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Vertical Pendulum Questions

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data

    A conical pendulum has length (L) 1.5m and rotates at 4ms-1. What is it's angle (θ) to the vertical?


    2. Relevant equations

    r = Lsinθ

    tanθ = v2/Lsinθg

    tanθsinθ = v2/Lg

    sinθsinθ/cosθ = v2/Lg

    sin2θ/cosθ = v2/Lg

    sin2θ + cos2θ = 1


    3. The attempt at a solution

    I've followed the list of equations given:

    L = 1.5m
    v: 4ms
    g: 10ms2 (In our workbook, we usually round it up to 10)

    r = 1.5sinθ

    tanθ = 42/1.5sinθg

    tanθsinθ = 42/1.5 x 10

    I get stuck here and I'm not sure how to continue. I get the impression that I've approached this the wrong way entirely :P

    Any help? It would be most appreciated.

    Thanks
     
  2. jcsd
  3. Jan 29, 2014 #2
    There's nothing wrong with your approach. Why don't you go ahead with it?
     
  4. Jan 29, 2014 #3
    Well, when I get to sin2θ + cos2 θ = 1, I assume you can cancel out the square because the square root of one is one. But if it's sinθ + cosθ = 1, I'm not really sure how to go about that. cos-1 x 1 = 0, and sin-1 x 1 = 90, so θ = 90? Even then, I'm not really sure how to write that.
     
  5. Jan 29, 2014 #4

    BvU

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    tanθ=sin/cos, so you get sin^2/cos. Write sin^2 = 1-cos^2 and you end up with a quadratic equation in cos.
     
  6. Feb 8, 2014 #5
    I'm still stuck on this question. Is anyone willing to help me? I'm still not sure how to go from sin^2 θ / cos θ = 106.7 to sin^2θ + cos^2 θ = 1.
     
  7. Feb 8, 2014 #6

    BvU

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    Screeech. That's me hitting the brakes. I suppose our posts 3 and 4 crossed, because otherwise I would have been sleepless since!

    You learned about Pythagoras with his simplest triangle ? Like 3^2 + 4^2 = 5^2 ? And you still post something like that, amounting to "therefore 3+4=5" ?

    Or, conversely: 3+4 = 7 so 3^2+4^2 = 49 ? I hope not!

    Did you read my post #4? If you did, for a while already you'd have in you notebook the relationship

    ##{1-\cos^2\theta\over\cos\theta} = {v^2\over Lg}##

    So you would definitely not have 106.7 on the right hand side, because that is ##{v^2 g\over L}##. Easily mistaken if typing things like tanθ = 42 /1.5sinθg without brackets!

    Equation solving skills make you see a simple quadratic equation in ##\cos\theta## here. If you don't see it, rewrite using ##x## instead of ##\cos\theta##.

    If you don't see it yet, multiply with ##x## on the left and on the right. Later check that ##x\ne 0##
     
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