Status: Unit 2, Chapter 3

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Status Unit 2, Chapter 3

• Vectors and Scalars
• Addition of Vectors Graphical Methods
• Subtraction of Vectors, and Multiplication by a
  Scalar
• Adding Vectors by Components
• Unit Vectors
• Vector Kinematics
• Projectile Motion
• Solving Problems in Projectile Motion
• Relative Velocity

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Section Two Problem Assignment

• Q3.4, P3.6, P3.9, P3.11, P3.14, P3.73
• Q3.21, P3.24, P3.32, P3.43, P3.65, P3.88

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Kinematic Equations for Projectile Motion (y up,
ax 0, ay-g -9.8m/s2)
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What angle gives the most range?

• This can be cast as a nifty calculus question!
  And the results may surprise you.
• What we want is an expression for x in terms of
  all the other variables.
• Then we take the derivative with respect to the
  angle to find the maxima.
• So lets get to it, we really just follow the
  illustrative problem from last lesson but keep
  results general.

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• From the calculus you will recall that finding a
  maximum or minimum of a function is done by
  setting the derivative with respect to the
  independent variable to zero.
• This corresponds to finding those points where
  the slope of the function is zero the very
  definition of an extreme point.

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http//www.lon capa.org/mmp/kap3/cd060.htm
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Properties of Trajectories w/ x0y00

• Maximum Range
• At q 45o
• Goes as velocity squared
• Goes inversely with acceleration
• For all other ranges
• Two initial angles.

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Problem Solving

• Information can be propagated forward and
  backwards using the equations of motion as long
  as a minimum amount of information is available.
• To keep yourself organized follow the usual
  steps
• 1) Draw a figure with a thoughtful origin and xy
  coordinate system
• 2) Analyze the x and y motion separately,
  remember they share the same time interval
• 3) Make the known and unknown table with ax0,
  ay-g, vx is constant, and vy0 at the apex.
• 4) Select the equations with some forethought.

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Finding Initial Variables Given Final Variables

• The examples last lesson used information about
  initial position velocity to find final
  location.
• The problem can be reversed. In the next example
  we use final information to derive initial
  parameters.
• Lets consider that very familiar situation of a
  hit softball or baseball
• A ball player hits a homer and the ball lands in
  the seats 7.5 m above the point at which the ball
  was hit. The ball lands with V36m/s, 28 degrees
  below the horizontal. Find the initial velocity
  of the ball.

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• Well at first glance this seems difficult. What
  are we after? Well we need Vox and Voy - from
  that we can find Vo and the direction of the
  initial velocity vector using Vo2 Vx2 Vy2 and
  tanq Voy/Vox.
• Remembering to keep the dimensions independent,
  lets work with the horizontal or x dimension
  first.
• Since there is no acceleration in the horizontal
  direction
• Vox Vx (36 m/s) (cos28o) 32 m/s.
• Now we are halfway there and can focus on
  y-direction.

• This looks dire but we actually do have Vy
  -(36 m/s) (sin28o) -17 m/s
• Which is more than enough to proceed.

Known Unknown
yo0 t?
y7.5m v0y?
g9.8m/s2 vy?
Known Unknown
yo0 t?
y7.5m v0y?
g9.8m/s2
vy-17m/s
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A Problem with Initial and final Information The
Air-mail Drop

• A plane traveling at an altitude of 200 m and a
  speed of 69m/s makes an air drop.
• At what distance x must the drop be made so
  that the package lands near the recipients?
• This is a mixed problem we know the initial speed
  and the final distance of the drop, from this we
  can get missing information.

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• So
• And the drop distance is just given by the
  horizontal distance corresponding to that time.

• Lets use the usual upy, right x coordinate
  system with the origin on the initial position of
  the plane.
• What we need is the time of the projectile motion
  which as usual is obtained from the vertical
  fall. Then our table takes the form
• And we can derive the missing time using the
  second projectile motion formula for the y
  direction

Known Unknown
yo0m t?
y-200m vy?
v0y0
g9.8m/s2
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• Lets make this a bit more difficult and for
  external reasons require that the drop occur 400
  m before the recipients.
• This means some vertical velocity down (since 400
  m is less that 440m) will be required to ensure
  the time taken is correct. What will that
  velocity be?

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• Well the table is clearly different now
• And doesnt have enough information. However we
  can get the time t from the horizontal
  information.

• And the table now is
• And the initial vertical velocity can be found
  from the 2nd equation

Known Unknown
yo0m
y-200m v0y?
t5.80s
g9.8m/s2 vy?
Known Unknown
yo0m t?
y-200m v0y?
g9.8m/s2 vy?
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As we expected the package required a downward
push.
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Projectile Motion Parabolic Motion

• It turns out in the absence of air resistance all
  projectile motion is simple parabolic motion.
• This can be show with combination of the x and y
  equations of motion through substitution for
  time.
• Setting x0y00 the equations are considerable
  simplified

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Parabolic Mirrors
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Relative Velocity

• As you can see from past discussions we need to
  move easily between reference frames.
• The tyke tossing a ball from the wagon was a good
  example.
• From the grounds frame of reference the ball
  clearly has two velocity components.
• From the girls reference frame there is only the
  vertical component.

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Relative Velocity

• Actually moving between frames gets confusing.
• This can be addressed with a rather prescriptive
  approach.
• Lets set it up with the rather easy example of
  two trains meeting each other and the extending
  to the more general situation.

VB-E
VA-E
Gloucestershire Warwickshire Railway Cotswolds,
England.
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• Suppose
• Train A has velocity 2m/s relative to the earth
  VA-E2m/s
• Train B has velocity -1m/s relative to the earth
  VB-E-1m/s
• Then from the trains points of view
• The earth has velocity of -2m/s relative to train
  A VE-A-2m/s
• The earth has velocity of 1m/s relative to train
  B VE-B1m/s
• And the following equations are true
• VB-AVB-EVE-A-1m/s-2m/s-3m/s
• VA-BVA-EVE-B2m/s1m/s3m/s
• Notice the nifty way the notation behaves, we
  have a sort of cancellation of inner subscripts
• Also VB-A-VB-A which is true for any vectors.

VB-E
VA-E
The axes points in the same direction for all
three frames.
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• Lets consider a person walking in a train at a
  single instant
• From vector addition its pretty clear that
• rPE rPTrTE
• If we simply take the time derivative of this
  equation
• vPE vPTvTE
• In words, this simply says that the velocity of
  the person with respect to earth equals the sum
  of the velocity of the person with respect to the
  train and the velocity of the train with respect
  to the earth.

• Because of commutativity we could have written
  this as
• vPE vPTvTE
• but the canceling of the inner products lends
  some preference to the former eq.

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• Lets consider the more complicated case of a
  boat in a current. Here well see the value of
  the subscripts.
• If we want to go straight across the river the
  boat will need to push at an angle with respect
  to water along the vector VBW.
• The water moves with respect to the shore
  according to VWS
• As a result the boat with respect to the shore
  will travel along VBS.
• Vectorially then,
• VBS VBW VWS
• Note for this equation
• The final vector subscripts correspond to the
  outer subscripts on the right side
• The inner subscripts are the same.

• This notational rule demonstrated above, and
  proven in the previous slide helps us move
  quickly and correctly between reference frames.

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Example Relative Velocity between Two Moving
Objects

• Both cars shown are traveling at 11m/s. What is
  the relative velocity of car 1 in the reference
  frame of car 2?

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• Let the velocity of car 1 in the reference frame
  of car 2 be denoted V12.
• We can also denote the velocity of car 1 relative
  to the earth as V1E.
• Likewise the velocity of the earth relative to
  care 2 is VE2.
• Using our rules we propose
• V12 V1E VE2
• The last vector is not something we have but can
  get with the identity
• VE2 -V2E
• Or
• V12 V1E V2E

• Which makes sense as it describes a vector moving
  at 45 degrees at car 2.
• The magnitude is simply given by the Pythagorean
  Theorem as

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Schedule

• Wednesday
• Well review the most important information and
  discus the test
• Start on Chapter 4 Dynamics!
• Friday
• No class, use the time to study for the test
• Monday Feb 12th
• Quiz 1
• Problem sets for Units 1 and 2 due.