Physics Circular Motion

 

Uniform circular motion  Angular displacement, angular velcit!, angular acceleratin, perid, "re#uenc! Centripetal acceleratin $!namic e#uatin, Centripetal "rce %inear vs circular mtin &e'tn(s %a' " )ravitatin *eig+t, *eig+ t, )ravit!, and satellite in circular c ircular rit

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 Lesson Outcomes At the end of the lesson, students should be able to: 1.

define angular displacement , angular velocity, angular acceleration, period  and  and frequency.

2.

state the relation between the linear  and  and circular parts of the motions.

3.

ap appl ply y New Newto ton’ n’ss univ univer ersal sal laws laws of grav gravit itat ation ion to de dete term rmin inee the weight of a body.

4.

use free-body diagrams to solve problems involving entripetal fores and aelerations.

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!ie a string to a stone and then swing it above your !ie head hori"ontally.

!he motion of the stone is an e#ample of irular motion motion..

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Uniform Circular Motion $t’s a motion of a partile around a irle or irular ar at onstant %uniform& speed. !he veloity is always direted tangent to the irle in the diretion of the motion.

v =  2πr  ! 'eriod !, is the time re(uired to travel one around the irle, that is to omplete one revolution. FAP 0015 PHYSICS I

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Uniform circular motion ) $n Fig. (a), at time t * the veloity is tangent to the irle at point O and at

Fig. (a)

a later time t  the  the veloity is a tangent at  point P . As the ob+et moves from O to P , the radius traes out the angle θ , and the veloity vetor hange the diretion.

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) $n Fig. (b), the veloity vetor at time t  is  is redrawn with its tail at O parallel to itself. ) !he angle β  between  between the two vetors indiates the hange in the diretion. ) ine the radii CO CO and  and CP   are perpendiular to the tangent at - and ' so it follows that

P 

α  + β  = *°  and  α  + θ  = *°,  !hus  β  = θ  FAP 0015 PHYSICS I

Fig. (b)

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) !he aeleration a, is the hange in ∆v in veloity divided be elapsed time ∆t, a = ∆v/∆t. )Fig. (c) shows two veloity vetors oriented at the angle θ , together with the vetor ∆v that represents the hange in the veloity vetors %v t * 0 ∆v& vt )  !he result resultant ant veloity vetor, v, has a new diretion after an elapsed time ∆t = t - t 0 Part of Fig. (b)

Fig. (c)

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)  Fig. (d)  shows the setor   of

Part of Fig. (b)

the irle COP. COP.   )  hen ∆t is very small the ar length OP   is straight line and e(uals to the distane v∆t   that traveled by the ob+et. )  $n this limit, COP   is an isoseles triangle with ape# angle θ . Fig. (d)

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ompare OP in Fig (d) with triangle in Fig (c). !hey are similar beause  both are isoseles triangles with ape# angles labeled θ  are same. !hus ∆v

v

=

v∆t 

r 

!his e(uation an be solved for

Fig. (c)

∆v  , ∆t

to show the magnitude of entripetal aeleration a aeleration  ac,

O

2

∆v v ac = = ∆t  r 

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Fig. (d) 06/08/15

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 Linear vs Circular   s θ  = r  ω = dθ  = 1  ds  = v dt  r   dt  r 

α = dω =  1  d v  = a dt  r   dt   r  FAP 0015 PHYSICS I

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 Dynamics of uniform circular motion hen an ob+et is moving in a uniform irular motion, there is an acceleration towards the enter of the irular path. %centripetal acceleration& !he magnitude of the aeleration is 2

ac = FAP 0015 PHYSICS I

v r 

=4

2

r 

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!o provide this aeleration, there must be a fore ats towards the center of the irular  path. ) !he fore is alled the centripetal force. ) !he magnitude of this fore an be nd

alulated motion. by using Newton’s 2 . law of

FAP 0015 PHYSICS I

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 Equations describing uniform circular motion

m a ∑ 5  ma  

∑5

 

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

=m

∑5

v

2

r 

=

mv

2

r  2



= m  r  06/08/15

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 adius and Centripetal !cceleration !cceleration ) !he bobsled tra6 ontained turns with radii of 33 m and 24 m. 5ind the entripetal aeleration at eah turn for a speed of 34 m/s, a speed that was ahieved in the two7man event. 8#press the answers as multiples of g  .9 m/s 2.

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UCM and Equilibrium" Conceptual #roblem$ ) A ar moves at a onstant speed, and there are three parts to the motion. $t moves along a straight line toward a irular turn, goes around the turn, and then moves away along a straight line. $n eah of the parts, is the ar in equilibrium

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 %peed and Centripetal &orce ) !he model airplane has a mass of *.* 6g and moves at a onstant speed on a irle that is parallel to the ground. !he path of the and its guideline lie airplane in the same hori"ontal plane, beause the weight of the plane is  balaned by the lift generated  by its wings. Find the tension  in the guideline %length 1; m& for speed of 1 and 39 m/s.

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 E'ample ) A 12**.* 6g ar rounded a orner of a radius r    4< m. $f the

oeffiient of stati frition µs  *.92, what is the greatest speed the ar an have in the orner without s6idding

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Example ) $f a lateral aeleration of 9. m/s 2  represents the ma#imum a   that an be attained without s6idding out of the the irular path, and if the ar is traveling at a c

onstant 4< m/s, what is a m minimum inimum radius of urve it an negotiate $f the driver rounding a flat with unban6ed urve with radius  . $f the oeffiient of frition between the tires and road is , what is the ma#imum speed v  at whih he an ta6e the urve without s6idding s

FAP 0015 PHYSICS I

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 (an)ed Curve

A vehile an negotiate a irular turn without relying on stati frition to provide the entripetal fore if the t he turn is ban6ed at an angle relative r elative to the hori"ontal.

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!o provide entripetal fore %without frition& : =  F  sin = = c  N   F  N  os   = = mg 

 F 

 F   N  sin θ   F   N  cos θ 

=

mv

2

r 

mv 2 / r 

tan θ  =

mg  v

2

rg 

5or a given speed, sp eed, v, the entripetal fore needed fo forr a turn of radius r an be obtained by ban6ing the turn at an angle θ, independent of the mass of vehile. hat would happen if a vehile moves at a speed muh larger or muh smaller than v FAP 0015 PHYSICS I

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GRAVI!

) Gravit"  is a fundamental fore in sense that annot be e#plained in terms of any other fore. ) 5undamental fores are: gravitational, eletromagneti and nulear fores. ) !hese fores seem to be responsible for everything that happens in the universe. ) Gravitational fores at between all bodies in the universe and hold together planets, stars and gala#ies of stars.

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)ravit!

&e'tn2s apple tree, 3rinit! Cllege, Camridge, 4ngland

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 A #ute ! Alert 4instein

*+ravity cannot be held responsible  for people falling in love,  ./ou can0t blame gravity for falling in love,

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 1ewton2ss Law of Universal +ravitation  1ewton2  

Newton proposed a fore law saying that every  particle attracts any other particle with a gravitational fore.  

8very partile of matter in the universe attrats every other partile with a fore that is directly  proportional   to the  product of the masses  of the  partiles and inversely proportional   to the square of the distance between them.

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Gravitational Attraction of #pherical $odies A uniform uniform sphere with a radius R radius R and  and mass M  mass M , and ob+et of mass m is brought near the sphere at the distane r distane r from the enter.  Newton showed that, the net fore e#erted by the sphere on the mass, m is the same as if all the masses of the sphere were onentrated at its enter this fore is,

 F   =

FAP 0015 PHYSICS I

GmM  r 2

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 1ewton2ss Law of Universal +ravitation  1ewton2 !he fore of gravity between any two point ob+ets of mass m1 an m is attractive and of magnitude

 F    = Gm12m2 r  where G is the universal gravitational onstant, G  >.>; # 1*711 Nm2/6g2 & 7gravity 7gravity forms ation7reation pair.

 

 Dependence of the +ravitational +ravitational &orce on %eparation Distance" r 

 F    =

Gm1m2 r 2

!he fore diminishes rapidly with the distane, but never ompletely vanishes. !hus, gravity is a fore of infinite range. FAP 0015 PHYSICS I

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%eight ) 'reviously we defined the weight of a body as the attractive  gravitational force e'erted on it by the earth. )  Now, we an broaden the definition as: the weight of the  body is the total gravitational force e'erted on the body by all other bodies in the universe . ) hen the body near the earth, we an neglet all other gravitational fores and onsider the weight as +ust the earth’s gravitational attration.

) At the surfae of the moon we an neglet all others fores and onsider the body’s weight to be gravitational attration of the moon, and so on. FAP 0015 PHYSICS I

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o, the weight of a body of mass, m, near the earth surfae,

GmM !  F   =  mg  =

2

r !

where M  where  M ! and r ! are the mass and radius of the earth respetively. ! so,   g   = GM  2 r !

 M e .39 × 1*> m

5or a body of mass, m, at a distane " distane " from  from the earth surfae,

 g  =

GM !

( " + r ! )

FAP 0015 PHYSICS I

2

o, eight = mg  =

GmM !

( " + r ! )

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 Mass in Circular Orbit 

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4ercise $etermine t+e average radius " rit " t+e mn arund t+e eart+ ased n its perid " rit

mv 2 r 

v2 = G r  = G r = G

=G

m $  r 

m $ m r 2 me .>; # 1*711 Nm2/6g2

m $ 

#     29 days  29 ? 24 ? 3>** s  2.4 ? 1*> s

v2 m $ # 2

  2π  r  v= # 

2 2

4π  r 

  r 3 = G

2 m$ #   

4π 2

=

(>.>; × 1* )  

−11

2 2   π  r  4 2 v = 2

# 



 

(

4π  

2

)

2

  m

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+ravitational and 3nertial Mass 

%e have had t&o definitions of mass'

)  !he property of an ob+et that resists hange in state of motion. ) Appears as the onstant in Newton’s seond law &     ma. $t is alled inertial mass. )!he property of an ob+et that determines the strength of the gravitational fore & 4 mg,  $t is alled gravitational mass. FAP 0015 PHYSICS I

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4ample 7 Find t+e acceleratin " gravit! n t+e sur"ace " t+e mn 7 3+e lunar rver +as a mass " --5 g *+at is its 'eig+t n t+e eart+ and n t+e mn9 :nte, t+e mass " t+e mn is  M m ; .5  10-- g and its radius is  Rm ; 1  106 m<

FAP 0015 PHYSICS I

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Cnceptual =uestin 7 >t+er t+ings eing e#ual, 'uld it e easier t drive at +ig+ speed arund unaned +ri?ntal curve n t+e mn t+an t drive arund t+e same curve n t+e eart+9 4plain

.5

FAP 0015 PHYSICS I

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 REASONING AND SOLU SOLUTION  TION  • The maximum safe speed with which a car can round an unbanked horizontal curve of radius r  is  is given by .

v

=

 µ  s rg 

•Since the acceleration due to gravity on the moon is roughly one sixth that on earth, the safe speed for the same curve on the moon would be less than that on earth. In other words, other beingaround equal, an it would be more difficult to drive at things high speed unbanked curve on the moon as compared to driving around the same curve on the earth.

FAP 0015 PHYSICS I

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Cnceptual =uestin 7  A stne is tied t a string and '+irled arund in a circular pat+ at a cnstant speed Is string mre liel!r t rea '+en Accunt t+e unt circle is +ri?ntal '+en it vertical9 Acc "r t !ur ans'er assuming t+e cnstant speed is t+e same in eac+ case

.

FAP 0015 PHYSICS I

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 REASONING AND SOLUTION  • When the string is whirled in a horizontal circle, the tension in the string, F !, provides the centripetal force which causes the stone to move in a circle. Since the speed of the stone is constant, and the tension in the string is constant. • When the string is whirled in a vertical circle, the tension in the string and the force, weightdepending of the stone contribute to the centripetal onboth where the stone is on the circle.

.8

FAP 0015 PHYSICS I

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Free d! diagrams Hri?ntal circle

mv

2

∑ F  =  F   cos φ  = r = *  F  =  F       sin φ  − mg  ∑  &

# 

 %

# 

@ertical @e rtical c circle ircle  At t+e tp " circle 2

∑ F  =  F  + mg  = mvr   %

# 

mv 2

 F #  =  

r 

− mg 

 At t+e ttm " circle

∑

 F  %

=  F #  − mg  =

mv 2

 F #  =  

r 

+ mg 

2

mv r 

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)  Now, however, the tension inreases and dereases as the stone traverse traversess the vertial irle. hen the stone is at the lowest point in its swing, the tension in the string pulls the stone upward, while the weight of the stone ats downward. !herefore, the entripetal fore is .

mv r 

2

=  F    − mg 

 Thus

 F # 

mv

=  

2

+ mg 

r 

# 

•  This tens tension ion is larger than in the horizontal ccase. ase.

 F #  =

mv

2

r  •  Therefore Therefore,, the string has a greater chance of breaking when the stone is whirled in a vertical circle. 

FAP 0015 PHYSICS I

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 Loop the Loop

!he rider who perform the loop7the loop tri6 6now that he must have a minimum speed at the top of the irle to remain on the tra6.

1

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1 .   5 N1

 −

m g  =

m v1

2

 

r  2.   5  N2 =

mv2

2

r 

3 .   5 N  N3 3 + m g  =

4.   5   N4 =

mv 4 2 r 

 

m v3 r 

2

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 At pint .,

2

mv +

 F  N

m v3

=  

 F  N  F  N

≥

m v3

2

v3

r 

≥

3

m g  =

r 

r 

2

−

* ≥

m g 

rg 

m g 

FAP 0015 PHYSICS I

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3+e rllercaster   At '+at minimum speed must a rller rller caster e traveling '+en '+en upside d'n at t+e tp " a circle s t+at t+e t +e passengers 'ill nt "all ut9 Assume Assume a radius " curvature "  m

