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Pendulum problem - there's got to be an easier way

  1. May 17, 2005 #1

    tony873004

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    Pendulum problem -- there's got to be an easier way!!

    What is the amplitude of a simple pendulum of length L = 0.60m if the bob has a speed v = 0.37 m/s when passing through the equilibrium position?

    I don't see why this is any different than figuring out the height a roller coaster will rise to, so... Work Energy theorum to get the max height of the pendulum. Then translate it into degrees.

    [tex]
    \Delta K=-\Delta U
    \]
    \[
    \Delta K=\frac{1}{2}mv^2,
    \quad
    \Delta U=mg(y_f -y_i )
    \]
    \[
    -mg(y_f -y_i )=\frac{1}{2}mv^2,
    \quad
    y_i =0
    \]
    \[
    -mg(y_f )=\frac{1}{2}mv^2
    \]
    \[
    (y_f )=\frac{\frac{1}{2}\rlap{--} {m}v^2}{-\rlap{--} {m}g}
    \quad
    \Rightarrow
    \quad
    y_f =\frac{\frac{1}{2}v^2}{-g}
    \]
    \[
    h=y_f ,
    \quad
    h=\ell (1-\cos \theta )
    \]
    \[
    \frac{\frac{1}{2}v^2}{-g}=\ell (1-\cos \theta )
    \quad
    \Rightarrow
    \quad
    \frac{\frac{1}{2}v^2}{-g}=\ell -\ell \cos \theta
    \quad
    \Rightarrow
    \quad
    \frac{\frac{1}{2}v^2}{-g}-\ell =-\ell \cos \theta
    \quad
    \Rightarrow
    \]
    \[
    \frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta
    \quad
    \Rightarrow
    \quad
    \theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell }
    \right)}{-\ell }} \right)
    \]
    \[
    \theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}\left( {0.37m/s}
    \right)^2}{-9.8m/s^2}-0.6m} \right)}{-0.6m}} \right)
    \quad
    \Rightarrow
    \quad
    \theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}0.37^2\rlap{--}
    {m}^{\rlap{--} {2}}/\rlap{--} {s}^{\rlap{--} {2}}}{-9.8\rlap{--}
    {m}/\rlap{--} {s}^{\rlap{--} {2}}}-0.6\rlap{--} {m}} \right)}{-0.6\rlap{--}
    {m}}} \right)
    [/tex]

    [tex]

    \theta =8.75 deg
    [/tex]

    Same answer as the back of the book. But there's got to be an easier way. The book gives no examples of this type of problem.
     
  2. jcsd
  3. May 17, 2005 #2
    Good job on solving the problem, but this problem isn't difficult, and the algebra is even pretty clean. You've made it a bit uglier than it needs be by not simplifying your expressions completely. Notice that your final expression can be written

    [tex]\cos{\theta} = 1 - \frac{v^2}{2gl}[/tex]

    Be happy that the problem is this straight-forward and get used to algebra. It only gets worse.

    --J
     
  4. May 18, 2005 #3

    tony873004

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    Thanks for the encouraging words :surprised

    if the -L didn't exist in the numerator, then
    I could see shoving the 1/2 downstairs as a 2 and
    I could see shoving the /-g downstairs as simply g, and neutralizing its minus sign with L's minus sign.

    [tex]\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta [/tex]

    giving me a modified formula of:

    [tex]\cos{\theta} = \frac{v^2-l}{2gl}[/tex]

    But your forumla
    [tex]\cos{\theta} = 1 - \frac{v^2}{2gl}[/tex]

    got rid of the -L in the numerator and introduced a "1-". Plugging in numbers, your forumla works, my original formula works, but my modified formula doesn't.

    Also, I would have thought there would be some pendulum formula to take care of this whole thing in 2 lines. Is this the right way to do this problem??
     
  5. May 18, 2005 #4
    Well, that's interesting. It looks like we both messed up the algebra! According to what you've written, the formula should simplify to

    [tex]\cos{\theta} = 1 + \frac{v^2}{2gl}[/tex]

    [tex]\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }= \frac{\frac{\frac{1}{2}v^2}{-g}}{-l}-\frac{l}{-l} = \frac{v^2}{2gl} + 1[/tex]

    Your -g should have actually been a +g. You got lucky with your original formula because cosine is even. A happy accident where you screw up, but your profs don't mark you off. :biggrin:

    --J

    Geh... LaTeX didn't work out so well. Too many nested \frac{}{}s. Gimme a sec.

    All better now.
     
    Last edited: May 18, 2005
  6. May 18, 2005 #5
    How did you modify the formula without taking the L.C.M.?
    [tex]\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta [/tex]
    It should be
    [tex]\cos{\theta} =\frac{v^2 + 2gl}{2gl}[/tex]
    which becomes
    [tex]\cos{\theta} =\frac{v^2}{2gl}+ 1[/tex]
     
  7. May 18, 2005 #6
    So we're agreed.

    Oh, and I just distributed the denominator to each term in the numerator right away.

    --J
     
    Last edited: May 18, 2005
  8. May 18, 2005 #7

    James R

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    Homework Helper
    Gold Member

    Here's an alternative approach, though it probably doesn't involve any less work. The motion of a simple pendulum is described by:

    [tex]\theta(t) = \theta_0 \sin (\sqrt{\frac{l}{g}}t)[/tex]

    where [itex]\theta_0[/itex] is the (angular) amplitude.

    The angular speed at any time is given by the time derivative of this equation, and from that you can work out an expression for the linear speed at any time. Use that formula to calculate the amplitude in your problem.
     
  9. May 18, 2005 #8
    g/l, not l/g. The argument of cosine must be unitless.

    And that equation assumes that [itex]\theta[/itex] is sufficiently small that [itex]\sin{\theta} \approx \theta[/itex], which isn't necessary for this problem. The original solution was better.

    --J

    Edit: Been too long since I last used TeX...
     
    Last edited: May 18, 2005
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