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Homework Help: Oscillations of a ruler on a cylinder

  1. May 14, 2010 #1
    1. The problem statement, all variables and given/known data

    You have ruler of length L and thickness 2d resting, in equilibrium , on a cylindrical body of radius r. Slightly unbalancing the ruler, and existing attrition between the surfaces prove that the ruler has a oscillatory motion of period:
    [tex] T = 2\cdot \pi\cdot \sqrt{\frac{L^2}{12\cdot g\cdot (r-d)}} [/tex]


    2. Relevant equations

    [tex]T=\frac{2\cdot \pi}{\omega}[/tex]

    [tex]\tau= F\cdot r\cdot \sin(\varphi)[/tex]

    3. The attempt at a solution

    I can't wrap my mind about the idea that the ruler won't immediately begin to fall. I can't figure out why the ruler would do a simple harmonic motion in the first place.
  2. jcsd
  3. May 14, 2010 #2


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    Hi benf.stokes! :smile:
    If the ruler doesn't slip, then when it tilts to the right, its centre of mass will be on the left of the point of contact. :wink:
  4. May 15, 2010 #3
    Hi tiny-tim :smile:. Thanks for the help
    I think I figured it out but I have three assumptions I'm not completely happy about and would like you to see if they're valid:

    Initially the center of mass of the ruler is at position [tex]y_{c} = R+d[/tex] and after the ruler is displaced the center of mass of the ruler is at

    [tex]y_{c'}=R\cdot \varphi\cdot \sin(\varphi)+(R+d)\cdot \cos(\varphi)[/tex]

    (I'm not really sure about the: [tex]R\cdot \varphi\cdot \sin(\varphi)[/tex], sometimes I think it should be [tex](R+d)\cdot \varphi\cdot \sin(\varphi)[/tex] but if I do so I don't obtain the correct result any more, is this right?).

    The frictional force in this case does no work since it's only role is to prevent slipping (I'm I correct?) and so the total energy of the ruler is conserved. And so:

    [tex]m\cdot g\cdot y_{c}= m\cdot g\cdot y_{c'}+\frac{1}{2}\cdot I\cdot \omega^2[/tex]
    (This is my last uncertainty: is it ok to not take into account the velocity of the center of mass of the ruler?)

    Then and using the small angle approximation :
    [tex]\sin(\varphi)\approx \varphi \ and \ \cos(\varphi)\approx\ 1-\frac{\varphi^2}{2}[/tex]

    [tex]m\cdot g\cdot (R-d)\cdot \frac{\varphi^2}{2}+\frac{1}{2}\cdot I\cdot (\frac{d\varphi}{dt})^2=0[/tex]

    And derivating this expression with respect to time yields:

    [tex]m\cdot g\cdot (R-d)\cdot \varphi\cdot \frac{d\varphi}{dt}+I\cdot \frac{d^2\varphi}{dt^2}\cdot \frac{d\varphi}{dt}=0[/tex]

    Which after some algebric manipulation yields:

    [tex]\frac{m\cdot g\cdot (R-d)}{I}\cdot \varphi + \frac{d^2\varphi}{dt^2}=0[/tex]

    Which is the equation for SHM with the desired period.

    [tex]I_{ruler}=\frac{1}{12}\cdot m\cdot l^2[/tex]
    Last edited: May 15, 2010
  5. May 15, 2010 #4
    The small angle approximation assumes that the ruler's center of mass moves only slightly, so neglecting the KE associated with the center of mass is fine (Though a more rigorous analysis is required to prove this point)

    If you want to work around the pitfall of neglecting the KE of the CoM, look at the forces and torques acting on the center of mass of the ruler instead, there the approximations should be a bit more straightforward.
  6. May 15, 2010 #5


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    Hi benf.stokes! :smile:

    (have a phi: φ :wink:)
    it's definitely Rφsinφ, not (R + d)φsinφ, because the distance along the bottom of the ruler is Rφ (because it's rolling).
    That's absolutely right :smile:, but

    why use energy, with its awkward squared terms (and yes, you do need the c.o.m. KE also), when you can use the much simpler torque equation? :wink:
  7. May 15, 2010 #6
    Thanks. I tried to use a simple torque equation first but I just couldn't find an appropriate axis
  8. May 15, 2010 #7


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    When you're dealing with rotating bodies, general rule is to use either the centre of mass or the instantaneous centre of rotation …

    and in this case the latter will eliminate the unknown reaction force. :wink:
  9. May 15, 2010 #8
    But won't the instantaneous center of rotation introduce a new component in the moment of inertia?
  10. May 15, 2010 #9


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    You'll need to use the parallel axis theorem to find the moment of inertia … is that what you meant?
  11. May 15, 2010 #10
    Yes. I tried that approach first but it the parallel axis theorem introduced an additional term I couldn't solve.
    Last edited: May 15, 2010
  12. May 15, 2010 #11
    Wait, I think I got it:

    The moment of inertia about the new axis will be:

    [tex]I_{ruler'}=\frac{1}{12}\cdot m\cdot l^2+m\cdot (R\cdot \varphi\cdot \sin(\varphi))^2[/tex]

    but since where're using the small angle approximation the second part will be very small compared to the first term?

    If this is so the torque equation makes complete sense and yields the desired result
  13. May 15, 2010 #12


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    (just finished watching the latest Dr Who :biggrin:)

    That's right! :smile:

    (except technically you left out an md2, but that will also be very small unless it's a horribly thick ruler :wink:)
  14. May 15, 2010 #13
    Yupi :cool: .Thanks you very much tiny-tim. Very helpful :smile:
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