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Question on rotational motion

  1. Nov 14, 2009 #1
    Solid cube of side 2a and mass M is sliding on a frictionless table with velocity v. It hits a small obstacle at the end of the table causing it to tilt. What is the minimum velocity to cause the cube to fall off the table?

    Totally no clue where to start I don't even know whats the condition for the cube to fall off....

    A bola basically consist of three stones at the end of three cord so when it is thrown in air they will spread out with constant 120 degree angles between the cord. Assume three stones with mass m and cord with length l. The hunter takes one stone and swing the bola so 2 stones move together in a horizontal circle of radius 2l and speed v0. Calculate the angular speed of the bola about the center of mass.

    Using COM 2mv0=3mvf so vf=[tex]\frac{2}{3}[/tex]v0 and v=l[tex]\omega[/tex] so [tex]\omega[/tex]=[tex]\frac{2}{3l}[/tex]v0
    However if I use conservation of angular momentum,
    L=I[tex]\omega[/tex]
    mrvf=[tex]\frac{4}{3}[/tex]lmv0=I[tex]\omega[/tex]
    [tex]\omega[/tex]=[tex]\frac{4}{9l}[/tex]v0

    Why are the angular velocity different?
     
  2. jcsd
  3. Nov 14, 2009 #2

    tiny-tim

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    Hi semc! :wink:
    A lower edge AB of the cube stops at the edge of the table. The cube was going so fast that the opposite lower edge starts to lift up, and the whole cube rotates about edge AB. How fast does it have to be for the rotation to go so far that the cube rotates right round, and goes off the table? :smile:
     
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