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Pendulum - Maximum Angle in 2nd half of swing

  1. Oct 18, 2015 #1
    A simple pendulum, 2.0 m in length, is released by a push when the support string is at an angle of 25° from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, to what maximum angle will it move in the second half of its swing?

    Let h1 be the elevation reached when potential energy = total energy transferred.
    mg(h1) = mgL(1 - cos θ0) + 1/2 (mV²)

    I am not sure if the approach is correct and how to proceed. My trig. background is not that strong at all.

    Please help. I believe the answer is 30 degrees.
     
  2. jcsd
  3. Oct 18, 2015 #2

    mfb

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    Staff: Mentor

    The approach is correct. You know everything apart from h1 (and m, but you can divide the equation by m to get rid of it), so you can solve that equation.
    With h1 and L you can find the angle.
    Alternatively, directly use the angle-dependent expression on the left side.
     
  4. Oct 18, 2015 #3
    Thanks for you feedback. I assume h1 = L ( 1 - cos 25 degrees) . Is that right?
    I have troubles to find the "V" . I don't think I should use the initial speed 1.2 m/s.
    Since it is asking for max angle for 2nd half of swing. Beside theta 0, do I need to set up theta 1 , another angle for 2nd half of swing? Confused.
    Need some guidance.
     
  5. Oct 18, 2015 #4

    mfb

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    Staff: Mentor

    Instead of 25 degrees, you have to use the angle you want to calculate there.
    Why not?
     
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