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Inertia of a bowling pin

  1. Oct 1, 2004 #1

    I need to know the moment of inertia of a bowling pin. I'd like to know how to calculate it. I would prefer an approximation formula according to the mass of the pin instead of an integration method.

    Thank you
  2. jcsd
  3. Oct 1, 2004 #2


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    The moment of inertia about what axis?
  4. Oct 1, 2004 #3
    Sorry about this lack of precision : around the horizontal axis.
  5. Oct 1, 2004 #4
    Where does the axis intersect with the pin? Any formula you develop here will be an approximation since the bowling pin has such an irregular shape that cannot be dimensionally defined (like, for example, a sphere can). If you know the length of the pin, you can assume it is a uniform horizontal bar, at which point you can use the formula for that, or derive it using calculus.
  6. Oct 4, 2004 #5
    The pin height = 38.1 cm
    The pin mass center height : 14.68m

    I think that it's possible to have a very good approximation of the bowling pin volume by using few spheres. In my application, the pins are represented by 6 spheres.
  7. Oct 7, 2004 #6
    Any idea about an integration method that I could use ?
  8. Oct 8, 2004 #7
    Wouldn't using the formula for a uniform bar be a better approximation? I forget exactly what the formula is, but you can find that online or in a book somewhere.
  9. Oct 14, 2004 #8
    I'm now using the moment of inertia of a cylinder on it's transversal axis :

    I = 1/12 * mh^2 + 1/4 * mr^2

    But since the mass of a pin is less uniformly dispersed than the one of a cylinder, I would have think that it's inertia would be lower.
  10. Oct 14, 2004 #9


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    Why not try three linked cylinders, or more.That is, build a bowling pin with cylinders -- that, in fact, is one way of simulating the z integral -- if z is the axis of symmetry. And, don't forget the other moments.
    Reilly Atkinson
  11. Dec 22, 2005 #10
    please help

    Can anyone explain whole physics of falling of ppin due to bowl - i mean forces on pins ,its rotations etc..
    Thanx in advance
  12. Dec 22, 2005 #11
    That's a complicated system! Here's a few factors I can think of. The pin till tip over when the force of impact on the pin, acting through the center of mass of the pin, displaces the center of mass over the edge of the base that's in contact with the alley. Given the approximate moment of inertia, that's easy enough to calculate. but the impact doesn't just go into that tipping. The pin slides horizontally when it's hit, and that depends on the energy of the impact, which dissipates some of that energy, and the coefficient of friction with the waxed alley. The pin may also rotate about it's vertical axis, also dissipating some of the impact energy and changing it's interaction with the alley from sliding to rolling friction. This will be determined by the direction of the ball's (or originally impacting other pin) travel, if the angle between point of contact and the pin's center of mass is not in line with the impactor's direction. And most importantly, the pin may be impacted by other pins, which might either add to or subtract from the motion imparted to it originally. Given the complex shape of the pin, other pins can hit it anywhere, but you can presume the ball can only contact a vertical pin at a certain height. Of course,if that pin was partially tipped by a prior collission, the ball can contact it in a broader region.

    This all goes to defy a clear real world analysis. Perhaps you'd be better off modeling table tennis? :-)
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