )

v r!

This is the change in elocit" with

respect to an interal of time

tt

= a / r

AVE=!

t

INS=d!

dt

* a = r (refers to !"#$%#!&"'acceleration)


4/37

s C&r*+,%r%#% 2-r


ANGULAR QUANTITIES

r

= 3

60 = 2 rad =1 rev

#eriod is Time for 1 reol$tion. %t is$s$all" in seconds orsecond&reol$tion

. PERIOD (T) /. FREQUENCY(,)A '$antit" that is the reciprocal of

#eriod .

%t is $s$all" in radians per secondsor reol$tion&second

f= 1 / T

T 0. ANGULAR FREQUENCY(,)This is the (aerage) ang$lar

elocit" for a reol$tion

f= !f f=

!/T


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CENTRIPETAL FORCE(FC)T

% N%! ,or% !"! !%#s !o %3%! "# o45%! ,ro++o6$ " s!r"&$! 7"! "# "*s%s !o $o "&r*'"r 7"!. T&s &s &r%!% !o !% "x&s o, ro!"!&o#

CENTRIPETAL ACCELERATION ("C)

A'so 8#o9# "s !% r"&"' "%'%r"!&o#. T&s &s !%"%'%r"!&o# "sso&"!% 9&! %#!r&7%!"' ,or% "#,o''o9 !% NSLM: &! "'so $o%s !o !% "x&s o, ro!"!&o#.


6/37


CENTRIPETAL FORCE (FC)

6T

ac

FC

r

+

NS"# F#%! ;Fx +" and $%

o$servation t&ere is no vertical'ove'ent ;F< =F#%! +"

FC +"

" 6T2>r FC +(6T2> r)

(T)Tangential elocit" is also the inear

elocit". %f the ang$lar elocit" is *nownthen

6T r

FC +r2

"C r(2->T)2

(-2r)>T2

%f the ang$lar fre'$enc" is *nown (!f) +

FC +-2,2r

FC (+-2

r) >T2

" r2

%f the period or fre'$enc" is *nown +

"C r(2-,)2

-2,2r


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%n -, the ang$lar elocit" /ang$lar acceleration at A

point or radial distance withinthat circle is constant.

6T

ac

+

12

?2

?1

??

6T

+616266

T, the linear or tangentialelocit" / ( acceleration )di3ers at A point or radialdistance within that circle.


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6F2 6O

2 @ 2"(x)

6F 6O@ "!

x xO @ 6O! @

"!2

L%"r Mo!&o# EB*"!&o#s Ro!"!&o#"' Mo!&o# EB*"!&o#s

x 6O! @

"!2

F2 O

2 @ 2?()

F O@ ?!

O @ O! @

?!2 O! @

?!2

is in radians



9/37

1 A oint on a +&eel rotatin, - rev/s and located . ' fro' t&e axis

exeriences +&at centrietal acceleration

r= .'

ac =

= - rev/s

ac = r

= (- rev/s)x(! rad/ 1 rev) = 1=-

r">sac = (. ')(1.!

rad/s)ac = 1020

'/s


10/37

A sled +it& 'ass = -3, rest on a &ori4ontal s&eet of frictionless iceIt is attac&ed $% a - ' roe to a ost set in t&e ice 5nce 6s&ed7 t&esled revolves 6nifor'l% in a circle aro6nd t&e ost If t&e sled 'a3es8ve co'lete revol6tions er 'in6te 9ind t&e force exerted on it $% t&e

roeTo Vie+

= (- rev/'in) : (! rad/1rev) :

(1'in/;. sec) =./2

T F

+r2T (2/ 8$)(/+)(=./2

r">s)2T .22 N

'=- 3,

r=- '

T T

Isolating thesled


11/37

UNIFORM CIRCULARMOTION U#&,or+ or&o#!"' C&r*'"r

Mo!&o#

9

6T

ac

6T

6T

ac

ac

< =

',

6T

ac

6T

ac

6Tac

6Tac

Tangential or inear elocit"(T) is constant and

perpendic$lar to the radial orcentripetal acceleration (a-)

!" #I$%

a&is

r

F'! #I$%(hal)

Fc or an*net force is not

drawn on the F4

;Fx F

+";F< =


12/37

2 A . 3, $loc3 in t&e 8,6re is attac&ed to a vertical rod $% 'eans of t+ostrin,s o+ 'an% revol6tions er second '6st t&e s%ste' '6st 'a3e in order t&att&e tension in t&e 6er strin, s&all $e 1- N

($)


13/37


($)


14/37


($)

(=. 8$)

= 3+


15/37

2 A . 3, $loc3 in t&e 8,6re is attac&ed to a vertical rod $% 'eans of t+ostrin,s o+ 'an% revol6tions er second'6st t&e s%ste' '6st 'a3e in order t&att&e tension in t&e 6er strin, s&all $e 1- N

($)


16/37

UNIFORM VERTICAL CIRCULAR MOTION

Tangential or Linear

Velocity (vT) is constant

and perpendicular to the

radial or centripetal

acceleration (aC)

W = mg

vT

ac

vT

vT

ac

ac

vT

ac

vT

ac

vTa

c

FRONT VIEWW = mg

W = mg

W = mgW = mg

W = mg

F

F

F

F

F

F

Effect of Weight is

present and its

reaction force (F) is

considered in the

analysis.

;F< F

+";Fx =


17/37

NON- UNIFORM VERTICAL CIRCULAR MOTION

Tangential or Linear

Velocity (vT) is not

constant and but still

perpendicular to the

radial or centripetal

acceleration (aC) which

also varies.

W = mg

vT

ac

vT

vT

ac

ac

vT

ac

vT

ac

vTa

c

FRONT VIEW

W = mg

W = mg

W = mgW = mg

W = mg

F

F

F

F

F

FEffect of Weight is

present and its

reaction force (F) is

considered in the

analysis.

Fc or anynet force is not

drawn on the F!

;F< F

+";Fx =


18/37

Vertical Circ. Motion

1. Tarzan (m=85 kg) tries to cross a river by singing !rom a 1"m long vine. #is s$ee% at t&e bottom o!

t&e sing (as &e 'st clears t&e ater) is 8 ms. Tarzan %oesn*t kno t&at t&e vine &as a breaking

strengt& o! 1""" +. ,oes &e make it sa!ely across t&e river-

ac

vT= 8 m/smT= 85 kg

r = 10 m

Tmax= 1000 N


19/37

UNIFORM CIRCULAR MOTION APPLICATIONS

ROAD CURVES DESIGN

FLAT CURVESTOP VIEW REAR VIEW

ris "radius of curvature#

axis

A FLAT !RVE" ROA" #A$ A %A&I%!% VELOIT' LI%IT IN

W#I# (ELOW T#I$ $PEE" T#E AR AN $AFEL' RO!N"

T#E !RVE WIT#O!T $)I""IN* FRO% T#E ROA"+ T,is can

-. caca.2 sing !% 3 N$L%

vTmax = 4


20/37


ROAD CURVES DESIGN

FLAT CURVES F

ris 5radi$s of c$rat$re6

+

$

N

a&is

ac

fis t&e (net) side frictional force actin,on t&e car It is t&e onl% force alon, t&e

xFaxis rovidin, t&e net force Fcentrietal force

N

+$

ac

NS"# ;F +" and $% o$servation t&ere isno vertical 'ove'ent ;F< =

;Fx +"xH (@)

@ , @+" ;F< =K(@)@ N =

, (+6T+"x2) >

r

N

+$ = s

sN (+6T+"x2) >

rs+$

(+

6T+"x

2) > rs$ (6T+"x

2

) > r

6T+"x2 sr$

6T+"xr


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ROAD CURVES DESIGN

ANED CURVES!" #I$% '$' #I$%

C4'#$ '!5 '$ $ 78$ (9)"'I:'I8; F!' 5F$; '$5!5 4'I7 %$ !'F'!


22/37


ANED CURVES

F

ris 5radi$s of c$rat$re6

+$N

N< N os

+$

a&is

ac

ac

N

+$

ac

Nx N s

NS"# ;F +" and $% o$servation t&ere isno vertical 'ove'ent ;F< =

;Fx +"xH (@)

@ Nx @+"

;F< =K(@)@ N<

=N s +" N os

+$

N (+$) > os

$6t

ac = vT'ax /r

(+$ > os ) s +"+$ !"# +"

$ !"# "

!"# ">$

!"# 1

(">$)

!"# 1(6T+"x2>

(r$)J


23/37

1 A Gat (6n$an3ed) c6rve on a &i,&+a% &as a radi6s of H. ' A carro6nds t&e c6rve at a seed of '/s s)2>(==,!)(2,!>s2)JW

12.2


24/37

2 T&e radi6s of a 9erris +&eel is '7 and it 'a3es one revol6tion in1 s9ind t&e aarent +ei,&t (Nor'al 9orce) of an . 3, assen,er at t&e

&i,&est O lo+est oints

+$

N

NT

+$

ac

a

c

r

A ferris wheel is a ertical circle moing atconstant speed (%F78 98T%-A-%8-A8 7T%7). Apparent weight meansthe e3ect of feeling light or hea" at certain

portions of the ride as the ferris wheel isoperated. This is d$e to the normal force

e:erted ;" wheels ca; in reaction to "o$rweight and the motion of the wheel

T 12s%

"C (-2)(+)J>(12s)2

= 8$( . +>s2) N

"C (-2r)>T2

"C 2.1 +>s2


25/37

NT

+$

ac +$

ac

NT

;F< +"s2X 2.1

+>s2

)NT 0=./0=

N

N

+$

ac

+$

N

;F< +"s2@ 2.1

+>s2

)

!% TOP o, !% F%rr&s%%'

!% OTTOM o, !% F%rr&s%%'


26/37

H A cord is tied to a ail of +ater7 and t&e ail is s+6n, in a verticalcircle +it& radi6s 1H'


27/37

T1W = mg

ac

vT

To get the $ini$u$ velocity% the tension

in the cord $ust also be the $ini$u$%which is &ero.

T1

W = mg

ac

Fy = may (+

!T1" W = "ma#

!T1" mg = "ma#

!(T1+ mg = "(mvT$ % &

a# = vT$% &

T1+ mg = (mvT$ % &

' + mg = (mvMIN$ % &

g = vMIN$% &

vMIN$

= g& = ()* m%$

(1), m = 1).$ m$

%$

vMIN= ).',m%


28/37

T&is +as discovered $% Sir Isaac Ne+ton

States t&at P Ever% article of 'atter in t&e 6niverse

attracts ever% ot&er article +it& a force t&at is directl%roortional to t&e rod6ct of t&e 'asses of t&e articleand inversel% roortional to t&e sJ6are of t&e distance$et+een t&e'Q

+1 +

+2

F12

F21 F2

F2

F1F1

Ill6stration

r

r 1 r2

91= F 91

92= F 92

912= F 921

Us$ N%9!o#Zs T&r L"9 o,

Mo!&o#

T +To eer" actionthere is alwa"s opposedan e'$al reaction, same inmagnit$de ;$t opposite in

direction.

UNIVERSAL LA OF GRAVITATION


29/37

UNIVERSAL LA OF GRAVITATION

Ronsider T+o 5$ects


30/37

MR

9#E= BM ME

RE2

G 0.0 x 1= 11N+2>8$2

FORCE OFGRAVITY

ME /. x 1=2

8$

(+"ss o, !%

%"r!)RE 0. x 1=0

+(r"&*s o, !%%"r!)

(0.0 x 1= 11 N+2>8$2)M (/. x 1= 2

8$)(0. x 1=0 +)2

9#E= (0 '/s) #

M [ (+"ss o, "#


31/37

1 T&e 'ass of t&e 'oon is a$o6t 127 and its radi6s is -7 t&at oft&e eart& Ro'6te for t&e acceleration d6e to ,ravit% on t&e 'oonUss6rface fro' t&is data

ME /. x 1=2

8$ (+"ss o, !% %"r!)RE 0. x 1=

0 + (r"&*s o, !%

%"r!)G 0.0 x 1= 11N+2>8$2 (U#&6%rs"') Gr"6&!"!&o#"'Co#s!"#!

M+ (=.=12)(/. x 1=2 8$) . x 1=228$

R+ (=.2/)(0. x 1=0 +) 1.// x 1=0+

,'= ?#'/ '

,'= (;;x1.F11N'/3,)(22x1.3,)/

(1-0-x1.;')$+ 1.0

+>s2


32/37

,'= ?#'/ '

$+ 1.2

+>s2

,E= ?#E/

E ? = ?

,E

E

/ #E

=,'

'

/ #'

,' = ,E(#'/#E)(E/')

1 T&e 'ass of t&e 'oon is a$o6t 127 and its radi6s is -7 t&at oft&e eart& Ro'6te for t&e acceleration d6e to ,ravit% on t&e 'oonUss6rface fro' t&is data

9ro' ,iven #'= (..12)#EO '=

(.-)E,E= 0 '/s

,' = (0 '/s)(..12#E/#E)(E/.-E)

,' = (0 '/s)(..12)(H)

ALTERNATIVE SOLUTION


33/37

At +&at oint $et+een t&e Eart& and t&e #oon is t&e ,ravitational6ll of t&e Eart& eJ6al in 'a,nit6de to t&at of t&e 'oon Ass6'e ano$ect +it& 'ass # in $et+een t&e eart& and t&e 'oon (Avera,e

distance $et+een Eart& O #oon 2Hx1.

')D = 2Hx1.'

SE S#

ME /. x

1=2 8$

MM .0x

1=22 8$

M


34/37



')D = 2Hx1.'

SE S#

ME /. x

1=2 8$

MM .0x

1=22 8$

MFES FSE

FES

FSE

FSM FMS

FSM FMS


35/37



')

ME /. x

1=2 8$

SE S#

MFSE FSM

FSE GMEM > RSE2FSM GMMM >

RSM2

FSE FSM

GMEM> RSE2

GMMM>RSM2

ME>RSE2 MM>RSM

2

GMEM>RSE2

GMMM>RSM2

MERSM2 MMRSE

2

RSM2 (MM>ME)RSE

2

RSM2 (=.=12)RSE

2

RSM

=.11=RSE


36/37



')

SE S#

MFSE FSM

RSM =.11=RSE

D = 2H x 1.'

D RSM @ RSE

.x1=+ =.11=RSE

@ RSE.x1=+ 1.11=RSE

(.x1=+)>1.11= RSE

RSE

.0x1=+

RSM

.x1=0+


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Satellite#otion

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