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Rotational inertia

  1. Dec 5, 2013 #1
    Hi,

    I would like to calculate the function of a rotating disc that has 2 spring slowing it down. (look at my pictures to understand)

    Let say I turn my disc 90 degrees clockwise and release it, it is going to oscillate a certain moment than stop.

    discrotating.png

    I want to plot that function and to have a sinus function decreasing to 0 after a "x" number of period.

    Thanks for your help,
    Jean-Philippe
     
    Last edited: Dec 5, 2013
  2. jcsd
  3. Dec 6, 2013 #2

    K^2

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    Doesn't work quite like that. A torsion spring pendulum is typically described by damped harmonic oscillator equation. For the equation of the following form.

    [tex]\frac{d^2x}{dt^2} + 2 \zeta \omega_0 \frac{dx}{dt} + \omega_0^2 = 0[/tex]

    And for initial conditions [itex]x(t) = x_0[/itex], [itex]x'(0) = 0[/itex], the general solution has the following form.

    [tex]x(t) = x_0 e^{-\zeta \omega_0 t} \left( cos(\omega t) + \frac{\zeta \omega_0}{\omega} sin(\omega t) \right)[/tex]

    Where the angular frequency is [itex]\omega = \omega_0 \sqrt{1-\zeta^2}[/itex]. To figure out the parameters [itex]\omega_0[/itex] and [itex]\zeta[/itex], you can follow prescription in the torsional harmonic oscillators article. But basically, you want [itex]0 < \zeta < 1[/itex] if you want decaying sinusoidal motion. Higher value will result in faster decay, but there is no way to say that it will make N oscillations. Each oscillation will be smaller than the last by a fixed ratio, but it never goes perfectly to zero. And, of course, once you figured out [itex]\zeta[/itex], you can find [itex]\omega_0[/itex] that gives you desired frequency of oscillations.
     
  4. Dec 6, 2013 #3
    This is my project, it is not a pendulum.

    2013_10_02_09_55_32.jpg
     
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