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Centripetal Acceleration/Rotational Motion HW Problem

  1. Dec 5, 2011 #1
    A block sits at the top of a smooth sphere of radius 1m. Suddenly, under the
    force of gravity, It begins to slide down the surface of the sphere until it leaves the
    surface. At what speed does it leave the surface?

    mgcosθ + N = mv^2/R
    But N =0
    v^2 = Rgcosθ
    V^2 = gcosθ
    speed = v = ( 9.8cosθ)^0.5
    Now,
    from work energy theorem,
    0.5mv^2 - 0 = mgR(1-cosθ)
    Rgcosθ = 2gR(1-cosθ)
    cosθ = 2(1-cosθ)
    3cosθ = 2
    cosθ = 2/3
    Now
    SPeed = v = (9.8 x 2/3)^0.5 = 2.556 m/s

    I am unsure if there's supposed to be a 1/2 in front of the mv^2/R on the first line of my work, which would change the final value to 6.32 m/s
     
  2. jcsd
  3. Dec 5, 2011 #2
    sorry i meant 3.62 for the new value lol
     
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