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Time Period of a Pendulum in Circular Motion

  1. Apr 1, 2005 #1
    Hi

    Here is a problem I need some help with:

    Find the time period of a simple pendulum (length l, mass m) when it is attached to the roof of a car moving in a circle with constant speed v and radius r.

    Here is what I did:

    IF the angle made by the string (inextensible) with the vertical at some instant is [itex]\theta[/itex] then the equations of motion in the direction along the length of the string and perpendicular to it are:

    [tex]F_{normal} = mg\cos\theta - \frac{mv^2}{r}\sin\theta[/tex]
    [tex]F_{tangential} = -ml \frac{d^2\theta}{dt^2} = mg\sin\theta - \frac{mv^2}{r}\cos\theta[/tex]

    Using the small angle approximation in the second equation, we get

    [tex]l\frac{d^2\theta}{dt^2} = -g\theta - \frac{mv^2}{r}[/tex]

    This gives [itex]T = 2\pi\sqrt{\frac{l}{g}}[/itex] whereas the answer in my book is,

    [tex]T = 2\pi\sqrt{\frac{l}{\sqrt{ g^2+(\frac{v^2}{r})^2 }}}[/tex]

    I would be grateful if someone could help me out with this (I know v^2/r has got to play a role but how??) at the earliest...
    Cheers
    vivek
     
    Last edited: Apr 1, 2005
  2. jcsd
  3. Apr 1, 2005 #2

    Doc Al

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    Staff: Mentor

    In an inertial frame, the period is [itex]T = 2\pi\sqrt{l/g}[/itex]. But in an accelerating frame, the effective "g" is different. This might help: https://www.physicsforums.com/showpost.php?p=377511&postcount=9

    (By the way, I think the answer given is incorrect.)

    Ah... I see you fixed the answer just as I was typing my response. That answer is correct. :smile:
     
  4. Apr 1, 2005 #3
    Hello Doc

    I understand what you just said and yes, I saw the link too. My book says the same thing...its just that I want to prove from first principles that the value of [itex]g_{eff}[/itex] to be used is what you and the book say. This should be possible using a differential equation approach I believe...isn't it?

    Thanks and cheers
    Vivek
     
  5. Apr 1, 2005 #4

    Doc Al

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    Sure. I'd start by writing the equations in terms of [itex]\theta[/itex] but measured from the equilibrium angle [itex]\theta_0[/itex], not from the vertical.
     
  6. Apr 1, 2005 #5
    I don't get it...could you please guide me? Whats the equilibrium angle not from the vertical?

    EDIT: What I am looking for is a pair of equations of motion in orthogonal directions, for the pendulum bob when the pendulum apparatus as a whole has nonzero acceleration (general case) not necessarily the present case of circular motion.
     
  7. Apr 1, 2005 #6

    Doc Al

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    My only point was that if you wish to derive the period of a pendulum using the usual small angle approximation, you must measure the angle from its equilibrium postion. Which is not vertical, if the apparatus is accelerating.
     
  8. Apr 1, 2005 #7

    krab

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    Look at your second equation. It means that for the bob to be in equilibrium, you must have [itex]\tan\theta=v^2/(rg)[/itex]. This gives you [itex]\theta_0[/itex]. So you were wrong to expand about [itex]\theta=0[/itex]; you must expand about [itex]\theta=\theta_0[/itex]. So for example, [itex]\sin\theta[/itex] becomes [itex]\sin\theta_0+\alpha\cos\theta_0[/itex] where [itex]\alpha=\theta-\theta_0[/itex] is the (small) angle deviation from equilibrium.
     
    Last edited: Apr 1, 2005
  9. Apr 1, 2005 #8

    Doc Al

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    Thanks for clarifying that, krab. (I guess I wasn't particulary clear!)
     
  10. Apr 1, 2005 #9
    Thanks a lot for the help krab and Doc Al. But I need some more help....

    According to what has been said, at equilibrium [itex]F_{tangential} = 0[/itex] gives [itex]\tan\theta_{0}=v^2/(rg)[/itex]. I understand this. Now how do I expand about [itex]\theta_{0}[/itex]. Should I do a Taylor expansion or is there some other way (from the way I think--the two methods should be equivalent save for some effort)...

    Thanks and cheers
    Vivek

    EDIT: I just saw that krab had discussed this already...thanks anway. I'll try this and post my work here for you folks to verify in a while...
     
  11. Apr 2, 2005 #10
    Hi again

    There are still a few more issues I want to be clear about. Here goes:

    1. The equation of motion I have written is with reference to an observer in the noninertial frame of the car. Is it only this observer who perceives the motion as simple harmonic (for small displacements about [itex]\theta_0[/itex] of course)? In principle, we are taking a snapshot of the system or rather a section view of the vertical plane which is actually rotating with the car. So what are the motions observed by (A) an observer in an inertial frame and (B) and observer in the rotating frame of the car?

    2. If I write the equations of motion with respect to an inertial frame G, then the acceleration of the bob in this frame is given by,

    [itex]\vec{a_{B/G}} = \vec{a_{B/C}} + \vec{a_{C/G}}[/itex] --(1)

    where [itex]\vec{a_{B/C}}[/itex] and [itex]\vec{a_{C/G}}[/itex] are respectively, the accelerations of the bob with respect to the noninertial frame of the car and that of the car with respect to the ground. Multiplying both sides by the bob's mass, I get an equation of motion in the inertial frame. Now, as I understand from the foregoing discussion, the equilibrium position is obtained when [itex]\vec{a_{B/C}} = 0[/itex] and not when [itex]\vec{a_{B/G}} = 0[/itex] (which in this case isn't possible either because of the nonzero term [itex]\vec{a_{C/G}}[/itex]. Is this okay?

    3. If it is okay, then equation (1) above should also give me an idea about the absolute motion of the bob (if desired of course) where

    [itex]\vec{a_{C/G}} = \frac{v^2}{R}\vec{e_R}[/itex]
    (along the radius of curvature of the car's circular path).

    4. So it is possible to treat this problem purely from an inertial frame point of view isn't it? I mean the equations I wrote did not explicitly mention that I was writing them in the noninertial frame but if I didn't, then I would get a +mv^2/r term (which if you will please check, I don't) giving the equilibrium position as [itex]tan\theta_0 = \frac{-v^2}{rg}[/itex] contrary to intuitive observation. What is the reason for this?

    Thanks and cheers,
    Vivek
     
  12. Apr 4, 2005 #11

    Doc Al

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    I would say that the inertial observer would see the bob executing SHM about it's "equilibrium" position as whirls in a circle along with the car. The noninertial observer (in the car) sees only the oscillation, not the circular motion.

    I'm not sure what you're saying. In the noninertial frame, the equilibrium position is when the forces (including the centrifugal force) add to zero; in the inertial frame, the the same position is found by setting the ("real") forces equal to ma, where a = the acceleration of the car.
    Your equation does imply a noninertial frame: you explicity include a centrifugal force term! From the inertial frame, you have to include all the acceleration terms that have a component tangential to the bob. That will produce a centripetal term (due to the motion of the car, of course) on the left side of your differential equation.
     
  13. Apr 9, 2010 #12

    Cleonis

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    As the other replyers remarked: starting point for the analysis is the angle of a plumb line in the moving car. The equilibrium angle of the swinging pendulum will be the angle of the plumb line.

    On my website there's a Java applet simulation, called Foucault rod, that can be used to explore the described setup.
    The Java applet evaluates the trajectory of a pendulum bob that is attached to a vibrating rod. The vibrating rod can be oriented at any angle relative to the plane of rotation, the bob can be positioned at any distance to the central axis of rotation, the number of pendulum swings during each circumnavigation can be varied over a range of values.

    Running the simulation illustrates the following: if there are a lot of vibration cycles per circumnavigation then the deviation from vertical is very small, in most cases negligably small.

    In the applet calculations the motion with respect to the inertial coordinate system is computed. Note that there is no need to introduce a centripetal force explicitly. All that is required is that the force exerted upon the pendulum bob is proportional to the displacement of the bob. That displacement - just as in the case of a plumb line - elicits the required centripetal force.

    Cleonis
    http://www.cleonis.nl
     
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