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Homework Help: Conical pendulum problem

  1. Dec 2, 2006 #1
    I tried to do this pendulum problem with force analysis, but I kept getting a wrong answer. Eventually I figured it out using torque, but I still have no clue why I was getting a wrong answer when I try to use f=ma. Any help appreciated!

    1. The problem statement, all variables and given/known data

    A conical pendulum, a thin uniform rod of length l and mass m, rotates uniformly about a vertical axis with angular velocity [tex]\omega[/tex] (the upper end of the rod is hinged). Find the angle [tex]\theta[/tex] between the rod and the vertical.

    3. The attempt at a solution

    Here's my original attempt:
    From vertical equilibrium, we have
    From the centripetal force, looking at each mass dm, we have
    [tex]dF\sin{\theta}=\omega^2 x \sin{\theta} dm[/tex], or
    [tex]dF=\omega^2 x (\frac{m}{l}) dx[/tex]
    Integrating from zero to [tex]l[/tex],
    [tex]F=\frac{1}{2} \omega^2 lm[/tex]
    equating with the F in the first equation, I got
    [tex]\cos{\theta}=\frac{2g}{\omega^2 l}[/tex], which is wrong.

    But I can get the right answer when I solve it with equilibrium/torque analysis in the rotating frame, which turns out to be [tex]\frac{3g}{2 \omega^2 l}[/tex]

    Can anyone point out what is wrong with my solution? It's been bothering me forever!!
    Last edited: Dec 3, 2006
  2. jcsd
  3. Dec 2, 2006 #2
    I'm having a hard time picturing the problem.
  4. Dec 2, 2006 #3
    It's just a basic conical pendulum made from a rod. Here's my attempt at a picture...

    Attached Files:

  5. Dec 3, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I assume that the m is a typo and that you meant to write:
    [tex]\cos{\theta}=\frac{2g}{\omega^2 l}[/tex]
    Why do you say that's wrong?

    How did you get this answer? (Again, I assume that the "m" is a typo.)
  6. Dec 3, 2006 #5
    Whoops, yeah that m was a typo :redface:

    But I know it's wrong because the book had answers at the back. (but it didn't have solutions)

    Here's what I did to get the "right" answer:

    Looking at the rotating reference frame of the rod, there is a centrifugal force and gravity. Their torque must sum to zero because the rod is in equilibrium in that reference frame.

    For each bit of mass dm, we have
    [tex]d\tau=(x sin{(\frac{\pi}{2}-\theta)})(\omega^2 x sin{\theta})dm[/tex], plugging in dm=(m/l)dx and simplify, I get
    [tex]d\tau=\frac{m\omega^2 x^2 sin{\theta}cos{\theta}dx}{l}[/tex]
    Integrating from zero to l,
    [tex]\tau=\frac{1}{3}m\omega^2 l^2 sin{\theta}cos{\theta}[/tex]

    The torque provided by gravity is simply mg(l/2)sin(theta). Equating with the centrifugal torque, I get
    [tex]\frac{1}{3}m\omega^2 l^2 sin{\theta}cos{\theta}=\frac{1}{2}mglsin{\theta}[/tex]
    Simplifying and solve for cos(theta), I found
    [tex]cos{\theta}=\frac{3g}{2\omega^2 l}[/tex]

    According to the book, this is the right answer. I just don't see why I couldn't get this result from my first way of doing it :frown:
  7. Dec 3, 2006 #6


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    Science Advisor
    Homework Helper

    I think your mistake is misunderstanding what happens on at the pivot.

    The rod is RIGID (it's not a string) so the reactions at the top can be in any direction, they don't have to be along the length of the rod. The fact that it's a pivot only means there can be no moment there.

    Your calcs for the forces on the rod due to gravity and rotation are correct but there is also an unknown sideways reaction force on the rod from the pivot. To find that, you have to take moments about somewhere. The easiest way is take moments about the pivot ...... but that's what you already did, to get the correct answer.
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