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Centre of inertia

  1. Jan 8, 2015 #1
    1. The problem statement, all variables and given/known data
    The centre of inertia (G) of a heterogeneous vehicle tyre has a distance D=0.1cm from its axis of rotation (the tyre's). The axis is referred to as Delta. The mass of the tyre is M=10kg and its radius is R=20cm. O is the center of the tyre (belongs to the axis)

    2. Relevant equations
    Find the mass of an object (dot) X attached to the tyre so that the centre of inertia G is identical to the axis of rotation.

    3. The attempt at a solution
    This is what I've done so far, but with no avail:

    I've considered G' as the centre of inertia of the whole group (the tyre and the object X), and G1 as the centre of inertia of X and "m" as its mass.

    Therefore: Vector OG' = (m*vector OG1 + m*vectorOG) / m + M
    I substituted G for O since we want it identical to the axis, which results into:
    Vector OG' = (m*vector OG1 + m*vecto OO) / m + 10
    Vector OG' = (m*vector OX) / m + 10 ( G1 = X itself)
    Thus: OG' = (m*20) / m + 10

    But I'm not sure where it's going to lead me. Please help me.
     
  2. jcsd
  3. Jan 8, 2015 #2

    haruspex

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    I presume you mean /(m+M), and that one of the m's in the numerator should be M, but I there's still something wrong with the vectors. How is your G' different from G?

    It seems to me the question is incomplete. It doesn't specify at what radius X is to be attached. Presumably it is at radius R.
     
  4. Jan 8, 2015 #3
    Yes, I meant (m+M) and M, sorry. The G' I considered is the centre of inertia of the entire group, ergo the tyre and the object itself. X is to be attached to the cover of the tyre, the latter is considered as a circle, therefore X is to be attached at the radius R, yes.

    I think the centre of mass equation I provided is correct, but I have no idea what to do next.
     
  5. Jan 8, 2015 #4

    haruspex

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    Ok, just realised that the question as posed uses G for two different things. I guess it should read:
    Find the mass of an object X attached to the tyre so that the centre of inertia G' is identical to the axis of rotation.
    Shouldn't you put G' = O?
     
  6. Jan 8, 2015 #5
    I'm not sure... I think it's a reference to the tyre's G. Could you help me in this case? (G=O)
     
  7. Jan 8, 2015 #6

    haruspex

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    No, I think the second "G" clearly means the mass centre of the system, tyre+X. And I don't like "is identical to the axis of rotation". Surely it should say "lies on the axis of rotation". (Note that the question as posed did not mention vectors. You introduced those. Seems to me the whole set-up can be considered planar.)
    So you have
    Vector OG' = (m*vector OG1 + M*vectorOG) /(m + M)​
    and you want G' to lie on the axis of rotation, so G'=O. What does that give you?
     
  8. Jan 9, 2015 #7
    This is much more clear... the thing is, the question didn't mention which centre of inertia, the tyre's or the system's. Anyway, this is what I got when I replaced G' by O:

    OO = (m*OG1 + M*OG) / (m+M)
    0 = (m*OX + 10*OG) / (m+10)
    0 = (m*20 + 10*0.1) / (m+10)
    m*20 + 1 = 0
    20m = -1
    m = -0.05kg but since a mass can't be negative I considered m = 0.05kg. Is this correct?
     
  9. Jan 9, 2015 #8

    haruspex

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    Numerically that's right, but you need to figure out why it went negative.
    The false step is here:
    0 = (m*OX + 10*OG) / (m+10)
    0 = (m*20 + 10*0.1) / (m+10)​
    You replaced two vectors by ... what?
     
  10. Jan 9, 2015 #9
    They're not vectors in that step, because if a vector A equals a vector B, then their distances are equal. That's what I did there, if the null vector equals an operation including vectors, their distances are equal. Am I right?
     
  11. Jan 9, 2015 #10

    haruspex

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    No, I'm referring to the way you replaced vectors OX and OG by 20 and 0.1.
     
  12. Jan 9, 2015 #11
    OX and OG are distances.

    If: vector 0 = (m*vector OX + M*vector OG) / (m+M)

    Then: 0 = (m*OX + M*OG) / (m+M)

    This is mathematically correct.
     
  13. Jan 9, 2015 #12

    haruspex

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    OK, I was interpreting OX and OG as vectors. If you are using them to stand for distances then the problem is in going from 'vector OX' to OX etc. It is not mathematically correct. What operation do you perform on a vector to get a corresponding distance?
    It might become clearer if you multiply out the first equation above to get the relationship between the two vectors.
     
  14. Jan 10, 2015 #13
    All right, this is what I get when I leave them as vectors:

    m*vector OX + M*vector OG = vector 0
    m*vector OX = -M*vector OG
    m*vector OX = M*vector GO
    m*OX = M*GO
    m*20 = 1
    m = 0.05kg

    Is this what you mean?
     
  15. Jan 10, 2015 #14

    haruspex

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    Not really. This step is not valid:
    On what basis do you simply throw away the minus sign?
    You want to get rid of the vectors. The correct procedure is to take the modulus of each side:
    m*vector OX = -M*vector OG
    |m*vector OX| = |-M*vector GO|
    |m| * |vector OX| = |M|*|vector GO|
    |m| |OX| = |M| |GO|
    Since masses are positive:
    m |OX| = M |GO|​
     
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