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Relationship Between Angular Velocity and the Spring Constant

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A mass attached to a spring which is mounted on a turntable. If you wanted to double the angular velocity, but keep the radius of the circle the same, by how much would you change the spring constant?

    2. Relevant equations

    [itex]\omega = \frac{v}{r}[/itex]




    3. The attempt at a solution

    I'm a bit confused on how these two are related.

    If the radius is staying the same & the goal is to double [itex]\omega[/itex], then the part of the equation which would change would be, v, the tangential velocity. I just can't figure out how k is related to tangential velocity. Can someone hel point me in the right direction? Thanks
     
  2. jcsd
  3. Feb 4, 2012 #2

    cepheid

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    For the mass to move in a circle, it must have a centripetal force acting on it. What's the equation for centripetal force? What must happen to the centripetal force if the velocity doubles, but the radius remains the same?

    Now, the centripetal force is just a requirement for circular motion. Something has to be physically present to provide that centripetal force, and if it can't, then the circular motion will cease. For example, with a planet orbiting a star, the force of gravity acts as the centripetal force that keeps the planet in its orbit, and if it is not strong enough given the speed of the object, that object will escape, instead of being in a bound orbit. Similarly, if you twirl a mass on a string in a circle around your head, the tension in the string is what acts as the centripetal force in this problem. If you twirl too fast, the motion will require more centripetal force than the string can provide, and it will snap.

    What is providing the centripetal force in THIS problem? The answer to that question will tell you how k comes into play.
     
  4. Feb 4, 2012 #3
    Well, I think the equation for the centripetal force would be [itex]\Delta r * k[/itex] , but I'm not sure if that is the formula you're looking for.

    If that is what you are referring to then the k would have to double because the [itex]\Delta r[/itex] is not changing.

    Am I on the right track?
     
  5. Feb 4, 2012 #4
    By what factor will the centripetal force increase if you double the angular velocity at constant radius?
     
  6. Feb 4, 2012 #5
    by two??
     
  7. Feb 4, 2012 #6

    cepheid

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    Yes, but you didn't directly answer the question I asked you:

    The answer is: the spring. The restoring force of the spring is what provides the centripetal force that keeps the mass moving in a circle in this situation.

    No. Doubling the angular speed at constant radius will not increase the centripetal force by a factor of two. To see why not, answer THIS question: what is the equation for centripetal force?
     
  8. Feb 4, 2012 #7
    [itex] k*x=m*r*\omega^{2}[/itex]

    and [itex]\omega^{2}= \frac{v^{2}}{r^{2}}[/itex]

    so i think the force, kx, is equal to [itex]\frac{m*v^{2}}{r}[/itex]

    If the above is true, then I can see how k is related to the velocity, but im not quite sure how to get v to double. Maybe multiply k by (1/2)??
     
  9. Feb 4, 2012 #8

    cepheid

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    Look at it from a reverse perspective: if v were to double, and r were to remain the same, what would happen to the centripetal force, mv^2 / r? By what factor would it increase? How would this increase in centripetal force be achieved by changing the spring constant?

    Alternatively, solve your equation for v. Since k is the only variable, and all the others are constants, you can see more easily what needs to be done to k in order to get v to double, when the equation is written in this form.
     
  10. Feb 4, 2012 #9
    Ok, now I see, it would take an increase of 4k to double the velocity with a constant radius. I was over thinking it...thanks for you help everyone
     
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