Physics 111: Lecture 3 Todays Agenda

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Physics 111 Lecture 3Todays Agenda

• Reference frames and relative motion
• Uniform Circular Motion

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Inertial Reference Frames
Cart on track on track

• 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.
• More about this later when we discuss forces.
• For now, just remember that we can make
  measurements from different vantage points.

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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 1Relative 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
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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 1solution

• 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 UCM?
Puck on ice

• Motion in a circle with
• Constant Radius R
• Constant Speed v v

y
v
(x,y)
R
x
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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!

y
v
(x,y)
R
?
x
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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
• 2?R R?c
• ?c 2?
• ??has period 2?.
• 1 revolution 2??radians

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Polar Coordinates...

• x R cos ?
• y R sin ?

1
sin
cos
0
?
3?/2
2?
?/2
?
-1
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Polar Coordinates...
Tetherball

• In Cartesian coordinates, we say velocity dx/dt
  v.
• x vt
• In polar coordinates, angular velocity d?/dt ?.
• ? ?t
• ? 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

• 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?/?

v
R
s
? 2? / T 2?f
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Recap

• 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
  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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Derivation
We know that and
v ?R
Substituting for v we find that
?
a ?2R
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Lecture 3, Act 2Uniform 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 2Solution
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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 due to its
  motion around the Earth?
• 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)

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Lecture 3, Act 3Centripetal Acceleration

• 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
  the reference frame of the Earth?
  (The radius of the
  Earth is 6.4 x 106 m.)
• (a) 0 m/s2
• (b) 8.9 m/s2
• (c) 9.8 m/s2

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Lecture 3, Act 3Centripetal Acceleration

• First calculate the angular frequency ?
• Realize that

RO
RO RE 300 km 6.4 x 106 m 0.3 x 106 m
6.7 x 106 m
300 km
RE
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Lecture 3, Act 3Centripetal Acceleration

• Now calculate the acceleration

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Recap for today

• Reference frames and relative motion. (Text
  2-1, 3-3, 4-1)
• Uniform Circular Motion (Text 5-2, also
  9-1)
• Look at Textbook problems Chapter 3 47, 49,
  97, 105