Time period of simple pendulum with varying mass

How do you find time period as a function of time for a simple pendulum that is in the form of a hollow sphere that is filled with mercury and there is a hole in the bottom through which the mercury is constantly falling at a fixed rate? I tried creating a function for the time period to check how it varies with the mass of mercury, but I had too many variables.

EDIT: This question was a concept based multiple choice question in my homework book. The option were time period becomes erratic, time period increases, time period decreases, time period first increases then decreases.

• Could you write your trial functions? It will help us point you in the right direction. Commented Oct 10, 2013 at 14:08

The center of mass of the system consisting of the bob and the mercury inside it shifts downward, effectively increasing the length. However, the final position of the center of mass after all the fluid has drained out is at the center of the bob, so the center of mass shifts up again after some time. Thus, the time period first increases due to the effective increase in length, then decreases since the center of mass shifts back to the center of the bob.

I might be misunderstanding something, so please correct me if that is true.

But I think you're missing a basic fact about pendulums.

Have a look at the equation of a basic pendulum on wikipedia. In the differential equation, the mass $m$ does not appear:

$$\ddot\theta - \frac g L \sin\theta = 0$$

Also have a look at all the derivations on there as well. In all cases, you simply divide the $m$ away somewhere along the line.

Even if you would make $m = m(t)$, you would still be able to divide by $m(t)$ without consequences at some point, proving that your pendulum's period depends only on the string length $L$ but not the bob mass $m(t)$.

So, the period $T$ of your pendulum will be roughly equal to

$$T \approx 2\pi\sqrt{\frac L g}$$

for small initial angles $\theta_0$, and for larger angles,

$$T = 4\sqrt{\frac L g }K\left(\sin\frac{\theta_0}{2}\right)$$

with $K$ the complete elliptic integral of the first kind.

This is of course true for an idealized pendulum. This is all not true for any real-world, physical pendulums that are subject to air drag, finite string elasticity, non-zero string mass, damping, etc. Your mercury might also flow out in a way that produces net forces, changing the situation entirely. However, I assumed you were not going into that kind of detail -- otherwise, we also need a lot more details to get to a decent description of your pendulum.

EDIT

Thanks to Kyle's answer, I realized that there is something I overlooked -- the length $L$ changes as the mass changes.

For any physical pendulum, the pendulum length $L$ is defined to be the string length plus the offset from the point where the string attaches to the bob to the bob's centre of mass. As the CoM changes with $\dot m \neq 0$, it follows that $L=L(t) = L(m(t))$.

Note that you have to be careful in deriving equations here. As I mentioned in Kyle's answer, interpreting Newton's second law

$$\mathbf{F} = d\mathbf{p}/dt$$

as

$$\mathbf{F} = \frac {d(m(t)\mathbf{v}(t))}{dt} \leftarrow \text{ this is NOT Newton's second law}$$

is wrong -- Newton's second law only applies to constant mass systems (see the wiki):

$$\mathbf{F} = m\frac {d(\mathbf{v}(t))}{dt} \leftarrow \text{ THIS is Newton's second law}$$

You must apply Newton's second law to a system of which the mass is constant. That means that for your pendulum, you should consider the entire system, so including the pendulum itself, the mass that is currently falling, all the mass that has already fallen to the ground, etc.

You end up with this:

$$\mathbf{F} + \mathbf{v_{\text{rel}}} \dot{ m} = m \dot {\mathbf{ v}}$$

where $\mathbf{v}_{\text{rel}}$ is the velocity of the mass that is leaving the system, relative to that system. See the wiki article on variable mass systems for a more complete coverage.

Important to note is that at the instant the mass leaves the pendulum bob, its $\mathbf{v}_{\text{rel}} = \mathbf{0}$, so it does not exert a force on the bob; only if it is pushed out will there be a net force.

Example: take a bowl of water. Weigh it. Punch a hole in the bottom at $t_0$. Do you expect the instantaneous weight at $t_0$ to change because of you opening the hole?

Let the water escape until $t_1$. Weigh the bowl with the hole open. Quickly close the hole, weigh it again. Do you expect these measurements to be different?

Repeat the (thought) experiment with a pressurized bowl, so that the water will be forced out. Will these measurements differ?

• The Lagrangian approach disagrees with you. Commented Oct 10, 2013 at 16:19
• Actually, Kyle's solution above proves that the varying mass will play a role in the motion of the pendulum. Could you check your solution? Commented Oct 11, 2013 at 2:29
• @PhyStu: see my comments on Kyle's answer. Commented Oct 11, 2013 at 5:28
• @PhyStu: and my latest edits :) Commented Oct 11, 2013 at 5:53

What happens is that the pendulum's effective length, $$\ell_e$$, increases for a time because the center of mass is moving further out as the mercury flows out of the pendulum bob. Since the period is proportional to the square root of the length, $$T\propto \sqrt{\ell_e}$$, then the larger effective length means a larger period. Eventually, the bob would be drained of the fluid and the effective length returns back to where it was at the beginning (when the bob was filled), so the period returns to what it used to be (decreases).

Modeling this using Newtonian mechanics might be a little difficult as the length of the pendulum and the mass of the bob are changing with time. An alternative would be to look at the Lagrangian approach in which you consider the energies (kinetic and potential): $$L=T-V=\frac{1}{2}m(t)\left(\dot{\ell}^2+\ell(t)^2\dot{\theta}^2\right)+mg\ell(t)\cos\theta$$ This leads to the equations of motion, \begin{align} \frac{\mathrm d}{\mathrm dt}\left[m(t)\dot{\ell}\right]-m(t)g\cos\theta&=0 \\ \frac{\mathrm d}{\mathrm dt}\left[m(t)\ell(t)^2\dot\theta\right]+m(t)g\ell(t)\sin\theta&=0 \end{align} Note that since we have $$m(t)$$ being differentiated with respect to time, we cannot simply "divide it out" the mass as suggested by other answers without making a further assumption:

• Since you're told that $$\mathrm dm/\mathrm dt=\alpha$$, then $$m(t)\sim m_0+\alpha t$$ where $$m_0$$ is the initial mass of the bob (fluid + shell). If you assume that $$m_0\gg\alpha t$$ (i.e., mass loss is negligible), then you can factor out the $$m$$.

Under this assumption, and using the small angle approximation, we have the differential equations: \begin{align} \frac{\mathrm d}{\mathrm dt}\left[\frac{\mathrm d\ell}{\mathrm dt}\right]-g\left(1-\frac{1}{2}\theta^2\right)&=0\tag{1}\\ \frac{\mathrm{d}}{\mathrm{d}t}\left[\ell(t)^2\frac{\mathrm{d}\theta}{\mathrm{d}t}\right]+g\ell(t)\theta&=0,\tag{2} \end{align} the latter of the two is a Sturm-Liouville equation. The solution of this depends on the actual function for the effective length, $$\ell(t)$$. One option would be to assume the center of mass moves linearly with time, $$\ell(t)=\ell_0+\beta t$$ (at least until $$\beta t\geq r$$, then it is just $$\ell_0$$ because this means the mass has drained fully from the bob). You could also integrate the volume to find the center of mass of fluid, $$r_\text{CoM}=\frac{\int z\rho\mathop{}\!\mathrm{d}V}{\int \rho\mathop{}\!\mathrm{d}V},$$ and add this to the initial length of the rod, $$\ell_0$$. Of course, assuming that $$\ell_0\gg r$$ would bring us back to the "uninteresting" case of the standard pendulum since the length would be roughly independent of time.

I have not tried proceeding further, but I imagine that, given appropriate boundary conditions, Eq. (1) can be solved; however, it could be the case that numerical methods are required.1

1. It may be the case that the Hamiltonian formalism provides a nicer pairing of functions, $$\mathrm dp/\mathrm dt=f(t,\,\theta)$$ and $$\mathrm d\theta/\mathrm dt=g(t,\,p)$$, for numerically integrating.

• I'm a few years late here, but I'm confused as to why your solution only considers the time dependence of the mass in one place in the equation. Assuming the change in mass is constant is one thing, but the mass at any given time is not. You do not have a second order DE with constant coefficients; you have a Sturm-Liouville equation. Commented Apr 12, 2018 at 13:00
• @SteelAngel does seem like quite the oversight. If I remember, I'll try fixing it tonight (at work now). Commented Apr 12, 2018 at 13:15
• @SteelAngel I believed I addressed your concerns. 9 months later, but better than never. Commented Jan 25, 2019 at 21:55
• This is much more reasonable of an explanation. I pity the person trying to solve this analytically. Commented Feb 7, 2019 at 14:48

I think the best place to start is a little thought experiment.

In place of a single ball as your pendulum bob, consider another compact shape. No change, right?

Let that shape be two balls stuck together, with the same mass as the original, with the wire split in two, one to each ball, swinging together. No change, right?

Now let the point where the balls are stuck together come undone, so the balls aren't really connected any more. No change, right?

Now the wire connecting to one of the balls is cut, and the ball falls. The other ball keeps swinging as before.

EDIT: I stand corrected, due to @Kyle's comment, not because the thought experiment is inconsistent, but because the situation of leaking fluid is a different problem. It strongly depends on the shape of the pendulum bob, that functions as a fluid reservoir. There are two factors to consider:

1. How the center of mass of the bob, and thus the effective length of the pendulum, changes as the fluid leaks out,

2. The angular moment of inertia of the pendulum bob + fluid.

Assuming (2) is small (as in the case of a spherical bob), as the fluid initially drains out, the center of mass lowers, effectively increasing the length and increasing the period. Then as the quantity of fluid drains away to nothing, the center of mass returns to the center of mass of the empty bob, which we assume is about where it was at the beginning. So the effective length and period return to about where they were at the beginning.

So the answer is, the period increases, and then it decreases back to roughly its original period.

The effect of the shape determines how much the center of mass moves. If the shape is a long vertical tube, one would expect the effect to be greatest. If the shape is a long essentially horizontal tube, the effect should be minimal. If the shape is a sphere, the effect should be intermediate.

• This is not true. See the Arun 2010 paper that I linked in my comments Commented Oct 12, 2013 at 0:33
• @Kyle: You are right. I edited the answer. Commented Oct 12, 2013 at 1:37
• See, isn't it a much more interesting problem!? Commented Oct 12, 2013 at 2:35
• @Kyle: It is. I like counterintuitive problems. My favorite is downwind-faster-than-the-wind DWFTTW, not to mention Robert H. Goddard and lenr-canr.org Commented Oct 12, 2013 at 14:16