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Rigid body motion

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data

    A sphere of mass M and radius R it`s given a velocity Vo on the base of a inclined plane (theta being the angle) and friction coefficient mu (you may assume that static and kinetic friction are equal) Find the position of the ball as a function of time

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 1, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi hellsnake! Welcome to PF! :smile:

    (have a theta: θ and a mu: µ :wink:)

    (and btw, is the sphere skidding, or is it rolling without slipping?)

    Use work done and conservation of energy …

    what do you get? :smile:
     
  4. Apr 1, 2010 #3
    It doesn'st tell you if it's slipping or not you have to see that. By the way I forgot you have to study the two cases when µ≤ 2/7 tan θ and µ≥ 2/7 tan θ. Without the equal. And when I do conservation of energy and work done by friction I get this differential equation
    (1/2)M.(Vcm)² + (1/2)Icm.w² + MgH - (1/2)MVo² = -Ffr X(t)
    But Vcm= (d/dt)X(t) and H=X(t)sin θ . Then I get that

    (1/2)M.(d/dt)X(t))² + (1/2)Icm.w² + MgX(t)sin θ - (1/2)MVo² = -Ffr X(t)

    Where:
    M= mass of the sphere
    Vcm= It's the velocity of the center of the mass
    Icm= It's the moment or inertia with an axis that pass through the center oh the sphere
    w= It's the angular velocity
    H= It's how high is the sphere measure from the floor
    Vo= It's the initial velocity
    Ffr= It's the force that the frcition does
    X(t)= It'sthe position of the ball measure from the point where it start his motion. And I choose the X axis parallel to the motion

    So if I read the equation I get that the energy on a arbitrary point of the sphere is equal to
    (1/2)M.(Vcm)² + (1/2)Icm.w² + MgH. Now this minus the energy in the initial moment(the sphere doesn't have rotational energy here) (1/2)MVo² . All of this is equal to the work done by friction -Ffr X(t)

    I think that I have to do another thing because this differential equation It's too hard for this course I think. However if you have the solution for this equation I appreciate it too

    Thanks alot in advance
     
  5. Apr 1, 2010 #4

    tiny-tim

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    Hi hellsnake! :smile:

    (have an omega: ω :wink:)

    You can simplify it slightly by writing Icm in terms of M and R, and writing Ffr = µMg.

    After that, you need a relation between V and ω … try the easier, rolling-without-slipping case first: V = Rω. :wink:

    (and after that, you'll need to find the condition on µ for rolling … presumably it's going to be µ ≥ 2/7 tanθ ! :rolleyes:)
     
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