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Rotating Rod - Please check answer.

  1. May 18, 2003 #1
    Here's the problem:

    A uniform rod of mass M and length L stands vertically on the edge of a frictionless table. A blow of impulse J is applied horizontally at the lower end of the rod.

    a) What are the initial horizontal velocity of the CM of the rod relative to the table, and the velocity of the lower end of the rod relative to the table?

    b) How far does the CM of the rod move horizontally in the time the rod rotates one revolution?


    w = angular velocty of rod.
    I = moment of inertia of rod
    Vcm = velocity of CM of rod
    Vtan = velocity of end of the rod

    OK, I place the origin at the bottom of the rod so that angular momentum is conserved.

    Iw = (L/2)MVcm
    [(1/12)ML^2]w = (L/s)MVcm

    Impulse in x direction is equal to the change of momentum in the x direction so: Vx = Vcm = J/M

    w = 6J/(ML) so Vtan = 3J/M => My textbook says it's 4J/M

    Plug w back into the equation above and you get.

    Vcm = J/(2M) => Textbook says it's J/M

    b) Using the above I get piL/6

    I've thought about the problem pretty carefully and can't see why my answers don't agree with the book.

    Can someone please check them.

    Thanks.
     
  2. jcsd
  3. May 18, 2003 #2
    You're very close -- I think you're just being a little inconsistent.

    You already said "Impulse in x direction is equal to the change of momentum in the x direction so: Vx = Vcm = J/M", and that's correct. So you already have the textbook answer for that part of the question. But then you went on and contradicted yourself a few lines later!

    As to the rotational motion, you say "I place the origin at the bottom of the rod so that angular momentum is conserved" but I don't think you actually did that. Actually, you placed your axis of rotation at the center of mass, not the bottom of the rod (and anyway, that's OK). Notice that you used the moment of inertia about the center of mass, (ML^2)/12. If you were figuring rotation about the end of the rod, you would have to use (ML^2)/3.

    You got w = 6J/ML which is correct, for rotation about the cm. And since Vtan=rw, and for rotation about the cm, r = L/2, Vtan=3J/M, just as you said.

    The only thing you really missed is that since your Vtan of 3J/M is relative to the cm, and the cm is moving along at a velocity of J/M relative to the table, the velocity of the lower end of the stick relative to the table is 3J/M + J/M = 4J/M.

    The rest is just arithmetic.

    (Keep 'em coming. My physics final is 9 days away & you're giving me a great review.)
    :smile:
     
  4. May 18, 2003 #3
    Thanks again Gnome,

    I should have seen that I solved for Vtan with respect to the CM of the rod and not the table and also that I already Vcm = J/M. I've been working on these problems all day gearing up for a test this Wednesday and I'm confusing myself a bit.

    I've posted another problem that I'm not quite able to work out.
     
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