A cute problem

  1. Here's a cute problem I came across recently.

    Suppose you have a rubber band with spring constant k, mass m and unstretched radius r. Now suppose you have a frictionless cone and the angle of the peak is [itex]2 \theta [/itex] (that is, if you project the shape of the cone onto a plane it looks like a triangle and the top angle is [itex] 2 \theta [/itex]. If you were to gently slide the rubber band down the cone (so it doesn't have any appreciable momentum, but you're not forcing it either) it will come to rest at some point on the cone where it will be at equilibrium (we're assuming that the cone is big enough so the rubber band doesn't go all the way to the bottom). What is the radius of the rubber band at this point of equilibrium?
  2. jcsd
  3. I would expect that it will be very close to r. My reasoning is that the mass of the rubber band is almost negligible and it is the gravitational force F=mg that must be equal to the force of the spring. I've been taught in general that F_spring=kx, but this equation will not hold here simply for geometric reasons.
  4. Well I haven't told you anything specific about the rubberband. Perhaps this is my very dense rubberband with a very small k. The idea is that you can get a closed form solution dependent only on the given parameters.
  5. Meir Achuz

    Meir Achuz 2,072
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    Draw a free body diagram of one half of the rubber band and...
  6. Fairly easy problem. Is there any friction?
  7. LURCH

    LURCH 2,507
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    The OP states that there is no friction, can we also asume no gravity? If there's no gravity, and "you're not forcing it either", then the rubber band comes to rest at r, yes?
  8. Answer

    Of course Lurch is right for the trivial case of no gravity, but assuming there is gravity I got the answer to be:

    R = r + [gm/(4k)]*Cot(theta)

    I am pretty sure this is right, but did someone else get something different.
    You have to be careful applying Hooke's Law. Circular symmetry causes force components in the rubberband to cancel, which introduces a sine in the integral equation:

    dF = 2k(r-R)Sin(theta)*Int[Sin(psi/2)*dpsi,0,2pi]*r-hat.

    this just has to be set equal to the gravity component:

    dF = gCos(theta)*Int[dm,0,M]

    Who's up for solving the oscillatory analog for appreciable momentum?
    Append the question to read:
    A rubberband is dropped from the tip of a cone with radius R1 and frictional coefficient u...............et. al.
    find the location (y), speed (y'), and acceleration (y'') for a given time t.
    I doubt there will be an accessible closed form solution but setting up the DE should be challenging enough (unless you use a Langrangian).
  9. Meir Achuz

    Meir Achuz 2,072
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    I get [itex]r-r_0=\frac{mg}{k\pi^2\tan\theta}[/itex], using the FBD
    for half the rubber band.
    Last edited: Oct 2, 2006
  10. [itex]r-r_0=\frac{mg}{2k\pi^2\sin\theta}[/itex]

    using potential energy change due to gravity = spring energy change
    Last edited: Jan 16, 2007
  11. Gokul43201

    Gokul43201 11,046
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    "Change" from the unstretched position? Since the unstretched position is not the equilibrium position (where the PE is at a local minimum), neither will this new position be - they are merely equipotential positions.

    Since the PE is parabolic, if this calculation is correct, the equilibrium position would lie midway between these 2 positions.

    PS: But I get [itex]~r-r_0=\frac{mg}{4\pi^2k\tan\theta}[/itex]
    Last edited: Jan 16, 2007
  12. as an exercise... what is the oscillation frequency when the rubber is displaced slightly?:biggrin:
  13. Great problem! I get the same answer for the equilibrium position as Gokul. For the oscillation period of small displacements, I get:

    [itex]T = \cot\theta\sqrt{m/k}[/itex]
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