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Setup

A ball of mass $m$ is attached to a string of length $L$. It is made to rotate at speed $\omega$ as a conical pendulum making an angle $\theta$ with the vertical.

Question

This post shows how the angle the ball makes is

$cos (\theta) = \frac g {L \omega^2}$

One would expect the angle $\theta$ to approach zero as $\omega$ approaches zero, but that is not what this solution shows. Can someone help me understand the above solution?

Thanks!

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  • $\begingroup$ do not divide by zero, and you get the solution to your doubts $\endgroup$
    – basics
    Commented Dec 29, 2022 at 22:33

3 Answers 3

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When the angle $\theta$ gets small then $$ \omega\to \sqrt{\frac{g}{L}} $$ which is the angular frequency of small amplitude swinging of a simple pendulum with length $L$. The angular velocity $\omega$ cannnot get smaller than this value. The small-angle conical pendulum motion is just independant equal-amplitude $x$ and $y$ direction swings that are 90 degrees out of phase: $$ x(t)= a \cos\omega t\\ y(t)= a \sin\omega t $$

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  • $\begingroup$ what allows you to conclude that $\omega$ cannot be less than $\sqrt{\frac{g}{L}}$? Is it simply because if that's true, $cos (\theta) = \frac g {L \omega^2}$ has no solution? $\endgroup$
    – eball
    Commented Jan 3, 2023 at 16:39
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    $\begingroup$ That's one way. The other is to understand how the conical pendulum is a special case of a simple pedulum that can swing in both the $y$ and $y$ directions. $\endgroup$
    – mike stone
    Commented Jan 3, 2023 at 17:06
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If $\omega$ is too small then the ball cannot “rotate upwards” (as asked in the linked question), i.e. the ball will execute the usual motion about the lowest point.

This question contains additional technical details, but roughly speaking there is a critical minimum frequency so that the rotating pendulum is stable about some non-zero $\theta$ angle. This is the gist of the linked question. As your manipulations show, if you are blow this critical frequency, there is no angle $\theta$ for which $\cos\theta=g/(L\omega^2)$ because the right hand side is greater than 1.

When there is no solution the effect of the rotation is simply to reduce (but not overcome) the effect of $g$. When $\omega^2$ is large enough, the centripetal force can overcome gravity and the pendulum can oscillate about some angle that is not at the bottom.

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  • $\begingroup$ can you clarify what you meant by "When there is no solution the effect of the rotation is simply to reduce (but not overcome) the effect of $g$." $\endgroup$
    – eball
    Commented Jan 3, 2023 at 16:42
  • $\begingroup$ also, can you confirm the following. (1) If $\omega<\omega_c$, then the ball will oscillate about $\theta=0$ (like a simple pendulum). (2) If $\omega>\omega_c$ then the ball will oscillate about $\theta>0$. $\endgroup$
    – eball
    Commented Jan 3, 2023 at 16:46
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    $\begingroup$ Yes to your second. To your first, see my answer to the linked question. $\endgroup$
    – ZeroTheHero
    Commented Jan 4, 2023 at 14:45
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the equations This

$$T\,\sin(\theta)=m\,\,r\,\omega^2=m\,L\,\sin(\theta)\,\omega^2\tag 1$$ $$T\,\cos(\theta)=m\,g\tag 2$$

if $~\omega\ne 0~$ then you obtain

$$T=m\,L\,\omega^2\quad,\cos(\theta)=\frac {g}{L\,\omega^2}$$

and if $~\omega=0~$ from equation (1)

$$T\,\sin(\theta)=0\quad\Rightarrow \theta=0$$

and from equation (2)

$$T=m\,g$$

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  • $\begingroup$ maybe you mean if $\theta \neq 0$, not $\omega $ $\endgroup$
    – user328248
    Commented Apr 23 at 3:48
  • $\begingroup$ I use this equation $~ T\,\sin(\theta)=m\,\,r\,\omega^2~$ for $~\omega=0~$ $\endgroup$
    – Eli
    Commented Apr 23 at 6:45

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