Review of Lecture 3

1
Review of Lecture 3

• Vectors Scalars
• Adding Vectors Geometrically
• Components of Vectors
• Unit Vectors
• Adding Vectors by Components
• Multiplying Vectors
• Multiplying a Vector by a Scalar
• Scalar (Dot) Product
• Vector (Cross) Product

2
Position Displacement

• Lets begin with the definition of a position
  vector which is a vector that extends from a
  reference point (usually the origin of a
  coordinate system) to a particle
• In unit vector notation we therefore have

3
Position Displacement

• Here we have graphically a position vector of

4
Position Displacement

• As you can see, if the particle moves, then the
  position vector also changes
• So if the particle moves from r1 to r2, then the
  displacement (vector) for the particle is given
  by

5
Position Displacement

• Written in unit vector notation we
  getorand finally as

6
Position Displacement

• Given an initial position vector ofand a
  later position vector ofwhat is the
  displacement ?r ?
• Is the magnitude of r2 different from r1?

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Position Displacement

• The initial position vector for a proton
  isand later is
• What is the protons displacement vector?
• To what plane is the displacement vector parallel?

8
Average Velocity Instantaneous Velocity

• We can see that if a particle moves through a
  displacement ?r in a time interval of ?t, then
  its average velocity vavg isor

9
Average Velocity Instantaneous Velocity

• Because ?t is a scalar, we can see that the
  vector vavg must be in the same direction as the
  vector ?r
• We can also rewrite vavg using unit vector
  notation as

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Average Velocity Instantaneous Velocity

• We also know that as ?t ? 0, we can express the
  instantaneous velocity at the time t as
• So in the limit as ?t ? 0, we have vavg ? v
• Most importantly, vavg takes on the direction of
  the tangent line at the time t in other words,
  v is always tangent to the particles path at
  the particles position

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Average Velocity Instantaneous Velocity

• Again going back to unit vector notation, we
  getwhich can be simplified down to

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Average Velocity Instantaneous Velocity

• In doing this we can see that
• The values vx, vy and vz are simply the scalar
  components of the vector v
• So we can find the scalar components of a vector
  v by simply differentiating its position vector r

13
Average Velocity Instantaneous Velocity

• This is shown diagrammatically in the following
  figure

14
Average Velocity Instantaneous Velocity

• Note one important point
• Unlike a position vector or a displacement
  vector, a velocity vector is not drawn from one
  point (here / A / F(x1, y1, z1)) to another
  point (there / B / F(x2, y2, z2))
• Rather, it shows the instantaneous direction of
  travel of the particle located at the tail of
  the vector the magnitude can be drawn to any
  arbitrary scale

15
Average Velocity Instantaneous Velocity

• The position of an electron is given bywith t
  in seconds and r in meters
• What is the function for the electrons velocity?
• At t 2.00 s, what is v in unit vector notation
  and as a magnitude direction (relative to the
  positive x axis)?

16
Average Acceleration Instantaneous Acceleration

• We can see that if a particles velocity changes
  from v1 to v2 in a time interval of ?t, then
  its average acceleration aavg isor

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Average Acceleration Instantaneous Acceleration

• And just as we did for the velocity case, if we
  shrink ?t to zero about some instant t, then the
  instantaneous acceleration is
• Important If the particles velocity changes in
  either magnitude or direction (or both), then
  there is an acceleration

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Average Acceleration Instantaneous Acceleration

• We can also rewrite a using unit vector notation
  asor

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Average Acceleration Instantaneous Acceleration

• This of course leads directly towhere the
  scalar components of the acceleration are

20
Average Acceleration Instantaneous Acceleration

• The scalar components can also of course be
  expressed as

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Average Acceleration Instantaneous Acceleration

• Review problem 4-15

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Average Acceleration Instantaneous Acceleration

• Again note
• Just as with the velocity vector, the
  acceleration vector doesnt point from here to
  there
• The vector shows the instantaneous direction of
  the acceleration of the particle located at the
  tail of the vector the magnitude can be drawn
  to any arbitrary scale

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Average Acceleration Instantaneous Acceleration

• Now lets compare the two figures that we have
  for the directions of the instantaneous velocity
  and instantaneous acceleration

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Average Acceleration Instantaneous Acceleration
25
Average Acceleration Instantaneous Acceleration

• A proton initially has a velocityand then 4.0
  seconds later has a velocityin meters per
  second
• For that 4.0 seconds, what is the protons
  average acceleration aavg in unit vector notation
  and expressed as a magnitude direction?

26
Projectile Motion

• A special case of 2-dimensional motion is when
  the particle has some initial velocity v0, but
  the acceleration is always the free-fall
  acceleration (-g), which is downward
• This case is called projectile motion in that the
  particle (a baseball, a cannonball, ) is
  projected or launched

27
Projectile Motion
28
Projectile Motion

• In studying projectile motion, we shall always
  assume (unless specified otherwise) that air
  friction is being ignored
• This assumption considerably simplifies the
  equations of motion
• Lets start by examining Figure 4-10

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Projectile Motion

• The projectile is launched with an initial
  velocityv0 v0xi v0yjwherev0x
  v0cos?,v0y v0sin?

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Projectile Motion

• During its flight, we know that the projectiles
  position vector r and velocity vector v are
  constantly changing, but the acceleration vector
  a is constant and always pointing directly
  downward
• We also know that there is no horizontal
  acceleration therefore the horizontal velocity
  of the particle is constant

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Projectile Motion

• This leads us to an important point
• In projectile motion the horizontal motion and
  vertical motion are completely independent of
  each other

32
Projectile Motion

• This means that projectile motion problems can be
  separated into two, single dimension problems
  one with zero acceleration (the horizontal
  problem) and the other with a constant downward
  acceleration (-g) (the vertical problem)
• We have of course seen these kinds of problems
  before, but lets now take a more detailed look

33
Projectile Motion(Horizontal)

• Because there is no acceleration in the
  horizontal direction, the horizontal velocity vx
  will always be just
• The horizontal position of the particle will
  therefore be

34
Projectile Motion(Vertical)

• The vertical motion is just what we examined
  previously when we looked at what happens when a
  ball is tossed directly upwards and then
  experiences free-fall
• So the equations in Table 2-1 apply where the
  acceleration is g in the downward direction and
  we substitute y for x

35
Projectile Motion(Vertical)

• We therefore have (equation 2-15)or

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Projectile Motion(Vertical)

• Similarly we can also get (equation 2-11)and
  (equation 2-16)
• As with a ball thrown vertically, when the
  (vertical) velocity goes to zero, the projectile
  is at the maximum height along its trajectory

37
Projectile Motion

• Now that we have the two independent equations
  (as a function of t), we should be able to create
  one equation that describes the projectiles path
  that is, an equation for y as a function of x
• We know x as a function of t, so lets just turn
  that equation around and find t as a function of x

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Projectile Motion

• To simplify things a little, we will choose to
  set our x0 0 we then get
• For the same reason, we will also choose to set
  our y0 0

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Projectile Motion

• Substituting for t we getwhich reduces down
  to

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Projectile Motion

• Notice that because g, ?0 and v0 are all
  constants, the equationis of the
  formwhich is the equation for a parabola

41
Projectile Motion

• Review problem 4-18

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Horizontal Range

• Lets define the horizontal range that the
  projectile travels to be the distance (along the
  x axis) that the projectile travels when it
  returns to the launch height

43
Horizontal Range

• Going back to our two independent
  equations,we can see that the maximum range
  occurs when y y0 0

44
Horizontal Range

• Lets set the distance traveled along the x axis
  to beand the height at that distance to be

45
Horizontal Range

• Substituting for t we getwhich reduces down
  to

46
Horizontal Range

• Using the trig identity
• we finally get

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Horizontal Range

• Note that the maximum value of sin 2?0 occurs
  when ?0 45º (which if you have fired many
  projectiles is what you would expect)
• Note that this equation does not hold true if the
  final height is not the launch height

48
The Effect of Air Resistance

• A complete study of the effects of air resistance
  is well past the level of complexity we are
  prepared to deal with in this course
• The amount of air resistance (e.g., the drag
  force) depends on
• the shape of the projectile,
• the density of the projectile, and
• the size of the projectile

49
The Effect of Air Resistance

• Depending on these factors, the air resistance
  may introduce either a linear or a
  non-linear(e.g., quadratic) drag force or both
• In either case, the drag force increases with
  velocity and eventually will match the force
  provided by the acceleration of gravity
• When this occurs, the object has reached its
  terminal velocity and will not further accelerate

50
The Effect of Air Resistance

• The effect of air resistance can be quite
  substantial as you can see from Table 4-1, the
  range, maximum height and total time of flight
  can be dramatically reduced

51
More on Acceleration

• It was noted earlier that a change in the
  direction of a particle causes an acceleration
  even if the magnitude of the particles velocity
  vector didnt change
• Why is that?

52
Uniform Circular Motion

• Suppose that we have a particle on the end of a
  string and we are swinging it in a circle at a
  constant (e.g., uniform) speed
• We would not normally think that the particle is
  accelerating as its speed is constant
• But the particles velocity (being a vector
  quantity) is not constant (because it is
  constantly changing direction)

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Uniform Circular Motion

• Given our definition of acceleration (a
  non-constant velocity), then it must be that the
  particle is accelerating
• Here we see the velocity and acceleration vectors
  for our particle moving at a uniform speed in a
  circle

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Uniform Circular Motion

• The velocity vector is always tangent to the
  circle in the direction of motion
• As a result, the acceleration is directed
  radially inward

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Uniform Circular Motion

• This inward directed acceleration is called
  centripetal (center seeking) acceleration and
  is given by the equationwhere a is the
  magnitude of the acceleration, r is the radius of
  the circle and v is the speed of the particle
• We will derive this equation shortly

56
Uniform Circular Motion

• The period of revolution (or simply the period)
  is the time it takes to go around the circle once
• The period is given by

57
Uniform Circular Motion

• Now lets derive the centripetal acceleration
  equation
• Here we see our particle at some instant it is
  moving at a constant speed v and is located at
  the point(xp, yp)

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Uniform Circular Motion

• We know that the velocity v is always tangent to
  the particles path
• That means that v is perpendicular to the radius
  r drawn from the origin to the particles position

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Uniform Circular Motion

• And using a little geometry, we can see that the
  angle ? between the velocity vector and the
  vertical line at p matches the angle the radius
  makes with the x axis

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Uniform Circular Motion

• The scalar components of v are shown here asvx
  and vy
• The velocity can also be written as

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Uniform Circular Motion

• Replacing sin ? with yp/r and cos ? with xp/r we
  get
• We know that we have to take the time derivative
  of the velocity to get the acceleration

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Uniform Circular Motion

• So we then get

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Uniform Circular Motion

• But we know that dyp/dt is nothing more than the
  velocity component vy similarly, we know that
  dxp/dt vx
• We also know from earlier that vx - v sin ? and
  vy v cos ?

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Uniform Circular Motion

• Making those substitutions we finally get

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Uniform Circular Motion

• To compute the magnitude of a we have

66
Uniform Circular Motion

• To compute the angle ? we have

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Relative Motionin One Dimension

• So far we have discussed the motion of a particle
  in 1 and 2 dimensions
• Now lets discuss how different observers in
  different frames of reference would see that
  particle
• We will begin by defining what we mean by a
  reference frame

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Relative Motionin One Dimension

• The easiest way to think of a reference frame is
  that its a physical object to which we attach our
  coordinate system
• For most purposes that is the ground, and
  sometimes we want an absolute reference
• Boire Field is at 42-46-54.347N / 071-30-53.206W
  (42.7817631 / -71.5147794) (estimated)

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Relative Motionin One Dimension

• Sometimes though our reference is relative
• Your honor, he was going 55 mph in a 35 mph
  zone
• Here the speed of the particle (the car in this
  case) was relative to the ground and also
  relative to the observer (the police officer
  who we assume was not moving)
• But what would happen if the officer were moving
  what would be observed then?

70
Relative Motionin One Dimension

• Lets create two reference frames A and B
• Frame A is not moving relative to the ground, but
  frame B is moving at a constant speed along the
  highway
• An observer at the origin of each frame Alex in
  Frame A and Barbara in Frame B each measure the
  position of a car moving down the highway

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Relative Motionin One Dimension

• We can see that the following is true

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Relative Motionin One Dimension

• This tells us that The position xPA of P as
  measured by A is equal to the position xPB of P
  as measured by B plus the position xBA of B as
  measured by A
• If we take the time derivative of that relation
  we get

73
Relative Motionin One Dimension

• We know that v dx/dt, so we then get
• This tells us that The velocity vPA of P as
  measured by A is equal to the velocity vPB of P
  as measured by B plus the velocity vBA of B as
  measured by A

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Relative Motionin One Dimension

• Now lets differentiate once again to get the
  acceleration of particle P as viewed in both
  frames of reference

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Relative Motionin One Dimension

• The result is
• This is because vBA is constant and therefore
  drops out when the differential is taken
• Observers in two different frames of reference
  (moving with a constant velocity relative to each
  other) will see that a moving particle has the
  same acceleration

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Relative Motionin One Dimension

• It is important to note that our frames of
  reference (A and B) are moving with a constant
  velocity relative to one another
• As such, these are called inertial frames of
  reference (from Newtons 1st Law the Law of
  Inertia)
• We would not get the same result if frame B were
  accelerating relative to frame A

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Relative Motionin Two Dimensions

• So far we have limited ourselves to relative
  motion in one dimension
• By extension this can of course be applied to 2
  (or 3) dimensions

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Relative Motionin Two Dimensions

• The relationship for positions gives us
• For velocities we get
• And finally for acceleration we get

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Next Class

• Homework Problems Chapter 421, 23, 59, 85, 88,
  97
• Read sections Chapter 5