4.4 Uniform Circular Motion

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4.4 Uniform Circular Motion
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Uniform Circular Motion

• Uniform Circular Motion (UCM) occurs when an
  object moves in a circular path with a constant
  speed
• An acceleration exists since the DIRECTION of the
  motion is changing
• This change in velocity is related to an
  acceleration!!!
• The velocity vector is always tangent to the path
  of the object

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Changing Velocity in UCM

• The change in the velocity vector is due to the
  change in direction
• The vector diagram shows
• Dv vf - vi

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Centripetal (Radial) Acceleration

• The acceleration is always perpendicular to the
  path of the motion
• The acceleration always points toward the center
  of the circle of motion
• This acceleration is called the Centripetal
  (Radial) Acceleration!!!

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Centripetal Acceleration, 2

• The magnitude of the centripetal acceleration
  vector is given by
• (4.15)
• The direction of the centripetal acceleration
  vector is always changing, to stay directed
  toward the center of the circle of motion

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Deriving Eqn 4.15 Centripetal (Radial)
Acceleration

• Motion in a circle from A to point B. Velocity v
  is tangent to the circle.
• As ?t ? 0, ?v ? ? v ?v is in the radial
  direction

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Deriving Eqn 4.15, 2

• a ? ac is (Radial ) Centripetal!!!
• Similar triangles
• ?

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Deriving Eqn 4.15, final

• As ?t ? 0, ?? ? 0, A ? B, (?r/?t) ? v
• Magnitude
• (4.15)
• Direction Radially inward!!
• Centripetal ? Toward the center
• ? Center-seeking
• Centripetal acceleration
• Acceleration toward the center!!
• aC ? v always!!

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Period Frequency

• A particle moving in UCM of radius r (speed v
  constant)
• Description in terms of period frequency
• Period (T) ? time for one revolution (time to go
  around once), usually in seconds. T t/n
• Frequency (f) ? number of revolutions (n) per
  second. f n/t ?
• T 1/f f 1/T

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Period Frequency, 2

• A particle moving in UCM of radius r
  speed v constant
• One Complete Circumference
• distance around 2pr
• time Period (T1/f)
• The speed of the particle would be the
  circumference of the circle of motion divided by
  the period
• Speed distance/time ?

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Period Frequency, final

• Therefore, the period is (4.16)
• And the frequency is (4.16a)
• Centripetal acceleration
• (4.16b)

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Example 4.9 The Centripetal Acceleration of the
Earth

• Calculate ac of the Earth, assuming it moves in a
  circular orbit around the Sun.
• Note that ac ltlt g

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Material for the Midterm

• Examples to Read!!!
• None
• Homework Problems to be solved in Class!!!
• Problem 28
• Problem 29