physics1-0608

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University Physics: Mechanics

Ch4. TWO- AND THREE-DIMENSIONAMOTION

ect!re "

Dr.-In#. Er$in Sit%&'!(http://zitompul.wordpress.com

)*+)

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",)Er$in Sit%&'!( University Physics: Mechanics

Uni%r& Circ!(ar M%ti%n

 A particle is in !ni%r& circ!(ar &%ti%n ifit travels around a circle or a circular arc

at constant (uniform) speed.  Although the speed does not vary, the

particle is accelerating  because thevelocity changes in direction.

The velocity is always directed tangent  to the circle in thedirection of motion.

The acceleration is always directed radially inward . ecause of this, the acceleration associated with uniform

circular motion is called a centri'eta( (!center see"ing#)acce(erati%n.

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Uni%r& Circ!(ar M%ti%n

The magnitude of this centripetal acceleration a is:$

2v

ar 

=

where r  is the radius of the circle and v  is the speed of theparticle.

%n addition, during this acceleration at constant speed, theparticle travels the circumference of the circle (a distance of&πr ) in time of:

2   r T 

v

π =

(centripetal acceleration)

(period)

with T  is called the period of revolution, or simply the period ,of the motion.

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Centri'eta( Acce(erati%n

r 

r 

v

v   ∆=

∆

r 

t v

v

v   ∆≅∆

r 

v

t 

va

2

≅

∆

∆=

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 An ob'ect moves at constant speed along a circular path in ahorizontal xy  plane, with the center at the origin. hen theob'ect is at x    *& m, its velocity is *(+ m/s) '.ive the ob'ect-s (a) velocity and (b) acceleration at y   & m.

Chec0'%int

1

) &

v +  24 &,s 31$

v )  24 &,s i1$

2v

ar 

=

2(4)

2=   28m s=

a  2 &,s) 31$

a$

v +$

v )$

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","Er$in Sit%&'!( University Physics: Mechanics

ighter pilots have long worried about ta"inga turn too tightly. As a pilot-s body undergoescentripetal acceleration, with the headtoward the center of curvature, the bloodpressure in the brain decreases, leading tounconsciousness.

hat is the magnitude of the acceleration, in g  units, of a pilotwhose aircraft enters a horizontal circular turn with a velocity ofv i   +i 0 1' m/s and &+ s later leaves the turn with a velocityof v f   *+i * 1' m/s2

111 1

2v

ar 

=  2   r 

T v

π =

  2v

r T 

π ⇒ =

2v

T 

π =

2 2(400) (500)v  = +

640.312 m s=

2640.312

48

π =

283.818 m s=

8.553 g =   29.8m s g   =

1

2  24 sT   =

E5a&'(e: 6i#hter Pi(%t

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 An Aston 3artin 45 4antage has a!lateral acceleration# of .67g . Thisrepresents the ma8imum centripetalacceleration that the car can attainwithout s"idding out of the circular path.

%f the car is traveling at a constantspeed of 9++ "m/h, what is theminimum radius of curve it cannegotiate2 (Assume that the curve isunban"ed.)

2v

ar 

=

2vr 

a⇒ =

2

2

(40m s)

(0.96)(9.8m s )=

170 m≈

• The re8!ire9 t!rnin# ra9i!s r  is

'r%'%rti%na( t% the s8!are %

the s'ee9 v 

• Re9!cin# v  y s&a(( a&%!nt

can &a0er  s!stantia((ys&a((er 

E5a&'(e: Ast%n Martin

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Re(ative M%ti%n in One Di&ensi%n

The velocity of a particle depends on the reference frame ofwhoever is observing or measuring the velocity.

or our purposes, a reerence ra&e is the physical ob'ect towhich we attach our coordinate system.

%n every day life, that ob'ect is the #r%!n9.

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Th%&

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Re(ative M%ti%n in One Di&ensi%n

uppose that Ale8 (at the origin of frame A) is par"ed by theside of a highway, watching car P  (the #particle#) speed past.arbara (at the origin of frame B) is driving along the highwayat constant speed and is also watching car P .

uppose that both Ale8 and arbara measure the position of

the car at a given moment. rom the figure we see that PA PB BA

 x x x= +

?The c%%r9inate % P  as &eas!re9 y  A is e8!a(

t% the c%%r9inate % P  as &eas!re9 y B '(!s

the c%%r9inate % B as &eas!re9 y  A@

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Re(ative M%ti%n in One Di&ensi%n

Ta"ing the time derivative of the previous e;uation, we obtain

( ) ( ) ( ) PA PB BA

d d d  x x x

dt dt dt  = +

?The ve(%city % P  as &eas!re9 y  A is e8!a( t%

the ve(%city % P  as &eas!re9 y B '(!s the

ve(%city % B as &eas!re9 y  A@

 PA PB BAv v v= +

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Re(ative M%ti%n in One Di&ensi%n

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E5a&'(e: Re(ative >e(%city

uppose that arbara-s velocity relativeto Ale8 is a constant v BA  1& "m/h andcar P  is moving in the negative directionof the x  a8is.

(a) %f Ale8 measures a constant v PA  *>5 "m/h for car P , what

velocity v PB will arbara measure252 km h BAv   =

 PA PB BAv v v= +

 P  moving in the negative direction

78 km h PAv   = −

 PB PA BAv v v= −   ( 78) (52)= − −   130km h= −

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E5a&'(e: Re(ative >e(%city

(b) %f car P  bra"es to a stop relative to Ale8 (and thus relative tothe ground) in time t   9 s at constant acceleration, what is

its acceleration aPA relative to Ale82

uppose that arbara-s velocity relativeto Ale8 is a constant v BA  1& "m/h andcar P  is moving in the negative directionof the x  a8is.

0,   78 km h , PAv   = −   10 st   =

0, PA PA

 PA

v va

t 

−=

  0 ( 78)7.8 km h s

10

− −= =

22.167 m s=

(c) hat is the acceleration aPB of car P  relative to arbaraduring the bra"ing2

0, PB PB

 PB

v va

t 

−=

  52 ( 130)7.8 km h s

10

− − −= =

22.167 m s=

0, AB PBv v

t 

−=

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Re(ative M%ti%n in T$% Di&ensi%ns

%n this case, our two observers are again watching a movingparticle P  from the origins of reference frames A and B, while

B moves at a constant velocity v BA relative to A. The corresponding a8es of these two frames remain parallel,

as shown, for a certain instant during the motion, in the ne8tfigure.

 PA PB BAr r r = +r r r

 PA PB BAv v v= +r r r

 PA PBa a=r r

The following e;uationsdescribe the position, velocity,and acceleration vectors:

$

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E5a&'(e: Re(ative >e(%city

 A plane moves due east while the pilot points the planesomewhat south of east, toward a steady wind that blows to thenortheast.The plane has velocity v PW  relative to the wind, with an airspeed(speed relative to wind) of &91. "m/h, directed at angle θ  southof east.The wind has velocity v 

WG

 relative to the ground with speed 71."m/h, directed &.? east of north.hat is the magnitude of the velocity v PG of the plane relative tothe ground, and what is θ .

$

$

$

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E5a&'(e: Re(ative >e(%city

 PG PW WGv v v= +r r r

215km h PW v   θ = ∠r

65km h 70WGv   = ∠ °r

km h 0 PG PG

v v= ∠ °r

sin sin 70 PW WG

v vθ   = − °   cos cos70 PG PW WGv v vθ = + °

215 sin 65 sin70θ × = − × °

16.50θ   = − °

215 cos( 16.50 ) 65 cos70= × − ° + × °

228.38 km h=

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E5ercise Pr%(e&s

9. A cat rides a mini merry@go@round turning with

uniform circular motion. At time t 9  & s, the

cat-s velocity is v 9  i 0 + 3 m/s, measured ona horizontal xy  coordinate system. At t &  1 s,

its velocity is v &  *i  * + 3 m/s.

hat are (a) the magnitude of the cat-s

centripetal acceleration and (b) the cats

average acceleration during the time interval

t & * t 92

1 1

1 1

&. A suspicious@loo"ing man runs as fast as he can along a moving sidewal"

from one end to the other, ta"ing &.1 s. Then security agents appear, andthe man runs as fast as he can bac" along the sidewal" to his starting

point, ta"ing 9. s. hat is the ratio of the man-s running speed to the

sidewal"-s speed2

     A    n    s      $    e   r :  (  a  )  1 .  &  A  7   m  /  s  & B  (  b  )  *  & i  *  & .  7  7  > '   m  /  s

  & .

     A    n    s      $    e   r :  9 .  7  >