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Amgular impulse

  1. Jun 3, 2012 #1
    How far above its center should a billiard ball be struck in order it roll without any initial slippage?
    Denote the ball's radius by R, and assume that the impluse delivered byt the cue is purely horizontal

    Book Solution:
    Since any static frictional forced supplied by the table is limited to μMg,the only sure way to avoid slippage is to create a situation in which zero frictional force is needed.

    Fδt=Mvo and h(Fδt)=Icωo

    ωo=vo/R

    Hope somebody can explain the underline sentence.
     
  2. jcsd
  3. Jun 3, 2012 #2
    Hi azizlwl! :smile:

    If you have friction acting, that means the force can cause slipping and this occurs till the ball starts pure rolling. If there is no frictional force acting(at the point where the ball touches the board, ofcourse), then the ball will begin pure rolling from the beginning, which is what you are solving for.

    This is a really interesting article about the physics behind billiard balls : http://www.real-world-physics-problems.com/physics-of-billiards.html

    I read it the first time I played the game, and I was really astounded :biggrin:
     
  4. Jun 3, 2012 #3

    Astronuc

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    The statement "the only sure way to avoid slippage is to create a situation in which zero frictional force is needed" implies that linear velocity at the radius of the ball is equal and opposite to the translational velocity of the center of the ball. If the center of the ball moves at v, then surface must move at -v = -ωr, where ω = angular velocity and r = radius.
     
  5. Jun 3, 2012 #4
    If no frictional force, won't there be slippage as in example below.
    http://img266.imageshack.us/img266/4816/p1010018me.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Jun 3, 2012 #5
    Yes, slippage can occur without friction, too. But in this case, the condition for no slipping is already given. You basically need that the center and the contact point should have equal and opposite velocity for pure rolling. So here, 0 frictional force will cause slipping. This is essentially what Astronuc said.


    If the ball were hit hard enough above or below the maximum 'optimum' height, the friction might be insufficient to prevent rolling. And we do not know anything about the maximum friction. But, if you hit however hard at this exact height that you are trying to find out, the ball will not be affected by frictional force, so it will always have pure rolling. To have this assurance that the ball will always perform pure rolling, friction is taken to be zero in your sum.
     
    Last edited by a moderator: May 6, 2017
  7. Jun 3, 2012 #6

    mfb

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    If you hit the ball with radius r at height h above the center with a momentum transfer p, if will
    - get the transversal momentum p and therefore the velocity v=p/m.
    - get an angular momentum of p*h (relative the the center of mass) and therefore rotate with angular velocity ω=p*h/(2/5mr^2).

    Now, to avoid slipping, you want pure rolling, which requires the angular velocity ω=v/r. Solving these equations for h gives the solution, assuming I did not mess anything up.
     
  8. Jun 3, 2012 #7
    Thanks. Pure rolling(from initial to end of δt). Instantenous ω=v/r. Frictional force=0. Fnet=Fapplied±0
     
    Last edited: Jun 3, 2012
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