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Need help setting up the problem.

  1. Apr 7, 2012 #1
    Hello,
    I was looking to get some help with setting up this problem so I may solve it. For some reason I just don't see what to do to work towards a solution.
    1. The problem statement, all variables and given/known data
    A 6kg collar is allowed to slide over a frictionless pole whose height above the ground obeys the parabolic equation y=8-(1/2)x^2. Attached to the collar is a k=30N/m spring. The spring is 1 m when unstretched and connected in a way that the spring will always start at the origin. If the collar started from rest at (0,8), how fast will it be traveling at x=2? Hint: Don't forget gravity!


    Any help to point me in the right direction would be fantastic. Thank you.
     
  2. jcsd
  3. Apr 7, 2012 #2

    tiny-tim

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    welcome to pf!

    hello qball1982! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)

    hint: try conservation of energy …

    kinetic energy + gravitational potential energy + elastic potential energy = constant …

    what do you get? :smile:
     
  4. Apr 7, 2012 #3
    Excellent! Thank you tiny-tim!

    I worked this out using the formula: K1+U1=K2+U2
    where U= mgy+1/2kx2

    Crunching the numbers I get :
    0+1205.4=K2+778.05
    427.35=K2
    427.35=1/2(6)v2

    then finally: v=11.935m/s

    Does this seem like the right path?
     
  5. Apr 8, 2012 #4

    tiny-tim

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    hi qball1982! :smile:

    (just got up :zzz:)

    yes that's the correct method :smile:

    (but we need to see your detailed calculations if you want us to check the actual figures … for example, did you use √(x2 + y2) in the spring PE?)
     
  6. Apr 8, 2012 #5
    Hello tiny-tim!

    I suppose exact calculations would help. Sorry about that. :wink:

    Spring displacement calculations:
    x1 = 8-1 = 7 meters
    x2 = √(22+62) -1 = 5.32 meters
    And of course the -1 in each is due to the springs unstretched length of 1 meter.

    Spring Uel Calculations:
    Uel1= 1/2(30N/m)(7m)2 = 735N/m
    Uel2= 1/2(30N/m)(5.32m)2 = 425.25N/m

    Gravitational Potential Energy Calculations:
    Ug1= (6kg)(9.8m/s2)(8m)=470.4
    Ug2= (6kg)(9.8m/s2)(6m)= 352.8

    To solve:
    K1+Uel1[/SUB+Ug1=K2+Uel2+Ug2

    0+ 735+ 470.4 =K2+ 425.25 + 352.8
    1205.2 = K2 + 778.05
    427.15 =K2
    427.15 = 1/2(m)(v2)
    427.15 = 3(v2)
    142.38 = (v2)
    v=√(142.38)
    v = 11.93 m/s

    I think it looks right but I am also wrong quite a bit too. :smile:

    Thanks again for all the help!
     
  7. Apr 8, 2012 #6

    tiny-tim

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    yes that looks ok :smile:

    just one thing, you could save time and make it look neater if instead of doing all this …
    you just work out Uel1 - Uel2 = 1/2(30N/m){(7m)2 - (5.32m)2} :wink:

    (and the same for Ug)
     
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