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Charged mass connected to spring, swung in circle in mag. field

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A spring with an unstretched length of 20 cm and a force constant of 100 N/m is attached to a 2-kg mass with a charge of 3.0 C. The mass is swung in a circle in a zero gravity environment, so that the spring is perfectly horizontal and is parallel to the radius of the circle, as shown at right. A vertical 1.5-T magnetic field permeates the entire region.
    [Broken]

    If the mass is moving at 5 m/s, what is the maximum radius of its motion?

    2. Relevant equations
    Centripetal Force: [itex]F= \frac{mv^{2}}{r}[/itex]
    Hooke's Law: [itex]F=kx[/itex] where k is the spring constant and x is elongation (how long the string is stretched)
    Magnetic Force: [itex]F = qvBsin \theta[/itex]

    3. The attempt at a solution
    I tried qvBsin(90 degrees) = [itex]qvB =F= \frac{mv^{2}}{r} + kr[/itex]
    I assumed the force on the spring itself is the tenion (the constant * how long it stretched)

    I'm not sure if I did it correctly...please help.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 18, 2013 #2

    mukundpa

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    What is x in F = kx ?
     
  4. May 18, 2013 #3
    Reply

    I have updated the information in the question.
     
    Last edited by a moderator: May 6, 2017
  5. May 18, 2013 #4

    mukundpa

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    is the elongation is equal to radius of the circular path?
     
  6. May 18, 2013 #5
    response

    Yes, that is what I assumed. But, I can't get the radius in real number. What did I do wrong? Do you have suggestions for what I should do?

    Should I change something about [itex]qvB =F= \frac{mv^{2}}{r} + kr[/itex] or do something different?
     
    Last edited: May 18, 2013
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