I encountered the following problem on an old exam for a university course I am in. It involves a conical pendulum with an elastic string:
I attempted a solution, and got 10000 for my answer, which was listed (a). However, I have some questions about the solution, namely, it doesn't seem to make physical sense to me. First, here is my solution:
Let $\theta$ be the angle between the string and the vertical. Then $T_{\text{horizontal}}=T\sin\theta$. The ball describes circular motion, with acceleration $r\omega^2$, thus
$$T\sin\theta = mr\omega^2.$$
From the geometry, we can see that $r=L\sin\theta$, $L$ being the length of the string. $L=L_0+\frac{T}{k}$, thus
$$T\sin\theta=m\bigg(L_0+\frac{T}{k}\bigg)\sin\theta\cdot\omega^2.$$
Since $\sin\theta$ does not equal $0$ between $0$ and $\pi/2$ rad ($0\text{ to }90^\text{o}$), we can divide both sides by $\sin\theta$,
$$T=m\bigg(L_0+\frac{T}{k}\bigg)\omega^2.$$
Solving for $T$,
\begin{align} T&=mL_0\omega^2+\frac{m\omega^2}{k}T \\ \Rightarrow T-\frac{m\omega^2}{k}T&=mL_0\omega^2 \\ \Rightarrow T\bigg(1-\frac{m\omega^2}{k}\bigg)&=mL_0\omega^2 \\ \Rightarrow T&=\frac{mL_0\omega^2}{1-\frac{m\omega^2}{k}} \\ \end{align}
Substituting the values from the question yields,
$$T=\frac{0.5\cdot1\cdot100^2}{1-\frac{0.5\cdot100^2}{10000}}.$$
While this is a listed answer, this doesn't make sense to me. It suggests that
(a) Gravity has no effect.
(b) When $\omega\approx 70.7$ rad/s, the tension is undefined. (Due to division by zero).
(c) When $\omega$ is greater than approx $70.7$ rad/s, the tension becomes negative.
Below is a graph of the equation (zoomed out a lot), from desmos.com:
This doesn't make any physical sense to me? Did I make a mistake somewhere? If so, why is this wrong and what is the correct solution? If I didn't make a mistake, then how does this make physical sense? Is the domain of $\omega$ limited somehow?