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Rolling Forces

  1. Apr 21, 2008 #1
    [SOLVED] Rolling Forces

    1. The problem statement, all variables and given/known data


    A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s and initial angular speed 0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.19. The kinetic frictional force fk acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed vcom has decreased enough and angular speed has increased enough, the ball stops sliding and then rolls smoothly.
    (a) What then is vcm in terms of ?
    ___________m·[tex]\omega[/tex]

    (b) During the sliding, what is the ball's linear acceleration?
    ___________m/s2

    (c) During the sliding, what is the ball's angular acceleration?
    ___________rad/s2

    (d) How long does the ball slide?
    ___________s

    (e) How far does the ball slide?
    ___________m

    (f) What is the speed of the ball when smooth rolling begins?
    ___________m/s



    3. The attempt at a solution

    I am utterly lost in this problem. I assume that since the ball is sliding the force of kinetic friction is applying a torque onto the ball so:

    T = FkR
    T = Fk(.11m)

    and Fk = uN so

    T = (.19)(9.8)(.11)m
    so
    T = .2048 * m

    now, I'm lost. Any pointers in the right direction?
     
    Last edited: Apr 21, 2008
  2. jcsd
  3. Apr 21, 2008 #2
    So far you're on the right path. The key is realizing when the ball stops slipping, and starts rolling. This is going to happen when the angular velocity can be directly related to the com velocity. Imagine a ball making one complete revolution, use this to relate com velocity to angular velocity when something isn't slipping. Use this velocity to find out how long it takes to decelerate to that point.
    Does that help?
     
  4. Apr 21, 2008 #3
    Ok so I got that angular acceleration during sliding is 42.3 rad/s by the fact that, T = I[tex]\alpha[/tex] and also equal to the equation above.

    I realize that when the ball stops slipping that the Acceleration of the Center of mass = [tex]\alpha[/tex] * R and that all those equations start to apply where Velocity of the Center of mass = [tex]\omega[/tex] * R . But how do I figure out the linear acceleration during the sliding?
     
  5. Apr 21, 2008 #4
    Stupid Question....sorry Wasn't thinking. Ok so I have the Linear Acceleration. But how do I figure out when the ball actually starts to roll and stop sliding?
     
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