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Speed of the rod as it leaves the rails?

  1. May 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails fig.1 that are the d=12.0 cm apart and L=45.0 cm long. The rod carries a current of I=48.0 A in the direction shown and rolls along the rails without sleeping. A uniform magnetic field of magnitude 0240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails?

    2. Relevant equations
    0+0+Fb *L*cos θ = (mv2)/2 +(Iω2)/2

    3. The attempt at a solution
    The rod gains energy from the rest, so begin kinetic energy and rotational energy are zero. The only source of energy rod gains is magnetic filed.
    I = (mr2)/2
    K=(mv2)/2
    w=v/r
    That is right side of equations becomes:
    (mv2)/2 + 1/2 (mv2)+1/2(1/2 mr2) * (v2/r2)

    Fb * L = 3/4 mv2
    Where
    Fb = I * d * B
    etc
    v=1.07 m/s

    Because the force move the rod parallel to the rails along x axis, the dot product will be:
    E=Fb * L * cos θ = Fb*L*1=FbL
    The question is how do we get the value of energy exerts magnetic field on the rod?
    That is force time Length of rails (Fb times L)?

    For the electric field is E = qFe
     

    Attached Files:

  2. jcsd
  3. May 2, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It looks like you are adding the kinetic energy twice here (highlighted in bold).
    The line afterwards is right again.
    Sure.

    I don't see an electric field.

    Slipping, not sleeping, by the way.
     
  4. May 3, 2015 #3
    Great!.

    Indeed I have added it twice mv2/2 (This is my type fault).
    • I was wondering about the energy was added in the left side Fb times L and which is likely the energy contribution of magnetic field.
    • "Sleeping" = "Slipping"
    Thanks a lot!
     
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