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A bead lies on a frictionless hoop of radius $R$ that rotates around a vertical diameter with constant angular speed $\omega$, as shown in the figure below.

enter image description here

As the title suggests, I am particularly concerned about the range of angular speeds $0 \lt \omega \lt \omega_0$ for which the fixed angle of the bead with respect to the vertical is $\theta = 0$ (basically at the bottom of the hoop).

In solving the problem, I first drew a free-body force diagram of the bead.

enter image description here

Applying Newton's 2nd Law,

$$\begin{align} N \sin \theta & = m \omega^2 R \sin \theta \\ N \cos \theta & = mg \\ \end{align}$$

Dividing the two equations and solving for $\omega$ gives

$$\omega = \sqrt{\frac{g}{R \cos \theta}}$$

Finally, substituting $\theta = 0$ to get $\omega_0$ or the maximum angular speed at which the bead stays fixed at the bottom,

$$\omega_0 = \sqrt{\frac{g}{R}}$$

So I've got everything up to that point covered. However, I still am having some trouble making sense of the bead's motion within the range $0 \lt \omega \lt \omega_0$. I'm fairly certain that as $\omega$ increases from 0 to $\omega_0$, the angle remains zero and the bead stays fixed at the bottom. But when I try to analyze it mathematically by rearranging the equation for $\omega$ to get

$$\cos \theta = \frac{g}{R \omega^2}$$

I noticed that if I try to start with an angular speed $\omega = 0$, I am dividing by zero and thus unable to solve for the angle $\theta$. And if I try to plug in a value of $\omega$ that is less than $\omega_0$, the resulting expression falls outside the domain of $\arccos$. I only have some intuition for when $\omega$ is zero (since hoop is stationary, the normal force simply counteracts force of gravity) and when $\omega = \omega_0$. Can someone please help me try to understand this? I think I'm just overthinking things.

TL;DR: Why does $\theta$ remain zero for angular speeds between zero and $\omega_0$

P.S. I haven't yet learned about the Lagrangian, hence my method of solution.

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  • $\begingroup$ If $\omega=0$ then $\theta=0$ thus your first equation is 0=0 and the second N=mg $\endgroup$
    – Eli
    Dec 5, 2022 at 7:21
  • $\begingroup$ if θ=0 thus Sinθ=0, so you can't divide your first equation by Sinθ. Your first equation turns to 0=0. The reason is when θ=0, the bead is spinning around itself and it is not rotating in a circle. $\endgroup$
    – Ebi
    Jun 19, 2023 at 14:57

1 Answer 1

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Using the equation

$θ = cos^{-1}\left(\frac{g}{Rω^{2}}\right)$

Where $g≈9.8$ meter/sec and R $=1$ meter

We get the graph enter image description here

As you increase the value of $ω$, $\left(\frac{g}{Rω^{2}}\right)$ gets closer and closer to $0$.

$cos^{-1}(0) = \frac{\pi}{2}$ radians, or 90 degrees, hence $y$ (angle) tends to the value $\frac{\pi}{2}$ for increasing values of $x$ (angular speed)

As you rotate the ring faster and faster, the bead will place itself horizontal to the center of the ring, making an angle of 90 degrees with the vertical

When angular speed is 0

$Nsinθ = mω^2Rsinθ$ ... (1)

$Ncosθ = mg$ ... (2)

By dividing (2) with (1) and simplifying

$cos θ = \left(\frac{g}{Rω^{2}}\right)$ when $ω≠0$

In the process of dividing we already make the assumption that $ω≠0$. But, using equation (1) we can easily see that $θ=0$ when $ω=0$ `

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  • $\begingroup$ Then what about values of the angular speed $\omega$ that are between 0 and $g/R \omega^2$? Surely $\omega \neq 0$ but the graph doesn't have values for $\theta$ in that range. $\endgroup$
    – Niko
    Dec 5, 2022 at 9:53
  • $\begingroup$ @Niko Not adding this to the answer, but in the case when $g/Rω^2$ is outside the $[0, 1]$ as you said the expression falls outside the domain of arccos. To intuitively think about this, imagine the ring rotating and the bead staying in the bottom as the angular speed is insufficient, and hence there is no angle made with the vertical. $\endgroup$
    – mark
    Dec 5, 2022 at 10:09
  • $\begingroup$ Yeah, that is also how I imagined it to be, but I was wondering why exactly that's the case, like if I can't solve for $\theta$ in that range, then is it safe to assume that the angle simply remains zero? For all I know, $\theta$ could be some sort of function that takes on a range of values within that range. $\endgroup$
    – Niko
    Dec 5, 2022 at 10:21
  • $\begingroup$ No that simply means that the initial equations we wrote do not hold true for this one, as the bead isn't moving in a circular path w.r.t the access of rotation. There is no angle, the radius is 0 $\endgroup$
    – mark
    Dec 5, 2022 at 10:41
  • $\begingroup$ Apologies for the late reply but after thinking about this problem for a while now, I came up with the following reasoning that seems to clear my confusion: It is not possible for $\theta$ to be non-zero for the range $0 \lt \omega \lt \omega_0$ as it will contradict the fact that the expression $g/R \omega^2$ falls outside the domain of arccos. Therefore, $\theta$ must be zero and furthermore we cannot divide (2) by (1) as doing so would mean division by zero since $\sin 0 = 0$. Is my reasoning viable? $\endgroup$
    – Niko
    Dec 20, 2022 at 13:46

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