Review of Lecture 8

1
Review of Lecture 8

• Potential Energy
• Work and Potential Energy
• Conservative Nonconservative Forces
• Path Independence of Conservative Forces
• Determining Potential Energy Values
• Gravitational PE
• Elastic PE
• Conservation of Mechanical Energy

2
Review of Lecture 8

• Reading a Potential Energy Curve
• Turning points Equilibrium points
• Work Done on a System by an External Force
• With Without Friction
• Conservation of Energy
• Power definition revisited

3
Systems of Particles

• We have discussed parabolic trajectories using a
  particle as the model for our objects
• But clearly objects are not particles they are
  extended and may have complicated shapes mass
  distributions
• So if we toss something like a baseball bat into
  the air (spinning and rotating in a complicated
  way), what can we really say about its
  trajectory?

4
Center of Mass

• There is one special location in every object
  that provides us with the basis for our earlier
  model of a point particle
• That special location is called the center of
  mass
• The center of mass will follow a parabolic
  trajectory even if the rest of the bats motion
  is very complicated

5
Center of Mass

• To start, lets suppose that we have two masses
  m1 and m2, separated by some distance d
• We have also arbitrarily aligned the origin of
  our coordinate system to be the center of mass m1

6
Center of Mass

• We define the center of mass for these two
  particles to be

7
Center of Mass

• From this we can see that if m2 0, then xcom
  0
• Similarly, if m1 0, then xcom d

8
Center of Mass

• Finally, if m1 m2, then xcom ½d
• So we can see that the center of mass in this
  case is constrained to be somewhere between x 0
  and x d

9
Center of Mass

• Now lets shift the origin of the coordinate
  system a little
• We now need a more general definition of the
  center of mass

10
Center of Mass

• The more general definition (for two particles)
  is
• Note that if x1 0 we are back to the previous
  equation

11
Center of Mass

• Now lets suppose that we have lots of particles
  all lined up nicely for us on the x axis
• The equation would now bewhere M m1 m2
  mn

12
Center of Mass

• The collection of terms in the numerator can be
  rewritten as a sum resulting in

13
Center of Mass

• This result is only for one dimension however, so
  the more generalized result for 3 dimensions is
  shown here

14
Center of Mass

• Noticing that xcom and xi, etc. are distances
  along the main axis of our coordinate system, we
  could just as easily switch to vector notation
• First recall that the position of mass mi using
  vector notion would be

15
Center of Mass

• Our center of mass equation using vector notation
  would therefore beremembering that M is the
  total mass of the system

16
Center of Mass

• As the number of particles gets to be large (as
  it would be for everyday objects like a baseball
  bat or a fighter jet), the easiest thing to do is
  to treat the object as a continuous distribution
  of matter
• The particles then become differential mass
  elements and the sums become integrals

17
Center of Mass

• Given continuous matter, the location of the
  center of mass becomes

18
Center of Mass

• Moving over to the integral form is nice, but now
  the problem is one of dealing with the
  non-uniformity of mass distribution in our
  everyday objects
• For this course, we will assume uniform objects
  objects with a uniform density

19
Center of Mass

• Lets rearrange that to be

20
Center of Mass

• If we now substitute for dm in the previous
  integrals, we get
• Now we are simply integrating over the volume of
  the object

21
Center of Mass

• From these equations you can see that if the
  object has a point, line or plane of symmetry,
  then things can get even easier

22
Center of Mass

• You know by intuition that the center of mass of
  a sphere is at the center of the sphere for a
  rod it lies along the central axis of the rod
  for a flat plate it lies in the plane of the plate

23
Center of Mass

• Note however that the center of mass doesnt
  necessarily have to lie within the object or have
  any mass at that point
• The CM of a horseshoe is somewhere in the middle
  along the axis of symmetry
• The CM of a doughnut is at its geographic
  center, but there is no mass there either

24
Center of Mass

• Going back to our baseball bat, the CM will lie
  along the central axis (the axis of symmetry)
• And it is the CM that faithfully follows the line
  of a parabola

25
Center of Mass

• Checkpoint 1
• Sample Problem 9-1
• Sample Problem 9-2

26
Checkpoint 1
27
Sample Problem 9-1
28
Sample Problem 9-2
29
Sample Problem 9-2
30
Newtons 2nd Law for aSystem of Particles

• We know from experience that if you roll the cue
  ball into another billiard ball that is at rest,
  that the two ball system will continue on away
  from you after the impact
• You would be very surprised if one or the other
  of the balls came back to you (putting English
  on the ball not withstanding)
• What continued to move away from you was the CM
  of the two ball system

31
Newtons 2nd Law for aSystem of Particles

• Remember that the CM is a point that acts as
  though all of the mass in the system were located
  there
• So even though we may have a large number of
  particles possibly of different masses, we can
  treat the assembly as having all of its mass at
  the point of its CM
• So we can assign that point a position, a
  velocity and an acceleration

32
Newtons 2nd Law for aSystem of Particles

• And it turns out that Newtons 2nd law holds for
  that point
• This looks just like the previous version of
  Newtons 2nd law, but we need to be careful in
  understanding the specific meaning of the terms

33
Newtons 2nd Law for aSystem of Particles

• Fnet is the net of all of the external forces
  acting upon our system of particles internal
  forces are not included (as they generally have
  no net effect)
• M is the total mass of the system (and is assumed
  to be constant)
• acom is the acceleration of the center of mass
  we cant say anything about what the acceleration
  might be of any other part of the system

34
Newtons 2nd Law for aSystem of Particles

• Circling back around to where we started from, we
  can break the one vector equation down into an
  equation for each dimension

35
Newtons 2nd Law for aSystem of Particles

• Now lets re-examine what is going on when we
  have the collision between the two billiard
  balls
• Once the cue ball has been set in motion, there
  are no external forces acting on the system
• Thus, Fnet 0 and as a consequence, acom 0
• This means that the velocity of the CM of the
  system must be constant

36
Newtons 2nd Law for aSystem of Particles

• So when the two balls collide, the forces must be
  internal to the system which doesnt affect
  Fnet
• Thus the CM of the system continues to move
  forward unchanged by the collision
• We can see the effect of this in many ways some
  not so obvious

37
Newtons 2nd Law for aSystem of Particles

• When a fireworks rocket explodes, the CM of the
  system does not change while the fragments all
  fan out, their CM continues to move along the
  original path of the rocket

38
Newtons 2nd Law for aSystem of Particles

• This is also how a ballerina seems to defy the
  laws of physics and float across the stage when
  doing a grand jeté

39
Newtons 2nd Law for aSystem of Particles

• In actuality, she is very much making use of the
  laws of physics
• By raising her arms and legs during the first
  part of the leap, she shifts her CM upwards

40
Newtons 2nd Law for aSystem of Particles

• As a result, her upper body and head follow
  nearly a horizontal path her CM must still
  follow a parabolic path however

41
Newtons 2nd Law for aSystem of Particles

• Checkpoint 2
• Sample Problem 9-3

42
Sample Problem 9-3

• What is the acceleration of the CM and in what
  direction does it move?

43
Sample Problem 9-3

• Solve for the net force then use Fnet Macom
• Assume that all of the mass is concentrated at
  the CM e.g., M m1 m2 m3

44
Linear Momentum

• Other casual definitions aside, momentum has
  one precise meaning in physics
• The linear momentum of a particle is a vector
  defined as

45
Linear Momentum

• Newton actually expressed his 2nd law of motion
  in terms of momentum
• The time rate of change of the momentum of a
  particle is equal to the net force acting on the
  particle and is in the direction of that force

46
Linear Momentum

• Because acceleration is the time derivative of
  velocity, we can also state the law as

47
Linear Momentum

• These two ways of expressing the force on a
  particle are entirely equivalent

48
Checkpoint 3

• Rank the four regions according to the magnitude
  of the force, greatest first
• In which region is the particle slowing?

49
Linear Momentumof a System of Particles

• Now suppose that we have a system of n particles,
  each with their own mass and velocity and
  therefore momentum
• The particles may interact with each other and
  there may be external forces acting on the system
  (e.g., the collection of particles)

50
Linear Momentumof a System of Particles

• The system as a whole has a total linear momentum
  which is defined to be the vector sum of the
  momenta of the individual particles, thus

51
Linear Momentumof a System of Particles

• Comparing this to equation 9-17 we can therefore
  see thatwhere M the total mass and vcom is
  the velocity of the center of mass
• The linear momentum of a system of particles is
  equal to the product of the total mass M of the
  system and the velocity of the center of mass

52
Linear Momentumof a System of Particles

• If we take the time derivative of the previous
  equation, we getwhich of course leads us
  straight to

53
Class Exercise

• A 2.0 kg toy car is going 0.50 m/s before the
  turn and 0.40 m/s after the turn
• What is the change ?P in the linear momentum of
  the car due to the turn?

54
Class Exercise

• We know that the velocities of the car before and
  after the turn takes place are
• Can we just subtract the velocity after the turn
  from the velocity before the turn?

55
Class Exercise

• The answer is No this is because the vectors
  are pointing in different directions
• So we have to say thatand do the vector math
  which gives us

56
Collisions

• We certainly have an intuitive feel for what the
  word collision means, but what is the specific
  definition given to a collision in physics?
• A collision is an isolated event in which two or
  more bodies (the colliding bodies) exert
  relatively strong forces on each other for a
  relatively short time

57
Collisions

• An interesting point to note is that the
  definition doesnt necessarily require the bodies
  to actually make contact a near miss (near
  enough so that there are relatively strong
  forces involved) will suffice

58
Collisions

• To analyze a collision we have to pay attention
  to the 3 parts of a collision
• Before
• During
• After

59
Collisions

• By using knowledge of various conservation laws,
  and looking at the system of particles before,
  during and after the collision, physicists can
  deduce much about what is going on in the system

60
Impulse Linear Momentum

• Lets start by examining a head-on collision
  between just two bodies of different masses
• We know that there will be a pair of 3rd law
  forces involved the force from body 1 on body 2
  and vice versa the force from body 2 on body 1
• And of course we know that these forces will be
  equal in magnitude but opposite in direction
• The forces are therefore F(t) and -F(t)

61
Impulse Linear Momentum

• These forces will change the linear momentum of
  the bodies the amount of change will depend on
  average value of the force as well as how long
  the force acts on the body (e.g., ?t)

62
Impulse Linear Momentum

• To state this qualitatively, lets go back to
  Newtons 2nd law in the form of

63
Impulse Linear Momentum

• Lets see what is happening to object R (on the
  right)

64
Impulse Linear Momentum

• We can rearrange the previous equation a bit to
  getwhere F(t) is a time varying force as
  shown in the following graph

65
Impulse Linear Momentum

• If we now integrate over the interval ?t we will
  get
• The left side evaluates to pf pi which is
  just the change in linear momentum

66
Impulse Linear Momentum

• The right side is a measure of the strength as
  well as the duration of the collision force and
  is defined as the impulse J of the collision
  (not to be confused with the SI units of joules)

67
Impulse Linear Momentum

• We can see from equation 10-3 that the impulse is
  simply the area under the force curve during the
  interval in question
• We can also see that
• This is called the impulse linear momentum
  theorem

68
Impulse Linear Momentum

• We can see that the impulse is just the area
  under the time-varying force curve over the time
  interval in question
• If we take the average value of the force
  instead, we will get the same result for the
  impulse

69
Impulse Linear Momentum

• We can see that in the plot to the right

70
Impulse Linear Momentum

• Checkpoint 4

71
Series of Collisions

• What happens when there are multiple collisions
  to consider?
• Lets imagine an object that is securely bolted
  down so it cant move and is continuously pelted
  with a steady stream of projectiles

72
Series of Collisions

• Each projectile has a mass of m and is moving at
  velocity v along the x axis (so we will drop the
  vector arrows and remember that we are talking
  about what is happening only along the x axis)
• The linear momentum of each projectile is
  therefore mv lets also suppose that n
  projectiles arrive in an interval of ?t

73
Series of Collisions

• As each projectile hits (and is absorbed by) the
  mass, the change in the projectiles linear
  momentum is ?p thus the total change in linear
  momentum during the interval ?t is n ?p

74
Series of Collisions

• The total impulse on the target is the same
  magnitude but opposite in direction to the change
  in linear momentum thus
• But we also know that

75
Series of Collisions

• This leads us to
• So now we have the average force in terms of the
  rate at which projectiles collide with the target
  (n/?t) and each projectiles change in velocity
  (?v)

76
Series of Collisions

• In the time interval ?t, we also know that an
  amount of mass ?m nm collides with the target
• Thus our equation for the average force finally
  turns out to be

77
Series of Collisions

• Assuming that each projectile stops upon impact,
  we know that
• In this case the average force is

78
Series of Collisions

• Suppose instead that each projectile rebounds
  (bounces directly back) upon impact in this
  case we have
• In this case the average force is

79
Series of Collisions

• Checkpoint 5

80
Conservation ofLinear Momentum

• Suppose we have a system that has no external
  forces acting upon it (the system is isolated)
  and no particles enter or leave the system (the
  system is closed)
• We then know that Fnet 0 which in turn means
  that dP/dt 0, or thatP constant (in a
  closed, isolated system)

81
Conservation ofLinear Momentum

• In other words
• If no net force acts upon a system of particles,
  the total linear momentum P of the system cannot
  change
• This very important result is called the law of
  conservation of linear momentum
• It can also be written as Pi Pf

82
Conservation ofLinear Momentum

• Because these are vector equations, we can derive
  a little more insight if we further examine what
  they mean along each dimension
• If the component of a net external force on a
  closed system is zero along an axis, then the
  component of the linear momentum of the system
  along that axis cannot change

83
Conservation ofLinear Momentum

• Checkpoint 4
• Sample Problem 9-6

84
Sample Problem 9-6

• A space hauler with an initial velocity vi 2100
  km/hr separates from its cargo container
• The cargo module has a mass of 0.2M where M is
  the initial mass of the hauler plus cargo module
• After separation, the hauler has a velocity of
  500 km/hr relative to the cargo module
• What is the velocity of the hauler relative to
  the sun?

85
Sample Problem 9-6

• We will assume that the system consists of the
  hauler cargo module and is closed
• Because it is closed, we know that conservation
  of momentum will hold and that Pi Pf
• Because this is a single-dimension problem, we
  can drop the vector notation

86
Sample Problem 9-6

• We know the initial linear momentum of the system
  it iswhere vi is the velocity of the
  combined hauler module relative to the sun
• Let vMS and vHS be the velocities of the ejected
  cargo module and the hauler respectively,
  relative to the sun

87
Sample Problem 9-6

• After the cargo module has been ejected, the
  total linear momentum of the system is
• We dont know what vMS is, but we can relate it
  to other factors as follows

88
Sample Problem 9-6

• We can now substitute vMS back into our equation
  for the final linear momentum
• Rearranging a little gives us

89
Sample Problem 9-6

• Finally, we can substitute in the known values
  and solve for vHS to get

90
Systems with Varying MassA Rocket

• So far we have always assumed that the mass of
  the system stayed constant
• There is an important class of problems where
  this is not so however Rockets
• Most of the mass of a rocket is fuel a rocket
  provides its thrust by burning its fuel and
  ejecting the fuel mass at a high velocity

91
Systems with Varying MassA Rocket

• As you can imagine, it would be difficult to deal
  with a system where the mass was changing as a
  function of time
• But thats really not necessary we already know
  how to solve this problem
• We do it by picking a system where the mass is
  not changing (remember the last problem???)

92
Systems with Varying MassA Rocket

• So the system we pick is one which includes the
  exhaust gas (mass) as well as the mass of the
  rocket (which is rapidly becoming emptied its fuel

93
Systems with Varying MassA Rocket

• Lets assume that our rocket is isolated out in
  space to make things a little easier
• We begin by looking at the rocket at some time t
  and assume it has some initial velocity v

94
Systems with Varying MassA Rocket

• Now we look at the rocket at some time dt later
  the velocity of the rocket is now v dv and the
  mass of the rocket is now M dM where the value
  dM is a negative quantity

95
Systems with Varying MassA Rocket

• We need to account for the exhaust however as it
  is part of our isolated system it has a mass of
  -dM and a velocity U

96
Systems with Varying MassA Rocket

• Since our system is isolated and closed, we can
  say that, according to the law of conservation of
  linear momentum, we know that Pi Pf
• This can be rewritten as

97
Systems with Varying MassA Rocket

• This equation can be simplified a bit if we
  consider the relative velocity of the exhaust to
  the rocket itselfor

98
Systems with Varying MassA Rocket

• Substituting that back and turning the crank a
  little we get
• Dividing each side by dt we then get

99
Systems with Varying MassA Rocket

• If we then replace dM/dt by -R (the rate at which
  the rocket loses mass) and note that dv/dt is
  just the acceleration of the rocket, we get the
  following (the 1st rocket equation)

100
Systems with Varying MassA Rocket

• The terms on the left hand side of the equation
  all depend solely on the design of the rocket
  engine
• Note that the left side of the equation must be a
  force as it is equal to mass times acceleration
• The term Rvrel is called the thrust of the rocket
  engine and we often represent it by T

101
Systems with Varying MassA Rocket

• To find the velocity of the rocket at any given
  time we have to integrate equation9-40
• Rearranging it a bit we get

102
Systems with Varying MassA Rocket

• Now we need to integratewhere ln means the
  natural logarithm, where vi and vf are the
  initial and final velocities respectively, and Mi
  and Mf are the initial and final mass of the
  rocket respectively

103
Systems with Varying MassA Rocket

• The result is (the 2nd rocket equation)
• Note an important point the change in velocity
  is directly proportional to the relative velocity
  (the exhaust velocity) the higher that is, the
  higher the change in velocity of the rocket itself

104
Systems with Varying MassA Rocket

• Now lets look at the 2nd term on the right
• Lets suppose that 90 of the rockets mass is
  ejected as exhaust this means that the ratio
  Mi/Mf is 101

105
Systems with Varying MassA Rocket

• The natural logarithm of 10 is about 2.3
• This means that (given our parameters) the rocket
  can never go faster than 2.3 times the exhaust
  velocity again a reason to design an engine
  with the highest possible high exhaust velocity
• It also encourages the design of multi-stage
  rockets, so that the final mass is as close to
  that of the payload as possible

106
Sample ProblemA Shuttle Launch

• The first couple minutes of a shuttle launch can
  be described very roughly as follows
• The initial mass is about 2 x 106 kg
• The final mass (after 2 minutes) is about1 x 106
  kg
• The average exhaust speed is about 3000 m/s
• The initial velocity is, of course, zero

107
Sample ProblemA Shuttle Launch

• If all this were taking place in outer space,
  with negligible gravity, what would be the
  shuttles speed at the end of this stage?
• What is the thrust during the same period and how
  does it compare with the initial total weight of
  the shuttle (on earth)?
• Mi 2 x 106 kg, Mf 1 x 106 kg,vrel 3000
  m/s, vi 0, ?t 2 minutes

108
Sample ProblemA Shuttle Launch

• Plugging the specified values into the 2nd rocket
  equation we get

109
Sample ProblemA Shuttle Launch

• To get the thrust we use the equation
• We know that 1 x 106 kg of fuel is used in the
  first 2 minutes of launch, therefore

110
Sample ProblemA Shuttle Launch

• To get the thrust we plug the specified values
  into the equation

111
Sample ProblemA Shuttle Launch

• The mass of the shuttle before launch was2 x 106
  kg so the shuttles weight was

112
Sample ProblemA Shuttle Launch

• The last problem was to determine the
  thrust-to-weight ratio at launch
• We now have both quantities so we plug them in
  and get

113
External Forces Internal Energy Changes

• Lets look at the ice skater in the picture
• She pushes against the railing with a force F at
  an angle ? to the horizontal
• Her initial velocity was zero and as she loses
  contact with the railing her velocity is v

114
External Forces Internal Energy Changes

• The external force clearly increased her kinetic
  energy, but things are different from previous
  examples
• Her body was not rigid clearly her arm behaved
  differently from the rest of her body

115
External Forces Internal Energy Changes

• The 2nd difference is this
• In previous examples, energy was transferred
  between an object and its environment (the
  system) by an external force
• In this case energy was transferred internally
  within the system via the external force F
  specifically, biochemical energy in the skaters
  muscles was transferred into kinetic energy of
  her body as a whole

116
External Forces Internal Energy Changes

• We want to relate the external force to the
  internal energy transfer
• We begin by imagining the skater as a particle
  (with her mass) located at her center of mass
• We then treat the external force as acting on
  that particle

117
External Forces Internal Energy Changes

• The horizontal component of the force accelerates
  the particle, resulting in a change ?K in its
  kinetic energy
• So we have

118
External Forces Internal Energy Changes

• To allow for more generality, we will also
  imagine that the force could also possibly change
  the potential energy of the skaters CM thus we
  have

119
External Forces Internal Energy Changes

• The left side of the equation is just Emch so we
  end up with
• Now lets consider the energy transfers within the
  skater (which is the system)

120
External Forces Internal Energy Changes

• Although there is an external force acting on the
  system, there is no energy transfer to or from
  the system
• Thus the total energy of the system didnt change
  ?E 0
• When the skater pushes off from the rail, there
  is a net decrease in the energy in her muscles
  lets call that ?Eint

121
External Forces Internal Energy Changes

• So we can write the energy equation as
  followsor

122
External Forces Internal Energy Changes

• But we know what Emch is, so substituting that in
  we get
• So we can see that the magnitude of the energy
  expended internally is exactly equal to the
  magnitude of the mechanical energy

123
External Forces Internal Energy Changes

• While we used a skater in the previous example,
  the concept applies to many commonplace systems
  lets consider another one a car being
  accelerated

124
External Forces Internal Energy Changes

• Here we consider the four forces from the tires
  on the pavement (two are shown but we will
  consider the net of all four to be F)

125
External Forces Internal Energy Changes

• In this case the internal energy comes from the
  fuel being burned in the engine
• There is a transfer of energy from the fuel to
  the kinetic energy of the car
• If F is constant, then for a given displacement
  d of the cars CM along a level road we can
  relate the change in ?K to the external force F
  (with ?U 0 and ? 0º) through equation 9-45

126
External Forces Internal Energy Changes

• Equation 9-45 holds regardless of whether the car
  is accelerating or decelerating if decelerating
  then the frictional forces are pointing to the
  rear and ? 180º
• Energy in this case is transferred from the
  kinetic energy of the cars CM to thermal energy
  in the brakes

127
Next Class

• Homework Problems Chapter 94, 12, 24, 32, 42,
  52
• Read sections Chapter 10