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Potential energy of a pendulum when the angle is small

  1. Apr 2, 2008 #1
    [SOLVED] Potential energy of a pendulum when the angle is small

    1. The problem statement, all variables and given/known data

    When a simple pendulum makes an angle with the vertical, its speed is v. (a) Calculate the total mechanical energy of the pendulum as a function of v and [tex]\theta[/tex]. (b) Show that when [tex]\theta[/tex] is small, the potential energy can be expressed as [tex]\frac{1}{2}mgL\theta^2=\frac{1}{2}m\omega^2s^2[/tex] (Hint: In part (b), approximate [tex]cos\theta[/tex] by [tex]cos\theta\approx1-\frac{\theta^2}{2}[/tex]

    Note: s is the displacement of the pendulum.


    3. The attempt at a solution

    OK...I solved part (a) with no problem, I was able to obtain the correct answer by adding the kinetic and potential energy. My problem lies with part (b). I am really confused by what exactly I must do, and where the squares of the angles are coming from. I know I somehow have to find a way to connect all of the variables together, but I can't see where the variables in the final equation would come from. I would appreciate just a tiny push in the right direction to coming up with the right answer.
     
  2. jcsd
  3. Apr 2, 2008 #2

    rock.freak667

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    For the while forget about where all the variables come in. Just focus finding an expression for the potential energy. At the lowest point of the oscillation, the pendulum has 0 p.e. Right? Draw the pendulum at a random displacement([itex]\theta[/itex] radians).At that random displacement draw a perpendicular line through the bob to meet the string to form a right angle triangle. (Not sure if I explained that too well) If you understood well, you can find the change in height and hence the potential energy.

    EDIT: here I drew a picture of what I tried to say.

    [​IMG]
     
    Last edited: Apr 2, 2008
  4. Apr 2, 2008 #3
    OK thanks...I figured out how to get the [tex]\frac{1}{2}mgL\theta^2[/tex] by substituting the approximate that was given. But I'm not sure about the other part, the part with the s and the omega squared. I know it has to come from the kinetic energy part, and by substituting omega times s for v. But what could the logic be for setting potential and kinetic energy equal to one another? Is it simply the fact that they are equal but occur at different places?
     
  5. Apr 2, 2008 #4

    rock.freak667

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    Do you remember how to derive the periodic time for a pendulum? When showed it moved with SHM by getting it in the form [itex]a=-\omega^2 x[/itex]? That is the [itex]\omega[/itex] you need to get.

    Remember your circle geometry? Are you able to see that the pendulum moves in the arc of a circle? If the angle is in radians and the radius the length of the string...How do you relate the two to the length of the arc?
     
  6. Apr 2, 2008 #5
    The length of an arc would be equal to [tex]L\theta[/tex] when [tex]\theta[/tex] is in radians, right?

    L would be the radius in this case.
     
    Last edited: Apr 2, 2008
  7. Apr 2, 2008 #6

    rock.freak667

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    yep the displacement [itex]s=L\theta[/itex]...Now substitute and it should work out.



    EDIT:

    Remember that at the lowest point, the p.e. is zero while k.e. is max. And at the max displacement, the opposite is true. So the p.e. and k.e. is interchanged as the motion goes from 0 displacement to max displacement.
     
    Last edited: Apr 2, 2008
  8. Apr 2, 2008 #7
    Okay...but I'm still unsure as to where I get the [tex]
    \frac{1}{2}m\omega^2s^2
    [/tex] from. Does it come from kinetic energy?

    [tex]\frac{1}{2}mv^2[/tex]

    [tex]v=s\omega[/tex], so substituting this I get

    [tex]\frac{1}{2}m(s\omega)^2[/tex]

    Again, that's the only way I can see to get the second part of that equation...but you mentioned acceleration using omega squared. I don't see where to make the subsitution you mentioned regarding the arc length.

    Thanks for the help by the way.
     
    Last edited: Apr 2, 2008
  9. Apr 2, 2008 #8

    rock.freak667

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    [itex]s=L\theta \Rightarrow \theta=\frac{x}{L}[/itex]

    [tex]

    E_p=\frac{1}{2}mgL\theta^2 \Rightarrow \frac{1}{2}mgL(\frac{x}{L})^2[/tex]


    Period of a pendulum is

    [tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]

    and

    [tex]\omega = \frac{2\pi}{T}[/tex]
     
    Last edited: Apr 2, 2008
  10. Apr 2, 2008 #9
    But what does x represent, is that the same as s? If so, I just simplified this to get

    [tex]\frac{1}{2}mg\frac{s^2}{L}[/tex]. I still have a g and I need to get [tex]\omega[/tex] as well.

    Sorry I'm a little slow on this.
     
  11. Apr 2, 2008 #10
    And what do I need the period for?
     
  12. Apr 2, 2008 #11

    rock.freak667

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    Yep x is the same as s.

    That is the relation between periodic time and omega...so you can just sub T into the equation with omega in it and get what omega is.
     
  13. Apr 2, 2008 #12
    Well in that case, wouldn't omega be [tex]\sqrt\frac{g}{L}[/tex]?

    I don't see how that could fit in to what I need at all.
     
  14. Apr 2, 2008 #13
    [tex]
    E_p=\frac{1}{2}mgL\theta^2 = \frac{1}{2}mg \frac{1}{L} L^2 \theta^2 = \frac{1}{2}m \frac{g}{L} s^2 = \frac{1}{2} m \omega^2 s^2

    [/tex]

    Because

    [tex]
    \omega = \sqrt{\frac{g}{L}}
    [/tex]
     
  15. Apr 2, 2008 #14
    That makes it all come together perfectly for me!! Thank you two so much for your kind help!! :)
     
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