Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Swinging periods of physical and simple pendelums

  1. Nov 18, 2013 #1
    Hey! Where is the error in my reasoning?:

    The acceleration of the centre of mass in a swinging physical pendulum in simple harmonic motion is given by:

    ##M \ddot{\vec{r_{cm}}} = \sum m_j \ddot{\vec{r_j}} = \vec{g} M##

    If ##x## is the coordinate distance measured along the swinging-arc of the CM, ##\theta## is the angular displacement from the vertical line thru ##x=0## and ##R## is the distance from the pivot and to the CM, and if we assume the angular amplitude is small:

    ##\ddot{x_{cm}} \approx -g \theta_{cm} = -\frac{x_{cm}}{R} g##

    Thus the period the CM is swinging with is identical to that of a simple pendulum.

    Why is this wrong? The acceleration of the centre of mass in a physical pendulum has the same form as that of the ball in the simple pendulum, so why can't I just ignore moment of inertia, torque and all that and just set up a normal DE?
     
    Last edited: Nov 18, 2013
  2. jcsd
  3. Nov 18, 2013 #2
    One cannot "ignore" Moment of Inertia and Torque in the motion of any pendulum of this type on a simple reason that its motion (including the motion of the CM) is not linear, but curvilinear; in this case - rotational.
    Thus, the original equation MUST HAVE moment of inertia, instead of just mass multiplied by the derivative of angular coordinate, instead of the linear one, on one side, and the Torque of the Gravitational Force, instead of just the Force itself, on the other side.
     
  4. Nov 18, 2013 #3

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    You can "ignore" the rotary inertia it in a sense, except the correct "length" parameter is the radius of gyration of the pendulum about the pivot, not the distance to its center of mass. Of course for a point mass, the two lengths are the same.
     
  5. Nov 19, 2013 #4
    What is the radius of gyratation? I assume it can be found by comparing the formulas for the angular frequency of physical and simple pendulums?

    Anyway, is there nothing incorrect about the logic in the OP? Because if I use that, I get the same angular frequency expression as for simple pendulums...
     
  6. Nov 19, 2013 #5

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    Google is your friend, but search for gyration, not gyratation. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook