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Equations of Motion of a Mass Attached to Rotating Spring

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m is attached to the end of a light spring of equilibrium length a, whose other end is fixed, so that the spring is free to rotate in a horizontal plane. The tension in the spring is k times its extension. Initially the system is at rest and the particle is given an impulse that starts its movement at right angles to the spring with velocity v. Write down the equations of motion in polar co-ordinates.

    2. Relevant equations

    Radial Force Fr= -k⋅r
    Radial acceleration ar = mv2/r
    Tangential Force Fθ = Torque = r⋅F

    3. The attempt at a solution

    I tried applying Newton's second law for both the radial and tangential components.
    In r : mar= -k⋅r ⇒ r = a⋅cos(ωt) (which is not the correct answer)
    In θ: 1- mr⋅aθ= r⋅F (got me nowhere)
    2- vθ= v/r=ω ; ω=√k/m (how do I recover θ(t) from it?)

    Thank you in advance
     
  2. jcsd
  3. Oct 27, 2016 #2

    TSny

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    Welcome to PF!
    This does not account for the equilibrium length of the spring.
    Does the expression on the right have the correct dimensions for an acceleration?
    In polar coordinates, the radial acceleration ar has two terms. For a review of velocity and acceleration in polar coordinates, see http://faculty.etsu.edu/gardnerr/2110/notes-12e/c13s6.pdf
    Force and torque do not have the same dimensions. So, a force cannot equal a torque.
     
  4. Oct 27, 2016 #3
    Thank you for your reply!

    Following your hints, here's what I ended up with:
    In ##r##: ##~m( \ddot r -r \dot \theta^2)=- k(r-a)##
    In ##\theta##: ##~\dot\theta=\frac{v}{r}=-\frac{k}{mv}(r-a) ## (which I got from the centripetal force ##m\frac{v^2}{r}=-k(r-a)##)

    What's left to do is plug in ##\dot\theta## in the equation in ##r## and find ##r(t)##
    Is there still something incorrect in my reasoning?
     
    Last edited: Oct 27, 2016
  5. Oct 27, 2016 #4

    TSny

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    OK, this looks good. It expresses ##\sum F_r = ma_r##.
    This is not quite correct. Note that ##r \dot{\theta}## represents the ##\theta## component of ##\vec{v}##.
    No, the correct relation is the relation that you got above: ##m( \ddot r -r \dot \theta^2)=- k(r-a)##.

    OK. To get the correct expression for ##\dot \theta## in terms of ##r## you can do either of the following:
    (1) Set up ##\sum F_\theta = ma_\theta##. Consult the link for the expression for ##a_\theta##.

    (2) Decide if there is any torque acting on the particle relative to the origin. If not, what quantity is conserved?
     
  6. Oct 27, 2016 #5
    OK, got it. Thank you so much for your help! Cleared things up.
     
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