Circulair Motion

8/14/2019 Circulair Motion

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Physics 111: Lecture 3, Pg 1

Physics 111: Lecture 3Physics 111: Lecture 3

Todays AgendaTodays Agenda

q Reference frames and relative motion

q Uniform Circular Motion


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Inertial Reference Frames:Inertial Reference Frames:

q A Reference FrameReference Frame is the place you measure from. Its where you nail down your (x,y,z) axes!

q An Inertial Reference Frame ( IRF ) is one that is notaccelerating . We will consider only IRF s in this course.

q Valid IRF s can have fixed velocities with respect to each other.

More about this later when we discuss forces. For now, just remember that we can make measurements

from different vantage points.

Cart ontrack on

track


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Relative MotionRelative Motionq Consider a problem with twotwo distinct IRFs:

An airplane flying on a windy dayAn airplane flying on a windy day ..

A pilot wants to fly from Champaign to Chicago. Having

asked a friendly physics student, she knows that Chicagois 120 miles due north of Urbana. She takes off fromWillard Airport at noon. Her plane has a compass and anair-speed indicator to help her navigate.

The compass allows her to keep the nose of the planepointing north.

The air-speed indicator tells her that she is traveling at120 miles per hour with respect to the air with respect to the air .


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Relative Motion...Relative Motion...

q The plane is moving north in the IRF attached to the air: V V p, a is the velocity of the plane w.r.t. the air.

Air Air

V V p,a


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q But suppose the air is moving east in the IRF attached tothe ground.

V V a,g is the velocity of the air w.r.t. the ground (i.e. wind ).

V V a,g

Air Air

V V p,a


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q What is the velocity of the plane in an IRF attached to theground? V V p,g is the velocity of the plane w.r.t. the ground.

V V p,g


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V V p,g = V V p,a + V V a,g Is a vector equation relating the airplanesvelocity in different reference frames.

V V p,g

V V a,g

V V p,a

Tractor



Lecture 3,Lecture 3, Act 1 Act 1Relative MotionRelative Motion

q You are swimming across a 50m wide river in which thecurrent moves at 1 m/s with respect to the shore. Your swimming speed is 2 m/s with respect to the water.You swim across in such a way that your path is a straightperpendicular line across the river. How many seconds does it take you to get across ?

(a)

(b)

(c)

50 3 29=

2 m/s

1 m/s50 m

35 2 50 =

50 150 =



Lecture 3,Lecture 3, Act 1 Act 1solutionsolution

q The time taken to swim straight across is (distance across) / (v y )

Choose x axis along riverbank and y axis across river y

x

q Since you swim straight across, you must be tilted in the water so thatyour x component of velocity with respect to the water exactly cancelsthe velocity of the water in the x direction:

2 m/s 1m/sy

x

1 m/s

2 1

3

2 2

= m/s



Lecture 3,Lecture 3, Act 1 Act 1solutionsolution

q So the y component of your velocity with respect to the water is

q So the time to get across is

y

x

3 m/s

50 3

29m

m ss=

50 m

3 m/s



Uniform Circular MotionUniform Circular Motion

q What does it mean?

q How do we describe it?

q What can we learn about it?



What is UCM?What is UCM?

q Motion in a circle with: Constant Radius R

Constant Speed v = |v v |

R

v v

x

y

(x,y)

Puck on ice



How can we describe UCM?How can we describe UCM?

q In general, one coordinate system is as good as any other: Cartesian:

(x,y) [position] (v x ,v y ) [velocity]

Polar: (R, ) [position] (v R , ) [velocity]

q In UCM: R is constant (hence v R = 0 ). (angular velocity) is constant. Polar coordinates are a natural way to describe UCM!Polar coordinates are a natural way to describe UCM!

R R

v v

x

y

(x,y)



Polar Coordinates:Polar Coordinates:

q The arc length s (distance along the circumference) isrelated to the angle in a simple way: s = R , where is the angular displacement . units of are called radians .

q For one complete revolution:2 R = R c c = 2

has period 2

.

1 revolution = 21 revolution = 2 radiansradians

R R

v v

x

y

(x,y)s



Polar Coordinates...Polar Coordinates...

x = R cos

y = R sin

/ 2 3 /2 2

-1

1

0

sincos

R R

x

y

( x ,y )



Polar Coordinates...Polar Coordinates...

q In Cartesian coordinates, we say velocity dx/dt = v . x = vt

q In polar coordinates, angular velocity d /dt = . = t has units of radians/second .

q Displacement s = vt .but s = R = R t, so:

R R

v v

x

y

s=

t

v = R

Tetherball



Period and FrequencyPeriod and Frequency

q Recall that 1 revolution = 2 radians frequency (f ) = revolutions / second (a) angular velocity ( ) = radians / second (b)

q By combining (a) and (b)

= 2 f

q Realize that : period (T) = seconds / revolution

So T = 1 / f = 2 /

R R

v v

s

= 2 / T = 2 f



Recap:Recap:

R R

v v

s = t

(x,y)

x = R cos( ) = R cos( t)

y = R sin( ) = R sin( t) = arctan (y/x)

= t s = v t s = R = R t

v = R



Aside: Polar Unit VectorsAside: Polar Unit Vectors

q We are familiar with the Cartesian unit vectors: i j k i j k

q Now introducepolar unit-vectors r r and :

r r points in radial direction points in tangential direction

R R

x

y

i i

j j

r r ^

^

^ ^

^

(counter clockwise)



Acceleration in UCM:Acceleration in UCM:

q Even though the speed speed is constant, velocityvelocity is not not constantsince the direction is changing: must be some acceleration !

Consider average acceleration in time t a av = v v / t

v v 2

t

v v 1v v 1v v 2

vv

R R




seems like v v (hence v v / t )points at the origin!

R R

q Even though the speed speed is constant, velocityvelocity is not not constantsince the direction is changing.

Consider average acceleration in time t a av = v v / t

v v




q Even though the speed speed is constant, velocityvelocity is not not constantsince the direction is changing.

As we shrink t , v v / t d v v / dt = aa

aa = d v v / dt

We see that aa pointsin the - R R direction.

R R


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q This is calledThis is called Centripetal Acceleration.q Now lets calculate the magnitude:

v v 2

v v 1

v v 1v v 2

v v

R R R R

v

v

R

R =Similar triangles:

But R = v t for small t

So:

v

t

v

R

=2 v

v

v t

R

=

av R

=2


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Centripetal AccelerationCentripetal Acceleration

q UCM results in acceleration: Magnitude : a = v 2 / R Direction : - r r (toward center of circle)

R aa

^


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Derivation:Derivation:

( )R R

a2

=

We know that and v = R

Substituting for v we find that:

av R

=2

a = 2R


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Lecture 3,Lecture 3, Act 2 Act 2 Uniform Circular MotionUniform Circular Motion

q A fighter pilot flying in a circular turn will pass out if thecentripetal acceleration he experiences is more than about9 times the acceleration of gravity g . If his F18 is movingwith a speed of 300 m/s , what is the approximate diameter of the tightest turn this pilot can make and survive to tell

about it ?

(a) 500 m(b) 1000 m

(c) 2000 m


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Lecture 3,Lecture 3, Act 2 Act 2 SolutionSolution

av R

g = =2

9

2

2

2

2

sm

8199

sm

90000

g 9v

R .

==

m1000 m819

10000 R = . D R m

= 2 2000

2km


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Example: Propeller TipExample: Propeller Tip

q The propeller on a stunt plane spins with frequencyf = 3500 rpm . The length of each propeller blade is L = 80cm .What centripetal acceleration does a point at the tip of a

propeller blade feel?

f

L

what is aa here?


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Example:Example:

q First calculate the angular velocity of the propeller:

so 3500 rpm means = 367 s -1

q Now calculate the acceleration. a = 2R = (367s -1)2 x (0.8m) = 1.1 x 10 5 m/s 2

= 11,000 g

direction of aa points at the propeller hub ( -r r ).

1 11

60 2 0105 0105 rpm s -1= =

rot x

s x

rad rot

rad smin

min. .

^


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Example: Newton & the MoonExample: Newton & the Moon

q What is the acceleration of the Moon due to its motionaround the Earth?

q What we know (Newton knew this also): T = 27.3 days = 2.36 x 10 6 s (period ~ 1 month)

R = 3.84 x 10 8 m (distance to moon) R E = 6.35 x 10 6 m (radius of earth)

R R E


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Moon...Moon...

q Calculate angular velocity:

q So = 2.66 x 10 -6 s -1.

q Now calculate the acceleration. a = 2R = 0.00272 m/s 2 = 0 .000278 g

direction of aa points at the center of the Earth ( -r r ).

127 3

186400

2 2 66 10 6 .

.rot day

x day

s x

rad rot

x = s -1

^


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Moon...Moon...

q So we find that a moon / g = 0.000278q Newton noticed that R E 2 / R 2 = 0.000273

q This inspired him to propose that F Mm 1 / R 2

(more on gravity later)

R R E

a moon g


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Lecture 3,Lecture 3, Act 3 Act 3

Centripetal AccelerationCentripetal Accelerationq The Space Shuttle is in Low Earth Orbit (LEO) about 300 km

above the surface. The period of the orbit is about 91 min .What is the acceleration of an astronaut in the Shuttle in thereference frame of the Earth?(The radius of the Earth is 6.4 x 10 6 m.)

(a) 0 m/s 2

(b) 8.9 m/s 2

(c) 9.8 m/s 2


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q First calculate the angular frequency :

q Realize that:



R O

300 km

R O = R E + 300 km

= 6.4 x 10 6 m + 0.3 x 10 6 m= 6.7 x 10 6 m

R E

1-s00115 .0

rot

rad 2 x

s60

1 x

rot

91

1==

min

min


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q Now calculate the acceleration:



a = 2

R

a = (0.00115 s -1 )2 x 6.7 x 10 6 m

a = 8.9 m/s 2


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Recap for today:Recap for today:

q Reference frames and relative motion. (Text: 2-1, 3-3, & 4-1)

q

Uniform Circular Motion (Text: 5-2, also 9-1)q Look at Textbook problems Chapter 3: # 47, 49, 97, 105Chapter 3: # 47, 49, 97, 105

Circulair Motion

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