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Object sliding through hollow rotating tube

  1. Jul 20, 2010 #1
    This is neither homework nor a question from an exam.

    I am an aging engineer and I am reviewing all of the calculus that I have forgotton. On p.410 of 'Calculus with analytic geometry,' 5th ed. by Howard Anton, Chap. 7 technology exercises, problem #15, paraphrased slightly to be (hopefully) a little more clear:

    Suppose that an initially horizontal hollow tube rotates in a vertical plane with a constant angular velocity of omega radians/sec about a horizontal axis, perpendicular to the length of the tube, through one end of the tube. The direction of omega is such that the tube initially rotates down from its horizontal starting point. Assume that an object within the tube, initially at rest at the pivot-point-end of the tube, is free to slide without friction through the tube while the tube rotates, and that the object's diameter is equal to the inside diameter of the tube so that the object can't rattle around at all as it slides down the tube. Let r be the distance of the object from the pivot point for all time t >= 0 as the object slides through the length of the tube (r=0 when t=0). It can be shown that

    r = g/(2(omega)^2)(sinh((omega)t)-sin((omega)t))

    during the period that the object is in the tube.

    But it can't be shown by me! I tried. Forget the actual problem - it's easy. I just wanna be able to derive the formula, or I won't be able to sleep.

    Let the origin be at the pivot point, the positive y direction up, the positive x direction toward the other end of the tube. Let theta be the angle between the tube and the horizontal and assume that the sliding object has mass m.

    A free body diagram of the object in the tube involves 2 forces: gravity, mg, pointing straight down, and what I'm calling a normal force, N, acting perpendicular to either the upper or the lower inside surface of the tube, depending on t and possibly omega. N can also be labeled as a force in the theta direction, F_theta, since it clearly has no radial component.

    When t is small but greater than zero, then the object has moved from its initial position but is not falling very fast, and so the upper inside surface of the tube will be moving downward at a faster speed than the falling object and will then push the object with a normal force perpendicular to that surface, forcing the object to slide parallel to the surface and down the tube. When t exeeds some value, then the object will be falling faster than the speed at which the tube is moving downward. Now the falling object is held up by the lower inside surface of the tube. This lower inside surface exerts a force on the object in a direction perpendicular to the surface, and the object is forced to slide without friction parallel to that surface and down the tube.

    It may also be true, though it isn't initially clear, that at low omega the upper inside surface never exerts a force on the object, and at high omega the lower surface never exerts a force.

    Let us consider the sum of the forces and the sum of the torques on the sliding object at a time after which the normal force has switched from the upper to the lower inside surface of the tube, since I suspect that this will lead to the formula in the text:

    sum of the forces in the y direction = Ncos(theta)-mg = ma_y = mdv_y/dt

    sum of the forces in the x direction = Nsin(theta) = ma_x = mdv_x/dt

    sum of the torques = mgrcos(theta)-Nr = I_object(alpha_object) = 0 since omega_tube = constant implies angular acceleration_tube =0 implies angular acceleration_object = 0 since the object is constrained to remain in the tube. So we have:

    dv_y/dt = g((cos((omega)t))^2-1)

    dv_x/dt = gcos((omega)t)sin((omega)t)

    Integrating twice with v_y(0) = 0, v_x(0)=0, x(0)=0, y(0)=0 gives:

    y= (g/(4(omega)^2))(sin((omega)t))^2-(g/4)t^2

    x= (g/(4(omega)))t-(g/(8(omega)^2))cos((omega)t)sin((omega)t)

    Now, r=(x^2+y^2)^1/2 should give the desired equation. Good luck with that.

    If there are simplifying assumptions that will produce the desired equation, they weren't mentioned in the problem statement.

    Help me Obi-wan Kanobi! You're my only hope.
     
  2. jcsd
  3. Jul 24, 2010 #2

    Filip Larsen

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    Gold Member

    Splitting up in x and y may be possible, but to me it seems easier to describe the radial acceleration directly as the sum of the centripetal acceleration and the gravity which, with the given initial conditions, result in the differential equation

    (1) [tex]
    \ddot{r} = \omega^2 r + g sin \omega t
    [/tex]

    The solution given by your book, i.e.

    (2) [tex]
    r = \frac{g}{2\omega^2} \left( \sinh\omega t - \sin\omega t \right)
    [/tex]

    is a solution to (1) so I assume it should be possible to derive (2) from (1) by integrating.
     
  4. Jul 25, 2010 #3

    Filip Larsen

    User Avatar
    Gold Member

    The solution to (1) can be found as the complete solution to the homogeneous equation plus one particular solution to the inhomogeneous equation.

    Armed with a mathematical handbook, the homogeneous equation

    (3) [tex]
    \ddot{r} - \omega^2 r = 0
    [/tex]

    can be seen to have the complete solution

    (4) [tex]
    r_h = A e^{\omega t} + B e^{-\omega t}
    [/tex]

    while a particular solution to (1) is found as

    (5) [tex]
    r_p = -\frac{g}{2\omega^2} \sin\omega t
    [/tex]

    which combined gives the total solution

    (6) [tex]
    r = r_h + r_p = A e^{\omega t} + B e^{-\omega t} -\frac{g}{2\omega^2} \sin\omega t
    [/tex]

    Inserting the initial conditions [itex]r(0) = \dot{r}(0) = 0[/itex] gives [itex]B = -A[/itex] and [itex]2A = g/(2\omega^2)[/itex] which inserted into (6) yeilds (2), namely

    (7) [tex]
    r = \frac{g}{2\omega^2} \frac{e^{\omega t} - e^{-\omega t}}{2} - \frac{g}{2\omega^2} \sin\omega t = \frac{g}{2\omega^2} \left( \sinh\omega t - \sin\omega t \right)
    [/tex]
     
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