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A point mass attached to a spinning spring.

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data
    A spring has mass Ms=120g and, if unstreched, length l=65cm. A boy attaches a point mass m=500g to one end of the spring. He holds the spring at the other end and spins it horizontally around his head, such that the point mass moves with speed v=12ms-1 and the spring is streched by 4.5cm. The effect of gravity can be neglected.
    A: Determine the force constant of the spring and list all forces that act upon the point mass.
    B: Find the total energy of the spring and point mass.

    2. Relevant equations
    arad=V2/r
    F=kx

    3. The attempt at a solution
    Forces on point mass:
    Tension from spring.
    I want to say centrifugal force pointing outwards, but I have a feeling I remember being told not to call it that?
    Gravity is told to be neglected, its a point mass and I have no info on friction so assume that can be neglected.

    Attempt at force on spring:
    Moving at constant velocity, so no linear acceleration. arad=v2/r. For the body as a whole, I'd think the r used here would be for the center of mass, which would be (rcm, spring*Mspring+rpoint*Mpoint)/Mtotal, which is 62.8cm from the middle. (Note: Rounded here, stored on calc for precise value in later calculations).
    (1/Mmass)*Fspring=0.5*0.045*k=v2/rcm
    k=(v2/rcm)/(lextention/Mmass)
    V=12, rcm=0.628, lext=0.045, mmass=0.5

    This gives over 2500 as the answer, which is obviously extremely wrong, and I'm really not sure what to do.





    For part B: Etot=Klinear+Krot+Uspring
    =0.5*Mtotal*12^2+0.5*I*ω2+0.5*k*lextension^2
    EDIT: Although thinking through this a bit more, I'm not sure I need both the Kinetic Energy terms...
    If that is correct:
    Not entirely sure on the moment of inertia - I know it's integrate of r^2 dm, but would I treat it as a line with an extremely high density at one end, sum the I of the spring with something?



    So, I'm stuck. Anyone able to point me in the right direction?
     
    Last edited: May 7, 2012
  2. jcsd
  3. May 8, 2012 #2
    It is a pseudo force, but since your frame of reference is not the stone itself, its effect can be considered.

    Ideas seem right upto here :smile:

    How did you get this? Isnt there another mass in play here?
     
  4. May 8, 2012 #3
    My logic was that the spring acting on itself was purely internal on that object so that the mass of the spring would only affect the whole system in that its mass would affect the centre of mass of the system. Where else does mspring come into play here?
     
    Last edited: May 8, 2012
  5. May 8, 2012 #4
    This is what i meant.
    Wouldn't it be 2*0.045*k=v2/r?

    Mspring doesn't effect the spring force equation as you said.
     
  6. May 8, 2012 #5

    Yeah, that is 2 - that typo only stayed as M, instead of 1/M, for that line, it went back to being a (1/0.5=2) the line below.
     
  7. May 8, 2012 #6
    Assuming your calculations are right, I think the answer for spring constant is correct.

    For part B, you need to consider the velocity of the spring and mass separately, as the velocity of the spring will depend on its distance(of COM) from the centre of the circle. Your original equation would be wrong since it considers the velocity of the spring as 12m/s. To stay on the safer side, its a better idea to calculate the moments of inertia separately too and plug them into the equation. Since the spring has uniform mass density it should be easy to calculate.
     
  8. May 8, 2012 #7
    So, treat the mass as just having linear velocity, the chain as the rotational, and the spring's potential?

    0.5*mmass*vmass2+0.5*Ichain*ω+0.5*k*lextension2
    0.5*0.5*122+0.5*0.6852*0.12*17.3^2+0.5*2500ish*0.045
     
    Last edited: May 8, 2012
  9. May 9, 2012 #8
    Yep.
     
    Last edited: May 9, 2012
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