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Find max. displacement of a spring

  1. Jul 25, 2011 #1
    1. The problem statement, all variables and given/known data
    A cap on a steel pin has a mass of m=500g and the pin has an angle of 50 degrees from the horizontal plane. The kinetic friction coefficient between the steel pin and the cap that glides down is uk=0,1. There is a spring at the bottom of the steel pin with spring konstant k=300 N/m. The distance travelled by the cap before it reaches the spring is d=1.16m.
    a) What speed does the cap have as it reaches the spring?
    b) What maximum change in length will the spring experience? (This is where I need help)

    [PLAIN]http://beborche.net/blandat/hylsan.jpg [Broken]



    2. Relevant equations
    From a) I know that the speed is v=4 m/s, and that the acceleration
    of the cap along the pin is 6.9 m/s².

    (g*sin(50) - uk*N = 7.51 - 0.31 = 6.9)


    3. The attempt at a solution
    Shouldn't one be able to put up the equation like this:
    ma = -kx
    and then solve for x?
    0.5*6.9 = -300x <=> x = - 0.5*6.9 / 300 = 0.0115meters = 1.15 cm

    The answers have a very different answer. I can state it here if you'd like, but does anyone know where I go wrong?
    Thanks.

    Edit: I have to use Mechanics laws and Force laws for this task.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 25, 2011 #2

    gneill

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    Staff: Mentor

    The cap hits the spring with a certain KE determined by its speed at impact. As the spring compresses, the cap continues to gain some KE from gravitational PE as it falls, but it also begins shifting KE into the spring in the form of spring PE, and it loses some KE to friction as it goes. When the KE of the cap is finally zero, the spring will be as compressed as it's going to get.

    You should be able to write an energy equation to balance all those terms.
     
  4. Jul 25, 2011 #3
    Hi and thanks for your reply. I forgot to mention that I have to solve the task both by using Energy laws, and by using Force laws and laws of mechanics. It is the force and mechanics solution I need help with...

    But yes, you're right it is solvable with energy laws.
     
  5. Jul 25, 2011 #4

    gneill

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    Staff: Mentor

    I see. The energy method is much simpler, but analyzing using forces is not too, too bad :smile:

    You should start by writing expressions for all the forces involved during the spring compression. I suggest taking an "x-axis" along the direction of motion. After this you'll have to construct a differential equation of motion from the forces which, with a little mathematical sleight of hand, is separable and integrable.
     
  6. Jul 25, 2011 #5
    Alright. Thank you for your guidance. I'll return with my results.
     
  7. Jul 25, 2011 #6
    Hi again. I'm getting stuck with the construction of the differential equation. Here are the forces I have identified:
    Fp = -kx
    wx = wsin[itex]\Theta[/itex] = mgsin[itex]\Theta[/itex]
    fk = ukN = ukmgcos[itex]\Theta[/itex]

    Now to the formation of the differential equation:

    Fnet = ma = m[itex]\frac{dv}{dt}[/itex] = // chain rule // = m[itex]\frac{dv}{ds}[/itex][itex]\frac{ds}{dt}[/itex] = mv[itex]\frac{dv}{ds}[/itex] = wx + Fp + Fk = mgsin[itex]\Theta[/itex] - kx - ukmgcos[itex]\Theta[/itex] <=>
    mv dv = mgsin[itex]\Theta[/itex] ds - kx ds - ukmgcos[itex]\Theta[/itex] ds <=>
    [itex]\int mv dv[/itex] = mgsin[itex]\Theta[/itex][itex]\int ds[/itex] - k[itex]\int x ds[/itex] - ukmgcos[itex]\Theta[/itex][itex]\int ds[/itex]
    m[v²/2][itex]^{0}_{4}[/itex] = mgsin[itex]\Theta[/itex][itex]^{s}_{0}[/itex] - k[s²/2][itex]^{s}_{0}[/itex] - ukmgcos[itex]\Theta[/itex][itex]^{s}_{0}[/itex]
    -8 = gsin[itex]\Theta[/itex]s - [itex]\frac{ks²}{2m}[/itex] - ukgcos[itex]\Theta[/itex]s

    after this I enter all values for all variables except for S. I put all numbers on the left hand side leaving 0 on the right hand side. Then I use "completing the square" method (or PQ-method) to solve it but it leaves me with a root of a negative number. Where am I going wrong? Is my differential equation correct? Please help.
     
    Last edited: Jul 25, 2011
  8. Jul 25, 2011 #7

    gneill

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    Staff: Mentor

    You've missed an S variable on the gsin(θ) term on your last line.
     
  9. Jul 25, 2011 #8
    I didn't miss it when I did ma calculations. Edited now! But can you see where I go wrong?
     
  10. Jul 25, 2011 #9

    gneill

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    Staff: Mentor

    When I solve for x in:

    [tex] -8\frac{m^2}{s^2} = g\;sin(\theta)\;x - \frac{k\;x^2}{2\;M} - \mu\;g\;cos(\theta)\;x [/tex]

    I get x = 0.175m and x = -0.152m.

    Quadratic formula suffices, with [itex]A = - \frac{k}{2\;M}[/itex]; [itex]B = g\;sin(\theta) - \mu\;g\;cos(\theta) [/itex]; [itex]C = 8\frac{m^2}{s^2}[/itex] .
     
  11. Jul 25, 2011 #10
    Credits to you! :P 0,175m is the correct answer. I dont understand a few things just yet but I'll dive into it and see if I get grasp it. Thanks a lot! and brb
     
  12. Jul 25, 2011 #11
    I understand everything except where m²/s² comes from. Unless they are the units for v², i.e. velocity squared? (In that case, why no units for gravity or mass?) thanks // Beborche
     
  13. Jul 25, 2011 #12

    gneill

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    Staff: Mentor

    The units on the 8 are indeed the units for a velocity squared. This is so because the '8' came from squaring the initial velocity. I like to carry my units along in my calculations so I can tell when I muck up the algebra! (When adding or subtracting items with units, the units have to agree or you've mucked up somewhere) Saves going through umpteen lines of derivation when the error could have been caught at about the second line!

    There are no units on the g or M terms because they are left in variable form -- the units would get plugged in when the numerical values are entered for solving the equation. You could have left the velocity term as vo2 and plugged the number in later.
     
  14. Jul 25, 2011 #13
    Alright, then I understand. Savvy :)
    And thanks for explaining.
     
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