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Central Hooke's Law Force

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle of mass ##m## is placed on a smooth table and attached to a fixed point ##O## on the table by a spring with spring constant ##k## and natural length ##l##.

    (i) Show that the particle can execute circular motion about ##O## with angular velocity ##\omega## provided ##\omega < k/m## , and calculate the resulting extension ##e## of the spring.

    (ii) Show that the frequency ##\Omega## of small oscillations of the particle about the circular motion computed in (i) is given by $$\Omega = \omega\sqrt{4 + \frac{l}{e}}$$


    2. Relevant equations
    $$F_s = -ke = -k(r-l) = -\frac{dU}{dr}$$
    $$m\ddot{r} = mr\omega^2 - \frac{dU}{dr}$$ (from Newton's 2nd or the Euler-Lagrange equation)
    $$r-l=e$$

    3. The attempt at a solution
    (i). In this case, ##\ddot{r}=0##, as ##r=##const. Thus,
    $$mr\omega^2 - k(r-l) = 0$$
    $$kl = kr - mr\omega^2$$
    $$\frac{k}{m}l = \frac{k}{m}r - \omega^2r$$
    $$\frac{k}{m}l= (\frac{k}{m} - \omega^2)r$$
    ##r>0## and ##\frac{k}{m}l > 0## (as ##k,m,l > 0##) so for this equation to hold (and therefore, for ##\ddot{r}## to equal ##0##, ##(\frac{k}{m} - \omega^2)>0##, so ##\omega<\sqrt{k/m}##.
    Plugging in ##r=l+e##.
    $$\frac{k}{m}l = (\frac{k}{m} - \omega^2)(l+e)$$
    $$\frac{k}{m}l = (\frac{k}{m} - \omega^2)l + (\frac{k}{m} - \omega^2)e$$
    $$(\ast)\:\frac{\omega^2l}{\frac{k}{m}-\omega^2} = e $$

    (ii). The radial equation of motion (plugging in ##r=l+e## once again) is
    $$m(l+e)\omega^2 - ke = m\ddot{e}$$
    $$m\omega^2l + m\omega^2e - ke = m\ddot{e}$$
    $$\ddot{e} + (\frac{k}{m} - \omega^2)e = \omega^2l$$
    $$e(t) = e_{hom}(t) + e_{part}(t)$$
    $$e(t) = A\cos((\frac{k}{m} - \omega^2)t-\delta) + \frac{\omega^2l}{\frac{k}{m} - \omega^2}$$, where ##A, \delta## are some undetermined constants from the solving of the differential equation.
    The (angular) frequency of ##e(t)## is then $$\Omega = \sqrt{\frac{k}{m} - \omega^2}$$
    Solving ##(\ast)## for ##\frac{k}{m} - \omega^2##, we find
    $$\frac{\omega^2l}{e} = \frac{k}{m} - \omega^2$$
    Thus,
    $$\Omega = \omega\sqrt{\frac{l}{e}}$$

    Apologies for the long post. Where exactly am I going wrong?
     
  2. jcsd
  3. Oct 9, 2014 #2
    I feel that you also need to consider the force balance in the theta direction, since, with the oscillatory motion, ##\omega## is varying with r.

    Chet
     
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