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How can you explain that a group of circular hoops of radius R, which are hung over a knife edge, (with different lengths) have the same frequency when brought in oscillation?

Let's say that the hoops are made of one big hoop and then cut into smaller pieces. Therefore, they still have the same radius but a different length because they were cut into smaller pieces. After that, they are hung up over a knife edge (picture).

Remarkable enough, the hoops have the same frequency of oscillation when brought out of their equilibrium position.

This is the test set-up for the oscillating hoops.

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    $\begingroup$ What do you mean by them all having the same radius $R$, but different "lengths"? $\endgroup$ – mike stone Mar 24 at 16:38
  • $\begingroup$ Consider adding a diagram to let people understand the question better $\endgroup$ – Davide Dal Bosco Mar 24 at 17:18
  • $\begingroup$ A picture is added. I think I need to have a differential equation to solve this, but I don't see possibilities for this. Maybe try moment of inertia? $\endgroup$ – Jasper Vis Mar 25 at 17:34
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    $\begingroup$ Unfortunately the picture gives no clear impression about what's going on. Just shooting from the hip, but with 99% probability, the answer is: because the rings have the same size (same resonant frequency) and they are weakly coupled via the knife (which causes them to almost keep their individual resonances, but exchange enough energy for synchronizing). $\endgroup$ – oliver Mar 25 at 17:38
  • $\begingroup$ It would appear from the picture that you are talking about segments of hoops, oscillating as physical pendulums. The problem is to find the center of gravity and the rotational inertia of such a segment. $\endgroup$ – R.W. Bird Mar 25 at 17:45
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My result is equivalent to that from JalfredP, but I considered just a segment of the hoop measured to an angle $θ_m$ on either side of the top position. I'm looking for the angular frequency of oscillation: $ω^2$ = mgL/I where L is the distance from the pivot point to the center of gravity, and I is the rotational inertia about the pivot point. For the position of the center of gravity above the center of curvature, the integral of (ρ dθ)R cos(θ) divided by the integral of (ρ dθ) gives h = R sin($θ_m$)/$θ_m$, where ρ is the mass/unit angle. Then L = R(1 – sin($θ_m$)/$θ_m$). For the rotational inertia: I is the integral of $[2R sin(θ/2)]^2$ρ dθ. Using the trig substitution $sin^2$(θ/2) = (1/2)[1 – cos(θ)], I = 4ρ$R^2$$θ_m$[1 – sin($θ_m$)/$θ_m$]. Using these expressions and m = 2ρ$θ_m$ yields $ω^2$ = g/(2R) which is independent of the arc length of the hoop segment.

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The reason is that the period will be dependendent on the radius of curvature $R$ but not on the length of the hoops, provided the oscillations are small and the hoop hangs at its center of curvature (i.e. with the same amount on mass on the two sides).

To prove this, we will go step by step: first we consider a full ring and solve its motion, then we generalize the results for fractions of a ring. It's going to be a long road but we will prove the final results eventually! Brace with me!

For oscillations, we need to use Newton's equation in angular form i.e.

$$I\alpha = \tau$$

where $\alpha$ is the angular acceleration around a given axis, $I$ the moment of inertia and $\tau$ the total torque. Let's solve this for the case of a ring.

The ring

Assume we have a ring of radius $R$, length $L=2\pi R$ and mass $M$. We model it as a thin ring lying the the $xy$ plane, whose center is in the origin and that oscillates around a fixed axis $\hat{a}=(0, 0, 1)$ passing through the point $p=(0, R, 0)$ and oriented in the $z$ direction.

First, we need

the moment of inertia

To compute the moment of inertia of the ring around the axis $\hat{a}$ we first compute the (easy) moment of inertia around the center $o=(0, 0, 0)$ of the ring and then use the parallel axis theorem to compute the one passing through the point $p$.

Our ring has a mass density given by $\lambda = M/ L=M/2\pi R$ and its moment of inertia around the center is $$I_o=\int dm h^2= \int \lambda dl h^2$$ where $dl=Rd\theta$ is the element of length of the ring in polar coordinates and $h$ is the distance of the infinitesimal mass element $dm=\lambda dl$ from the axis. Because it is a ring, we have $h=R$ always so that $$I_o=\lambda R^2\int dl = \lambda R^3\int_{0}^{2\pi} d\theta$$ which of course is

$$I_o=2\pi \lambda R^3$$

finally, we use $\lambda = M/2\pi R$ and get

$$I_o=MR^2$$ which is our moment of inertia around $o$. To get the one around $p$ we use the parallel axis theorem and find

$$I_p=I_o+MR^2=2MR^2$$

Next, we need

the total torque

We choose again $p$ as a reference point for the torque so that the fixing force keeping the pendulum fixed has no net momentum. This means that the total torque is given by gravity only, which points in the negative $y$ direction and acts on the center of mass. For a full ring the center of mass is always at a distance $R$ from the point $p$ so that the total torque is given by

$$\tau_T=-R\;Mg\sin(\phi)$$

where $\phi$ is the angle between gravity and the position vector of the center of mass as it oscillates.

Putting everything together

Newton equations then reads, using $\alpha=\ddot{\phi}$:

$$I_p \ddot{\phi}=-RMg\sin(\phi)$$ or substituting the moment $I_p$ $$2MR^2 \ddot{\phi} =- RMg\sin(\phi)$$

in the approximation in which $\phi\approx 0$ and $\sin(\phi)\approx \phi$ we get an harmonic motion $$2MR^2 \ddot{\phi} =- RMg\phi$$

whose solution for the period is $$T=2\pi\sqrt{{2MR^2\over RMg}}=2\pi\sqrt{{2R\over g}}$$ so that the period only depends on $g$ and on the radius $R$.

More in general, for a pendulum like this, the period will always be given by

$$T=2\pi\sqrt{{I_p \over r_{cmp}Mg}}$$ with $r_{cmp}$ the distance of the center of mass from the point $p$

The general case of a hoop

We use what we learnt to compute the general case. We now have a fraction of a ring (a hoop) of radius $R$ and defined by its arc $\gamma$ (the angle the hoop forms) so that its length is $L=R\gamma$ and its mass density is $\lambda(\gamma)=M/R\gamma$.

If $\gamma=2\pi$ we have a full ring, if $\gamma=\pi$ a half ring, if $\gamma=\pi/2$ a quarter of a ring and so on..

What we need to prove is that our final result (the period) does not depend on $\gamma$ i.e. it is the same for all kind of hoops!

To find its period we need the new moment of inertia and the new position of the center of mass.

As far as the distance from the center of mass is concerned, that is easy and we can integrate the position of the center of mass from the origin, using in polar coordinates $y=Rcos(\theta)$ (due to the way in which we defined $\gamma$ as the angle with the y axis!) and integrating again 2 times from $0$ to $\gamma/2$

$$y_{cm}(\gamma)=2{1\over M}\int_0^{\gamma/2}\lambda Rd\theta R cos(\theta) = 2{1\over M} {M \over R\gamma} R^2 \int_0^{\gamma/2} d\theta$$ which yields

$$y_{cm}={2 R\over \gamma}sin(\gamma/2)$$

so that the distance of $p$ from the CoM is (considering that $p$ is in $(0, R, 0)$ and the cm in $(0, y_{cm}, 0)$ ) $$r_{cmp}=R-y_{cm}=R\left(1-{2 \over \gamma}sin(\gamma/2)\right)$$ and if $\gamma=2\pi$ we recover that $r_{cmp}=R$ as expected for the full ring.

For the moment of inertia, again, we compute $I_o$ around the center which is now given by the same integral as before, except we only compute it from $0$ to $\gamma/2$ and multiply by two (because of the symmetry around the $y$ axis), so that

$$I_o(\gamma)= 2 \lambda R^3\int_{0}^{\gamma/2} d\theta$$

giving $$I_o(\gamma)=2{M\over R\gamma} R^3 \gamma/2$$ i.e.

$$I_o(\gamma)= MR^2 $$ as before.

Unfortunately this is the moment around $o$ and we need the one around $p$. To find that we use the parallel axis theorem is a different way i.e .we do it in two times to connect the moment of inertia of the CM with the one we found and then the results to the one we need. In other words, we first compute the moment wrt the CoM using the distance between the origin and the CoM and then the one we look for, $I_p$, using the distance between the CoM and $p$.

We first find the moment around the center of mass $I_{cm}$ as

$$I_{cm}(\gamma)=I_o(\gamma)-My_{cm}^2$$

and then we find the moment of inertia around $p$ using the newly found $I_{cm}(\gamma)$ and using the distance $r_{cmp}=R-y_{cm}$ i.e.

$$I_p(\gamma)=I_{cm}(\gamma)+Mr_{cmp}^2$$

This leads to

$$I_p(\gamma)=I_{cm}(\gamma)+M(R-y_{cm})^2=I_o(\gamma)-My_{cm}^2+M(R-y_{cm})^2$$ which we rewrite as

$$I_p(\gamma)=I_o(\gamma)-My_{cm}^2+MR^2+My_{cm}^2-2MRy_{cm}$$ and finally, simplyfying and using $y_{cm}={2 R\over \gamma}sin(\gamma/2)$

$$I_p(\gamma)=I_o(\gamma)+MR^2-2MRy_{cm}=2MR^2-4{MR^2\over \gamma}\sin(\gamma/2)$$

which eventually turns to

$$I_p(\gamma)=2MR^2\left(1-{2\over\gamma}\sin(\gamma/2)\right)$$

and again for $\gamma=2\pi$ we recover the moment of inertia of a ring.

Final result!

After all this work, we can look at the period of oscillation which, remember, is given by $$T=2\pi\sqrt{{I_p\over Mgr_{cmp}}}$$ using $$I_p=2MR^2\left(1-{2\over\gamma}\sin(\gamma/2)\right)$$ and $$r_{cmp}=R\left(1-{2 \over \gamma}sin(\gamma/2)\right)$$

and that is

$$T=2\pi\sqrt{{ 2MR^2\left(1-{2\over\gamma}\sin(\gamma/2)\right) \over Mg R\left(1-{2 \over \gamma}sin(\gamma/2)\right)}}$$

which simplifies to

$$T=2\pi\sqrt{{ 2R \over g}}$$

exactly as in the ring case as the $\gamma$-dependent factor $\left(1-{2\over\gamma}\sin(\gamma/2)\right)$ simplifies on both sides.

So the final result is that the period does not depend on anything except the radius of curvature $R$. Also this result is mass-independant so you could to that also with hoops of different mass/materials.

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