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Period of small oscillations for a pendulum

  1. Nov 21, 2015 #1
    1. The problem statement, all variables and given/known data A pendulum consists of a light rigid rod of length 250 mm, with two identical uniform solid spheres of radius of radius 50 mm attached one on either side of its lower end. Find the period of small oscillations (a) perpendicular to the line of centres and (b) along it.


    2. Relevant equations Equation I have come across is

    db205bb67d7f7f685b18b19b451ac429.png



    3. The attempt at a solution From the parallel axis theorum the moment of inertia of a pendulum and sphere is mr2+2/5mr2. As there is two spheres I added on another 2/5 mr2. The answers for both parts are (a) 1.011 s and (b) 1.031 s. I have gotten answers close to, but not quite these answers, ie 1.2 s or 1.3 s. I feel like this is an easy question and that I'm after doing something stupid, so any guidance would be greatly appreciated.
     
  2. jcsd
  3. Nov 21, 2015 #2

    BvU

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    Hello mro, :welcome:

    Just so we are really talking about the same pendulum:

    upload_2015-11-21_21-24-45.png
    you didn't tell us what m is and I don't understand how this is a pendulum. Where is the axis of rotation ?
    How can it oscillate along the "line of centres"? and perpendicular to it ?
     
  4. Nov 21, 2015 #3
    Hi BvU,
    We weren't given m, and it doesn't make much sense to me either. When I read the question initially, that was the picture I had in my head. However, from reading it again, I think they're attached on the same end? Just simply because it said "one on either side of its lower end," not ends. To be honest, the book is very poorly worded at times. As regards to m, could it not be cancelled with the lower line m in the equation? Maybe it's total mass of the system? I apologise for the vagueness of the question by the way, it's not very clear.
     
  5. Nov 21, 2015 #4

    gneill

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    @BvU I think the two cases have the spheres at the same end of the rod, the difference being the plane of oscillation:

    Fig1.png
     
  6. Nov 21, 2015 #5
    @gneill Hi, thanks for the diagram, that's the picture I had in my head. Am I on the right track with my equations or is there something I'm forgetting about?
     
  7. Nov 21, 2015 #6

    gneill

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    I think you're on the right track, but you should be careful about your moment of inertia expressions. Be sure to carefully account for the displacement distance in the parallel axis theorem for each case.

    I just did a quick calculation, and I suspect that they may have forgotten to include both spheres when they calculated their "given" answers. Of course, it's also possible that I've mucked up, too! So let's see what you get (show your work), and in the meantime I'll go over my own work.

    Edit: Nope, I mucked up :oops: I'm seeing their results now.
     
  8. Nov 21, 2015 #7
    Ok, just want to double check something: for R in the period equation, is that L/2? Or do I have to include the radius of the spheres in that R too? Thanks for all your help so far by the way, really appreciate it! :)
     
  9. Nov 21, 2015 #8

    gneill

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    The R in your formula is the distance from the center of mass of the physical pendulum to the pivot point. See:

    Hyperphysics: Physical Pendulum
     
  10. Nov 21, 2015 #9
    I seem to be making a mess of things :H I'm inputting 0.125 m for the moment of inertia of the pendulum and 0.05 m for the spheres, and I am not getting the answers out. I'm definitely doing something really stupid and I'll be so annoyed when I figure out what it is!
     
  11. Nov 21, 2015 #10

    gneill

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    I'm not sure how you're coming up with a value for the moment of inertia alone, since the sphere mass was not specified. The mass cancels out eventually in the period formula though.

    Why don't you share the details of your work? Maybe someone can spot where it goes awry.
     
  12. Nov 21, 2015 #11
    https://scontent-lhr3-1.xx.fbcdn.net/hphotos-xat1/v/t34.0-12/12282819_10204148176339731_1402560635_n.jpg?oh=e147dab589d46be63cf24a1a026741fc&oe=56535411
    I apologise for this being sideways but this was my calculations part.
     
  13. Nov 21, 2015 #12

    gneill

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    This problem seems to hinge on obtaining the correct moment of inertia for the given setup, so let's focus on that.

    I'm not understanding the moment of inertia you've used in your period formula. Can you type it out and explain the variables? If r is the radius of a sphere, then what is the variable for the offset a la parallel axis theorem?
     
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