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Sliding to Rolling motion problem.

  1. Dec 28, 2006 #1
    1. The problem statement, all variables and given/known data
    A bowling ball is released without spin. How far does it travel before its motion is pure, rolling motion? No data is given, it asks for an order of magnitude estimate.

    2. Relevant equations
    k = mv^2/2
    k = Iw^2/2 + mR^2w^2/2
    I = 2MR^2/5
    torque = Ialpha
    torque = fk(R)
    Work = fk(d)
    V = Rw (w = omega, condition for rolling motion)

    3. The attempt at a solution
    The ball may slide a bit before it starts to roll at all. It will then roll and slide, then, eventually, only roll. There will be energy lost to friction. At the point the ball begins to roll (which is where V = Rw)

    Krolling - Ksliding = -fk(d) or

    Iw^2/2 + MR^2w^2/2 - Mv^2/2 = -fk(d)

    Kinetic friction exerts a torque on the ball:

    Ia = fk(R), and since Icm for a sphere is 2MR^2/5 and w = at

    fk = 2MRt/5, substitute this in the above and simplify:

    2Rw^2/5 + Rw^2/2 - V^2/2 = -2dt/5

    and I think I just took a wrong turn some where :uhh:.

    A gentle (or not so gentle) prod in the right direction would be very welcome right now.

    Thanks so much,
  2. jcsd
  3. Dec 28, 2006 #2

    Doc Al

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    Staff: Mentor

    Try thinking of it this way: The ball starts with pure translational motion. The friction slows the translational motion, and starts the ball rotating. So write equations for the translational motion and for the rotational motion, due to the friction. The ball will slow down (but increase rotational speed) until the speed is just right to stop slipping.

    You'll find that the time it takes to begin rolling without slipping depends on how fast it was going when released. You'll just have to take a guess at what a reasonable speed for that would be.
  4. Dec 29, 2006 #3
    I'm still not sure I have this right. When the ball is finally rolling along, I should have

    Krolling = Iw^2/2 + mR^2w^2/2, and using I = 2MR^2/5 and w = v/r, I end up with

    Krolling = 7Mv^2/10

    I can get the acceleration from the torque:

    sumof t = Ialpha = umgR

    which simplifies to

    a = 5ug/2

    and I would like to use v^2 = v0^2 -2as to get s, that would be nice.

    v0 I can just estimate, but how do I get v (the speed of the rolling CM)?

    Last edited: Dec 29, 2006
  5. Dec 29, 2006 #4
    Urgh.. Defeat... I just can't figure out how to get v. I can get a bunch of equations, but I don't think I know enough math to solve them, or even know if they can be solved :frown:

    The best I can do is this:

    using a = 5ug/2, and assuming we know the average velocity, then the time is t = v_avg/a, that should be close to the time it takes to convert to rolling friction, right?

    So the distance is x = v_avg(t) = v_avg^2/a = v^2(2/5ug).

    I found a figure for average velocity of v = 4.60 m/s from some bowling website, and using 0.100 (waxed wood on snow, from my physics book. I used this because I read that they oil the lanes, and this is the closest I could find)

    x = 4.6^2(2/(5(0.100)(9.8))) = 9.4 m.

    So, I would say order of magnitude of 10^1 m.

    I have to say this all seems like a cheat to me, and since a bowling lane is about 18.3 m, this really seems high.

    Is this even on the right track?

  6. Dec 29, 2006 #5

    Doc Al

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    Staff: Mentor

    This is true, but you won't need it.
    OK. But realize that this is the acceleration of the ball's surface with respect to its center. What's the translational acceleration of the ball's center of mass? (Hint: Apply Newton's 2nd law twice--for rotation and for translation.)

    As I said earlier, two things are going on: The rotation is speeding up; the translation is slowing down. At some point, the ball's surface speed with respect to the center will just equal the ball's translational speed. That's the point where the motion becomes rolling without slipping. Solve for that point by setting up the equations for rotational and translational speed.
  7. Dec 29, 2006 #6
    But I had to multiply alpha (angular acceleration) by R (radius of ball) to get the linear acceleration (a) in that equation. Doesn't that give me the translational acceleration of the center of mass?

    Maybe I don't understand what you mean by surface speed. I will try this some more, now that I have had some sleep.

    Thanks, Doc Al. You are really great.

  8. Dec 29, 2006 #7
    It's amazing how dense I can be. Ok, I think I have it, thank you. But I am not that great at estimates and order of magnitude stuff. My estimates are bad, in fact :eek:

    Here's what I did:

    The sliding translational motion, that would have to be:

    f = -ma

    and the rotational motion should be:

    fR = I(alpha)

    where f is force of kinetic friction.

    If T is the time it takes to 'get rolling', so to speak, then

    fRT = Iw (w is final angular velocity)

    The condition for rolling motion is w = vf/R, so

    fT = Ivf/R^2

    Returning to f = - ma, since a = (vf-v0)/t

    fT = -m(vf-v0)

    where vf, v0 are final and initial velocities of the center of mass.

    Using I = 2mR^2/5 as before, I get this really cool relation:

    vf = (5/7) v0

    To get the distance, I have an average v of 4.6, m/s, so say it starts out at 5.0 m/s. It seems to me that bowling balls don't slow down that much, and here I think I can use energy:

    -fd = mvf^2/2 - mv0^2/2

    and using a coefficient of friction of 0.100 and v0 = 5.0 m/s, I get about 6 m for the distance to complete rolling, or about 1/3 the distance of the typical bowling lane.

    Well, this is the same order of magnitude as before, 10^1, but I feel much better.

    I hope this is right (??).

    Thanks again,
    Last edited: Dec 29, 2006
  9. Dec 29, 2006 #8

    Doc Al

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    Staff: Mentor


    Just to make it clear--probably not necessary, since you've obviously got it--here's another way to view the condition for rolling without slipping: The instantaneous speed of the point of contact of ball with respect to the floor must be zero. (That's what "not slipping" means, of course.) When you caculate alpha (using Newton's 2nd law for rotation) you are finding the angular speed of the ball about its center; when you mulitply that by R, you are find the linear speed of the ball's surface about its center. In order for the ball not to slip, the speed of the surface with respect to the center must exactly cancel the speed of the center with respect to the floor. Make sense?
  10. Dec 29, 2006 #9
    :blushing: Oh Wow... thank you, you made my year.

    Yes, that makes sense. I think you are also hinting that I could have found an easier way, using this relationship. I will look for that.

    This rolling stuff was very confusing, this problem (and you) helped a lot!

    Thank you, thank you, thank you, as always.
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