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Rotational motion and conservation of energy problem

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data

    The sliding block has a mass of 0.800 kg, the counterweight has a mass of 0.460 kg, and the pulley is a hollow cylinder with a mass of 0.350 kg, an inner radius of 0.020 m, and an outer radius of 0.030 m. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.820 m/s toward the pulley when it passes through a photogate.

    p10-43.gif

    (a) Use energy methods to predict its speed after it has moved to a second photogate, 0.700 m away.


    2. Relevant equations

    KET = (1/2)mv2
    KER = (1/2)Iw2

    I for a hollow cylinder = (1/2)M(R12 + R22)

    w = omega = angular velocity

    3. The attempt at a solution

    I know work is equal to the change in kinetic energy so I go:

    W = ((1/2)(M1+M2)Vf2 + (1/2)IwF2) - ((1/2)(M1 + M2)VI2 + (1/2)Iw2 + M2gs)

    and W = -f * s = - (coefficient of friction)M1gs

    s = .7 as given in the problem...

    So I substitute that expression for work into the first equation, solve for VF but the website I'm using is telling me I'm getting the wrong answer... what am I doing wrong?
     

    Attached Files:

  2. jcsd
  3. Feb 18, 2009 #2

    Tom Mattson

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    That's not right. There is no reason that the work done by friction must equal the change in gravitational potential energy of the object. The Work-Energy Theorem states that the net work done on a system equals the change in total kinetic energy of the system. So we have:

    [tex]W_{total}=\Delta K_{total}[/tex]

    [tex]W_f+W_g=\Delta K_{block}+\Delta K_{pulley}+\Delta K_{counterweight}[/tex]

    [tex]-Fs+m_2gs=\frac{1}{2}m_1\left(v_2^2-v_1^2\right)+\frac{1}{2}I\left(\omega_2^2-\omega_1^2\right)+\frac{1}{2}m_2\left(v_2^2-v_1^2\right)[/tex]

    You'll need to relate [itex]\omega[/itex] to [itex]v[/itex], but other than that this should be a piece of cake. You know everything except [itex]v_2[/itex].
     
  4. Feb 18, 2009 #3
    Ok, isn't friction force equal to the coefficient of friction times the normal force though? ..and the normal force of the block = M1g since it isn't moving vertically?

    Thanks for the help by the way, besides that everything is clear now.
     
  5. Feb 19, 2009 #4

    Tom Mattson

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    Yes on both counts.

    No problem!
     
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