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Net Force & CM

  1. Aug 19, 2011 #1
    Here's a dumb question I came up with that I'm not seeing the answer to.

    For any system of particles, F_ext = M a_cm. Consider a uniform rod lying on a frictionless surface:

    -----------^
    ----------------
    --------------v

    Equal and opposite forces are exerted on the rod where the ^ and v are. The net external force is zero, so the center of mass, which is at the center of the rod, does not move. But obviously it would move if you applied those forces, right? The whole thing would rotate around a point between the two points of application.

    I'm probably making a really stupid mistake in thinking about this, so some help would be appreciated.
     
  2. jcsd
  3. Aug 19, 2011 #2
    Why do you think that they have a different point of application? Both force vectors act at the center of mass.

    -------^ N
    -------|--------
    -------v Fg

    Or are you referring to some other forces, not the weight and the normal?
     
    Last edited: Aug 19, 2011
  4. Aug 19, 2011 #3
    Nope, no weight and normal. This is a top-down view and I apply those two forces myself by, say, tapping the rod there.
     
  5. Aug 19, 2011 #4
    If you only give two taps at those points ,the final result will be a rotation about the center of mass along with a translational motion of the center of mass. Since you have a torque the rod will gain angular momentum .The motion of the rod exactly when the forces are applied is probably a mathematical mess.
     
  6. Aug 19, 2011 #5
    Where does the translational motion of the CM come from? The net force is zero, so the acceleration of the CM should be zero as well, right?
     
  7. Aug 19, 2011 #6

    vela

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    Right. The center of mass wouldn't move, and the rod would rotate about it. That might not match up with what you intuitively think will happen, but that's what happens.
     
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