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A light elastic string with elastic modulus $\lambda$ and natural length $a$ is fixed at one end at a point $O$, and a particle of mass $m$ is attached at the other end. The particle moves in a horizontal circle with centre $O$ at a constant angular speed $\omega$.

The tension between the particle and the point $O$ can be expressed, in polar coordinates, as $$T=\frac{\lambda(r-a)}{a}.$$

Using Newton's $2$nd law we get the equation of motion as $$m\ddot{\mathbf{x}}=-T\mathbf{e}_r.$$ Since $\dot{\theta}=\omega$ we have $\ddot{\theta}=0$ and also $\ddot{r}=\dot{r}=0$, which simplify the equation of motion to $$r(\lambda-ma\omega^2)=\lambda a.$$

From here, we consider two cases, either $\lambda$ is positive or non-positive.

If $\lambda$ is positive then we get $\lambda<ma\omega^2$ which is actually a restriction for the circular motion to be maintained.

If $\lambda$ is non-positive then we have $\lambda\le ma\omega^2$.

What is the physical interpretation of $\lambda\le ma\omega^2$? Will we have $\ddot{r}=\ddot{\theta}=0$?

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Typically, you are going to be looking at cases where $$\lambda>>ma\omega^2\tag{1}$$ so that the string is very stiff and doesn't stretch much. So, you will have $$\frac{r}{a}=\frac{1}{1-\frac{ma\omega^2}{\lambda}}$$ If $\lambda$ is small, so that relationship 1 is not satisfied, the string cannot be assumed to exhibit linear elastic behavior, and a more accurate equation must be used to represent the string behavior (which does not assume that the tension is a linear function of the strain). Certainly, you can't have the radius being less than the unextended length "a" when the string is under tension, so negative values of $\lambda$ are excluded. Materials don't get shorter when you put them under tension.

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  • $\begingroup$ Thank you very much for your answer. May I ask why the statement "the string cannot be assumed to exhibit linear elastic behavior" holds? Is it because of a theorem or a proposition? $\endgroup$ – johnny09 Apr 5 '16 at 17:16
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    $\begingroup$ Linear elastic behavior is typically just an approximation for small strains. It's like the first term in a Taylor series expansion. For larger deformations, the quantity we use for strain $\Delta L/L$ does not give a good approximation for the observed ratio of the stress to the elastic modulus. In fact, the elastic modulus itself depends on strain, and there are numerous approximations to strain which all become equal in the limit of small strains, but are all different at large strains. For more information, study large deformation material behavior. $\endgroup$ – Chet Miller Apr 5 '16 at 18:49

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