This question was given to me during a lecture. The axis of rotation is into the page, such that the pendulua all swing to the left and the right (and I think it would be obvious to note we're disregarding any collisions of the pendulua).

enter image description here

My thought process is as follows:

  • These are all simple pendula. Since this is the case, we can derive expressions for the resonant frequency and therefore the period of each.

$$ \omega = \sqrt{\frac{mgl}{I}}$$ $$T = \frac{2*\pi}{\omega} = 2\pi \sqrt{\frac{I}{mgl}}$$

  • For pendula $1$ and $2$ to have the same resonant frequency, they must have the same $l$ and $I$, as $m$ and $g$ are already implied to be equivalent.

  • However, I can clearly tell that $l_1 \ne l_2$, just by inspection. I then think about the note that they all have the same center of mass, which may lend to say that $l_1 = l_2$, but I don't see how that's true. I could be convinced that if $l$ is defined as the length from the pivot point to the center of mass of the pendulous object, then $l_1 = l_2$, but most sources I find say $l$ is the length of the string. Maybe it's assuming the object is a point mass.

  • Now, I'm not very good yet on moments of inertia, but I reckon that the moments of inertia for $1$ and $2$ is respectively:

$$I_1 = (M/2) \ R^2 + (M/2) \ R^2 = MR^2$$ $$I_2 = (M/2) \ R^2 + (M/2) \ R^2 = MR^2$$

  • Now, if this is true, then I would agree $I_1 = I_2$. I'm beginning to accept that these must indeed be point masses, as it would make sense that we add them like this to render this result, but it just confuses me how these are point masses with different shapes. I also would've thought that $r$ in this consideration of moment of inertia to not be $R$. I suppose I'm confused about to what $r$ in $mr^2$ refers to, but I thought it refers to the radius from the axis of rotation, but the $R$ arrows don't seem to point to it. Can anyone shed some light to justifications for why pendula $1$ and $2$ have the same $w$?

closed as off-topic by John Rennie, Kyle Kanos, stafusa, Bill N, Qmechanic Nov 14 '17 at 16:22

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  • 2
    $\begingroup$ Isn't the "I" in the period formula the moment of inertia about the pendulum pivot? $\endgroup$ – DJohnM Nov 7 '17 at 17:10
  • $\begingroup$ I believe so, yes. Did I just make a really stupid mistake? $\endgroup$ – sangstar Nov 7 '17 at 17:11

Only the 1st pendulum is "simple" (as indicated), the others are "compound" pendula.

The arrows marked $R$ are only indicating a length. They are not indicating the axis of rotation.

Your calculations of $I_1, I_2$ are incorrect. The $r$ which you must use in $I=\Sigma mr^2$ is the distance of each mass from the axis of rotation. However, $I_1=I_2$ is correct.

In your formula $l$ is the distance between the pivot and the centre of mass, which is probably intended to be the same in all cases. (Note that $l$ is not necessarily the same as $r$ in the calculation of $I$.)

I think it must be assumed that the blue lines are rigid rods which are rigidly connected, otherwise there are other modes of oscillation - ie rotation about the point of attachment of the vertical rods with the masses or the horizontal rod.

Yes you must assume that the masses are point masses, except for #3, because no dimensions have been given.

  • $\begingroup$ Right, so if I understand you correctly (which I think I do since I accepted this answer): where I got caught up is mistaking $l$. $l$ is actually the distance from the pivot to the axis of rotation, and so it is the same in both cases, and the $r$ I use is the distance from each mass from the axis of rotation, which means $I_1 = I_2$ is equivalent in this case. $\endgroup$ – sangstar Nov 9 '17 at 17:50
  • $\begingroup$ Not quite : the pivot is at the axis of rotation. $l=L$ is the distance of the CM from the axis of rotation. Yes, this is the same in all cases. And $I_1=I_2 \ne MR^2$. $\endgroup$ – sammy gerbil Nov 9 '17 at 20:04
  • $\begingroup$ Hmm.. is the axis of rotation at the midpoint of either masses pendula $1$ and $2$ and going in and out of the page? And does that therefore mean the pivot is there as well? That doesn't seem to make sense though if $l=L$ is the distance from the CM (centre of mass?) and axis of rotation, since CM and axis of rotation would be both at the centre of mass of both pendula $1$ and $2$. My intuition was that the pivot is at the very top of the pendula, so perhaps the axis of rotation is there as well? But then the masses in $1$ and $2$ don't have the same $l$? $\endgroup$ – sangstar Nov 9 '17 at 20:39
  • $\begingroup$ Ah, wait: am I correct in saying the pivot and axis of rotation is at the top of the pendula, and the CM of both those objects in $1$ and $2$ are both roughly on that dotted blue line? Which would make the statement $l = L$ is the distance of the CM from the axis of rotation true? And the moment of inertia $I_1 = I_2 = ML^2$? $\endgroup$ – sangstar Nov 9 '17 at 20:40
  • $\begingroup$ You are correct about the positions of the pivot and axis of rotation (top, black line) and of the CM (bottom, dotted blue line). However, the moments of inertia are not $ML^2$. Use the formula in my answer. $\endgroup$ – sammy gerbil Nov 9 '17 at 20:47

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