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Cube-rod pendulum

  1. Feb 15, 2016 #1
    Hello
    I am studying for my exam and found an interesting exercise that i'm trying to solve

    1. The problem statement, all variables and given/known data

    A pendulum consists of a thin rod of length ℓ and
    mass m suspended from a pivot ∇ in the figure to the
    right. The bob is a cube of side L and mass M, attached
    to the rod so that the line of the rod extends through
    the center of the cube, from one corner to the
    diametrically opposite corner (dashed line).
    . upload_2016-2-15_13-23-45.png
    (a) Locate the distance of the center of mass
    from the point of support.
    (b) Find the moment of inertia I of the (entire)
    pendulum about the pivot point.
    (Hint: obviously it is too hard to find the
    moment of inertia of a uniform cube about an
    arbitrary axis through its center of mass by
    integrating directly, so there must be some simple trick…)
    (c) Write down the equation of motion in terms of I and any
    other relevant parameters.

    (d) Find the frequency of small oscillations.

    2. Relevant equations


    3. The attempt at a solution
    I don't know if this is correct :
    a/ The center of mass of the cube is at a distance L√3/2 to the corner where the rod is fixed. So the distance between the two centers of mass is L√3/2+l/2.
    So the center of mass of the whole system is at a distance m/(M+m)*(L√3/2+l/2) to the cube center of mass and at a distance d=m/(M+m)*(L√3/2+l/2) - L√3/2 to the corner ?

    b/ For the moment of inertia of the cube about an axis through the cdm and a corner, I found that it is equal to the moment of inertia about an axis that passes through the center of mass and the center of one face =ML²/6 and for the rod about the corner I=ml²/3
    so Itot=ml²/3+ML²/6

    c/ If we take φ to be the angle made by the pendulum and the vertical (passes through the pivot) we have to use I*d²φ/dt²=τ with τ the torque
    I don"t know what is the expression for the torque here considering that for this pendulum the rod has a mass ?
    Thanks
     
  2. jcsd
  3. Feb 15, 2016 #2

    haruspex

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    As I read the question, you are supposed to treat the rod as massless. Other than that, I agree with your answer ro a).
    How do you arrive at your answer to b)? I agree with it, just interested in how you got there?
     
    Last edited: Feb 15, 2016
  4. Feb 15, 2016 #3
    Using cosine direction, and the angle between the axis passing through a corner-com and an edge is 55°
    So for c/ he torque is just -mgl sin(φ) ? But what if the rod has a mass ?
     
    Last edited: Feb 15, 2016
  5. Feb 15, 2016 #4

    haruspex

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    I just realised that although I agree with the result you got for that, it isn't actually what b asks for. They want the MoI about the pivot point.
    So you need the MoI about an axis that's through the cdm and orthogonal to a long diagonal, and then you need to use the parallel axis theorem.
    Where/how are you using that?
    Where l is the length of the rod? Haven't you forgotten something?

    Adding in terms for MoI and torque from the rod is pretty straightforward, but let's get the question as posed sorted first.
     
  6. Feb 16, 2016 #5
    Oh I thought it was about the corner, i don't know why ..
    But the moI that I found is about a diagonal, and the rod is aligned with the diagonal, isn't it the same then ?
    if not, i don't think i understand your explanations...

    I took the origin about the com
    The moment of inertia Ixx Iyy Izz is ML2/6 and 0 everywhere else
    So the moment of inertia about an arbitrary axis is
    I= nIn = cos(a)2Ixx+cos(b)2Iyy+cos(c)2Izz
    unit vector n going from origin to the corner n=(Cos(a), cos(b), cos(c))
    No need to use the angle because we have I= (cos(a)2+cos(b)2+cos(c)2) ( ML2/6) and we know that for direction cosines : (cos(a)2+cos(b)2+cos(c)=1
     
  7. Feb 16, 2016 #6

    haruspex

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    Well, not quite. If you knew the MoI about an axis through the corner and perpendicular to the diagonal through the corner then that would do. You could get from there to the answer using the parallel axis theorem. But note that "through the corner" is not precise enough. You have to specify the axis direction.
    Yes, that's excellent. I just had not been able to understand why you seemed to care about the angle before.
    Anyway, that gives you MoI about any axis through the c.o.m., so in particular about an axis through the c.o.m. perpendicular to the diagonal. So what do you have to do to get the MoI about the pivot point?
     
  8. Feb 16, 2016 #7
    Add Md² to the moment of inertia where d is the distance between the com and the pivot, that is √3L/2+l ?
     
  9. Feb 16, 2016 #8

    haruspex

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    Yes.
     
  10. Feb 17, 2016 #9
    For the torque tou told me that I forgot something. The force is applied at the center of the cube so it would be -mg(l+sqrt(3)L/2) sin(theta) ?
     
  11. Feb 17, 2016 #10

    haruspex

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    Yes.
     
  12. Feb 17, 2016 #11
    thank you for your help :smile:
     
  13. Feb 17, 2016 #12

    TSny

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    Adding the effect of the mass of the rod shouldn't cause too much of a headache.
     
  14. Feb 17, 2016 #13
    If we add the effect of the mass of the rod, for the torque Is it just going to be -(M+m)g(l+√3L/2) sinθ ?
     
  15. Feb 17, 2016 #14

    TSny

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    No, the torque can be calculated by letting the total weight act at the center of mass of the system.
     
  16. Feb 17, 2016 #15
    So I just use what I found in a/ !
    Thank you for the answer !
     
  17. Feb 17, 2016 #16

    TSny

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    Yes. You will need the distance from the pivot to the CM of the system.
     
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