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How hard must you hit a rod for it to swing up 180 degrees?

  1. Oct 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A thin rod, 0.79 m long, is pivoted such that it hangs vertically from one end. You want to hit the free end of the rod just hard enough to get the rod to swing all the way up and over the pivot.
    How fast do you have to make the end go?

    2. Relevant equations
    P(change in potential energy)=K(change in rotational kinetic energy) = 1/2(I)(Wv)^2
    P=mgh (where h is change in height)
    I=1/12(mL^2) (the rotational inertia of a thin rod)
    V(rotational coordinate omega)=s/r (where s is distance traveled around circle and r is radius)
    Wv(rotational velocity)=(change in V)/(change it time)
    Wv=(Vt (tangential velocity))/r
    Ac (centripetal acceleration)=(V^2)/r
    *the change in height of the thin rod is the change in the center, not the tip
    *there could be more but I think I covered everything

    3. The attempt at a solution
    https://goo.gl/rsq8QO <---- url link to a picture of my work
    I know the answer is not 5.56 m/s or 3.78 m/s
     
  2. jcsd
  3. Oct 28, 2015 #2
    The rod is rotating about one end not the center for which your moment of inertia is for. Do you know the parallel axis theorem?
     
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