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Rolling forces

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Ball starts to slide with initial speed v0 = 6.5 m/s on horizontal surface. After what time will ball start to roll? Kinetic friction between the ball and the surface is 0.3.
    v0 = 6.5 m/s
    [tex]\mu[/tex] = 0.3
    I = 1/2 mr^2
    2. Relevant equations
    I have solved this problem but I'm not sure if is correct, so please check....

    3. The attempt at a solution

    Fk = N [tex]\mu[/tex]
    ma = mg[tex]\mu[/tex]
    a = g[tex]\mu[/tex]

    This is the acceleration which is opposite to direction of the ball so it is slowing it down.
    So velocity when it starts to roll must be equal to the "rolling velocity" [tex]\omega[/tex] r (what is correct name in English?)

    v(t) = v0 - at = [tex]\omega[/tex] r
    [tex]\omega[/tex] = [tex]\frac{v0 - g*mu*t}{r}[/tex]

    Ek = Fk*s + (1/2) I[tex]\omega[/tex]^2
    When I loose the masses and substitute [tex]\omega[/tex]:

    1/2 v0^2 = g*mu*(v/t) + (1/5) * (v0 - g*mu*t)

    And that is one equation with one unknown (t).
    But equation is pretty big and I suppose I got something wrong...
  2. jcsd
  3. Jun 22, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Careful: What's the rotational inertia of a ball?

    This is good for the translational acceleration. What about rotation?

    That's the condition for "rolling without slipping".

    Not exactly. The translational speed decreases while the rotational speed increases. Write expressions for both speeds as a function of time and solve for when they meet the condition [itex]v = \omega r[/itex].
  4. Jun 22, 2008 #3
    There's several things wrong here. You solved for the correct equation for velocity as a function of time, but you don't set that equal to omega*r. You need to find the equation for omega as a function of time.

    Once you have velocity and omega as functions of time, you can use the no slip condition to isolate t. (the rolling velocity is called the angular velocity in english, represented by omega)

    Edit: Totally stole my thunder, doc.
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