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Simple harmonice motion

  1. Jul 13, 2013 #1
    1. The problem statement, all variables and given/known data

    The 20cm long wrench swings on its hook with a period of 0.90s. When the wrench hangs from a spring of spring constant 360 N/m , it stretches the spring 3.0 cm. What is the wrench 's moment of inertia about the hook ?

    2. Relevant equations

    torque =I alpha , Fsp =kx

    3. The attempt at a solution
    I put R to be (14+.03)
    1. The problem statement, all variables and given/known data




    3. The attempt at a solution
    We need to find I and I = 1/2mg
    Consider Fsp = mg when the wrench is at the lowest point .
    -kx=mg and m=`kx/g but I dun know how to find the I as the pivot is not on the rotation and should I consider the wrench as a rod for the moment of inertia ?
     
    Last edited: Jul 13, 2013
  2. jcsd
  3. Jul 13, 2013 #2

    haruspex

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    If you could treat the wrench as a 20cm uniform rod pivoted at one end you wouldn't need the information about the period of swing. If its radius of inertia is k, what would its period be?
     
  4. Jul 14, 2013 #3
    period = 2 pi * square root ( k/2g) ?
     
  5. Jul 14, 2013 #4

    haruspex

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    Not quite. Why the factor 2 on the g? (I should have stated k as the radius of inertia about the pivot point, not the mass centre, but if you took it as about the mass centre the answer would be quite different.)
    Anyway, once you have that equation correct, you can use it to deduce k, yes? What else do you need to find out in order to calculate the moment of inertia about the pivot point?
     
  6. Jul 14, 2013 #5
    I put that wrong and that should be T=2 pi * square root( K/g) but how could the period can help to solve this problem ?
     
  7. Jul 14, 2013 #6

    haruspex

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    From that equation, T tells you k. And the moment of inertia is mk2, right? So what's the other quantity that you need to determine?
     
  8. Jul 14, 2013 #7
    I got the k , radius is 0.20 but I wonder why the moment of inertia is mk^2 and not 1/3mK^2+ mK^2 and I am not good at the concept of inertia . thanks so much
     
  9. Jul 14, 2013 #8

    haruspex

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    You're confusing two things. If the radius of inertia (about some point) is k then the moment of inertia (about that point) is mk2. That's the definition of radius of inertia.
    For a uniform rod length 2l, the moment of inertia about its mass centre is ml2/3, so its radius of inertia is l/√3. About one end, the moment of inertia is 4ml2/3, so its radius of inertia is 2l/√3.
    In the present problem, we do not know and do not care whether the wrench is uniform, or whether the pivot point is at one extreme end (it surely isn't).
     
  10. Jul 15, 2013 #9
    you mean the I = Icm + Md^2 is used when the pivot is any point off the center mass except the end right ? Therefore , I had to use the rod of moment of inertia which is I= 1/3mL^2 and the length is found by the period = 2 pi square root of ( L/g) . IS that corrected ?
     
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