Ball on a rotating spring

Problem

A ball of $2\;\text{kg}$ is tied to a spring ($K = 2000\;\text{N/m}$, rest length $\ell_0 = 1\;\text{m}$) which rotates with period $T = 0{,}5\;\text{s}$ about a pivot. Find the length of the spring during the rotation.

Solution

Data. $m = 2\;\text{kg}$, $K = 2000\;\text{N/m}$, $\ell_0 = 1\;\text{m}$, $T = 0{,}5\;\text{s}$.

Angular velocity: $\omega = \frac{2\pi}{T} = \frac{2\pi}{0{,}5} = 4\pi\;\text{rad/s}$

Setup. During the rotation the spring stretches: the elastic force provides all of the centripetal force. The radius of the motion coincides with the length $\ell$ of the spring, so: $K(\ell - \ell_0) = m\omega^2 \ell$

Numerical substitution ($m\omega^2 = 2 \cdot (4\pi)^2 = 2 \cdot 157{,}9 = 315{,}8$): $2000(\ell - 1) = 315{,}8\,\ell$ $2000\,\ell - 2000 = 315{,}8\,\ell$ $(2000 - 315{,}8)\,\ell = 2000 \quad\Rightarrow\quad 1684{,}2\,\ell = 2000$ $\ell = \frac{2000}{1684{,}2} = 1{,}19\;\text{m}$

Result. $\ev{\ell = \frac{K\,\ell_0}{K - m\omega^2} = 1{,}19\;\text{m}}$

Note: if $K$ were too small compared with $m\omega^2$, the denominator would become negative and the equation would give $\ell < 0$: the spring would not be able to sustain the rotation (it would break and the ball would fly off).

Links

Topics: Dinamica Concepts: Moto circolare uniforme · Forza centripeta · Forza elastica e legge di Hooke · Velocità angolare Skills: Impostazione simbolica Objects: Molla