Physics Project on circular motion

7/23/2019 Physics Project on circular motion

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Circular Motion



1 Kinematics of Uniform Circular Motion

Uniform circular motion: motion in a circle of

constant radius at constant speedInstantaneous velocity is always tangent to

circle.



1 Kinematics of Uniform Circular Motion

Looking at the change in velocity in the limit that the

time interval ecomes infinitesimally small! we see that

"#$1%

&his acceleration is called the centripetal! or radial!

acceleration! and it points towards the center of thecircle.



Example 1

' 1#( kg all at the end of a string is revolving uniformly in a

hori)ontal circle of radius (.*(( m. &he all makes +.(( revolutionsin a second. ,hat is its centripetal acceleration-

aR =v2

r

v =2π r

T=

2(3.14)(0.600 m)

(0.500 s)= 7.54 m/s

aR = v

2

r= (7.54 m/s)

2

(0.600 m)= 94.7 m/s2



Example 2

&he moons nearly circular orit aout the earth has a radius of aout

/0!((( km and a period & of +2./ days. 3etermine the acceleration

of the moon toward the earth.

r = 3.84x108 m, T = (27.3 d)(24.0 h/d)(3600 s/h) = 2.36x106 s

v =2π r

T → a =v2

r =4π 2r2

T2r =4π 2r

T2 =4π 2(3.84x108 m)

(2.36x106 s)2 = 0.00272 m/s2



+ 3ynamics of Uniform Circular Motion

4or an o5ect to e in uniform circular motion! there

must e a net force acting on it.

,e already know theacceleration! so can

immediately write the force:

"#$1%

,e can see that the force

must e inward y

thinking aout a all on a

string:



+ 3ynamics of Uniform Circular Motion

&here is no centrifugal force pointing outward6

what happens is that the natural tendency of theo5ect to move in a straight line must e

overcome.

If the centripetal force vanishes! the o5ect flies

off tangent to the circle.



Example 3

7stimate the force a person must e8ert on a string attached to a (.1#(

kg all to make the all revolve in a hori)ontal circle of radius (.*((m. &he all makes +.(( revolutions per second "&9(.#(( s%.

(ΣF)R = maR = mv2

r

= m(2π r/T)2

r

= (0.150 kg)4π 2(0.600 m/0.500 s)2

(0.600 m)

=14 N



/ ighway Curves! ;anked and Unanked

,hen a car goes around a curve! there must e

a net force towards the center of the circle ofwhich the curve is an arc. If the road is flat! that

force is supplied y friction.




If the frictional force is

insufficient! the car will

tend to move more

nearly in a straight line!

as the skid marks show.




's long as the tires do not slip! the friction is

static. If the tires do start to slip! the friction iskinetic! which is ad in two ways:

1. &he kinetic frictional force is smaller than the

static.+. &he static frictional force can point towards

the center of the circle! ut the kinetic frictional

force opposes the direction of motion! making

it very difficult to regain control of the car and

continue around the curve.




;anking the curve can help keep

cars from skidding. In fact! forevery anked curve! there is one

speed where the entire centripetal

force is supplied y the

hori)ontal component ofthe normal force! and no

friction is re<uired. &his

occurs when:



Example 6

A 1000. kg car rounds a curve o a flat road of radius

50. m at a speed of 50. km/h (14 m/s). Will the car

follow the curve or it will it skid! Assume (a) the

pavement is dr" and the coefficient of static friction is

µs#0.$0% (&) the pavement is ic" and µs#0.'5.

FN = mg = (1000. kg)(9.8 m/s2) = 9800 N

(ΣF)R = maR = m v2

r= (1000 kg) (14 m/s)

2

(50 m)= 3900 N

(a) (Ffr )max = µ sFN = (0.60)(9800 N) = 5900 N

Since this is greater than 3900 N, the car will

follow the curve.

(b) (Ffr )max = µ sFN = (0.25)(9800 N) = 2500 N

Since this is less than 3900 N, the car will not

follow the curve, so it will skid.



Example 7

(a) or a car traveling with speed v around a curve

of radius r determine a formula for the angle atwhich a road should &e &anked so that no friction is

reuired. (&) What is the angle for an e*presswa"

off+ramp curve of 50 m at a design speed of 50

km/h!

ΣFR = maR → FNsinθ =mv2

rΣFy = may = 0→ FNcosθ - m g = 0→ FN =

mg

cosθ

FNsinθ =mv2

r→

mg

cosθ sinθ =

mv2

r→mgtanθ =

mv2

r

tanθ =

v2

rg

For r = 50 m and v =14 m/s, we have

tanθ =(14 m/s)2

(50 m)(9.8 m/s2)= 0.40→θ = 22o



=onuniform Circular Motion

If an o5ect is moving in a circular

path ut at varying speeds! it

must have a tangential component to its acceleration as

well as the radial one.



=onuniform Circular Motion

&his concept can e used for an o5ect moving

along any curved path! as a small segment of thepath will e appro8imately circular.



# Centrifugation

' centrifuge works y

spinning very fast. &hismeans there must e a

very large centripetal

force. &he o5ect at '

would go in a straightline ut for this force6 as

it is! it winds up at ;.



Example 9,he rotor of an ultracentrifuge rotates at 50000

rpm (revolutions per minute). ,he top of a 4.00

cm long test tu&e is $.00 cm from the rotation a*isand is perpendicular to it. ,he &ottom of the tu&e

is 10.00 cm from the a*is of rotation. -alculate

the centripetal accelerations in gs at the top

and the &ottom of the tu&e.

At top, 2π r = (2π )(0.0600 m) = 0.377 m per revolution

50,000 rpm = 833 rev/s so T =1

833 rev/s=1.20x10-3 s/rev

v =2π r

T=

0.377 m/rev

1.20x10-3 s/rev= 3.14x102 m/s

aR =v2

r =(3.14x102 m/s)2

0.0600 m =1.64x106 m/s2 =1.67x105 g's

At the bottom, v =2π r

T=

(2π )(0.1000 m)

1.20x10-3 s/rev= 523.6 m/s

aR =v2

r=

(523.6 m/s)2

0.1000 m= 2.74x106 m/s2 = 2.80x105 g's









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Physics Project on circular motion