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Potential energy, and use of datum line

  1. Feb 27, 2015 #1

    MacLaddy

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    1. The problem statement, all variables and given/known data

    The mechanism in the attached photo contains two masses, each treated as particles. When released from rest, the 4-kg block falls and the 6-kg sphere rises. Initially, the angle between the two rigid, massless linkages is θ = 60°

    What is the velocity of the sphere when θ = 180°?


    2. Relevant equations

    $$T_0 + V_0 = T + V$$


    3. The attempt at a solution

    I have two questions about my attempt at this problem. One is regarding my use of the datum line, and if I am using it correctly. The other is regarding the final velocity of the falling 4kg block. I am using the assumption that it will be zero, as it needs to change directions, but I am not sure if that is right. I have seen discussions on here in the past about the change in speed of a piston vs. the crank, but I don't recall what the final verdict was.

    Anyhow, here is my attempt. Please let me know if I am on the correct track.

    [itex]T_0 + V_0 = T + V[/itex]

    Labeling the 4kg mass as "A", and the 6kg as "B", and using the pivot point just under the B mass as my datum line. Also assuming that [itex]T_0=0[/itex], as nothing is initially moving.

    [itex]W_Ay_0 + W_By_0 = \frac{1}{2}m_Av_A^2 + \frac{1}{2}m_Bv_B^2 + W_Ay + W_By[/itex]

    Assuming [itex]T_{A}=0[/itex]

    [itex]W_Ay_0 + W_By_0 = \frac{1}{2}m_Bv_B^2 + W_Ay + W_By[/itex]

    Solving for [itex]v_B[/itex] yields

    [itex]v_B=\sqrt{2\frac{W_Ay_0 + W_By_0 - W_Ay - W_By}{m_B}}[/itex]

    [itex]W_Ay_0=4kg(9.81 m/s^2)(-0.3m)=-11.772 kg*m^2/s^2[/itex]

    [itex]W_Ay=4kg(9.81 m/s^2)(-0.6m)=-23.544 kg*m^2/s^2[/itex]

    [itex]W_By_0=6kg(9.81 m/s^2)(0.3\sin(30)m)=8.829 kg*m^2/s^2[/itex]

    [itex]W_By=6kg(9.81m/s^2)(0.3m)=17.658 kg*m^2/s^2[/itex]

    Plugging everything in yields
    $$v_B = 0.99 m/s \approx 1 m/s $$

    Look about right?

    Any advice is, as always, greatly appreciated.

    Thanks,
    Mac
     

    Attached Files:

  2. jcsd
  3. Feb 27, 2015 #2

    BvU

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    Looks just fine.

    No urgent advice, but plugging in is best postponed until the end (you never know what cancels, you run less risk of errors and you can check dimensions).
    I.e. at the end ## {1\over 2} l\; \Delta m\; g = {1 \over 2} m_2 \;v^2 \rightarrow v^2 = g \; { 1\over 10}##
     
  4. Feb 27, 2015 #3

    MacLaddy

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    Thanks BvU. Good advice on waiting, I typically get impatient and end up with rounding errors.
     
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