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I need help with understanding of Inertia of a slender rod and plate?

  1. May 25, 2012 #1
    Hello,

    Before I get barraged with comments about this being a homework question it isn't. It is about helping me understand inertia. I am trying to understand how to chose which form of equation for the rod and which form of equation for the plate in the diagram attached.

    Am i right in saying that as the rods are attached at their ends we use 1/3*m*L^2 but had they been fixed in the middle we would have to use 1/12*m*L^2.

    As for the plate I am confused about which one you would choose as it has no rotational motion about the end or the center. It has purely translational motion. So how do you know whether to chose 1/12*m*L^2 or 1/12*m(a^2+b^2)

    Any help in understanding would be much appreciated.

    Thanks

    MM
     

    Attached Files:

  2. jcsd
  3. May 25, 2012 #2

    tiny-tim

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    hello mm391! :smile:

    (try using the X2 button just above the Reply box :wink:)
    (you probably know this, so i'll put it in small type … moment of inertia only applies to rigid bodies, so you have to deal with the pair-of-rods and the plate completely separately)

    not exactly

    you can always use either the centre of mass or the centre of rotation

    fixed as in the diagram, you can use either (try both, you'll find they give the same result in the end)

    fixed at their centres, the centre of mass and the centre of rotation are the same, so there's no choice! :biggrin:
    not following you :confused:

    ω = 0, so why would you want to use moment of inertia at all? :wink:
     
  4. May 25, 2012 #3
    Well I am not sure how to say much more without making it into a previous homework question. We had this question about 3 months ago and needless to say I didn't do very well. I am now revising for exams. The question asks us to derive an equation for natural frequency for the system, ωn
     
  5. May 25, 2012 #4

    tiny-tim

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    show us your work, if you like :smile:
     
  6. May 25, 2012 #5
    I=1/3*ml^2
    Angular K.E = Iω^2
    Linear K.E = 1/2*mv^2
    Potential Energy = mgh
    ƩmIrod=2(1/3*mL^2)

    ƩK.E =1/2*mv^2+Iω^2

    ∴ ƩK.E =1/2*mv^2+1/3*ml^2ω^2

    Then I get a little confused?
     
  7. May 25, 2012 #6

    tiny-tim

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    yes, the KE of the plate is 1/2 Mv2

    the KE of each rod is either 1/2 Ic.o.rω2 or 1/2 mvc.o.m2 + 1/2 Ic.o.mω2 (same thing :wink:)
     
  8. May 25, 2012 #7
    But now I am stuck. I am not sure how to finish the revision exercise. I was told that the system conservative so d/dt(E+V)=0 which I can't get my head round either.
     
  9. May 25, 2012 #8

    tiny-tim

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    you now need an equation relating v and h :wink:
     
  10. May 25, 2012 #9
    Sorry for not understanding but I still don't see it:

    I have everything down that could be related.

    m*g*h + 1/2*mv^2+2(1/3*ml^2ω^2)=0

    m*g*h=1/2mv^2

    g*h=1/2v^2

    V=√(2gh)
     
  11. May 25, 2012 #10

    tiny-tim

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    that's physics!

    i'm talking geometry!! :wink:

    every problem like this has geometrical constraints … these give you the extra equation(s) you need to solve the problem!

    call the angle " θ ", and find h and v as a geometrical function of θ and dθ/dt :smile:

    EDIT: (just noticed) i mean the h in your mgh,

    not the (different) h in your diagram! :wink:
     
    Last edited: May 25, 2012
  12. May 25, 2012 #11
    I will have a look in the morning and then I will hopefully post the answer or anymore problems.

    Thanks Tiny-Tim
     
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