Rotating spring

An elegant case: an object tied to a spring that rotates about a pivot on a smooth horizontal plane. Here it is the elastic force that provides the centripetal force, and the peculiarity is that the radius of the motion coincides with the length of the spring, which in turn depends on how much the spring stretches — a small game of mutual dependence to be solved.

During rotation the spring stretches to a length $\ell$ (greater than the rest length $\ell_0$). The elastic force $K(\ell - \ell_0)$ is directed towards the pivot and acts as the centripetal force. Since the radius of the motion is precisely $\ell$:

$K(\ell - \ell_0) = m\omega^2\ell$

This is a linear equation in the unknown $\ell$: collecting terms,

$\ell = \frac{K\,\ell_0}{K - m\omega^2}$

Key formula — Rotating spring

$K(\ell - \ell_0) = m\omega^2\ell$

When the spring "gives way"

If $K$ is too small compared with $m\omega^2$ (i.e. $K \leq m\omega^2$), the denominator $K - m\omega^2$ vanishes or becomes negative and the equation gives $\ell \to \infty$ or $\ell < 0$: the spring cannot sustain the rotation. Physically, the spring would stretch without limit or break, and the object would fly off. There is a critical rotation speed beyond which the system has no steady-state solution.

The numerical application is worked out in Ball on a rotating spring.

Links

Topics: Dynamics Concepts: Centripetal force · Elastic force and Hooke’s law Skills: Symbolic set-up Objects: Spring

Related exercises: Ball on a rotating spring · Why circular motion is accelerated · Maximum speed on a flat curve