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Rotational force compared to spring constant?

  1. Dec 29, 2004 #1
    Maybe I'm just losing it but I can't seem to find a way to reduce this equation to the terms it requires:

    A weight is suspended on the end of spring that is stiff enough to have no perceptible sag or bend and a equilibrium length of b. If the system is undergoing steady circular motion in which one rotation takes time T, create an equation to calculate the length of the spring in terms of b,T, and k (spring constant).

    I started out by making (M*4*pi^2*b)/T^2=k(L-b) where L is the current Length of the spring while under rotation and M is Mass.

    If I solve for L I get L=(M*4*pi^2*b+b)/T^2*k, but am unable to get rid of the M for Mass. I suspect that it could be canceled out by the mass component of k, but then that would not leave a k in the equation.

    Am I way off on this? Can it be expressed in only those three terms? Any help would be much appreciated.
     
  2. jcsd
  3. Dec 29, 2004 #2

    Gokul43201

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    Was there a picture accompanying this ?

    What I picture is an initially vertical spring hanging from the ceiling, with the mass now executing horizontal circles.

    In this case you have :

    [tex]\frac{M \cdot 4 \pi^2 x^2}{xT^2} =kx \cdot \frac{\sqrt{b^2 + x^2} - b}{\sqrt{b^2 + x^2}} [/tex]

    where x is the radius of the circle, so [itex]\sqrt{b^2 + x^2} [/itex] is the number you want.

    Clearly, you've pictured this differently, so please throw in an explanation of the geometry.
     
    Last edited: Dec 29, 2004
  4. Dec 29, 2004 #3

    Andrew Mason

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    This is a complicated problem. Is the equilibrium length b with or without the mass suspended from it? Assuming it is, just a few observations that may be of some help:

    1. the vertical force is the vertical component of kx = mg where x is the extension of the spring. Length of spring, L = x+b
    2. the horizontal force is the horizontal component of kx.
    This provides the centripetal force:
    [tex]F_c = m\omega ^2R = \frac{4\pi^2R}{T^2}[/tex]
    where R = Lsinz where z is the angle from vertical.
    3. the ratio of horizontal force to vertical force is the tangent of the angle z and is: [itex]tanz = \omega^2R/g[/itex]. This means [itex]cosz=g/\omega^2L[/itex]

    Good luck.

    AM
     
  5. Dec 30, 2004 #4
    Clarification

    Sorry I tried to quickly summerize the problem, but I guess i didn't make it very clear. Her is a link to the problem online.
    http://www.faqs.org/docs/Newtonian/Newtonian_183.htm

    Now unless I'm mistaked the wording seems to suggest that the relevent forces are all acting on the horizontal plane, with gravity being cancled out by the verticle stiffness of the spring. Is that how you guys read it?

    With only foces in the horizontal plane it seems like it would be a simple matter to equate the spring force to the rotational force. But I wasn't able to get rid of Mass as a componant. Mabey I am making it to simple? should I be factoring other planes of force? again sorry for the confusion let me know you impressions of the problem or any glaring oversights I've made. Thanks.
     
  6. Dec 30, 2004 #5

    Gokul43201

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    Okay, this is not what I had pictured (nor do I suspect, had AM).

    Yes, all you have to do is equate forces along the horizontal plane.

    [tex] Mv^2/l = k (l-b) ~~[/tex]

    Now use the fact that

    [tex]T = \sqrt{M/k}= 2\pi l /v [/tex]

    to substitute for M, v in terms of T, in the first equation.
     
    Last edited: Dec 30, 2004
  7. Dec 30, 2004 #6

    Andrew Mason

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    Ok. This is a much simpler problem since gravity is not an issue. It may be simpler to work in angular velocity and then convert to the period using [itex]\omega = 2\pi/T[/itex]. You seem to have the centripetal acceleration as proportional to b. It is actually proportional to L. In other words, try starting with:

    [tex]K(L-b) = M\omega^2L[/tex]

    I do agree, however, that M has to be a factor in the expression for L.

    AM
     
  8. Jan 3, 2005 #7
    Once more into the breach

    Ok so at least I know I was on the right track. Thanks both of you guys for responding so quickly after my 2nd post, yes your right L should be the radius, and good call Gokul43201 on the T=sqrt(M/k).
    If I could trouble you for a little more of your time however, if I solve Gokul43201's equation for M and substitute it in my original equation it not only cancels out M, but T and k as well. This leaves me with something like: L=b/(4*pi^2-1).
    This doesn't seem to fulfill the requirements the original problem was asking for especially when you look at part 2, where it asks you to analyze the problem for specific values of T. Any further thoughts on this problem? I'm having some trouble picturing what it is driving at in the second part much less defining it mathematically. Once again any tips or pointers would be appreciated.
     
  9. Jan 3, 2005 #8

    Andrew Mason

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    The physical interpretation of this system is that it has a maximum period which Gokul correctly gave you. But that is derived from the equation of motion for the system, which is [tex]K(L-b) = m\omega^2L[/tex]. Work out the expression for L from this.

    As

    [tex]T \rightarrow 2\pi \sqrt{\frac{m}{k}}[/tex]

    what happens to L?

    AM
     
    Last edited: Jan 3, 2005
  10. Jan 20, 2005 #9
    Thanks

    Sorry to be so long responding, I got involved with something else. That finally makes sense, Thank you for all your patence.
     
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