Announcement:

1

• Announcement
• LABS start this week !
• Homework 2 due Fri. (Sept. 15) by 5.00 PM on
  webassign Problems from Chapter 3 and 4.

2
Physics 151 Lecture 6Todays Agenda

• Todays topics
• Finish circular motion
• Discuss relative motion
• text sections 4.4-4.6

3
Review Uniform Circular Motion (UCM)
See text 4-4

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

4
Polar Coordinates...

• In Cartesian co-ordinates 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
frequency (f) ?/2 ?
period (T) 1 / f 2?/?
5
Lecture 6, 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 ?

6
Acceleration in UCM
See text 4-4

• 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
7
Acceleration in UCM

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

But ?R v?t for small ?t
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Centripetal Acceleration
See text 4-4

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

a
R
?
9
Lecture 6, 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) 20 m
• (b) 200 m
• (c) 2,000 m
• (d) 20,000 m

10
ExampleNewton 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
a 0.00272 m/s2
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Moon...

• So we find that amoon / g .000278
• Newton noticed that RE2 / R2moon .000273

• This inspired him to propose that FMm ? 1 / R2
• (more on gravity later)

12
Lecture 6, ACT 2Uniform Circular Motion
A satellite is in a circular orbit 600 km above
the Earths surface. The acceleration of gravity
is 8.21 m/s2 at this altitude. The radius of the
Earth is 6400 km. Determine the speed of the
satellite, and the time to complete one orbit
around the Earth.

• Answer
• 7,580 m/s
• 5,800 s

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Lecture 6, ACT 3Uniform Circular Motion
A stunt pilot performs a circular dive of
radius 800 m. At the bottom of the dive (point B
in the figure) the pilot has a speed of 200 m/s
which at that instant is increasing at a rate of
20 m/s2. What acceleration does the pilot have
at point B ?
14
Lecture 6, ACT 3The Pendulum

• Which statement best describes
• the motion of the pendulum bob
• at the instant of time drawn ?
• the bob is at the top of its swing.
• which quantities are non-zero ?

q 30
1m
B) vr 0 ar 0 vq 0 aq ? 0
C) vr 0 ar ? 0 vq 0 aq ? 0
A) vr 0 ar 0 vq ? 0 aq ? 0
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Lecture 6, ACT 3The PendulumSolution
O
q 30
NOT uniform circular motion Is circular motion
so must be ar ? 0 Speed is increasing so aq not
zero
1m
At the top of the swing, the bob temporarily
stops, so v 0.
aq
ar
C) vr 0 ar ? 0 vq 0 aq ? 0
Animation
What are components of v and a at q 0 ?
16
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.
• More about this later when we discuss forces.
• For now, just remember that we can make
  measurements from different vantage points.

Animation
17
Text Ch 4, sect 6
Lecture 6 ACT 4 Relative Motion

• Consider an airplane flying on a windy day.
• A pilot wants to fly from New Haven to Bradley
  airport. She knows that Bradley is 120 miles due
  north of New Haven and there is a wind blowing
  due east at 30 mph. She takes off from New Haven
  Airport at noon. Her plane has a compass and an
  air-speed indicator to help her navigate. She
  uses her compass at the start to aim her plane
  north, and her air speed indicator tells her she
  is traveling at 120 mph with respect to the air.

After one hour she is (A) at Bradley (B)
east of Bradley (C) southeast of Bradley
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See text Ex. 4.9 and 4.10
Lecture 6 ACT 4 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
See example 4-9,10 (Boat Crossing a River)
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See text Ex. 4.9 and 4.10
Lecture 6 ACT 4 Relative Motion

• But 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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Lecture 6 ACT 4 Relative Motion
See text Ex. 4.9 and 4.10

• 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...
See text Ex. 4.9 and 4.10

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

Va,g
Vp,a
Vp,g
The north component of vp,g is just her air
speed. So she gets far enough north. The wind
takes her further east. Answer is (B) due east.
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Lecture 6, ACT 5Relative 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?

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Lecture 6, ACT 5Solution

• 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

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Lecture 6, ACT 5Solution
Answer (c)
25
Recap for today

• Circular Motion (Text Ch. 4.4)
• Relative Motion (Text Ch 4.6)
• Reading assignment for Wed.
• Read about Forces Ch 5.1-3
• Homework 2 due Fri. (Sept. 15) by 5.00 PM on
  webassign
• Problems from Chapter 3 and 4