1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Centrifugal force and elastic deformation

  1. Jan 24, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider a spring of natural length L_0 with constant k which rests on a horizontal frictionless surface. The spring is attached at one end to a fixed post and at the other end to a mass m. Suppose the spring is rotating around the post in a circle with angular velocity w. What is the new length of the spring?

    The above problem is simple enough, and I have found a solution to it. Now consider the following related problem: you have a circular elastic band of natural radius R_0, mass m, and spring constant k. It is rotating on frictionless surface about its center with angular velocity w. What is the new radius of the rubber band?

    I am trying to use my ideas from the first problem to solve the second problem.

    2. Relevant equations

    F=-Δx
    F_c=mw^2R

    3. The attempt at a solution

    First I will show my solution to the first problem. It is as simple as setting the centripetal force equal to the restoring force of the spring: F_s=kΔx=F_c=mw^2L, where Δx=L-L_0. Solving I find,

    L=(kL_0)/(k-mw^2)

    As for the second problem, the main difficulty is how to model an elastic band. One idea is to think about it as the sum of many small springs attached to eachother at the center of the circle, and extending to the boundary of the circle, with a small mass element on the other end (in other words, each mass element dm of the band can be thought of as attached to it's own spring).

    I am not really sure how to proceed this way, so another idea is to use Young's modulus and think about the stress/strain of the rubber band as it is deformed by the centrifugal force.

    Another idea is to use energy. The rotational energy of the band is 1/2Iw^2, where I=mR^2 (the band can be thought of as a hoop). Not sure about the elastic potential energy of the band.
     
    Last edited: Jan 24, 2016
  2. jcsd
  3. Jan 24, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Consider an element subtending a small angle ##d\theta## at the centre. If the tension in the band is T, what is the radial force on the element?
     
  4. Jan 24, 2016 #3
    Drawing a circle and doing using some geometry we find the radial force is F_r=2Tsin(dθ/2), which is also equal to the centripetal force on the element: F_c=dm*w^2*R. Thus it suffices to find the tension T, presumably in terms of the spring constant k. I'm not sure how to do this...
     
    Last edited: Jan 24, 2016
  5. Jan 24, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Use a small angle approximation for sine. In the limit it will be exact.

    You need to relate the overall extension to the tension etc.
     
  6. Jan 24, 2016 #5
    Clearly the tension T is a function of the radius R, the natural radius R_0, and k. Naively I guess that T=k(R-R_0) and doing the math I get

    R=2*pi*kR_0/(2*pi*k-mw^2)

    Interestingly this is of a similar form to the answer I gave to the first problem.

    Now my only assumption is T=k(R-R_0), which seems reasonable, because it is the analogous to Hooke's law for a spring, however it could be off by a constant factor.
     
  7. Jan 24, 2016 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If you allow for a rapid spin, and hence a large increase in radius, I get that r is the solution of a quadratic.
    For modest spins, the linear approximation is ##\Delta r=\frac{r_0\rho\omega^2}{2\pi k}##.
    Yes, it must be so.
     
  8. Jan 24, 2016 #7
    I don't believe that r is the solution of a quadratic. I actually made a mistake in post #3 which I later corrected. I originally wrote F_c=dm*w*R^2 and corrected it to F_c=dm*w^2*R. When I used the first formula I got a solution with a quadratic as well, which leads me to believe you did not catch my mistake.

    I believe the correct derivation is as follows: as you said, we can use the small angle approximation, so the radial force on the small mass element dm is F_r=T*dθ, which equals the centripetal force F_c=dm*w^2*R. So T*dθ=dm*w^2*R, and summing over all mass elements we get T*2pi=mw^2R. Substitute T=k(R-R_0) and the result follows immediately.

    By the way, what is the justification for T=k(R-R_0)?
     
  9. Jan 24, 2016 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, sorry, that was wrong. It should be ##T=2\pi k(r-r_0)##. The increase in length is ##2\pi (r-r_0)##.
    No, you're right. (Not having a good day! Need to be more careful with my scribblings.)
    Putting that together I get the rather surprising ##r=\frac{4\pi^2kr_0}{4\pi^2k-m\omega^2}##.
    Note the consequence as omega increases.
     
  10. Jan 24, 2016 #9
    Right that was my other question, as w increases, r becomes negative. What is the interpretation of this?
     
  11. Jan 24, 2016 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It's not so much that it becomes negative... on the way it would become infinite. That is, for any given k and m there is a critical rate at which the radius will just keep increasing. Not surprising, when you think about it.
     
  12. Jan 24, 2016 #11
    It's sort of surprising to me...I mean what happens when you go past that critical rate?
     
  13. Jan 24, 2016 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Consider r >> r0. Required tension for the centripetal acceleration is proportional to r; actual tension is also almost proportional to r. So there comes a point at which getting wider to increase the tension is self-defeating.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Centrifugal force and elastic deformation
  1. Elastic deformation (Replies: 6)

  2. Centrifugal Forces (Replies: 3)

  3. Centrifugal Force (Replies: 4)

  4. Centrifugal force (Replies: 3)

Loading...