Physics 151: Lecture 32 Today

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Physics 151 Lecture 32 Todays Agenda

• Topics
• The Pendulum Ch. 15
• Potential energy and SHM

2
Simple Harmonic MotionReview
See text 15.1 to 15.3

• The most general solution is x Acos(?t ?)
• where A amplitude
• ? frequency
• ? phase constant
• For a mass on a spring
• The frequency does not depend on the amplitude
  !!!
• The oscillation occurs around the equilibrium
  point where the force is zero!
• Energy is a constant, it transfers between
  potential and kinetic.

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Lecture 32, Act 1Simple Harmonic Motion

• You have landed your spaceship on the moon and
  want to determine the acceleration due to gravity
  using a simple pendulum of length 1.0 m. If he
  period of the pendulum is 5.0 s what is the
  value of g on the moon ?
• a) 1.3 m/s2.
• b) 1.6 m/s2.
• c) 0.80 m/s2.
• d) 0.63 m/s2.
• e) 2.4 m/s2.

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General Physical Pendulum
See text 15.5

• Suppose we have some arbitrarily shaped solid of
  mass M hung on a fixed axis, that we know where
  the CM is located and what the moment of inertia
  I about the axis is.
• The torque about the rotation (z) axis for small
  ? is (sin ? ? )
  
  
  ? -Mgd -MgR??

z-axis
R
?
x
CM
d
Mg
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Lecture 32, Act 2Physical Pendulum

• A pendulum is made by hanging a thin hoola-hoop
  of diameter D on a small nail.
• What is the angular frequency of oscillation of
  the hoop for small displacements ? (ICM mR2 for
  a hoop)

pivot (nail)
(a) (b) (c)
D
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Lecture 32, Act 2 Solution

• The angular frequency of oscillation of the hoop
  for small displacements will be given by

Use parallel axis theorem I Icm mR2
mR2 mR2 2mR2
pivot (nail)
cm x
R
m
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Torsion Pendulum
See text 15.5

• Consider an object suspended by a wire attached
  at its CM. The wire defines the rotation axis,
  and the moment of inertia I about this axis is
  known.
• The wire acts like a rotational spring.
• When the object is rotated, the wire is twisted.
  This produces a torque that opposes the rotation.
• In analogy with a spring, the torque produced is
  proportional to the displacement ? -k?

See figure 13.15
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Torsion Pendulum...
See text 15.4

• Since ? -k??? ? I???becomes

Similar to mass on spring, except I has taken
the place of m (no surprise)
See figure 13.15
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Lecture 32, Act 3Period

• All of the following pedulum bobs have the same
  mass. Which pendulum rotates the fastest, i.e.
  has the smallest period? (The wires are identical)

C)
B)
A)
D)
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Lecture 32, Act 5Period

• Check each case.

The biggest is (D), the smallest moment of inertia
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Energy in SHM
See text 15.3

• For both the spring and the pendulum, we can
  derive the SHM solution using energy
  conservation.
• The total energy (K U) of a system undergoing
  SMH will always be constant!
• This is not surprising since there are only
  conservative forces present, hence energy is
  conserved.

Animation
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SHM and quadratic potentials
See text Fig. 15.6

• SHM will occur whenever the potential is
  quadratic.
• Generally, this will not be the case
• For example, the potential betweenH atoms in an
  H2 molecule lookssomething like this

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SHM and quadratic potentials...
See text Fig. 15.6

• However, if we do a Taylor expansion of this
  function about the minimum, we find that for
  smalldisplacements, the potential IS
  quadratic

U?(x) 0 (since x0 is minimum of potential)
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SHM and quadratic potentials...
See text Fig. 15.6
U(x) U?? (x0) x? 2 Let k U?? (x0)
Then U(x) k x? 2
U
U
x0
x
x ?
SHM potential !!
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Recap of todays lecture

• Chapter 15 Pendula
• Simple Pendulum
• Physical Pendulum
• Torsional Pendulum
• Next time
• Damped and Driven Oscillations