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Rotational Inertia of a Rod Falling and Slipping

  1. Jul 3, 2004 #1
    A rod of mass M and length L, initially standing, falls in such a way that the center of mass experiences only vertical motion. What is the moment of inertia of the rod?

    I picture the rod as having both translational and rotational motion since it's rotating about an axis through the CM which is moving vertically downward. I can also picture the rod as rotating about the end touching the surface, which is moving horizontally. Now, to find the moment of inertia, I need to pick a rotation axis. Which one of these axes should I pick?
     
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  3. Jul 3, 2004 #2

    AKG

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    It doesn't matter which one you pick as far as I can tell, it's simply that the moment of inertia about one axis is different from the moment of inertia about another.
     
  4. Jul 3, 2004 #3

    Doc Al

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    The moment of inertia is a property of the rod, not of its motion. I don't understand the point of this question.
     
  5. Jul 3, 2004 #4
    I asked this since I needed some insight on the following problem: A thin uniform stick of mass M and length L is positioned vertically, with its tip on a frictionless table. It is released and allowed to slip and fall. Determine the speed of its center of mass just before it hits the table.

    There is a picture that comes with this problem which basically depicts the rod standing and just before it hits the table, and from it I've concluded that the center of mass does not displace horizontally, but vertically only.

    Anywho, I can solve this problem using conservation of mechanical energy. When I set up the equation, the rotational kinetic energy term involves the moment of inertia and this of course led me to my original question.
     
  6. Jul 3, 2004 #5

    Doc Al

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    Since all the forces on the rod act vertically, the center of mass must fall straight down.
    Right. But I assume you can calculate the rotational inertia of a rod about its cm. The trick is to relate the speed of the cm to the rotational speed of the stick.
     
  7. Jul 3, 2004 #6
    Yes of course! Hmm...Somehow [itex]\omega \propto v[/itex], but that's all I can think of. This is giving me a headache. Argh...
     
  8. Jul 4, 2004 #7

    well, here's how [itex]\omega \propto v[/itex], to get you started

    [itex]y[/itex] will be the distance that the stick has fallen

    therefore:
    [tex]\cos \theta = \frac {\frac{L}{2}-y}{\frac {L}{2}}[/tex]

    which gives:
    [tex]\dot{y}=\frac{L}{2}\sin\theta \ \dot{\theta}[/tex]

    or, in your [itex]\omega \propto v[/itex] terms:
    [tex]v=\frac{L}{2}\sin\theta \ w[/tex]

    then use this relation with energy considerations to solve for the motion.
    btw, [itex]I=\frac{1}{12}ML^2[/itex]

    if i remember correctly, this was an example in my K&K book at MIT
     
  9. Jul 4, 2004 #8
    Got the answer already, but thanks anyways.
     
  10. Jul 4, 2004 #9

    Doc Al

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    Almost. It should be:
    [tex]\sin \theta = \frac {\frac{L}{2}-y}{\frac {L}{2}}[/tex]
    Which gives:
    [tex]\dot{y}= - \frac{L}{2}\cos\theta \ \dot{\theta}[/tex]
    (The minus sign is just an artifact of how the angle is defined.)

    This makes sense, since when the cm hits the floor, the angle is zero and [itex]\omega[/itex] is maximum.
     
  11. Jul 4, 2004 #10

    well, that depends on how you define [itex]\theta[/itex]. Since I used [itex]I=\frac{1}{12}ML^2[/itex], [itex]\theta[/itex] will be the angle that the stick makes with the vertical as it is falling as measured from the center, not the angle it makes with the horizontal as measured from the floor. in this case, [itex]\theta[/itex] will be maximum at the end of the problem.
     
  12. Jul 4, 2004 #11

    Doc Al

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    D'oh! You're perfectly correct. My bad. :uhh:

    (But [itex]\omega[/itex] is the same no matter how you define [itex]\theta[/itex].)
     
    Last edited: Jul 4, 2004
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