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Rotational Motion

  1. Jan 4, 2008 #1
    A rod of mass M and length L is connected to a pivot at one of its ends. The pivot is hanging from a horizontal ceiling and the rod is set parallel to the ceiling (its also horizontal). The force that will cause it to rotate once it is released is gravity and what I need to find is the initial angular acceleration of the rod and the initial linear acceleration of its right end.

    For T (torque) and @ (angular acceleration), if we were to mathematically use T = Fd for d = rsin(X), and Net T = I@, the @ found would be the same at any point on the rod, so it would be acceptable to use L as the length, L/2 as the length, and so forth to find @ while the force acting on the rod, gravity, is perpendicular to it. Therefore F = Mg and T = Mgrsin(X), in which sin(X) equal to 1. Therefore T = Mgr. Using that and T = I@, then @ = T/I = Mgr/(1/3*M*L^2). Now all that is needed is to plug in for r, which could supposedly be the length for any point along the rod. But if I were to plug in random values of r, for 0 < r < L, then I would get a different @ for different lengths. Is it customary to use the center of mass for r?
     
    Last edited: Jan 4, 2008
  2. jcsd
  3. Jan 4, 2008 #2
    Note: I didn't very carefully read your post but I'm sidetracked by this

    They never made you do that thing in elementary school where everyone joins hands in a line and rotates around a person at the end? Then the people at the end non-moving end just walk casually while the people on the outside sprint?

    What's that tell you?


    Edit2!: I actually got the wrong result out of that myself. The angular quantities are the same along the rod, yes. The linear quantities will differ, so you're gonna go through like the below poster says and get the torque along the rod, and hence the angular acceleration and THEN the LINEAR acceleration is what you will find that's different as you go along the rod.

    DOUBLE NOTE: Why d=rsin(X)?
    EDIT: Yah I should read these posts completely first. Coincidentally since the problem wants it initially and the sin(x)=1 that ends up not messing you up, though I think it's still incorrect as I mentioned.
     
    Last edited: Jan 4, 2008
  4. Jan 4, 2008 #3

    Dick

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    It's not just customary, it's mandatory. The sum of all of the gravitational forces acting on the parts of the rod is equivalent to a single force of magnitude Mg located at the center of mass of the rod. In rotational problems like this you can't just stick the force anywhere you like.
     
  5. Jan 4, 2008 #4
    That's what he needs to do for T=F*d, it's mg, and d=the CM, right?

    He may've done that part wrong but that's not what he was quite specifically asking at the end, right?

    If I jacked up my explanation lemme know so I can chunk it
     
  6. Jan 4, 2008 #5

    Dick

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    Right, I think. I took the question to be what r to use to compute torque, T=F*r*sin(X). Where X is the angle between the force and the line to the axis of rotation. I think the poster is confusing r*sin(X) with some sort of distance. The only thing in your post I'd clarify is that the angular acceleration is certainly constant along the rod, it's rigid. The linear acceleration isn't constant.
     
  7. Jan 4, 2008 #6
    Yah, clarify is polite speak for "you tried hard to make a point but it was the wrong one!" >_> but now no one will ever know!
     
  8. Jan 4, 2008 #7
    uhhh....so when considering a force that acts on all parts of the object in rotation, its best to use the center of mass as the point of application?
     
  9. Jan 4, 2008 #8

    Dick

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    If it acts on all parts proportional to mass, like gravity, yes. It's not only 'best', it's the only right thing to do.
     
  10. Jan 4, 2008 #9
    I see, I see. Interesting. Thanks; just in case, when dealing with conservative forces, the center of mass should be considered as the point application?
     
  11. Jan 4, 2008 #10

    Dick

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    Don't jump to conclusions. For example if a charged body is in a uniform electric field, then the analogous conclusion is that the force can be treated as acting at the 'center of charge', which can be quite different from the center of mass if the mass is distributed differently from the charge.
     
  12. Jan 4, 2008 #11
    hmmm...I see once more. Thanks.
     
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