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Simple pendulum: Determine the velocity

  1. Aug 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A homogeneous bar with length 0.6 m and mass m = 2 kg is fixed to a wall via a hinged connection in the vertical plane. At the end of the bar a constant force F acts of 150 N. The bar is released from the vertical equibrilium position. Determine the velocity (of the COG) and the angular velocity of the bar when the bar reaches a horizontal postition.

    The answer from the book is: [itex]\omega [/itex] = 26.4 rad/s and [itex]v_{COG}[/itex] = 7.9 m/s

    2. Relevant equations
    Work moment : [tex] M = I \theta [/tex]
    Rotational energy:

    [tex] \frac{1}{2} I \omega ^2 [/tex]

    3. The attempt at a solution
    [tex] M \theta =\frac{1}{2} I \omega ^2 =\frac{1}{2} (\frac{1}{3}ML^2) \omega ^2 [/tex]

    [tex] 150 \cdot0.6 \cdot \pi/2= 1/2 \cdot 1/3\cdot 2 \cdot 0.6^2 \cdot \omega^2 \rightarrow \omega =\ 34.32 rad/s [/tex]

    What mistake am I making?
     
    Last edited: Aug 5, 2015
  2. jcsd
  3. Aug 5, 2015 #2

    DEvens

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    I see no gravity here.
     
  4. Aug 5, 2015 #3

    andrevdh

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    Potential energy
     
  5. Aug 5, 2015 #4
    The gravity is not mentioned in the exercise but let's add it:

    [tex] M \theta =\frac{1}{2} I \omega ^2 + mg\Delta h \rightarrow\ \omega = 23.5\ rad/s [/tex]

    Why is my answer not agreeing with the one from the book?
     
    Last edited: Aug 5, 2015
  6. Aug 5, 2015 #5

    haruspex

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    The question is not clear. Which way is F acting initially, and which way does it act later?
    It says the force is constant, not that its magnitude is constant. That suggests its direction is constant.
     
  7. Aug 6, 2015 #6
    Yeah you're right the question is posed wrongly.. I found out I get the right answer if I assume the force acts horizontally.
     
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