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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:

enter image description here

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:

enter image description here

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?

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Interpretation of equations must be consistent with the physical reality. Your graph shows values of $T$ and $\omega$ which are -ve; the former is not realisable, the latter is not meaningful.

The tension in the string is given by $T\cos\theta=mg$. The minimum tension is $mg$ when $\theta=0$ and it grows infinitely large as $\theta \to \frac12\pi$.

If the string were a spring or a rod it could be in compression, which is negative tension. This could be realised if the mass rotated above the point of suspension. However, with the spring or rod in compression then there cannot be any centripetal force to keep the mass moving in a circle.

Therefore $T$ cannot be negative and $\theta$ cannot be greater than $\frac12\pi$.

Rearranging your equation, the angular frequency of circular oscillations is given by
$$\omega^2 = \frac{T(\theta)}{mL(\theta)}=\frac{g}{L_0\cos\theta+\frac{mg}{k}}$$

The smallest possible value of $\omega$ occurs at $\theta \approx 0$ and is $\sqrt{\frac{g}{L_1}}$, which is the same as for small oscillations of a simple pendulum of the non-rotating length $L_1=L_0+\frac{mg}{k}$.

The largest possible value of $\omega$ occurs for $\theta=\frac12\pi$ and is $\sqrt{\frac{k}{m}}$, which is the same as for oscillations of the elastic string.

So yes, the value of $\omega$ is restricted to the range $\sqrt{\frac{g}{L_1}} \le \omega \le \sqrt{\frac{k}{m}}$.

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  • $\begingroup$ Great answer. So my final answer was valid, except it's domain is limited? $\endgroup$ – hddh Jun 20 '17 at 4:20
  • $\begingroup$ I suspected as much, especially since $\omega=v/r=v/L\sin\theta$, which meant that there was still a hidden dependence on $\theta$, which has a limited domain. Also, doesn't the answer mean that the presence of gravity, or any forces acting along the vertical, has no effect on the expression for $T(\omega)$? $\endgroup$ – hddh Jun 20 '17 at 4:24
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I think gravity has its effect because here $T\cos(\theta)=mg$. So we have $\theta=\cos^{-1}\Big(\frac{mg}{T}\Big)$. Now you have pointed out $T=10^4 N$. We can also calculate the inclination angle ($\theta$), plugging the value of $m=0.5$kg and $g=10 m/s^2$ as, $$\theta=\cos^{-1}\bigg(\frac{0.5 \times10}{10^4}\bigg)=\cos^{-1}(0.0005) \sim 90^o$$ Which means the angular velocity $\omega$ is surprisingly high!!! And I think that will create a huge tension on the spring and which is true ($T=10^4$N, which create $10^4 m/s^2$ on $1$kg mass!). So Hooke's law is not valid here and you can not use the expression for final length of spring as $L=L_0+\frac{T}{k}$.

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  • $\begingroup$ I do not think this answers the question. The angular velocity is given in the question as $(100 rad/s$ or $70.7 rad/s)$ so it is not surprisingly high. Why is Hooke's Law not valid? The question says that it is valid. You cannot change the conditions of the question. $\endgroup$ – sammy gerbil Jun 19 '17 at 20:35

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