Physics 2211: Lecture 7 Todays Agenda

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Physics 2211 Lecture 7Todays Agenda

• Uniform Circular Motion

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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?

• 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...

• 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 8, Act 1Uniform 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 8, Act 1Solution
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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 8, Act 2Centripetal 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 8, Act 2Centripetal 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 8, Act 2Centripetal Acceleration

• Now calculate the acceleration

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Recap for today

• Uniform Circular Motion