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Falling meter stick on a frictionless surface

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    The question I'm working on says "A special, frictionless, thin meter stick is initially standing vertically on the floor. If the meter stick falls over, with what angular velocity will it hit the floor?"


    2. Relevant equations
    v_rot=rw, I=1/12mL^2, KE=1/2Iw^2


    3. The attempt at a solution
    I know that the stick is going to be spinning about its center of mass, assumably in the center of the stick. I also know that the cm will be falling straight down. When I set mgh=1/2Iw^2 I get stuck trying to solve for w without being given the mass of the stick.
     
  2. jcsd
  3. Oct 19, 2009 #2

    rock.freak667

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    The mass of the stick as at the geometric centre. The point where it touches the ground, that acts like a pivot. So yes you must use mgh=1/2 Iω2

    But the 'I' in this case is not about the centre but about one end.
     
  4. Oct 19, 2009 #3
    Thanks. But I forgot to add that the problem also says "assume that the end in contact with the floor experiences no friction and slips freely." I assumed they meant that the end on the floor would be moving in the opposite direction and with equal velocity as the top of the stick, hence the rotation about the cm. Do I need to use I=1/3mL^2?
     
  5. Oct 19, 2009 #4
    And what about h? Do I set it equal to the cm, 1/2m?
     
  6. Oct 19, 2009 #5

    rock.freak667

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    Well it still means it rotates about the end, slips freely and no friction means that in your energy equation, you don't need to account for friction.

    So yes I=(1/3)mL2

    so what in mgh=(1/2)Iω2, what does I change to? and what is h in terms of L? (the centre of mass is at the geometric centre of the stick)
     
  7. Oct 19, 2009 #6
    ok, so substituting back in, I get w=sqrt(3g).

    Thanks
     
  8. Oct 19, 2009 #7

    rock.freak667

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    Yep...I was almost going to tell you it should be ω=√(3g/L) when I remembered it was a metre stick :redface: . Anyhow, it's still correct.
     
  9. Oct 31, 2009 #8
    FYI....In case anyone is reading this thread for help with their own homework, I thought about the problem one last time before I turned it in and changed my answer. After receiving my homework back, with the correct answer, the actual formula would be sqrt((24gh)/(l^2)). The way to think about it is since the stick is standing on a frictionless surface, once the top of the stick moves past vertical there is an x-component of its weight pushing the bottom sideways. Therefore, the stick rotates about its center (I=1/12MR^2) with the cm falling straight down. Be careful, because this problem didn't indicate an initial velocity for the top of the stick, which means that it was allowed to just fall over not pushed over.

    Hope this helps clarify the wonderful world of physics.
     
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