Physics 211: Lecture 3 Todays Agenda

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Physics 211 Lecture 3Todays Agenda

• Reference frames and relative motion
• Uniform Circular Motion
• Polar coordinates
• Centripetal acceleration!

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Lecture 3, Act 1Which statement is true

• A) If an objects speed is constant then its
  acceleration must be zero.
• B) If an objects acceleration is zero then its
  speed must be constant.
• C) If an objects speed is constant then its
  velocity must be constant.
• D) An objects velocity is always in the same
  direction as its acceleration.

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Inertial Reference Frames

• A Reference Frame is the place you measure from.
• Its where you nail down your (x,y,z) axes!
• An Inertial Reference Frame (IRF) is one that is
  not accelerating.
• We will consider only IRFs in this course.
• Valid IRFs can have fixed velocities with respect
  to each other.
• but not relative accelaration
• see what follows

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Relative Motion

• Consider a problem with two distinct IRFs
• An airplane flying on a windy day.
• A pilot wants to fly from Champaign to Chicago.
  Having asked a friendly physics student, she
  knows that Chicago is 120 miles due north of
  Urbana. She takes off from Willard Airport at
  noon. Her plane has a compass and an air-speed
  indicator to help her navigate.
• The compass allows her to keep the nose of the
  plane pointing north.
• The air-speed indicator tells her that she is
  traveling at 120 miles per hour with respect to
  the air.

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Relative Motion...

• The plane is moving north in the IRF attached to
  the air
• Vp, a is the velocity of the plane w.r.t. the
  air.

Air
Vp,a
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Relative Motion...

• But suppose the air is moving east in the IRF
  attached to the ground.
• Va,g is the velocity of the air w.r.t. the
  ground (i.e. wind).

Air
Vp,a
Va,g
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Relative Motion...

• What is the velocity of the plane in an IRF
  attached to the ground?
• Vp,g is the velocity of the plane w.r.t. the
  ground.

Vp,g
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Relative Motion...
Tractor

• Vp,g Vp,a Va,g Is a vector equation
  relating the airplanes velocity in
  different reference frames.

Va,g
Vp,a
Vp,g
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Lecture 3, Act 2Relative Motion

• You are swimming across a 50m wide river in which
  the current 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 straight perpendicular line across
  the river.
• How many seconds does it take you to get across
  ?(a) (b)(c)

50 m
2 m/s
Test problems
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Lecture 3, Act 1solution
y
Choose x axis along riverbank and y axis across
river
x

• The time taken to swim straight across is
  (distance across) / (vy )

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

1 m/s
y
2 m/s
m/s
x
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Lecture 3, Act 2solution

• So the y component of your velocity with respect
  to the water is
• So the time to get across is

m/s
m/s
50 m
y
x
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Uniform Circular Motion

• What does it mean?
• How do we describe it?
• What can we learn about it?

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What is Uniform Circular Motion?
Puck on ice

• Motion in a circle with
• Constant Radius R
• Constant Speed v v
• It happens in a plane

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How can we describe UCM?

• In general, one coordinate system is as good as
  any other
• Cartesian
• (x,y) position
• (vx ,vy) velocity
• Polar
• (R,?) position
• (vR ,?) velocity
• In UCM
• R is constant (hence vR 0).
• ? (angular velocity) is constant.
• Polar coordinates are a natural way to describe
  UCM!

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Polar Coordinates

• The arc length s (distance along the
  circumference) is related to the angle in a
  simple way
• s R?, where ? is the angular displacement.
• units of ? are called radians.
• For one complete revolution (?c)
• 2?R R?c
• ?c 2?
• ??has period 2?.

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Relating Polar to Cartesian Coordinates
1
sin
cos
0
?
3?/2
2?
?/2
?
-1
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Velocity of UCM in Polar Coordinates
Tetherball

• In Cartesian coordinates, we say velocity dx/dt
  v.
• x vt (if v is constant)
• In polar coordinates, angular velocity d?/dt ?.
• ? ?t (if w is constant)
• ? has units of radians/second.
• Displacement s vt.
• but s R? R?t, so

y
v
R
s
???t
x
v ?R
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Period and Frequency of UCM

• Recall that 1 revolution 2? radians
• frequency (f) revolutions / second
  (a)
• angular velocity (?) radians / second
  (b)
• By combining (a) and (b)
• ? 2? f
• Realize that
• period (T) seconds / revolution
• So T 1 / f 2?/?

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Recap of UCM

• 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

v
(x,y)
R
s
???t
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Aside Polar Unit Vectors

• We are familiar with the Cartesian unit vectors
  i j k
• Now introducepolar unit-vectors r and ?
• r points in radial direction
• ? points in tangential direction

(counter clockwise)
y
R
?
j
x
i
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Acceleration in UCM

• Even though the speed is constant, velocity is
  not constant since the direction is changing
• If the velocity is changing, there must be some
  acceleration!
• Consider average acceleration in time ?t
  aav ?v / ?t

v2
R
v1
??t
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Acceleration in UCM

• Even though the speed is constant, velocity is
  not constant since the direction is changing.
• Consider average acceleration in time ?t
  aav ?v / ?t

R
seems like ?v (hence ?v/?t ) points at the origin!
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Acceleration in UCM

• Even though the speed is constant, velocity is
  not constant since the direction is changing.
• As we shrink ?t, ?v / ?t dv / dt a

a dv / dt
R
We see that a points in the - R direction.
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Acceleration in UCM

• This is called Centripetal Acceleration.
• Now lets calculate the magnitude

?v
v1
v2
But ?R v?t for small ?t
v2
R
So
v1
?R
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Centripetal Acceleration

• UCM results in acceleration
• Magnitude a v2 / R
• Direction - r (toward center of circle)

R
a
?
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Useful Equivalent
We know that and
v ?R
Substituting for v we find that
?
a ?2R
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Centripetal Acceleration

• If an object is undergoing UCM, it must be
  accelerated by a centripetal acceleration of
  magnitude a v2/R w2R
• Object feels acceleration pushing in direction
  of center of circle
• merry-go-round
• What kind of forces give rise to circular motion
• Direct contact (ball on end of rope)
• Gravity, electric force
• magnetism

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Fermilab

• Protons and anti-protons undergo UCM bent by
  magnets that provide centripetal acceleration

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galaxies

• Stars undergo ucm around the center of a galaxy
  which may be a black hole

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Lecture 3, Act 3Uniform Circular Motion

• A fighter pilot flying in a circular turn will
  pass out if the centripetal acceleration he
  experiences is more than about 9 times the
  acceleration of gravity g. If his F18 is moving
  with 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, Act 3Solution
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Example Propeller Tip

• The propeller on a stunt plane spins with
  frequency f 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
what is a here?
L
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Example

• First calculate the angular velocity of the
  propeller
• so 3500 rpm means ? 367 s-1
• Now calculate the acceleration.
• a ?2R (367s-1)2 x (0.8m) 1.1 x 105 m/s2
  11,000 g
• direction of a points at the propeller hub (-r ).

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Example Newton the Moon

• What is the acceleration of the Moon that
  provides its UCM motion around the Earth?
• What keeps Moon in circular orbit? (gravity)
• What we know (Newton knew this also)
• T 27.3 days 2.36 x 106 s (period 1 month)
• R 3.84 x 108 m (distance to moon)
• RE 6.35 x 106 m (radius of earth)

R
RE
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Moon...

• Calculate angular velocity
• So ? 2.66 x 10-6 s-1.
• Now calculate the acceleration.
• a ?2R 0.00272 m/s2 0.000278 g
• direction of a points at the center of the Earth
  (-r ).

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Moon...

• So we find that amoon / g 0.000278
• Newton noticed that RE2 / R2 0.000273
• This inspired him to propose that FMm ? 1 / R2
• (more on gravity later)
• Gravity is the source of this centripetal
  acceleration

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Lecture 3, Act 4Geostationary Orbit

• A satellite in geostationary orbit is always
  directly above the same spot on the earths
  equator (most communication satellites are of
  this type).
• Compare the centripetal acceleration of such a
  satellite as to the centripetal acceleration a
  person on the surface of the earth ap.
• A) as gt ap
• B) as ap
• C) as lt ap

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Lecture 3, Act 4Geostationary Orbit

• A satellite in geostationary orbit is always
  directly above the same spot on the earths
  equator (most communication satellites are of
  this type).
• Compare the centripetal acceleration of such a
  satellite as to the centripetal acceleration a
  person on the surface of the earth ap.
• A) as gt ap
• B) as ap
• C) as lt ap

a w2R
Same forboth
bigger for satellite
There are lots of these satellites
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Recap for today

• Reference frames and relative motion. (Text
  3-1)
• Uniform Circular Motion (Text 3-3)