1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determination of moment of inertia

  1. Mar 15, 2014 #1
    1. The problem statement, all variables and given/known data

    I've been trying to find out what is the period os this kind of pendulum decribed here: http://www.eng.uah.edu/~wallace/mae364/doc/Labs/mominert.pdf [Broken]

    The thing is, i've came to the same result shown in equation (11) but my reasoning it's different. I would even say that there is an error on this website on equation 5: in this equaition where is the sin of theta?? It can be omitted, only if the angle between the force vector and position vetor is 90º, but it's not the case clearly!

    Am I thinking well? In my reasoning I had also to the aproximattion of cos theta ≈ cos of phi ≈ 1 and in this site things are really simplified just because of that lack of sin of theta...

    If you're interested i can also post my reasoning but I'd like to get some help on this equation 5 first...

    Thanks a loot!


    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 15, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    At best there is a cosΘ (which is very close to 1), but the suspension points are still 2a apart on the equilibrium line, so the F point a little more outward than drawn in the figure.
     
  4. Mar 15, 2014 #3
    the thing is: what goes to torque equation is Tx right? Beacause Ty and Mg/2 cancell one another
     
  5. Mar 16, 2014 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The torque equation would be numbered (5) ? That is for the z component of the torque vector ##\tau = \vec r \times \vec F## (I suppose you will learn that vector version much later on, for now: it is for the torque about a vertical axis going through C). The expression is correct, the drawing is not. F should point more outwards.

    I can't find much wrong with (6) for the magnitude of F.

    For the torque around the x-axis (front view, fig.1, from C towards you) you indeed have
    (mg/2 a/2) + (mg/2)(-a/2)=0.
     
  6. Mar 16, 2014 #5
    But suppose you're viewing from end view. Then, the forces are above...

    So, what really makes torque isnt Tx??

    Then we make Ty=Mg/2, and so on...
     

    Attached Files:

  7. Mar 16, 2014 #6

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    From the pdf I gather the x axis is (in the front view) pointing towards the viewer. Take C as the origin. If the y axis is pointing to the right, then the z axis is pointing up. ##I\ddot \theta## is then the z component of the torque ##\tau##
     
  8. Mar 17, 2014 #7
    I get it now... But i still have that cos (phi) that i have to aproximate to 1 right?
     
  9. Mar 17, 2014 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Where is that ##\cos \phi##??
     
  10. Mar 17, 2014 #9
    Ty = Mg/2 = T cosϕ

    therefore: T = mg/2cosϕ

    therefore: Tx = mg sin ϕ / 2 cosϕ
     
  11. Mar 17, 2014 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ah, my mistake. Your T is the tension in the wires, not a torque as I was stupily reading. You are absolutely right. And yes, assume ##\phi## is so small that the cosine is 1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted