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Homework Help: Harmonic motion

  1. Apr 12, 2005 #1
    A 5 kg ball hangs from a 10 m strong. The ball is swung horizontally outward 90 degrees from its equilibrium position. Assuming the system behaves as a simple pendulum, find the maximum speed of the ball during its swing.


    what would I have to do to figure this problem out?

    [tex]\omega = \sqrt{\frac{g}{l}}[/tex]
    [tex] \omega = \sqrt{\frac{9.8}{10}}[/tex]
    [tex] \omega=0.99rad/s[/tex]

    [tex]\omega r =v[/tex]
    [tex]0.99rad/s* 10m =v[/tex]
    [tex]v=9.9m/s[/tex]

    I'm not getting the answer of 14, what am I doing wrong?
     
  2. jcsd
  3. Apr 12, 2005 #2

    dextercioby

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    HINT:Use the law of conservation of total mechanical energy.

    Daniel.
     
  4. Apr 12, 2005 #3
    yes conservation of energy is always better than mechanics when it comes to fussy math equations. Think of the change in gravitation potential energy.
     
  5. Apr 12, 2005 #4

    dextercioby

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    And BTW,[itex]v=\omega R [/itex] could work in this case if u knew the maximum angular velocity...

    Daniel.
     
  6. Apr 12, 2005 #5
    okay, i used [tex]gh=.5v^2[/tex] and got the answer I was looking for

    As for [tex]v=\omega R[/tex], isn't that what [tex]\omega=\sqrt{\frac{g}{l}}[/tex] is? what is omega in that previous equation if it isnt angular velocity?
     
  7. Apr 12, 2005 #6

    dextercioby

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    Nope,angular velocity is a very complicated function (something involving elliptic functions "cn" and "dn"),because the linear approximation [itex] \sin \vartheta\simeq \vartheta [/itex] would not be correct...

    Daniel.
     
  8. Apr 12, 2005 #7
    I read in the book that omega in [tex]\omega=\sqrt{\frac{g}{l}}[/tex] is angular frequency. How is that different from angular velocity?
     
  9. Apr 12, 2005 #8

    dextercioby

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    Angular velocity is

    [tex] \omega (t)=:\frac{d\vartheta (t)}{dt} [/tex]

    and angular frequency is

    [tex] \omega =:\frac{2\pi}{T} [/tex]

    These 2 #-s (denoted the same :yuck:) are equal only for a uniform circular motion .The bob from a mathematical pendulum (not even in the linear approximation) doesn't have a uniform circular motion,ergo the two "animals" are different.


    Daniel.
     
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