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Terminal velocity problem

  1. Apr 16, 2006 #1
    A bar of negligible resistance and mass of 17 kg is pulled horizontally across frictionless parallel rails, also of negligible resistance, by a massless string that passes over an ideal pulley and is attached to a suspended mass of 190 g. The uniform magnetic field has a magnitude of 600 mT, and the distance between the rails is 33 cm. The rails are connected at one end by a load resistor of 34 m[tex]\omega [/tex].What is the magnitude of the terminal velocity reached by the bar? Answer in units of m/s.

    First I drew a free body diagram on the weight, and found that [tex] F_g= Mg= F_m= ILB [/tex]
    so I= Mg/LB
    Then I used Ohm's Law.
    I= E/R= [tex] 1/R * d\phi /dt [/tex]
    and flux= BA
    so [tex] d\phi/dt [/tex] = B (dA/dt)= BLV
    then I= BLV/R , where V is the terminal velocity
    so Mg/LB = BLV/R
    Solving for V gives MgR/L^2 *B^2.
    Then I plugged in my numbers
    V= (.19)(9.8)(.034)/(.33^2)(.6^2)
    and I got 1.61 m/s
    The second part is: What is the acceleration when the velocity v= 1.1 m/s?
    I set 1.1 =(.19)(.034)g/(.33^2)(.6^2) and got 6.68 m/s.. which isn't right.. can someone please help?
  2. jcsd
  3. Apr 17, 2006 #2

    Andrew Mason

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    Where does the 17kg mass of the bar fit into this? The dropping weight provides the force that accelerates both itself and the bar.

  4. Apr 17, 2006 #3
    so would I just do
    1.1= (.19+17)(.034)g/(.33^2)(.6^2) ?
    Thats the only place that I think the mass can go.
  5. Apr 17, 2006 #4

    Andrew Mason

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    You are using a condition for v that applies only where the acceleration is 0.

    Go back to the beginning:

    [tex]Mg - T = Ma[/tex] and

    [tex]T - F_m = m_{bar}a[/tex] so:

    [tex]Mg - F_m = (m_{bar} + M)a[/tex]


    [tex]a = (Mg - F_m)/(m_{bar} + M)[/tex]


    [tex]a = (Mg - B^2L^2v/R)/(m_{bar} + M)[/tex]

    Just plug in the numbers.

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