Physics 2211: Lecture 13 Todays Agenda

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Physics 2211 Lecture 13Todays 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 Uniform Circular Motion?

• Motion with
• Constant Radius R
• (Circular)
• Constant Speed
• (Uniform)

y
v
(x,y)
R
x
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How can we describe Uniform Circular Motion?

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

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...
Conversion from polar to Cartesian coordinates

• x R cos ?
• y R sin ?

1
0
?
3?/2
2?
?/2
?
-1
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Polar Coordinates...

• In Cartesian coordinates, we say velocity dx/dt
  vx.
• x vxt
• 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
• (a) frequency (f) revolutions / second
• (b) angular velocity (?) radians / second
• By combining (a) and (b)
• ? (rad/s) 2? rad/rev x f (rev/s)
• ? 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
• Now introduce polar unit-vectors and
  
• points in radial direction
• points in tangential direction
  (counterclockwise)

y
R
?
x
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Acceleration in Uniform Circular Motion

• 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

R
??t
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Acceleration in Uniform Circular Motion

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

R
seems like (hence ) points at
the origin!
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Acceleration in Uniform Circular Motion

• Even though the speed is constant, velocity is
  not constant since the direction is changing.
• As we shrink ?t,

R
We see that points in the direction.
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Acceleration in Uniform Circular Motion

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

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

• Uniform Circular Motion 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 13, 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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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

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Lecture 13, 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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Problem Motion in a Circle

• A boy ties a rock of mass m to the end of a
  string and twirls it in the vertical plane. The
  distance from his hand to the rock is R. The
  speed of the rock at the top of its trajectory is
  v.
• What is the tension T in the string at the top of
  the rocks trajectory?

v
T
R
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Motion in a Circle...

• Draw a Free Body Diagram (pick y-direction to be
  down)
• We will use SF ma (surprise)
• First find SF in y direction
• SF mg T

y
mg
T
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Motion in a Circle...

• SF mg T
• Acceleration in y direction
• ma mv2 / R
• mg T mv2 / R
• T mv2 / R - mg

v
y
mg
T
F ma
R
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Motion in a Circle...

• What is the minimum speed of the mass at the top
  of the trajectory such that the string does not
  go limp?
• i.e., find v such that T 0.
• mv2 / R mg T
• v2 / R g
• Notice that this doesnot depend on m.

v
mg
T 0
R
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Lecture 13, Act 3Motion in a Circle

• A skier of mass m goes over a mogul having a
  radius of curvature R. How fast can she go
  without leaving the ground?

(a) (b)
(c)
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Problem Rotating puck weight.

• A mass m1 slides in a circular path with speed v
  on a horizontal frictionless table. It is held
  at a radius R by a string threaded through a
  frictionless hole at the center of the table. At
  the other end of the string hangs a second mass
  m2.
• What is the tension (T) in the string?
• What is the speed (v) of the sliding mass?

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Problem Rotating puck weight...
T

• Draw FBD of hanging mass
• Since R is constant, a 0.
• so T m2g

m2
m2g
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Problem Rotating puck weight...
N
T m2g

• Draw FBD of sliding mass

m1
2nd Law T m1a m1v2 / R
m1g
But, T m2g
m2g m1v2 / R
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Recap for today

• Uniform Circular Motion
• Next Work and Energy -- Read 6.1 in Tipler