Schedule

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Schedule

• Today, April 23
• Angular Momentum Torque for Systems and Rigid
  Bodies, (11-4, 11-5, 11-6, 11-7)
• Wednesday, April 25
• Conservation of Angular Momentum and Special
  Topics (11-7, 11-8, 11-9, 11-10)
• Friday, April 27
• Static Equilibrium (12-1, 12-2, 12-3, 12-4)
• Quiz 4 Review
• Monday, April 30
• Quiz 4 / 50 points / Crib Sheet
• Wednesday, May 2
• Catch-up / Final Review
• Wednesday, May 9 noon-150PM
• Final / 100 points / Open book

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Formal Definition of Torque and Angular Momentum
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• What is the angular momentum of a particle of
  mass m moving CCW with speed v in a circle of
  radius r? (See the figure)
• Choose the center of the circle as the origin.

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Angular Momentum and Torque for a System of
Particles

• Well for a single particle weve found the vector
  products definitions tr x F and lr x p are
  consistent with our earlier discussion for motion
  about a fixed axis.
• The next step is to explore the consequence for
  systems of particles. This is exactly what we did
  for linear motion and the equations Fma and
  pmv.
• The system could be a loose collection of
  individual particles or a rigid body. The total
  angular momentum L will be the sum of the
  individual angular momentum l1, l2, l3.

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Some comments on

• Torque is the change in angular momentum
• L and t must be calculated about the same origin.
• Valid for a fixed or moving point in any inertial
  reference frame.
• Valid only about the center of mass for an
  accelerating reference fame.
• Proven on page 284 of the text
• This is why we can say KEKECMKERot for a hoop
  rolling down a plane.

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L and t for a Rigid Body

• We start from our general result and apply to the
  case of rotation of a rigid body about a fixed
  axis.
• Well replicate the result found in Chapter 10
  for this special case.
• Focus will be the component of angular momentum
  along rotation axis which we call Lw

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Application to the Atwood Machine

• In Chapter One we used a free-body diagram to
  analyze the two weight pulley system, which is
  known as Atwoods Machine.
• We can also use angular momentum and torque to
  analyze the system.

q
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q
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The larger the moment of inertia the slower the
system. If I0 we recover the result of Chapter
4.
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Limitations of

• Angular momentum neednt be in the same direction
  as w.
• Equality requires that the rotation axis be along
  an axis of symmetry through the center of mass.
• Actually it can be shown that every rigid body no
  matter how oddly shaped has three axes about
  which LIw holds. These are called principal
  axes

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L and w Not in the Same Direction

• Consider this asymmetric assembly which is a
  massless rod at an angle f with respect to the
  rotating axis with attached masses m1 and m2.
• The angular momenta shown are for m1 approaching
  you and m2 receding.
• The direction of the sum of their momenta, L,
  clearly does not coincide with the direction of
  w.

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Rotational Imbalance

• If the system rotates with constant w, the
  magnitude of the angular momentum will not change
  but the direction will.
• In fact as the rod masses rotate around the
  axis, so will L.
• The figure has L in the plane of the screen. In a
  small time interval dt the rod will rotate an
  angle dqwdt. Accordingly, m1 will move out of
  the screen and m2 in.
• This a means L will move into the screen. The
  change in L or dL/dt must also point INTO the
  screen.

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• But by StdL/dt, a torque (also into the page)
  must be applied to the system.
• The torque must come from the system bearings at
  the end of the axle. The associated force must be
  in the w-L plane to give the correct change dL.
• This means the force is rotating at angular
  velocity w leading to wobbles.

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• The wheel at the right is similar and would
  wobble.
• Note that at rest the wheel is balanced
  (statically balanced).
• Dynamic balancing (L parallel to w and so t0)
  would require weights m3 and m4.

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What t is Needed for Balance?

• The torque needed to keep the system turning can
  be given in terms of I, w, and f.
• The component of L perpendicular to the z axis is
  Lsin(f-90)Lcosf.
• Looking down the z axis the changing vector
  appears at left.And dL(Lcosf)dq,
• But the angle changes by dqwdt, so dLLcosfwdt
• Taking the derivative tdL/dtwLcosf
• Our next step is to find L.

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Note the dependence on Io, w, and f For instance
at f0 or 90, L lines up with w and no torque is
needed! Here we see the power of the vector
approach to the problem