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What does it mean for a particle to be free?

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    From an inertial reference frame S, the vector position of a particle of mass
    m1 = 1kg is given by r1(t)=(tx-hat - t^2y-hat)m.
    The vector position of a particle m2=2m1 is given by r2=(t)=(tx-hat +t^3y-hat)m

    Is particle #1 free along the y-direction? Explain



    2. Relevant equations
    If the particle is free along the y-direction then the y's would equal to 0.




    3. The attempt at a solution

    r1= -t^2y
    r2=+t^3y
    =not free?
    (I am actually unsure if my definition of free is true)
     
  2. jcsd
  3. Dec 7, 2008 #2

    Fredrik

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    "Free" means "unaffected by forces", or equivalently that the potential is constant. In this case, it means you have to check if it's accelerating in the y-direction or not.
     
  4. Dec 7, 2008 #3
    Particle #1's motion along the y-direction is -t^2.
    Would this mean that it is not free?
     
  5. Dec 8, 2008 #4

    Fredrik

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    That's its position at time t. What's its acceleration at time t?
     
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