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Rotational kinematics using energy

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    A 45.0-cm diameter wheel, consisting of a rim and six spokes, is constructed from a thin rigid plastic material having a linear mass density of 25.0 g/cm. This wheel is released from rest at the top of a hill 52.0m high.


    a.) How fast is it rolling when it reaches the bottom of the hill?
    b.) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?


    2. Relevant equations

    MEi = MEf
    [itex]\omega[/itex] = V / R


    3. The attempt at a solution

    mghi = .5m*v2 + .5I*(v2 / R2)
    v2 = (2*mghi) / (m + (I/R2))

    I = mR^2 + 6/3 *mR^2
    = 3mR^2

    v^2 = 2mgh / (m + (3mR^2 / R^2))
    = .5 * ghi



    I get v = 15.96
    but that is wrong?
     
  2. jcsd
  3. Sep 18, 2011 #2
    Nevermind. For those of you who are having trouble, figure I as a numerical number.

    So I = m*R^2 + 6/3 * m*R^2

    Note that the R is the radius of the wheel, hence, R is the same for both parts of the equation. However, m in the first part is the mass of the WHEEL (so you'll take the circumference of the rim and multiply it by the linear density) and m in the second part is the mass of each stoke (so you'll take one stoke's length and multiply it by the linear density).

    You should get I=.235863

    Then you have the equation

    v^2 = (2*m*g*h) / (m + (I/R^2))

    in both parts, the m is the mass of the total system (the 6 stokes and the rim, it should be like 6.909kg). The h is the initial height (52m) and the R is the radius of the rim. Take the square root and you'll get v.

    For part 2, it won't change.
     
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