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Bead sliding on a rotating rod

  1. Oct 9, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A bead of mass m slides under gravity on a smooth rod of length l which is inclined at a constant angle ##\alpha## to the downward vertical and made to rotate at angular velocity ##\omega## about a vertical axis. The displacement of the bead along the rod is r(t).

    A)Find the Lagrangian and the equation of motion of the bead, and obtain the general solution.
    B) Show that if the bead is released from rest at r=0, it falls off the rod after a time interval $$t = \frac{1}{\omega \sin \alpha} \cosh^{-1} \left( 1 + \frac{l \omega^2 \sin^2 \alpha}{g \cos \alpha}\right)$$

    2. Relevant equations
    Lagrangian L = T - V

    3. The attempt at a solution
    Write the kinetic and potential energy of the bead. I get: $$T = \frac{m}{2}(\dot{r}^2 + r^2 \omega^2)\,\,\,\,\,\,\,\text{and}\,\,\,\,\,\,V = -mgr\cos \alpha$$

    Using Lagrange's Eqns, this gives ##\ddot{r} - \omega^2 r = g \cos \alpha## so ##r = C_1 \exp(\omega t) + C_2 \exp(- \omega t) - g \cos \alpha/\omega^2##

    If the bead is released at the origin, set t=0 as well for convenience then ##r(0) = \dot{r}(0) = 0##. Solve these to obtain ##C_1, C_2## to get $$r = \frac{g \cos \alpha}{\omega^2} \left( \frac{\exp(\omega t) + \exp(-\omega t)}{2}\right) - \frac{g \cos \alpha}{\omega^2}$$

    It will fall off when r=l, so solving gives $$t = \frac{1}{\omega} \cosh^{-1} \left(1 + \frac{\omega^2l}{g \cos \alpha}\right)$$

    I'm not really sure where the sin term comes in. The question states that the rod is inclined at a constant angle ##\alpha## to the downward vertical. But the rod rotates so I don't see how this is possible.
    Many thanks.
     
  2. jcsd
  3. Oct 9, 2013 #2
    How did you get the second term in kinetic energy?
     
  4. Oct 9, 2013 #3

    CAF123

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    Hi voko,
    I believe this term describes the tangential velocity of the bead which is present because the rod is rotating in the plane, and the bead is constrained to move along the rod.
     
  5. Oct 9, 2013 #4
    But why is the tangential velocity ## r \omega ##? ## r ## is the distance along the rod, not from the axis of rotation.
     
  6. Oct 9, 2013 #5

    CAF123

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    I assumed the axis of rotation was at the origin, is is not?
     
  7. Oct 9, 2013 #6
    This seems reasonable, yet ##r## is still not the radius of rotation.
     
  8. Oct 9, 2013 #7

    CAF123

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    I set the problem up with one end of the rod at the origin and the other end at a radial distance ##r=l## from the origin with the rod at an angle ##\alpha## to ##-\hat{y}##. In this case, the position of the bead along the rod and the distance of the bead from the axis of rotation are the same, no?

    Perhaps I misinterpreted the problem statement. As noted in the OP, the rod is at a constant angle ##\alpha## to ##-\hat{y}##. But it rotates so ##\alpha## is not the same at all times.
     
  9. Oct 9, 2013 #8
    The very first sentence in the formulation states that the angle is constant. It cannot be both constant and "not the same".

    The only sensible explanation I can think of is that the rod and the vertical axis are always coplanar and at a constant angle, but the plane rotates about the axis.
     
  10. Oct 9, 2013 #9

    CAF123

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    What about if the rod is in 3D space, with one end attached to the origin and the other at a distance r=l from the origin, pointing downwards below the xy plane. Then the angle between the rod and the negative y axis is always alpha, while it rotates, creating a sort of cone shaped outline as it rotates.
     
  11. Oct 9, 2013 #10
    I think this is what I described earlier. What is the distance from the axis at any point of the cone, given its distance from the vertex is ##r##?
     
  12. Oct 9, 2013 #11

    CAF123

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    It is rsinα.
     
  13. Oct 9, 2013 #12

    CAF123

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    Thanks voko, I got it. I originally envisaged the problem in 2D, which is why I had the confusion with the angle.
     
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