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Conservation of Angular Momentum w/ Spring

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Relaxed spring is sitting on a horizontal surface. A block is attached at one of its ends is kicked with a horizontal velocity, v1, given to it. The block will move and stretch. Find the distance, x, the spring will stretch. X is in meters.

    l0=1 m
    m=1 kg
    k=1 N/m
    v1=1 m/s


    2. Relevant equations

    F=-kx T=2∏√(m/k) w=√(k/m) w=v/l0 kx=mrw2 KE=(1/2)mwI


    3. The attempt at a solution
    Plugging in the given and solving for the variables I get w=1 rad/s, T=2∏, l0=1 m. The equation that I get to relate the initial and final lengths is mw2(l0+x)=mwl02

    I am having hard time understanding this material. I would really appreciate some help. Thanks.
     

    Attached Files:

  2. jcsd
  3. Apr 29, 2012 #2
    Hi. Consetvations of angular momentum and energy make sense, do they?
     
  4. Apr 29, 2012 #3
    yes it makes sense. similar to the idea of conservation of momentum, P=m*v.I also understand kinetic energy initial will equal kinetic energy final.
     
  5. Apr 29, 2012 #4
    Energy 1/2 mv1^2 = 1/2 mv2^2 + 1/2 k x^2
    Angular momentum mv1l0 = mv2(l0+x)
    They would make sense to you.
    Regards.
     
  6. Apr 29, 2012 #5
    Solving for x in the energy equation, I get x=1/v2

    Substituting 1/v2 in the angular momentum equation, I get v2=0. When I go back to solve for x, I get x=0.

    Is this correct???
     
    Last edited: Apr 29, 2012
  7. Apr 30, 2012 #6
    Hi.

    You should get quadratic equation from these equation. Do good practice in mathematics.

    Regards.
     
  8. Apr 30, 2012 #7

    gneill

    User Avatar

    Staff: Mentor

    Query: Do they want the extension when the spring is 90° to its initial position, or when the spring is at maximum extension?

    The reason I ask is there's no guarantee that the angle that the velocity V2 will be perpendicular to the spring when the "true anomaly" (the angle between the initial position and the current position) is 90°. In fact if I were to guess, I'd put that occurrence at 180° from the initial position.
     
  9. May 7, 2012 #8
    Sorry I did not notice your post. The extension is at 90 degrees.
     
  10. May 7, 2012 #9
    Sweet Springs thank you for your help!

    The final equation I get is x^3+2(x^2)-2=0, which gives me 0.839m as the final answer
     
  11. May 7, 2012 #10
    Question regarding this problem, how come I cannot use ma=m(v^2/r)=kx to solve this equation?
     
  12. May 8, 2012 #11
    Yes I also have this question.. Why cant we use ma=(mV^2)/r=kx
     
  13. May 9, 2012 #12
    Hi.
    By deleting v2, I have got the equation
    y^4 + 2y^3 + y^2 - 2By - B = 0 where B = mv1^2 / kl0^2 = 1, y=x/lo

    Interesting. Show me your result please.
    Regards.
     
    Last edited: May 9, 2012
  14. May 9, 2012 #13
    I did not use the m(v^2/r)=kx. I was curious as to why one could not use this relationship. m(v^2/r) is to be used in an uniform circle and this system would not produce an uniform circle, thus one would yield an incorrect result.

    Thank you for all of your help sweet springs.
     
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