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Rotational Energy, Rolling downhill

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A solid steel sphere of radius 10 cm and a mass of 1.5 kg rolls down a 1.25 m incline that makes an angle of 20 degrees with the horizontal. Calculate the linear velocity of the following points relative to the ground, when it reaches the bottom of the incline.
    a) The center of the sphere
    b) A point at the top edge of the sphere
    c) A point at the bottom of the sphere


    2. Relevant equations

    mgh = .5Iw2+.5mv2

    I = 2/5(mr^2)

    3. The attempt at a solution

    Ok so using the equation above, energy, I substituted in the values, using wr = v1 for tangential velocities, putting W = v1/r for the angular velocity, But the velocity (v2) of the .5 mv2, is different than the tangential, as the velocity of the kinetic energy (normal) is V2, V2 = .5*V1, which is derived from the fact that a point on a sphere goes twice the distance compared to the center of mass, the point on the spheres velocity is V1, v2 is the speed of the Center of mass.

    Substituting we get

    mgh = .5(.4*m*r^2)(.5V1/r)2 + .5*m*V12

    which reduces down into

    gh = .1(v1^2/4) + .5v1^2, and I get 2.70 m/s for the linear speed of the center of mass... is this right?

    Also would the answers for b and c be the same?
     
  2. jcsd
  3. Jan 9, 2012 #2

    Delphi51

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