Mechanics prob: energy conservation with a swing pls help... ?

My answer for the lowest point works. 

But I'm having trouble doing the highest point one..? 

Explanation with answer pls :/

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  • 3 weeks ago
    Best answer

    Is this a trick question?  The ball *starts* at the highest point in its swing, and the speed is zero there!

    The potential energy at a distance h below the rest position, relative to that starting position, is -mgh, right?  So conservation of energy and the work-energy theorem say that all of that P.E. lost goes to kinetic energy:

    mgh = (1/2)mv²

    v = √(2gh)

    At the bottom of both circles, h = L = 1.080 m, so the speed is:

    v = √(2 * 9.81 * 1.080) ~~ 4.60 m/s

    At the top of the small circle, h = d - r = d - (L - d) = 2d - L = (2*0.756 + 1.080) m

    h = 0.432 m

    At that position, the speed is:

    v = √(2 * 9.81 * 0.432) ~~ 2.91 m/s

  • 3 weeks ago

    108 - 75.6 = 32.4 = r (from image)

    108 - 2×32.4 = 43.2 cm (distance below 

    original starting point)

    From energy conservation:

    m/2×v^2 = mgh

    v = (2gh)^.5 = (19.6×.432)^.5 = 2.91m/s

    Check: (19.6×1.08)^.5 = 4.6 m/s

    Update if you need more, and please choose 

    a best answer from among those you get.

    • Zabedul3 weeks agoReport

      Ohh I see, but then how is it that at the top of of the small circle we're equaling Mechanical Energy = PE? Aren't  we supposed to equal Mechanical Energy = PE + KE since there's kinetic energy at the top as well? 

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