Review of Lecture 10

1
Review of Lecture 10

• Collisions
• Impulse Linear Momentum
• Momentum Kinetic Energy in Collisions
• Inelastic Collisions in 1 Dimension
• Elastic Collisions in 1 Dimension
• Collisions in 2 Dimensions

2
Rotational Motion

• So far, we have been dealing mainly with
  translational motion
• Now we will move on to rotational motion the
  motion of something that is spinning about an
  axis
• Specifically we will examine the rotation of a
  rigid body around a fixed axis

3
Rotational Variables

• We have at the right an arbitrary rigid body
  which is rotating about an arbitrary axis
• This axis is called the rotation axis or axis of
  rotation

4
Rotational Variables

• You can see that any point within the body will
  describe a circle as it moves around the rotation
  axis
• Every point within the body will move through the
  same angle during the same time period

5
Rotational Variables

• We also have a reference line in our diagram
  which is normal to the rotation axis and rotates
  with the body (e.g., it is embedded within the
  body)

6
Rotational Variables(Angular Position)

• The angular position of the line is the angle of
  the line relative to some fixed reference
  direction which we take as the zero angular
  position
• This could be the x axis, the y axis or some
  other direction fixed in the xy plane

7
Rotational Variables(Angular Position)

• Looking down from above now, we can see that the
  angular position is being measured w.r.t. the
  positive x axis

8
Rotational Variables(Angular Position)

• The angleis in this case expressed in
  radians as s is the length of the arc described
  by the line of length r rotating through the
  angle ?

9
Rotational Variables(Angular Position)

• If you havent worked in radians before, then
  here is a quick overview
• There are 2p radians in a complete circle
• Thus 1 rad 57.3º 0.159 revolution
• Unlike using degrees, we do not reset our angle
  in radians to 0 (zero) when we pass the zero
  angular position, e.g., twice around a circle
  would be 4p radians

10
Rotational Variables(Angular Position)

• Note also that there are no units for an angle
  measured in radians
• This is because the definition of an angle
  measured in radians is simply the ratio of 2 real
  numbers (with whatever units they may happen to
  have)

11
Rotational Variables(Angular Displacement)

• Just as we can have a linear displacement, we can
  also have an angular displacement which is
  defined as

12
Rotational Variables(Angular Displacement)

• By convention, angular displacements are taken to
  be positive when the rotation is in the
  counterclockwise direction
• Remember the phrase Clocks are negative

13
Rotational Variables(Angular Velocity)

• Carrying the linear analogies one step farther,
  we can define an angular velocity as the angular
  displacement made over some period of time ?t

14
Rotational Variables(Angular Velocity)

• The average angular velocity is
  thereforeand is usually measured in
  radians-per-second or revolutions-per-second (or
  revolutions-per-minute a.k.a. rpm)

15
Rotational Variables(Angular Velocity)

• To get the instantaneous angular velocity we
  simply make our ?t very small and in the limit we
  get

16
Rotational Variables(Angular Acceleration)

• Lets take this one last step and introduce
  angular acceleration
• Just as we could have average and instantaneous
  angular velocity, so too can we have average and
  instantaneous angular acceleration
• And we will define them in just the same fashion
  as we did their linear counterparts

17
Rotational Variables(Angular Acceleration)

• Lets start with average angular acceleration
  which is defined as the change in angular
  velocity over some time period

18
Rotational Variables(Angular Acceleration)

• And similarly for the instantaneous angular
  acceleration we have
• Note Angular acceleration is typically measured
  in radians/s2 or sometimes revolutions/s2

19
Rotational Variables

• Sample Problem 10-1
• Checkpoint 1

20
Are Angular Quantities Vectors?

• Recall that we have often dispensed with the
  vector notation when we have a particle moving in
  a straight line along an axis
• We can do this because we dont really need a
  vector a plus or minus sign will suffice to
  tell us which direction the particle is moving

21
Are Angular Quantities Vectors?

• We can do the same with an object that is
  rotating around a fixed axis it can be rotating
  either clockwise or counterclockwise

22
Are Angular Quantities Vectors?

• We represent the angular velocity as a vector ?
  pointing along the axis of rotation
• This is quite different from any vectors we have
  previously used the vector in this case doesnt
  point in the direction of motion but rather
  points at right angles to it

23
Are Angular Quantities Vectors?

• We use a right-hand rule to define the direction
  for the angular velocity
• Curl your right hand around the rotating object
  such that your fingers point in the direction of
  rotation
• By definition then, your thumb points in the
  direction of the angular velocity

24
Are Angular Quantities Vectors?
25
Are Angular Quantities Vectors?

• Angular acceleration a is also a vector
  quantity and it follows the same rules
• Note that angular displacement however generally
  cannot be represented as a vector
• This is because in order to be a vector, a
  quantity must follow all of the rules of a vector

26

• In particular, a vector must follow the rules of
  vector addition which says that the order of the
  vectors being added doesnt make any difference
• However, angular displacements fail this test
• Lets see why

27
Rotation with ConstantAngular Acceleration

• When we were working with constant linear
  acceleration we had a nice set of equations that
  allowed us to solve for various quantities given
  a certain set of known values
• The same happens in the case of constant angular
  acceleration and in fact the equations look very
  similar

28
Rotation with ConstantAngular Acceleration

• What has been done in Table 10-1 (page 247) is
  that the variables for linear displacement,
  velocity and acceleration have simply been
  replaced with their angular equivalents
• So for example

29
Rotation with ConstantAngular Acceleration

• Checkpoint 2
• Sample Problem 10-2

30
Relating the Linear andAngular Variables

• While all of the particles of a rigid rotating
  object move with the same angular velocity, the
  linear velocity of a specific particle will vary
  with its perpendicular distance r from the axis
  of rotation
• Lets start with the angular position

31
Relating the Linear andAngular Variables

• Looking at the picture to the right, we can see
  that a particle at a distance of r from the axis
  will move a distance s along a circular arc given
  a change in the angle ?

32
Relating the Linear andAngular Variables

• Thus the relationship between linear distance
  (along the arc of the circle in this case) and
  the angle of rotation is

33
Relating the Linear andAngular Variables

• Note that the angle ? must be measured in radians
  in this case (where one revolution is equal to 2p
  radians)
• Differentiating the position equation with
  respect to time will give us the velocity
  equation

34
Relating the Linear andAngular Variables

• Notice that the linear velocity vector is always
  tangent to the circle
• Given that the angular speed is constant we can
  see that the linear speed is also constant

35
Relating the Linear andAngular Variables

• If the linear speed of a point on a rotating body
  is constant, then the body is undergoing uniform
  circular motion
• The period of rotation is given by

36
Relating the Linear andAngular Variables

• Substituting back into the equation relating
  linear velocity to angular velocity we get

37
Relating the Linear andAngular Variables

• Now lets differentiate the linear velocity
  equation to get the linear acceleration of our
  rotating object

38
Relating the Linear andAngular Variables

• Remember that the acceleration we just calculated
  is the tangential acceleration (thus the t
  subscript)
• We already know the radial component of the
  acceleration from our study of uniform circular
  motion

39
Relating the Linear andAngular Variables

• We can see that the radial acceleration component
  is present whenever the angular velocity is not
  zero

40
Relating the Linear andAngular Variables

• The tangential acceleration component however is
  present only when the angular acceleration is not
  zero

41
Relating the Linear andAngular Variables

• Checkpoint 3
• Sample Problem

42
Sample Problem

• An astronaut is riding in a centrifuge with a
  radius r 15 m
• At what constant angular speed must the
  centrifuge rotate if the astronaut is to have a
  linear acceleration of magnitude 11g?

43
Sample Problem

• Since the angular speed is constant, we know that
  the angular acceleration is zero
• So we also know that the tangential component of
  the linear acceleration is also zero

44
Sample Problem

• The only component of acceleration that is left
  is the radial acceleration which iswhere ar
  11g

45
Sample Problem

• Rearranging a bit we get

46
Sample Problem

• What is the tangential acceleration of the
  astronaut if the centrifuge accelerates at a
  constant rate from rest to the angular speed we
  just calculated in 120 seconds?

47
Sample Problem

• We know that the tangential component of angular
  acceleration is given byand that

48
Sample Problem

• Combining these two equations gives uswhere
  ?2 2.68 rad/s, ?1 0, t2 120 s,t1 0 and r
  15 m
• Doing the calculations gives us the resultat
  0.034 g

49
Kinetic Energy of Rotation

• If we want to calculate the kinetic energy of a
  rotating object (say a circular saw blade) we
  would be tempted to start with the familiar
• But we know from our previous study of rigid
  bodies that this will only give us the kinetic
  energy of the center of mass which, in the case
  of a circular saw blade, is zero (we hope!)

50
Kinetic Energy of Rotation

• Lets instead decompose our saw blade into lots
  of small masses (particles) which we know will be
  moving at different velocities (depending on
  their perpendicular distance from the axis of
  rotation)

51
Kinetic Energy of Rotation

• This series of terms can be expressed
  aswhere mi is the mass is the ith particle
  and vi is the linear speed of the particle and
  the sum is taken over all of the particles in the
  body

52
Kinetic Energy of Rotation

• We already know that the values vi will not be
  the same for all particles so lets make that
  substitution now
• This term is called the rotational inertia or the
  moment of inertia of the object

53
Kinetic Energy of Rotation

• So we get as a result
• Now lets compare this with our previous
  definition of kinetic energy

54
Kinetic Energy of Rotation

• Linear

• Rotational

RotationalInertia
55
Kinetic Energy of Rotation

• Note that while the formulas are somewhat
  different, they both describe kinetic energy
• The difference is only that in one case the
  object is moving in translation and in the other
  case the body is rotating
• Clearly it is also possible that a body is moving
  in both ways

56
Kinetic Energy of Rotation

• Checkpoint 4

57
Calculating the Rotational Inertia

• We know that rotational inertia depends on the
  distribution as well as the velocity of masses
  that are rotating around a fixed axis
• So for small numbers of simple masses, we can
  calculate the rotational inertia directly

58
Calculating the Rotational Inertia

• For objects with a continuous mass distribution,
  we need to move from the summation to an
  integralwhere we integrate over the mass of
  the object

59
Calculating the Rotational Inertia

• Lets take a simple example a thin rod rotating
  about the long axis (figure (c) in Table 10-2)
• Imagine a thin slice of therod it would look
  like aflat plate
• Also assume it hasuniform density ?

60
Calculating the Rotational Inertia

• A mass element in the plate would have a
  volumewhere rdr is the infinitesimal change
  in radius (at radius r), d? is the change in
  angle and dl is the thickness

61
Calculating the Rotational Inertia

• We can therefore express the mass element dm
  aswhich results in

62
Calculating the Rotational Inertia

• We can break this down into three integrals as
  follows (where I have inserted the limits over
  which you need to evaluate the integrals)

63
Calculating the Rotational Inertia

• This results in

64
Calculating the Rotational Inertia

• We know that the density of the rod is uniform,
  so the mass is simply
• Factoring that into the previous equation we get
  the following for the moment of inertia of a thin
  rod about its long axis

65
Calculating the Rotational Inertia

• Lets imagine two masses m separated by a distance
  L and rotating about a point halfway between them

66
Calculating the Rotational Inertia

• We have

67
Calculating the Rotational Inertia

• Now lets change the system so that the axis of
  rotation is through one of the masses

68
Calculating the Rotational Inertia

• We have

69
Calculating the Rotational Inertia

• From this we can see that even though the
  configurations of the two systems are identical,
  they have different moments of inertia just
  because the axis of rotation was changed

70
Parallel-Axis Theorem

• For objects that have a continuous mass
  distribution we can always use the
  equationto find the moment of inertia of any
  object about a specific axis by evaluating the
  integrals at the appropriate limits

71
Parallel-Axis Theorem

• But there is a shortcut we can use if we already
  know the moment of inertia of the object around
  an axis that passes through the center of mass
  and if that axis is parallel to the (new) axis in
  question

72
Parallel-Axis Theorem

• Given these constraints, the moment of inertia
  around the new axis iswhere h is the
  perpendicular distance from the new axis to the
  axis passing through the center of mass, and M is
  the mass of the object

73
Parallel-Axis Theorem

• Lets go back to our example of two balls of mass
  m separated by a massless rod of length L

74
Parallel-Axis Theorem

• We know that the moment of inertia is

75
Parallel-Axis Theorem

• Now lets change the system so that the axis of
  rotation is through one of the masses but this
  time we will calculate the new moment using the
  parallel-axis theorem

76
Parallel-Axis Theorem

• We get

77
Parallel-Axis Theorem

• Lets do another problem of this type
• We can see from table10-2 that a hoop
  ofdiameter R spunaround any axishas a moment
  ofinertia I ½ MR2

78
Parallel-Axis Theorem

• Now lets move the rotation axis to a new
  location as shown by the dotted line
• Using the parallel axis theorem we see that

79
Parallel-Axis Theorem

• Checkpoint 5
• Sample Problem 10-7

80
Torque

• When you try to swing a door, its clear that it
  is easier to move the door if you apply the force
  farther away from the axis of rotation (e.g., the
  hinges)
• Its also clear that the application of the force
  is most efficient if the force is applied
  perpendicular to the plane of the door

81
Torque

• Lets look at an object that is rotating about a
  fixed axis O
• A force F is applied at point P which is at a
  position r relative to the axis O
• Note also that the force F is applied at an
  angle F relative to the vector r
• For simplicity we also assume that the force F
  is in the plane of the screen

82
Torque

• Lets decompose the force into its components
  relative to the vector r
• The radial component is labeled Fr (FcosF)
• The tangential component is labeled Ft (FsinF)

83
Torque

• It should be clear that the ability to rotate the
  door depends only on the tangential component of
  the force
• And as we noted earlier, the farther away the
  force is applied from the axis of rotation, the
  easier it is to rotate the object

84
Torque

• So the ability to rotate an object using an
  external applied force depends on the tangential
  component of the applied force, and the distance
  that force is applied from the axis of rotation

85
Torque

• We therefore define torque to be the product of
  these two values
• Torque comes from the Latin word meaning to
  twist

86
Torque

• By rearranging things a bit in the previous
  equation we can equivalently see thatwhere
  r? is the perpendicular distance from the
  rotation axis O to an extended line running
  through the vector F at point P

87
Torque

• The extended line is called the line of action
  and the value r? is called the moment arm
• Clearly if the force is applied completely
  tangentially, then the moment arm is just r

88
Torque

• A word of caution
• You can see that the SI units for torque are Nm
  (force times distance) which also are the units
  for work
• Work can also be expressed in joules(1 J 1 Nm
  ), but torque is never expressed that way

89
Torque

• In the next chapter we will spend more time with
  torque and examine its properties as a vector
• For the moment however, it is sufficient to just
  treat it as a scalar
• If an object rotates counterclockwise (as seen
  from a position above the axis of rotation), then
  the torque is positive and vice versa (remember
  that clocks are negative)

90
Torque

• Also note that torques obey the superposition
  principle, e.g. when several torques act on a
  body, the net torque (or resultant torque) is the
  sum of the individual torques

91
Torque

• Checkpoint 6

92
Newtons 2nd Law for Rotation

• We know that Newtons 2nd law is
• We also know there are formulas that relate the
  angular and linear equivalents for displacement,
  velocity and acceleration

93
Newtons 2nd Law for Rotation

• So we should be able then to come up with an
  rotational equivalent to Fnet ma and we can
  it iswhere tnet takes the place of Fnet, I
  takes the place of the mass m, and a (the angular
  acceleration) takes the place of the linear
  acceleration a

94
Newtons 2nd Law for Rotation

• The proof of this is straightforward and is shown
  on page 257
• Note that the angular acceleration must (as
  always) be measured in radians/sec2

95
Newtons 2nd Law for Rotation

• Checkpoint 7
• Sample Problem 10-8

96
Sample Problem 10-8

• We have a disk with massM 2.5 kg and a
  radiusR 20 cm mounted on a fixed, frictionless
  axle
• A block of mass m 1.2 kg hangs from a massless
  string which is wrapped around the disk several
  times
• Find the acceleration of the block, the angular
  acceleration of the disk and the string tension

97
Sample Problem 10-8

• As usual, we begin by drawing the free body
  diagrams
• Starting with the block we see that our force
  equation turns out to be
• We have the acceleration a in this equation, but
  we cant solve for it yet as we dont know the
  value of T

98
Sample Problem 10-8

• Moving on to the disk, we can see that the torque
  on the disk is(since the disk is turning
  clockwise, the torque is negative)

99
Sample Problem 10-8

• But we also can express the torque in terms of
  the moment of inertiaand we know thatfor
  a flat plate rotating as this one is

100
Sample Problem 10-8

• Combining all of these we getwhich is fine
  except that that dont know what a is (so that we
  can solve for T)

101
Sample Problem 10-8

• But wait we can relate a to the tangential
  component of the linear acceleration at (which
  also just happens to be a in this case)
• Rearranging that we get

102
Sample Problem 10-8

• Plugging that back into our previous equation we
  get
• We can now substitute this result back into our
  equation for the motion of the block

103
Sample Problem 10-8

• When we do that we get
• Now that we know a we can solve for T and get

104
Sample Problem 10-8

• Finally, we can solve for a and get

105
Sample Problem 10-8

• As a final check lets see what happens when M
  0

106
Work and RotationalKinetic Energy

• Recall from Chapter 7 that the work done on an
  object can be expressed as the change in
  translational-related kinetic energy
• Suppose that the change in the kinetic energy is
  the only change in the overall energy of the
  system thus

107
Work and RotationalKinetic Energy

• For motion confined to a single axis (lets say
  the x axis) we havewhich reduces towhen
  the force is constant and the displacement goes
  from xi to xf

108
Work and RotationalKinetic Energy

• And finally, the power is
• Now lets consider what happens in the rotational
  case

109
Work and RotationalKinetic Energy

• In this case suppose that the change in the
  rotational kinetic energy is the only change in
  the overall energy of the system thus

110
Work and RotationalKinetic Energy

• The corresponding equation for work in the
  rotational case iswhich reduces towhen
  the torque is constant and the angular
  displacement goes from ?i to ?f

111
Work and RotationalKinetic Energy

• And finally, the power is
• Note Table 10-3 (on page 260) which provides some
  corresponding relations between translational and
  rotational motion

112
Work and RotationalKinetic Energy

• So now lets pull all of this together into one
  last sample problem

113
Sample Problem 10-11

• We have a sculpture consisting of a thin hoop of
  mass m and radius R 0.15 m and a thin radial
  rod also of mass m and length L 2.0 R arranged
  as shown
• Suppose the sculpture can rotate about the x axis
  as shown

114
Sample Problem 10-11

• In terms of m and R, what is the sculptures
  moment of inertia I about the rotation axis?

115
Sample Problem 10-11

• We begin by realizing that we can break the
  problem down into components the hoop and the
  rod which can be solved for separately
• The individual results can then be added together
  for the final result
• So lets start with the hoop

116
Sample Problem 10-11

• From Table 10-2h we can see that the moment of
  inertia for the hoop about its diameter is
• And also from Table 10-2e we see that the rod has
  a moment of inertia about its center of mass of

117
Sample Problem 10-11

• We assert that the rotation axis of the rod is
  parallel to the rotation axis of the hoop
• As a result, we can use the parallel-axis theorem
  to calculate the moment of inertia of the rod
  about the hoops rotational axis which results
  inwhere h is the distance between the rods
  COM and the rotation axis of the hoop, and m is
  the mass of the rod

118
Sample Problem 10-11

• This results in
• Substituting in L 2.0 R we get

119
Sample Problem 10-11

• Adding the rods and the hoops moments together
  we get

120
Sample Problem 10-11

• Assume that the sculpture starts from rest in the
  position shown and is given a very gentle nudge
  to cause it to rotate (the rotation due entirely
  to gravity)
• What is the angular speed ? when the sculpture is
  inverted?

121
Sample Problem 10-11

• To solve this we need to do two things
• We need to relate the sculptures angular speed ?
  to the sculptures kinetic energy K at the bottom
  of the rotation, and
• We need to relate the kinetic energy K at the
  bottom of the rotation to the sculptures total
  mechanical energy

122
Sample Problem 10-11

• Lets start with the formula for the conservation
  of mechanical energy
• We know that
• We also know that as the sculpture rotates, the
  kinetic energy is related to its angular speed
  by

123
Sample Problem 10-11

• Now lets attack the potential energy side
• We know thatwhere ?y represents the change in
  position of the sculptures center of mass
• So we get

124
Sample Problem 10-11

• Now we need to know what the sculptures COM is
  doing
• For this we will use the fact that we can
  represent our two objects (the rod and the hoop)
  as point masses at their respective centers of
  mass

125
Sample Problem 10-11

• The COM of the hoop is at y 0 and the rods COM
  is at h (which we know is 2R)

126
Sample Problem 10-11

• So we get as our initial COMand we can see
  by symmetry that when the sculpture is inverted
  the COM will be

127
Sample Problem 10-11

• So from this we can see that
• Pulling everything together we get

128
Sample Problem 10-11

• Substituting in I 4.83R2 and ?ycom -2R and
  then solving for the angular speed ? we get

129
Next Class

• Homework Problems Chapter 104, 15, 33, 41, 55,
  65
• Read sections Chapter 11