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Angular Momentum - Bowling ball

  1. Oct 11, 2008 #1
    [solved] Angular Momentum - Bowling ball

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

    A bowling ball is thrown down the alley with speed [tex] v_0 [/tex]. Initially, it slides without rolling, but due to friction, it begins to roll. Show that it's speed when it rolls without sliding is [tex] 5/7 v_0 [/tex]


    2. Relevant equations

    [tex] \tau = \vec R \times \vec f[/tex] where f is the force friction.

    [tex] \vec L = \vec R \times M\vec V[/tex]

    [tex] \vec v_f = \vec a t +\vec v_0 [/tex]

    3. The attempt at a solution

    I don't understand what is really going on conceptually when it goes from sliding to rolling.
    I know that when it slides, it's kinetic friction, and when it slides, it's static friction.
    So, there must be something that makes the bowling ball go from kinetic to static friction.

    What that is, is where I go lost.

    So I drew my FBD's. I set the origin along the axis of friction.
    As it slides, there's gravity, the normal and kinetic friction in the usual places.
    As it rolls, it's the same as above, but with static friction force and an arrow showing angular momentum.

    Initially, it had no angular momentum because [tex] \vec L_0 = I_o \omega[/tex] , but [tex]\omega = 0 [/tex] initally ([tex] \omega [/tex]is angular velocity)
    Thus, angular momentum would just be [tex] \vec L = \vec R \times M\vec V[/tex].

    I know there's torque being applied due to friction which would be equal to rf*sin(theta).

    Thereafter, I just wrote a bunch of equations down and I tried to see what would happen if I tried to solve for the force of friction using torque, but that leads no where. I tried going through linear energy and linear momentum, but that leads no where since I don't know under what condition it would make the bowling ball start rolling.
     
    Last edited: Oct 11, 2008
  2. jcsd
  3. Oct 11, 2008 #2

    arildno

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    The trick is to recognize that the ball's angular momentum, calculated with respect to the (moving) contact point remains CONSTANT throughout the motion.
    The reason for that is three-fold:
    1. The velocity of the the moving contact point is always parallell to the velocity of the ball's centre, under the fair assumption that that is, indeed, the centre of mass of the ball.

    2. The friction acts AT the contact point, and hence, would not contribute to the angular momentum as calculated with respect to the contact point.
    3. Gravity, which can be considered acting at the centre of mass, acts in a parallell direction to the distance vector between the contact point and the mass centre, and hence, it generates 0 torque about the contact point.
     
    Last edited: Oct 11, 2008
  4. Oct 11, 2008 #3

    arildno

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    In order to solve the problem:
    1. Calculate initial angular momentum wrt. to the contact point.
    2. Calculate final angular momentum wrt. to the contact point.
    3. Set these results equal to another, and solve for final linear velocity in terms of v0
     
  5. Oct 11, 2008 #4
    Thank you so much.

    I honestly did think about that case.
    But something made me think that it meant that it wasn't moving.
     
  6. Oct 11, 2008 #5

    arildno

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    In general, conservation of POSITION is dealt with within the statics sub-branch of mechanics, rather than within kinematics.
    :smile:
     
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