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Angular Momentum of a billiard ball

  1. Nov 7, 2007 #1
    To get a billiard ball to roll without sliding from the start, the cue must hit the ball not at the center (that is, a height above the table equal to the ball's radius, r) but at exactly at a height of 2r/5 above the centre. Prove the result.

    For a ball to roll without slipping, it must have tangential velocity twice that of its velocity of the centre of mass.
    so tangengial speed = ωr

    at the point where the wheel contact the surface must be at rest
    vcm - ωr = 0
    vcm = ωr

    L=Iω = Ip/mr
    p = Lmr/I

    so for conversation of momentum
    mvcmi = mvcmf + Lmr/I
    Δvcm = Lr/I

    I for a sphere = (2/5)mr^2
    where r is where the ball needs to hit

    Δvcm = L/(2/5)mr
    (2/5)mΔvcm/L = r
    (2/5)Δp/L = r
    (2/5)r = r

    these last few steps dont seem right?
    are they? am i even correct up to there?

    thanks
     
  2. jcsd
  3. Nov 7, 2007 #2

    Doc Al

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    Staff: Mentor

    Maybe I'm just braindead tonight, but I'm having a bit of trouble following your reasoning. Think in terms of forces. The cue applies a force at some distance above the center. That force creates an acceleration of the center of mass and an angular acceleration about the center of mass. Set the condition for zero acceleration of the contact point (same as rolling without slipping) and solve for that distance.
     
  4. Nov 7, 2007 #3

    Doc Al

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    Staff: Mentor

    It's perfectly OK to think in terms of an impulse that produces a velocity of the center of mass as well as a angular velocity. Perhaps that's closer to how you are thinking. (But when I see "conservation of momentum", I get confused. You're hitting the ball!)

    And don't use the same letter (r) to stand for the radius of the sphere and the position of the strike.
     
    Last edited: Nov 7, 2007
  5. Nov 7, 2007 #4
    F = ma
    will this force be equal to the torque from the force?
    F = ma = T = Iɑ
    ma = Iɑ

    I for a sphere = (2/5)mr^2

    ma = ɑ(2/5)mr^2
    a = ɑ(2/5)r^2

    rolling without slipping so:
    vcm = ωr
    vcm/t = ɑr
    vcm/rt = ɑ
    and
    a = vcm/t

    a = ɑ(2/5)r^2
    vcm/t = (vcm/rt)(2/5)r^2
    r = 5/2 = 2.5

    is this the answer im supposed to get?

    thanks
     
  6. Nov 7, 2007 #5

    Doc Al

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    Yikes. Don't set torque = force!!!

    You have two equations from Newton's 2nd law:
    [tex]F = m a[/tex]

    [tex]\tau = I \alpha[/tex]

    Let the strike be placed at a distance "d" above the center. Rewrite the torque equation:
    [tex]F d = I \alpha = 2/5 m r^2 \alpha[/tex]

    Combine those two equations (with the condition of rolling without slipping) to solve for d.
     
  7. Nov 7, 2007 #6
    F = ma
    T = Iɑ
    I for a sphere = (2/5)mr^2
    Fd = Iɑ = a(2/5)mr^2

    rolling without slipping so:
    vcm = ωr
    vcm/t = ɑr
    vcm/rt = ɑ
    and
    a = vcm/t

    mad = ɑ(2/5)mr^2
    ad = ɑ(2/5)r^2
    dvcm/t = (vcm/rt)(2/5)r^2
    d = (2/5)r

    there we go!

    thanks
     
  8. Nov 8, 2007 #7

    Doc Al

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    Good. Here's how I would do it:

    Starting with Newton's law for translation:
    [tex]F = m a[/tex]

    And rotation:
    [tex]F d = I \alpha = 2/5 m r^2 \alpha = 2/5 m r^2 (a/r) = 2/5 m r a[/tex]

    Multiply the first by d and set equal to the second:
    [tex]m a d = 2/5 m r a[/tex]
    [tex]d = 2/5r[/tex]
     
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