Suppose, a ideal pendulum which has a pendulum lenghth $L$ and a bob of mass $m$, another one whose bob has same mass and same effective length. But the second one's bob is hollow and and the hollow is full of water.

Will the periodic time $T$ of the two pendulums be same?

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    $\begingroup$ The friction of the water with the internal walls of the blob, and/or the mechanical oscillations experienced by the water, might be two elements making the periods differ. But both should be small effects for typical pendulums (with small angles such that motion is periodic). So they are probably the same in real conditions. Being theoretically rigorous though, they would differ. $\endgroup$ – rmhleo Oct 14 '16 at 7:27
  • $\begingroup$ can you define 'effective length' since I have a feeling that the answer to your question depends on that $\endgroup$ – Gonenc Oct 14 '16 at 9:38
  • $\begingroup$ Edited.. I was wondering if the pressure exerted by water on the upper side of the bob will increase $T$? $\endgroup$ – Mockingbird Oct 14 '16 at 11:58
  • $\begingroup$ No, the pressure of water on any side of the bob would not affect T. $\endgroup$ – lesnik Oct 14 '16 at 12:48
  • $\begingroup$ Are you sure about it? Is it because on every side pressure is same? But is really it is same? $\endgroup$ – Mockingbird Oct 14 '16 at 13:11

If we can neglect friction with air, the formula for the period of a pendulum will be:

$T= 2\pi \sqrt {\frac {l}{g}}$

If $g$ is costant, as in this case, the period of the pendulum will only depend on the lenght of the string.

Because the center of mass of both bobs lays in the middle, the effective lenght (lenght of the string + distance of CM from the top of the bob) will be the same, ergo the two pendula will have the same period.

It is also interesting to see how the period of the pendulum changes when we ad different amounts of water in the bob.

  1. At first, with no water, the center of mass of the bob lays in the middle (distance of CM from the top of the bob= Radius). The period is T.

2.When we star adding water, the CM of the system will go below the original one (distance of CM from the top of the bob= Radius +y ).The new period of the bob will be $T_f>T$. The period will rise as long as we keep adding water below the original CM of the bob. When the water fills in half the bob, the period will reach a maximum. As we add more water, the period will start to decrease again.

  1. Eventually,when we fill the bob (completly) with water, the CM of the system will lay in the same positon of the beginning. At this point, the effective lenght will be the same and so will be the period.
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    $\begingroup$ nicely written answer! $\endgroup$ – Kosala Oct 14 '16 at 19:38

I am interpreting your "effective length" as the length of the pendulum here.

If that is the case, my answer is no. The two periods would not be the same.

The reason is that there when we write an oscillator's energy as $\frac{1}{2}mv^2+\frac{1}{2}kx^2$, we can write the oscillator's period as $2\pi\sqrt{\frac{m}{k}}$. Since water can only translate without rotating under the assumption that there is no friction, the rotational inertia is $0$ for the water. Thus "$m$" is smaller and the period is shorter for the blob with water.

Hope I helped.

  • $\begingroup$ The hollow bob is **full **of water. $\endgroup$ – Farcher Oct 14 '16 at 7:11
  • $\begingroup$ @Farcher Sorry, when I answered this question it was "some" water so I think it was not full. I think we just have to worry about the friction now. $\endgroup$ – Junkai Dong Oct 14 '16 at 7:14
  • $\begingroup$ That is what I thought. So all you need to do is edit your answer to make it the correct one. The difficulty in answering the question is not knowing at which level it was set. $\endgroup$ – Farcher Oct 14 '16 at 7:20
  • $\begingroup$ The reason I asked this question is water acts pressure on the surrounding wall of the hollow. If so, why this pressure wouldn't change the effective '$g$' reacting on the bob and somehow change $T$? $\endgroup$ – Mockingbird Oct 14 '16 at 8:18

My answer is also "no", but for different reasons.

Let's consider two identical pendulums, same mass, same size, both has a spherical cavity filled with water, but in in one case the water is liquid, in the other is is freezed (pressure inside will be very high, but who cares). Even in this case periods will be different.

If the speed of center of mass is of pendulums is $V$, will the kinetic energy of pendulums be the same? The answer is no, because the ice rotates together with pendulum, but the water does not.

So the speed of 'water' pendulum will be higher than the speed of 'ice' pendulum, and the period of the 'water' pendulum will be smaller.


Consider the following example. Two pendulums, both are spheres of radius $r$ and mass $m$. Both are attached to a fixed point with a "rope" of zero length and are swaying around this point. But in one case the sphere is solid, in the other the sphere consist of a thin weightless shell filled with water. Let's assume there is no friction between water and sphere.

In first case (solid sphere) the momentum of inertia of this sphere is: $$I_1 = m*r^2 + 1*m*r^2/5 = 7/5 * m*r^2$$

Pendulum period would be: $$T_1 = 2*\pi*\sqrt{7/5*r/g}$$

But if the sphere is filled with water, the water does not rotate together with the sphere. It's momentum of inertia would be: $$I_2 = m*r^2$$ and the period: $$T_2= 2* \pi * \sqrt{r/g}$$ The period is smaller in case of "water filled" pendulum.


As you mentioned,the pendulum is ideal.One property of ideal pendulum is that

  • The bob is suppossed to have a point mass.

Another point to be noted is that both have same effective length.

Since $T$ is proportional only to $\sqrt{l}$ ($g$ is assumed to be constant here),The time periods are equal.

I have asked a question regarding the definition of ideal pendulum.

It is the answer.

An ideal pendulum, as in the question you refer to, is one without friction, air resistance, a point-mass bob - i.e. the bob is not a real massive object but just a point, etc. It's defined that way to make the maths simpler. Real, actual pendulums only behave in approximately the same way.

The ideal pendulum can, however, have any length or bob-weight, so it does not have any special proportions.

Why the bob is not rigid body?

It is because the $I_{cm}$ of bob requires the knowledge about the size and shape of the bob-which is not given in the question. And the Time period of a compound pendulum is dependent upon the $I_{cm}$ of the bob.So the question is impossible to solve if I assume that the bob is a rigid body.

  • $\begingroup$ not necessarily because his bob is not a point mass anymore! rather it is a rigid body $\endgroup$ – Gonenc Oct 14 '16 at 9:37
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    $\begingroup$ @gonenc,answer edited. $\endgroup$ – user132865 Oct 14 '16 at 10:13

The simple version

I think the time period will be same, in both cases.

Think of the pendulum's bob as the system. When "deriving" the expression for time period, we consider the bob to be the system, with the forces on it as $mg$, $T$ etcetera. We assert that: $$ m\ell^2\dfrac{\mathrm d^2\theta}{\mathrm dt^2} = -mg\ell\sin\theta \approx -mg \cdot \theta$$ Giving, $$T \approx 2\pi\sqrt{\dfrac {\ell}g}$$ All along, we involved only the forces that were applied on the bob, as a whole, externally. We didn't think of the internal forces. Now if we replace the rigid spherical bob by a hollow bob filled with fluid, we will have to consider "the hollow sphere, along with the fluid" as the system. We would then arrive at the same result. All along, we have assumed small oscillations.

Digging deeper

The thing is, all the above process is an approximation. But, importantly, I have also indirectly approximated that the fluid is a rigid body.$^1$ It is perfectly reasonable, if the hollow was completely filled, and the fluid was incompressible. But the question says water, which is compressible.

So what would happen is as the bob reaches the bottommost part of its trajectory, the water would be compressed due to the centrifugal force, thus shifting the centre of mass of the "bob system" lower for some time. This itself will vary so complicatedly that I cannot even begin to calculate the time period (good luck on that). But I predict that the time period will increase, seeing that for short periods $\ell$ effectively increases. For larger amplitudes, this effect will be even larger.

Note that, if the question had said "an incompressible fluid" instead of water, then whatever the situation (the bob should be completely filled though), the time period would be exactly the same.

I am open to discussion on this answer and any errors you might spot. I would like to know from the experts if my answer is indeed correct (just upvote! :P). Thank you.

$1$: I have also approximated that the bob is a point mass, but it doesn't really matter here, because both our bobs will be identical in all respects, except the ability of the fluid to be deformed. The essential difference has already been discussed!


protected by Qmechanic Oct 14 '16 at 11:18

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