FAP0015 Ch05 Circular Motion

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Uni orm circular motionAngular displacement, angular velocity, angular

acce erat on, per o , requency

Centripetal acceleration

Dynamic equation, Centripetal force

Newtons Law of Gravitation

Weight, Gravity, and satellite in circular orbit

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Lesson OutcomesAt the end of the lesson, students should be able to:

1. defineangular displacement,angular velocity,angularacceleration,periodandfrequency.

. s a e e re a on e ween e near an c rcu arpar s o

the motions.

.

the weight of a body.

4. use ree-bod dia rams to solve roblems involvin

centripetal forces and accelerations.

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Tie a string to a stone and then swing it above your

ea or zonta y.

The motion of the stone is an exam le ofcircular motion.

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Uniform Circular Motion

Its a motion of a particle around a circle or circular arcat constant (uniform) speed.

The velocity is always directed tangent to the circle inthe direction of the motion.

T

r2v

=

Period T, is the time

required to travel once

a ou e c c e, a ocomplete one revolution.

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Uniform circular motion

In Fig. (a), at time t0 the

ve oc ty s tangent to t e

circle at point Oand atFig. (a)

velocity is a tangent at

oint P.

As the ob ect moves from O to P the radius traces out theangle , and the velocity vector change the direction.

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In Fig. (b), the velocity vector at time tis redrawn

w t ts ta at para e to tse .

The angle between the two vectors indicates the

c ange n t e rect on.

Since the radii CO and

CP are perpendicular to

the tangent at O and P soP

=+ 90

==+

Thus,an

Fig. (b)

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The acceleration a, is the change in v in velocityv e e e apse t me t,a = v t.

Fig. (c) shows two velocity vectors oriented at the angle

, toget er w t t e vector v t at represents t e c angein the velocity vectors (vt0 + v)= vt T e resu tant ve oc ty vector, v,

has a new direction after an= -

Fig. (c)Part of Fig. (b)

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. g. s ows e sec or othe circle COP.

en s very sma e

arc length OP is straight line

that traveled by the object.

,

isosceles triangle with apex

angle .

Fig. (d)

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Compare COP in Fig (d) with triangle

n g c . ey are s m ar ecauseboth are isosceles triangles with apex

an les labeled are same. Thus

t=vv Fig. (c)

rv

This equation can be solved for ,

v

to show the magnitude of centripetal

acceleration ac, O

rtac vv =

=Fig. (d)

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Linear vs Circular

s=

r

rdtrdt

=

==

ad1d===

v

rdtrdt

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Dynamics ofuniform circular motion

When an object is moving in a uniform circularmotion, there is an acceleration towards the center of

t e c rcu ar pat . cen r pe a acce era on

The ma nitude of the acceleration is

v 22

rc

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To provide this acceleration, there must be a

path.

T e orce s ca e t e centr petal orce.

calculated by using Newtons 2nd. law of

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circular motion

Fc = mac

vmvmF

22

== rr

2

C

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Radius and Centripetal Acceleration

The bobsled track contained turns with

radii of 33 m and 24 m. Find the

centripetal acceleration at each turn fora speed of 34 m/s, a speed that was

ac eve n e wo-man even .

Express the answers as multiples of g

= 9.8 m/s2.

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Solution T e magn tu e can e o ta ne rom t e re at on rac v= Since the radius ris in the denominator on the right side,

22

e acce era on s sma er w en r s arger

For radius =33 and 24 m the centripetal acceleration is

3.6gm/s35m33

33For 2 =====r

a,r c

m/s34 22v

The acceleration approaches to zero when ris very large. UCM

. gm sm42

or =====r

a,r c

along the arc of an infinitely large circle entails no acceleration,because is just like at constant speed along straight line.

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UCM and E uilibriumConceptual Problem:

A car moves at a constant speed, and there arethree parts to the motion. It moves along a

straight line toward a circular turn, goes around

the turn, and then moves away along a straightline. In each of the parts, is the car in

equilibrium?

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Answer T e o ect s n equi ibrium w en t as ero acce eration.

As the car approaches the turn, both the speed anddirection of motion are constant.

Thus the velocity vector does not change and there is noacceleration.The same is true when the car moves awayfrom the t rn. For these arts of motion the car is inequilibrium.

As the car goes around the turn, the direction of the travel,characteristic of UCM.

Because of this acceleration the car is not in equilibrium

. An object that is in UCM can never be in equilibrium.

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Speed and Centripetal Force

The model airplane has amass of 0.90 kg and moves ata constant spee on a c rc e

that is parallel to the ground.The ath of the air lane andits guideline lie in the samehorizontal plane, because the

balanced by the lift generatedby its wings. Find the tension

for speed of 19 and 38 m/s.

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Solution Since the plane flies on a circular path, it experiences acentripetal acceleration. that is directed towards the center.

This acceleration is produced by a net force which is equalto the tension T. Because T in the guideline is the only force

pu ng e p ane nwar mus e e cen r pe a

force.Thus,Fc= T = mv2/r,

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Example . g ca ou e a co e o a a us r= .

the coefficient of static friction s = 0.82, what is thereatest s eed the car can have in the corner without

skidding?Solution:

mv2

The sum ofF in x-direction for the force of static friction

and is

rcpxssx

The sum of y-components of force is

mgWNmaWNF yy ==== Thus

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orth s

22

===

v

r

v

mm

SubstituteNand solve forv,

m/s19)m/sm)(9.8145)(082(Thus

2 ==v

rgr sss

The mass of the car has been eliminated. Thus all cars, heavy

or light have the same maximum speed.The speed it depends on s, thus the dry road allows greater

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Example

If a lateral acceleration of 8.9 m/s2 represents the

ax u c a ca e a a e w ou g ou

of the circular path, and if the car is traveling at a,

can negotiate? If the driver rounding a flat with

unbanked curve with radius R. If the coefficient of friction between the tires and road is s, what is themaximum speed v at which he can take the curve

w ou s ng

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Solution

(a) We have, ac = 8.9 m/s and v = 45 m/s.

= 2 = 2 2, = c, . .

(b) Acceleration v2/R toward the center of the curve

s,

vertical acceleration, Thus,Fs = m v2/R, The normal

force mg = 0.

We haveFs = sFN,, = smg, which is constant anddetermine the cars maximum speed. v = ( gR)1/2, ifs = 0.91, R = 230 m, Then v = 45 m/s.

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Banked Curve

A vehicle can negotiate a circular turn without relying

turn is banked at an angle relative to the horizontal.

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To provide centripetal force

(without friction) :

r

mvsinFF Nc ==

mcosF =

mr/mv

cosFsinFN

2

=

rgtan

2v=

For a given speed, v, the centripetal force needed for a turn of

radius r can be obtained by banking the turn at an angle ,

independent of the mass of vehicle.What would happen if a vehicle moves at a speed much larger or

much smaller than v?

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GRAVITY

Gravity is a fundamental force in sense that cannot beexplained in terms of any other force.

Fundamental forces are: gravitational, electromagnetic and

nuclear forces. These forces seem to be responsible for everything that

happens in the universe.

and hold together planets, stars and galaxies of stars.

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Newtons Law of Universal Gravitation

Newton proposed a force law saying that every

particle attracts any other particle with a gravitational

force.

Every particle of matter in the universe attracts

every other particle with a force that is directlyproport ona to t e pro uct o e masses o t e

particles and inversely proportional to the square of

.

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Gravitational Attraction of Spherical Bodies

A uniform sphere with a radiusR and massM, and object of

mass m is brought near the sphere at the distance rfrom the

cen er.

Newton showed that, the net

orce exer e y e sp ere on

the mass, m is the same as if all

the masses of the s here wereconcentrated at its center this

force is, GmM2 rF =

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Newtons Law of Universal Gravitation

The force of gravity between any two point objects of

1 2

mm2

r

F =

where G is the universal gravitational

constant, G = 6.67 x 10-11Nm2/kg2

F-gravity forms action-reaction pair.


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Dependence of the Gravitational Force

,

2

21 mGmF =

The force diminishes rapidly with the distance, but never completelyvanishes. Thus, gravity is a force of infinite range

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Weight Previously we defined the weight of a body asthe attractive

gravitational force exerted on it by the earth.

Now, we can broaden the definition as: the weight of the

body isthe total gravitational force exerted on the body by

.

When the body near the earth, we can neglect all other

earths gravitational attraction.

At the surface of the moon we can ne lect all others forces

and consider the bodys weight to be gravitational attractionof the moon, and so on.

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So, the weight of a body of mass, m, near the earth surface,

2

e

e

r

GmMmgF ==

whereMe and re are the mass and radius of the earth respectively.

= 24

2so,

e

e

rg= e

.

re= 6.38 106 m

For a body of mass, m, at a distance h from the earth surface,

eGM eGmM

( )2erh + ( )2,

erh +

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Mass in Circular Orbit

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Gravitational and Inertial Mass

We have had two definitions of mass:

T e property o an o ect t at res sts c ange n state

of motion.

ppears as e cons an n ew on s secon aw

F =ma. It is called inertial mass.

e property o an o ect t at eterm nes t e

strength of the gravitational force

= mg.It is called gravitational mass.

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Example

Find the acceleration of gravity on the surface oft e moon.

The lunar rover has a mass of 225 kg. What is itswe g t on t e eart an on t e moon

no e, e mass o e moon s m = . x gand its radius isRm = 1.74 x 10

6 m.]

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Solution:For the moon the acceleration of gravity is

22211

m s62.11035.71067.6

=

==xGMm

261074.1 Rmm

This is about 1/6 of thegon the earth.

NsmkgmgW 2210/81.9225

2

===

n t e eart t e rover s we g t was

2

On the Moon, its weight was

.

As expected 1/6 its earth weight.

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Conce tual uestion

,

easier to drive at high speed around

un an e or zon a curve on e moon

than to drive around the same curve onthe earth? Explain.

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REASONING AND SOLUTION

The maximum safe speed with which a car can round

.

rgv s=Since the acceleration due to gravity on the moon isroughly one sixth that on earth, the safe speed for the same

.other words, other things being equal, it would be moredifficult to drive at high speed around an unbanked curve

on the moon as compared to driving around the same curveon the earth.

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Conce tual uestion

a circular path at a constant speed. Is stringmore likel to break when the circle ishorizontal or when it vertical? Account for to

your answer assuming the constant speed isthe same in each case.

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REASONING AND SOLUTION

When the string is whirled in a horizontal circle, the

e o e g, T, ov e e ce e a

force which causes the stone to move in a circle.,

tension in the string is constant.

,tension in the string and the weight of the stone both

contribute to the centri etal force, de endin on

where the stone is on the circle.

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Now, however, the tension increases and decreases asthe stone traverses the vertical circle. When the stone isat the lowest point in its swing, the tension in the string

pulls the stone upward, while the weight of the stone

mgFvm T2

= Thus mgvmF2

T +=r

2

This tension is larger than in the horizontal case.

rFT =

e e o e, e g a a g ea e c a ce o ea gwhen the stone is whirled in a vertical circle.

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Loop the Loop

The rider who performthe loop-the loop trick

know that he must have

a minimum speed at

e op o e c rc e o

remain on the track.

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2

F1 1N1r

mg. =

F2

2

N2 r

mv

. =

r

mvmg.

23

N3F3 =+

mv.

2

4F4 =

r

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FAP0015 Ch05 Circular Motion