How Does Angular Velocity Affect Spring Extension in Circular Motion?

In summary, the conversation discusses a particle of mass ##m## attached to a fixed point ##O## on a smooth table by a spring with spring constant ##k## and natural length ##l##. It explores the possibility of the particle executing circular motion about ##O## with angular velocity ##\omega## and the resulting extension ##e## of the spring. The conversation also examines the frequency ##\Omega## of small oscillations of the particle about the circular motion, which is given by the equation $$\Omega = \omega\sqrt{4 + \frac{l}{e}}$$ The solution involves solving the radial equation of motion and considering the force balance in the theta direction."
  • #1
pixatlazaki
9
1

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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  • #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
 

Related to How Does Angular Velocity Affect Spring Extension in Circular Motion?

1. What is Hooke's Law?

Hooke's Law is a scientific principle that states that the force required to extend or compress a spring is directly proportional to the distance the spring is stretched or compressed.

2. What is the formula for Hooke's Law?

The formula for Hooke's Law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the distance the spring is stretched or compressed.

3. What is the significance of the negative sign in Hooke's Law formula?

The negative sign in the formula indicates that the force applied to the spring is in the opposite direction of the displacement of the spring. This means that when the spring is compressed, the force is directed outwards, and when the spring is stretched, the force is directed inwards.

4. How is Hooke's Law used in real life?

Hooke's Law is used in many real-life applications, such as in the design of suspension systems, measuring tools, and shock absorbers. It is also used in the study of elasticity and materials science.

5. What is the difference between Hooke's Law and Central Hooke's Law?

Central Hooke's Law is a modified version of Hooke's Law that takes into account the presence of a central force acting on the spring. This central force can be caused by the weight of the spring itself or other external forces acting on the spring. The formula for Central Hooke's Law is F = -kx + k'y, where k' is the spring constant for the central force and y is the displacement caused by the central force.

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