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Troublesome Question

  1. Jan 2, 2004 #1
    Hey everybody, this here was one of the questions Cambridge university gave to students applying there these past few years, i'm having some trouble figuring out how it works, can anybody help me out? Cheers:)

    Consider two balls, one small and one large, the smaller ball sitting on top of the larger. Assuming the small ball begins to slide down the side of the large ball, at what point will the two balls cease to be in contact with one another?

    Christian K.

    Attached Files:

  2. jcsd
  3. Jan 2, 2004 #2
    Depending on the material and the size, the typical answer will be at the equator of the largest ball.
    Things that would change this would be, whether the balls were sticky enough in material that the smaller ball held to the larger for a little longer, or if they were both slippery enough (and large enough) that the smaller ball was flung off the larger, instead of rolling to the equator and then straight down.
  4. Jan 2, 2004 #3

    Doc Al

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    Staff: Mentor

    Sounds like homework to me. :smile:
    Here's a hint. (Assume the larger ball is fixed in place.) As the small ball slides down, it speeds up. At some point the available centripetal force (What provides that force?) will not be enough to keep it pressed against the surface: find that point.
  5. Jan 2, 2004 #4
    Homework? hehe, not really. So the centripetal force varies according to the forumla mv²/r, and i need to try to find the point where the vertical component of the rolling ball becomes greater than (mv²/r)?

    Christian K.
  6. Jan 2, 2004 #5
    2 cents


    There is a Normal force acting between the two balls. When the top ball is sitting directly on top of the lower ball thye Normal force is equal to the weight of that ball in whatever gravitational field is present. As the top ball slides off the one under it, the Normal force will decrease until it is zero. At this point is when the ball leaves the sphere, which is not at the equator.

    Here's a brief synopsis for you....hope it helps:

    There is a

    tengential component: m g sinq = m at = m (q '' r)............(1)

    and a

    Normal component: mg cosq - N= m an= m ((q ')2 r).............(2)

    Since we know the Normal force (N) is zero when the top ball leaves the lower one, we have:

    g sinq = q '' r............(3)


    g cosq = (q ')2 r..........(4)

    The masses (m) have been cancelled from each side of each eq'n.

    If we now play a game with derivatives to express the angular velocity as a function of angle rather than time we have:

    q ''= dq '/d t= (dq '/dq) (dq /dt) = (dq '/dq) q '

    Integrate this eq'n to get: g (1-cosq) = (1/2) r (q ')2 ........(5)

    Eliminate (q ')2 from eq'ns 4 and 5 and you get:

    g (1-cosq) = (1/2) g cosq

    Solve this for the angle cosq and you have:

    cosq = 2/3........(6)

    Solve for q to find the required angle as: q = 48.2 deg's.

  7. Jan 2, 2004 #6

    Doc Al

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    Staff: Mentor

    The required centripetal force is given by mv²/r; the only force available is gravity. What's the centripetal component of the weight?
  8. Jan 2, 2004 #7

    Doc Al

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    Staff: Mentor

    Re: 2 cents

    Yikes!:smile: While perfectly correct, no need for the heavy artillery.
    You need to find the point where the radial component of the weight just balances the required centripetal force:
    Put this in terms of the change in Potential energy:
    Solving for θ:
  9. Jan 2, 2004 #8

    Nicely said and done!

  10. Jan 2, 2004 #9
    oh, i understand what is going on, but how do you get to 2mgR(1-cos(theta))?
  11. Jan 2, 2004 #10

    Doc Al

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    Staff: Mentor

    R(1-cosθ) is the height the ball drops when it moves an angle θ along the bottom sphere. (This ignores the size of the smaller ball; instead of R, one should use R + r... but it drops out anyway.)
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