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Not a clue Springs

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Ok sorry for the horrible drawing i freehanded it on paint.

    The 5lb collar is released from rest at A and travels along the smooth guide. Determine the speed of the collar just before it strikes the stop at B. The spring has an unstretched length of 12in. k=2 lb/in

    please note that from point C to point A it is a 1/4 circle of radius 12in
    From C to B, it is horizontal.


    2. Relevant equations



    3. The attempt at a solution

    I could do this problem if it was traveling from just C to B but the curve has me lost.

    I attempted to find my Y and i got y=(144-x^2)^1/2+10
    I solved that for x and got x=(144-(y-10)^2)^1/2
    I did this to find the hypotenus or the s length of the spring. would it be correct for my s to be (x^2+y^2)^1/2. i feel like im way off hlep
     

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    Last edited: Apr 8, 2009
  2. jcsd
  3. Apr 8, 2009 #2

    Doc Al

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    Not sure what you are trying to do. What are y, x, and s?

    Hint: Use energy conservation. Find the initial and final stretch of the spring.
     
  4. Apr 8, 2009 #3
    s is the length of spring. the initial stretch is 22in, and the final is 12in, but how does that help
     
  5. Apr 8, 2009 #4

    Doc Al

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    It will allow you to figure out the elastic potential energy stored in the spring. Be sure to take the unstretched length of the spring into account.
     
  6. Apr 9, 2009 #5
    I attempted to find my Y and i got y=(144-x^2)^1/2+10
    I solved that for x and got x=(144-(y-10)^2)^1/2


    am i correct is saying that my s would the hypotenus using the x and y above.

    I would then plug it into V(e) = .5ks^2 for elastic potential energy

    I believe I also have to use V(g) = Wy

    My main problem is that my y value is a peacewise function. From point B-C it is a different function than from C-A

    Note the Picture attached above
     
  7. Apr 9, 2009 #6

    Doc Al

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    Sorry, but I still have no idea what you are trying to calculate here.


    No. All you need to find is the initial and final stretch of the spring from its unstretched length--the curved section is irrelevant. Your previous post had all the data needed to find the spring PE.

    Yes, once you find the amount that the spring is stretched, that's how you find the elastic PE.

    If you mean gravitational PE, then yes, you definitely need that.

    So what? Energy is conserved. All you care about it is the total energy (elastic, gravitational, and kinetic) at points A and B.
     
  8. Apr 9, 2009 #7
    Ok I think i got it, i thought it would be much more complicated, my book is confusing

    ok heres what i did,

    V(A) = V(e) - V(g) = .5ks^2 - Wy = -3292 k=2lb/in s=10in W=5lb*32.2 ft/s^2 <-- should this be in/s^2 since k is in lb/in...... & y=22in

    T(A) = .5mv^2 = (5/2)v^2

    And V(C) & T(C) = 0 because the spring is no longer stretcher and the y = o

    so i got v = 36.29 in/s

    Are my units correct.
     
  9. Apr 9, 2009 #8
    Last edited by a moderator: May 4, 2017
  10. Apr 9, 2009 #9

    Doc Al

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    Why the minus sign?
    I would put all quantities into standard units, distances in feet. (k = 2lb/in = 24lb/ft.) Note that 5 lbs is the object's weight, not its mass, so no need to multiply by g when finding the gravitational PE. (What's the object's mass?)

    The collar starts from rest at point A.

    The elastic and gravitational PEs are both zero at point B (not point C, at least according to your diagram). The KE is not zero.
     
  11. Apr 10, 2009 #10
    I subtracted Wy because of a similar example in my book but i guess i read into it wrong

    ok so my new mass is 2.26 slug

    ok so my T(a) = 0, but my T(B) is now 1.134v^2

    plug and chug and i got a velocity of 11.98 ft/sec, that just seems wrong, too fast

    note the new pic
     

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  12. Apr 10, 2009 #11

    Doc Al

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    How did you get that?

    As a sanity check for the speed, how fast would something be moving if it were simply dropped from that same height? (That speed must be less than the speed of the collar in this problem, since there's no spring pulling it down.)
     
  13. Apr 11, 2009 #12
    1 pound = 0.0310809502 slug

    5lb = .1554045... oops calc error

    ok so i got at point A
    V(g)=9.157366, V(e)=9.96, T=0

    Point B
    V(g)=0, V(e)=0, T=.77v^2

    v=4.96 ft/sec I hope
     
  14. Apr 12, 2009 #13

    Doc Al

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    That's better.

    Your V(g) is close to my value: mgh = (5)*(22/12) = 9.167
    Show your calculation for V(e).

    Show your calculation for T.

    Did you do the sanity check I suggested?
     
  15. Apr 12, 2009 #14
    Point A

    Vg = .554*32.2*1.83 = 9.157366<-- Different than urs Doc
    Ve = .5*24*.83 = 9.96

    Point B

    T = .5*1.554*v^2 = .777v^2
     
  16. Apr 12, 2009 #15
    oops Ve = .5 *24*.83^2 = 8.2668

    v = 4.74
     
  17. Apr 13, 2009 #16

    Doc Al

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    You must have a typo in here, since the left and right sides don't match.
    Since mg is given as 5 lb, mgh = 5*(22/12) = 9.167
    Since mg = 5; m = 5/g = 5/32.2 = 0.1553
    The 0.83 (really 0.8333...) should be squared.

    Wrong mass.
     
  18. Apr 13, 2009 #17
    Sorry for the typo, i meant 1.554

    I thought that the mass was 1.554 slug, and 5lb was the force
     
  19. Apr 13, 2009 #18

    Doc Al

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    You're off by a factor of ten. The mass is 0.1554 slugs. (See my previous post.)
     
  20. Apr 13, 2009 #19
    man i keep making stupid mistakes

    V(g) = .1554*32.2*1.83333333333 = 9.1738
    V(e) = .5*24*.8333333333333333^2 = 10
    T(B) = .5*.1554*v^2 = .0777v^2

    Therefor v = 15.71 ft/sec
     
  21. Apr 13, 2009 #20

    Doc Al

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    Redo this one.
     
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