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Rotational force caused by a rod through the radius of a cylinder

  1. Jan 12, 2009 #1
    I am playing with math and physics during my off-time at work, and I am wondering how to calculate a specific force, or if I am even looking at it correctly. Here it is.

    A rod through a cylinder's radius so that if it is off to one side causes rotational force equal to the difference in foot poundsof the two sides seems simple enough. But what happens once the cylinder has rotated enough so that the end of the rod is touching the same surface that the cylinder is resting on? How do I calculate that force. I am assuming that the cylinder has a hole through it just larger than the rod to allow motion. Where is the force applied? Thanks in advance for the answers :)
  2. jcsd
  3. Jan 14, 2009 #2
    I find your description a bit unclear, so let me ask. You have this cylinder lying down on a surface, a rod is inserted "in the side of it", ie. in radial direction so that it's not symmetrical but only points out on one side of the cylinder. And then you rotate it from one side til it hits the surface on the other drawing out the path of a half circle?
  4. Jan 14, 2009 #3


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    This problem was posted before, but I don't recall the exact wording required to find that thread or the answer.

    Assume a rod goes through a pair of frictionless holes or perhaps even better, a frictionless tube across the diameter of the cylinder. Assume that the initial condtion is that the end of the rod rests on the same surface as the cylinder. What are the forces, and could any movement result from these forces?

    In the previous case, there wasn't any maximum length given to the rod, so a very long rod could be well extended, resting on the surface at a near horizontal angle. Another critical point would be when the rod was vertical (assume the rod has a rounded end and ther's no rolling resistance). I'm not sure if the rod was considered to have any mass at all in the previous thread.
  5. Jan 14, 2009 #4
    Yes Jeff has my basic description covered quite well. Frictionless hole through the diameter with a rod inserted into that hole. basically tip the cylinder onto one of its flats, then insert a rod through it from left to right, directly through the middle of the cylinder. then tip it back up onto its round side and you've got a good idea of how the rod placement should be. The only thing other than that to remember is that the rod is free to slip out of the hole to a certain distance. Now, with this information, how would I calculate what kind of resistance to continued rotation would be caused when the protruding end of the rod strikes the surface that the cylinder is rolling on?

    Add: I am trying to calculate as if the rod is weightless, but with 1lb of weight at either end located exactly on the ends of the rod. My theory is that the weight of the rod is pushing down and sideways with a downward force equal to a larger amount for angles closer to horizontal, and vice versa for the sideways push. Am I anywhere near How I should be thinking on this? And how does this linear force translate to rotational force?
    Last edited: Jan 15, 2009
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