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I'm having trouble wihh this problem I got in an exam and I got it wrong. In the problem is you have a frictionless table and a nail in the centre where a spring is attached. At the end of the spring there is a mass attached to it. The problem doesn't have numbers, $K$ is the springs constant, $L$ is its natural length, $m$ is the mass of the object attached.

The question is: where is the position of the mass so that the movement is a uniform circular motion?

I feel like this problem is very vague. I have trouble understanding and imagining how can such movement occur. So far i have this:

\begin{equation} K(L-r)=mv^2/r \end{equation}

I'm almost sure this can't be the solution because I never specify the uniformity of the motion, but I don't know how to either. Still, if I consider that the elastic force only acts in the $r$ axes, it's like I'm saying it's uniform, right? Because there's no other force component that can change velocity's magnitude...

Still I'm almost sure I got it wrong because it seems too simple.

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To have a perfectly circular motion, the mass needs to be at an equilibrium position such that the centripetal force is equal to the spring force.

Using Hooke's law, if $x$ is the spring's elongation from equilibrium, then, the spring's force is $-Kx$.

Using centripetal equations of motions, the centripetal force is $\frac{mv^2}{r}=m\omega^2r$ where $\omega$ is the rotational speed. Here, $r=L+x$ for a stretched spring. Applying Newton's law we get:

$Kx = m\omega^2(L+x)$ and then,

$$x = \frac{m\omega^2L}{K-m\omega^2}$$

So, if the mass stands at $L+x$, the motion is perfectly circular.

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  • $\begingroup$ The spring force is the force causing the centripetal acceleration and it is not an equilibrium situation because the mass is accelerating. $\endgroup$
    – Farcher
    Dec 15, 2016 at 19:16
  • $\begingroup$ Indeed the mass is not in static equilibrium, but what I meant was in equilibrium in the radial direction. $\endgroup$
    – fgoudra
    Jan 5, 2017 at 16:56
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Anything kept on a table with a spring attached to it does not start to move on its own. So you have to give it an initial push perpendicular to the radius vector. That would elongate the spring and the spring will try to pull it back with a force. Hence the actual length of the spring would be $(r+x)$ and not $(r-x)$ as you have stated.

Also, there should be no tangential force because you state that the motion is uniform, hence its speed should not change.

And I think that the expression would be

$Kx=\frac{mv^2}{r+x}$

Where r is the length of the spring and x is the extension in the spring.

You see, if $x$ remains constant, the centripetal force will be constant and you can imagine the spring to be a string, as in string the tension remains constant for uniform circular motion and it happens in real life and thus is not vague. So similar stuff can also happen with springs!

( I hope you wont get this wrong next time in your exam :-P )

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The spring needs to apply a constant inward radial force on the mass.
To do that the spring must be stretched to a length $L+x$ and not be compressed to a length $L-x$.

The inward radial force on the mass due to the spring, $Kx$, produces the centripetal acceleration of the mass, $m(L+x) \omega ^2$, and Newton,s second law $F=ma$ connects the two quantities.

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