How Does Angular Velocity Affect Spring Extension in Circular Motion?

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SUMMARY

The discussion focuses on the relationship between angular velocity and spring extension in circular motion, specifically for a particle of mass ##m## attached to a spring with spring constant ##k## and natural length ##l##. It establishes that circular motion is possible if the angular velocity ##\omega## is less than ##\sqrt{k/m}##, leading to the extension of the spring given by the equation $$e = \frac{\omega^2l}{\frac{k}{m}-\omega^2}$$. Additionally, the frequency of small oscillations about this circular motion is derived as $$\Omega = \omega\sqrt{4 + \frac{l}{e}}$$, indicating the dynamics of the system under oscillatory conditions.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with spring mechanics and Hooke's law
  • Knowledge of circular motion dynamics
  • Basic differential equations for oscillatory motion
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  • Explore the implications of varying angular velocity on spring extension
  • Learn about the stability of oscillatory systems in physics
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Homework Statement


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 $$\Omega = \omega\sqrt{4 + \frac{l}{e}}$$

Homework 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)
$$r-l=e$$

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?
 
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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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