2
$\begingroup$

I'm trying to understand the following problem: A mass is attached to a spring and is set in uniform circular motion. We need to solve for how much the string is stretched, given the frequency of rotation, the unstretched length of the spring, the spring constant, and the mass of the object.

I would set up this problem like so (ignoring the directionality of the spring force):

$$F_s = kx = ma_c = m\frac{v^2}{r}=m\frac{(2\pi rf)^2}{r}=mr(2\pi f)^2$$ so $$kx = mr(2\pi f)^2$$

where k is the spring constant, m is the mass, r is the radius of curvature, f is the frequency, and x is the amount by which the spring is stretched.

Speed (v) is $2 \pi r\over T$ where T is period and $\frac{1}{T} = f$.

So then, since the radius of curvature is the unstretched length of the spring plus the amount it's stretched by, I'd substitute $r = r_0 + x$ where $r_0$ is the unstretched length of the spring. Then, we can solve for x, which gives

$$x = \frac{r_0(2\pi f)^2}{\frac{k}{m} - (2\pi f)^2}$$

As far as I can tell, this is the mathematically correct answer. However, this of course diverges at

$$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

which doesn't make physical sense. Furthermore, its answer approaches $r_0$ for values greater than the divergent f. I don't understand this behavior at all.

What would make intuitive sense is for the spring to be stretched more and more as the frequency gets larger. It won't be a linear relationship, but it shouldn't be this mess.

Is this some limitation of the model? Did I make a mistake in the setup? Did I make a math mistake? There's something in here I'm not getting. Any insight would be greatly appreciated.

Edit: (Wrong, see edit 2) Okay, I just realized I could simplify further:

$$x = \frac{mr_0 (2\pi f)^2}{k} - r_0$$

This gets rid of the divergence -- thank goodness. But there is still a zero point at $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$, which doesn't make sense. As I said above, the sensical thing would be for it to not have that $r_0$ on the end, so $x(f)$ just a regular parabola. Did I just make a math error?

Edit2: My simplification was wrong. The first equation is correct, and it diverges at the natural frequency.

$\endgroup$
  • $\begingroup$ Strange. $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ is the 'natural' frequency of a mass $m$ bobbing up and down from a $k$ spring. It doesn't appear to have anything with the problem, which you seem to have solved correctly. $\endgroup$ – Gert Dec 5 '17 at 19:22
1
$\begingroup$

Your "simplification" is incorrect. Your 1st equation is correct : $$x=\frac{r_0}{\frac{k}{m\omega^2}-1}$$ where $\omega=2\pi f$. This diverges $(x \to \infty)$ as $\frac{k}{m\omega^2} \to 1$ from above.

What this equation is telling you is that the spring is not stiff enough for circular motion to take place above the critical frequency $\omega_0=\sqrt{\frac{k}{m}}$.

enter image description here

For a fixed angular frequency $\omega$, the required centripetal force $F=m\omega^2 r$ is proportional to the radius $r$ (red lines in diagram). The slope of this line is $m\omega^2$. The spring force $F=kx=k(r-r_0)$ is also linear, with a slope of $k$ (blue line). Where the two lines cross, the spring force is sufficient to provide the required centripetal force at this radius.

As the angular frequency $\omega$ increases the centripetal force (red line) gets steeper and the crossing point moves further to the right. The spring stretches and the radius of the circular orbit increases. As $m\omega^2 \to k$ the red line becomes parallel to the blue line. The crossing point moves out to infinity, and the spring stretches without limit or until it breaks. When $m\omega^2 > k$ the two lines never cross, because the red line is steeper than the blue line. Bound motion (whether circular or elliptical) is no longer possible.

$\endgroup$
  • 1
    $\begingroup$ It would be good to mention what happens physically: when the rotational frequency is above the critical frequency, the spring is stretched without bound (or until it breaks). $\endgroup$ – Mark H Dec 5 '17 at 22:56
0
$\begingroup$

The thing you need to consider when thinking about your answer is that $F=-k(x-x_0)$ is only an approximation when the deviations from the rest length are small. In fact, as deviations grow larger, there are other terms, so it could well be written as: $$F=a(x-x_0)+b(x-x_0)^2+c(x-x_0)^3+.....$$

What you got here is the resonant frequency of such a system. Theoretically, it should be infinite at resonant frequency, but in reality there are inevitable imperfections in the spring, higher order terms to consider, etc. which all cause damping and while the stretch is maximum at said frequency, but it is not quite infinite.

$\endgroup$
0
$\begingroup$

which doesn't make physical sense.

Why not? You're trying to equate two forces. One of them (spring force) is linear with distance. The other (centripetal force) is quadratic with distance and frequency. As the required frequency increases, there is some point past which you cannot make them equal at any $x$. As $x$ increases, the required centripetal force increases faster than linear, so it never catches up. An ideal spring in this situation could not maintain rotation faster than that frequency.

Your second equation doesn't appear to be equivalent. If $f$ is zero, then $x$ should be zero. In that other equation, when $f = 0$ then $x= -r_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.