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Rotation Question -Spring

  1. Feb 19, 2017 #1
    1. The problem statement, all variables and given/known data
    Overhead view of a spring lying on a frictionless surface and attached to a pivot at its right end.The spring has a relaxed lenght of ##l_0=1.00m## and a negligible mass.A small 0.100 kg disk is attached to the free end at left.That disk is then gvena velocity ##\vec {v_0}## of magnitude ##11\frac {m} {s}## perpendicular to the springs lenght.The disk and spring then move around the pivot.
    (a)When the streching of the spring reaches its maximum value of ##0.100l_0##, what is the speed of disk,
    (b)What is the spring constant ?
    Adsız.png

    2. Relevant equations
    Energy conservation.

    3. The attempt at a solution
    From newtonian equations we know that ##fx=\frac {mv^2} {r}## so from that we can obtain ##1.1k=v^2##

    Then I wrote energy conservation

    İnitally it has a speed ##11\frac {m} {s}## and later time it will have some velocity v' and theres also spring potantial energy

    Here I stucked,Do I have to add rotational energy of spring-disk system ? I think I should but I got wrong result.
    The equation will be ##\frac 1 2m(v_0)^2=\frac 1 2m(v')^2+\frac 1 2kx^2##
    here
    ##x=0.1m##
    or ##\frac 1 2m(v_0)^2=\frac 1 2mI(w)^2+\frac 1 2kx^2## but its same as upper equation.

    Any help ?
     
  2. jcsd
  3. Feb 19, 2017 #2

    kuruman

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    When you write F = kx for the magnitude of the spring force, x is not the length of the spring. It is the extra stretch (or compression) of the spring.
    Because you have a point mass, you can either write the rotational energy or the translational energy. As you have seen, they give the same result.
     
  4. Feb 19, 2017 #3
    I know I wrote k0.1=0.1v^2/1.1
     
  5. Feb 19, 2017 #4

    kuruman

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    Yes, I see now that you canceled the 0.1 for the mass m with the 0.1 for the displacement x.

    When you say you got the wrong result, what result did you get and how? Do you know what the "correct" result is?
     
  6. Feb 19, 2017 #5
    ##1.1k=(v')^2##
    ##\frac 1 2m(v_0)^2=\frac 1 2m(v')^2+\frac 1 2kx^2##here
    ##x=0.1m##

    so we can cancel ##\frac 1 2## then
    ##0.1kg (11\frac {m} {s})^2=0.1kg(v')^2+k(0.1m)^2##
    ##12.1J=0.11k+0.01k##
    ##12.1J=0.12k##
    ##k=100.83 \frac {N} {m}##

    answer is ##k=210\frac {N} {m}##
     
  7. Feb 19, 2017 #6

    kuruman

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    You conserved mechanical energy and that gave you one equation. Is there anything else that is also conserved? That is where you get the second equation you need.
     
  8. Feb 19, 2017 #7

    Angular momentum will be conserved cause ##τ=0## so ##L_f=L_i##
    ##1.1mv'=1.m(v_0)##

    from that ##v'=10\frac {m} {s}## and ##k=210\frac {N} {m} ##

    but still didnt understand why ##kx=\frac {mv^2} {r}## is not correct ?
     
  9. Feb 19, 2017 #8

    kuruman

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    Because it does not guarantee you that there is no radial component of the velocity. The best way to tackle problems of 2d motion involving central forces (forces that depend on r only) where angular momentum is conserved is to write the mechanical energy as
    $$ME = \frac{L^2}{2mr^2}+\frac{1}{2}mv_r^2+V(r)$$ where ##L## is the conserved angular momentum, here ##L=m v_0 l_0##, ##r=l_0+x## and ##V(r)=\frac{1}{2}k x^2##. The first term in the equation is the rotational energy of the mass and is also known as "the centrifugal potential."

    The condition for maximum extension of the spring is obtained by solving the ME conservation equation for ##v_r##, and then setting ##v_r = 0##. This results in $$\frac{k}{m}=\frac{v_0^2(2l_0+x)}{x(l_0+x)^2}.$$
     
  10. Feb 19, 2017 #9
    I see I guess,Is that means if theres radial veloctiy the force will be different then ##kx=\frac {mv^2} {r}## ?
     
  11. Feb 19, 2017 #10

    kuruman

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    When you say "the force" which force do you mean? There is only the force, the spring force, which provides both a centripetal acceleration and a radial acceleration. In that case, Newton's second law in the radial direction is $$m\frac{d^2x}{dt^2}=\frac{mv_{\theta}^2}{l_0+x}-kx$$
    Here, the left side of the equation, mass times radial acceleration, is zero instantaneously at maximum extension (or compression) of the spring. When the spring goes from zero to maximum extension, it does so in a way that conserves angular momentum. The mass goes around while executing radial oscillations. It is not in a circular orbit, which is what your initial solution assumes.
     
  12. Feb 19, 2017 #11
    I tried to meant force equation or the moton equation.Ok.Now I understand the whole idea.Thanks a lot :)
     
  13. Feb 19, 2017 #12
    Or wait I hate this but
    It gives my equation when left side is zero ? sorry I am tired but I have to ask
     
  14. Feb 19, 2017 #13
    Nevermind maybe I understand later.thanks again
     
  15. Feb 19, 2017 #14
    Or wait its not a general equaiton (##kx=\frac {mv^2} {r})##its true for instantenously thats why we cant apply ?
     
  16. Feb 19, 2017 #15

    kuruman

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    Yes.
     
  17. Feb 19, 2017 #16
    Finally,I understand no more questions.Thanks again :)
     
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