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Let a circular hoops of radius $r$, is hanging on nails in a wall. Can I consider this as simple pendulum so the frequency $\omega = \sqrt{\frac{g}{L}}$?

On the other hand If I consider that pendulum as physical pendulum the frequency would be $\omega = \sqrt{\frac{mg r}{I}}$ where $I$ is the moment of inertia. In the first system, mass was not dependent on the frequency but it does for the physical pendulum. If I do any physics problems for the circular hoops, I see differences in the time period.

So what is the main key to distinguish the oscillator from the two system?

EDIT:

What about this question? A circular hoop is hanging on nails in a wall. The radius is r. I want to find the frequency. If I use the first one I get $$\omega = \sqrt{\frac{g}{r}}$$ But If I use the second equation I get $$\omega = \sqrt{\frac{mg r}{I}} = \sqrt{\frac{mgr}{2 m r^2}} = \sqrt{\frac{g}{2r}}$$

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    $\begingroup$ The moment of inertia is proportional to the mass, so $m$ really drops out of the second formula. Is that what you are asking? $\endgroup$
    – CuriousOne
    Sep 19, 2015 at 22:22
  • $\begingroup$ Hi Sabbir and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$
    – John Rennie
    Sep 20, 2015 at 6:24
  • $\begingroup$ But can't I ask some questions to clear my basic understanding? $\endgroup$
    – user58143
    Sep 20, 2015 at 8:32

2 Answers 2

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I'm not entirely sure what the question is, but I think that I do see the source of confusion. The first equation that you gave is for the specific case of a pendulum with a fixed mass located at a distance 'L' away from the point of rotation. The second equation you have is for the more general case where the pendulum has some distribution of mass along its entire length. In this case, you have to first calculate 'I', the moment of inertia, which you can think of as breaking the pendulum into small, itty-bitty sections of mass, and then summing all of the itty-bitty sections of mass while weighting the contribution of each bit of mass by by r^2, the distance of each section of mass from the rotation point squared.

If you consider the special case where all of the mass of the pendulum is at one point located a distance 'L' from the rotation point, then the second equation should reduce to the first equation.

Now, as for the apparent confusion over the fact that the frequency given by the first equation doesn't depend on the total mass but the second equation apparently does depend on mass, the answer to your confusion is that, actually, the second equation doesn't depend on the total mass, either. It just seems that it does because you see an 'm' in the equation. Remember that, in actuality, the moment of inertia 'I' also depends on the total mass since it is effectively a weighted sum of all of the mass elements of the pendulum. If you double the mass of the pendulum, you double the value of 'm' in the second equation but you also double the value of the moment of inertia 'I' in the second equation, so the frequency given by the second equation doesn't depend on changes in the total mass (provided the distribution of the mass remains the same). Therefore, both the first equation and the second equation show the same behavior with respect to changes in the total mass and there is no paradox.

EDIT: Concerning your additional edit, here's my response: You have to remember the conditions under which the first equation applies. The first equation does NOT say that the frequency of any pendulum is Sqrt(g/r). If you look at your textbook, it should say that the frequency of a pendulum is Sqrt(g/r) IF it is a pendulum in which all the mass is concentrated at a point at a distance 'r' from the point of rotation. Your hoop obviously doesn't satisfy that condition, so if you use the first equation then the answer is a rough approximation at best. The answer that you got from your second equation is the correct one.

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  • $\begingroup$ Can you please check the edit? $\endgroup$
    – user58143
    Sep 19, 2015 at 23:13
  • $\begingroup$ @SabbirHasan: I arrived at the same conclusion while you were editing. The edit answers your question: you need to take the extra inertia that the inertial moment provides to correctly calculate $\omega$. Well done! $\endgroup$
    – Gert
    Sep 19, 2015 at 23:51
  • $\begingroup$ @Samuel Weir it was a Physics Gre question and the first answer satisfies there. I do not understand the basis. All the answers are available there. Please check it and compare with my question .grephysics.net/ans/9277/74 $\endgroup$
    – user58143
    Sep 20, 2015 at 8:35
  • $\begingroup$ @SabbirHasan: I think that the reasoning of the "Official Solution" is wrong. The "Official Solution" assumes that the first equation you showed applies to this problem, but as I explained earlier it does not. You have to use the second equation since the mass is not all concentrated at one point. But since the question is asking for a comparison of two pendulums of the same shape which only differ in size scale, and since both the first and second equations have the same 1/Sqrt(r) dependence on r, the "Official Solution" gets the right answer for the scaling despite using the wrong equation. $\endgroup$
    – user93237
    Sep 20, 2015 at 18:48
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I like Samuel Weir’s answer because it’s true that $\omega$ is mass invariant in both cases.

But the calculated values for $\omega$ aren’t the same.

1. If we consider the hoop to be a point mass ($m$) at distance $R$ from the nail, then:

$\omega = \sqrt{\frac{g}{R}}$.

2. If we consider the inertial moment:

$\omega = \sqrt{\frac{mg R}{I}}$

For a hoop hanging from a nail in the wall we can apply the parallel axis theorem, so that:

$I=2mR^2$, so that for that case:

$\omega=\sqrt{\frac{g}{2R}}$.

So the first approach overestimates $\omega$ considerably. It has to be considered a simplification and should not be used for accurate $\omega$ predictions.

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  • $\begingroup$ it was a Physics Gre question and the first answer satisfies there. I do not understand the basis. All the answers are available there. Please check it and compare with my question .grephysics.net/ans/9277/74 $\endgroup$
    – user58143
    Sep 20, 2015 at 8:35

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