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Very Simple Work/Energy Pendulum Problem

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A 2.0kg bob of a pendulum is held at an angle of 55 degrees from the vertical. When released it reaches a maximum speed of 3.5m/s. What is the length of the pendulum arm?


    2. Relevant equations
    Ek = 1/2mv^2
    Ep = mgh

    3. The attempt at a solution

    1/2 * 2 * 3.5 ^ 2 = 12.25 J
    12.25 = 2(9.8)h
    h = 0.625

    That's as far as I can get. I know that the length of the pendulum arm is 0.625 plus "length - 0.625" but I'm stuck there. Help would be much appreciated :)

    Thanks in advance!
     
  2. jcsd
  3. Jan 15, 2012 #2

    Simon Bridge

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    Draw the picture - h is the vertical distance between the start position and the bottom of the arc - and you know the angle of the arc is 55 degrees.

    Construct the obvious triangle and use trig.
     
  4. Jan 15, 2012 #3
    Could you please specify what the triangle would look like and which trig functions to use? I know that h is the vertical distance between the start position and the bottom of the arc and that I should use trig on a constructed triangle but I'm a little brain dead as of now and totally stonewalled.
     
  5. Jan 15, 2012 #4

    Simon Bridge

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    Did you draw the picture?
    There's only like, two possible triangles.

    Hint: you want the one where the third side is horizontal.
    Draw it and stare at it - keep staring at it: what is the length of the hypotenuse? What is the length of the vertical side? How are these related?
     
  6. Jan 15, 2012 #5
    I did draw the picture; there's a big triangle of which the three sides are the vertical, the string, and an imaginary horizontal line, and a small one of which the three sides are h, the imaginary horizontal, and a hypotenuse roughly where the pendulum bob would swing. I've been staring at it for a good half hour before making a topic... maybe my algebra or trig is rusty? Sorry Simon, I'm not seeing anything.
     
  7. Jan 15, 2012 #6

    Simon Bridge

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    You will kick yourself...
    attachment.php?attachmentid=42718&stc=1&d=1326662485.png
    Write the expression for the cosine of 55 using this triangle, and solve for r.
     

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  8. Jan 15, 2012 #7
    Thanks for the diagram! That's what I had down, and I tried to use sine law but I can only get to sin35 = (r - 0.625) / r

    I hate having people spell things out for me but I am just not getting this :(
     
  9. Jan 15, 2012 #8

    Simon Bridge

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    That should do it! You just used the complimentary angle.
    I was suggesting you do:[tex]\cos(55)=\frac{r-h}{r}[/tex](remember SOH CAH TOA?)
    But yours is exactly the same.

    Now solve for r.

    You can do algebra right?
    This is an equation of the form: ax = x - b
     
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