Simple pendulum (with a twist) 2

Relevant wiki: Pendulums

The mass does not affect the period, but the position of the center of mass does.

• When the cylinder is filled, the COM lies in the middle of the cylinder.

• When the cylinder is half-full, the COM lies slightly more than one-quarter above the bottom.

• When the cylinder is empty, the COM lies in the middle of the cylinder.

Thus the effective length of the pendulum is greater between the initial and final situations, making the period temporarily longer.

Specifically, if $t = 1$ is the initial situation and $t = 0$ is the situation where the cylinder is empty; $m$ is the mass of the cylinder and $M$ is the mass of the water; $h$ is the height of the cylinder; and $\ell$ is the distance from the fulcrum to the end of the cylinder; then the effective length of the pendulum is $\ell - \frac h2 \frac{m + Mt^2}{m + Mt}.$ The maximum value occurs when $t = \left(\sqrt{1 + \frac{M}{m}}-1\right)\ \frac{m}{M}.$ In the limit $M << m$ , this reduces to $t \approx \tfrac12$ . In the limit $M >> m$ , it becomes $t \approx \sqrt{m/M}$ .

Relevant wiki: Pendulums

The COM (Center of Mass) of the bob first moves down and then returns to the initial position (once fully empty). Since, T is proportional to the square root of the length at which the COM exists (from the hinge); therefore, Time period shall first increase then decrease returning back to almost the initial value (as rod may have some mass).

I just figured that a heavier bottle will mean more force, but at the same time more inertia. Since the water is leaking out, it means that the inertia on the bottle would decrease faster than the force on the bottle (considering the size of the bottle, the leak would lead to a significant decrease on weight over time). This would lead to a momentary increase in the period of the pendulum, but since both are constantly decreasing, it would reach a point in which the loss of water would be unable to make up for the loss and the period would start to decrease until it reaches a point in which the force is zero.

We note that the period $T$ increases with the pendulum length $L$ , the distance between the pivot to the center of mass of the pendulum. Let the masses of the rod and water be $m_r$ and $m_w$ , and the lengths of the rod and cylinder be $L_r$ and $L_w$ respectively. Assuming that the cylinder has negligible mass. And let the portion of water flows out be $x \in [0,1]$ . Then the pendulum length for an $x$ is given by:

$\begin{aligned} L(x) & = \frac {(1-x)m_w \left(L_r + xL_w + \frac {(1-x)L_w}2\right) + \frac {m_rL_r}2}{(1-x)m_w+m_r} \\ & = \frac {(1-x)m_w \left( 2L_r + (1+x)L_w\right) + m_rL_r}{2\left((1-x)m_w+m_r\right)} \end{aligned}$

For large $x$ or $x \approx 1$ or the cylinder is almost full, we can assume $m_w \gg m_r$ . Then,

$\begin{aligned} L(x) & \approx \frac {(1-x)m_w \left( 2L_r + (1+x)L_w\right)}{2(1-x)m_w} = L_r + \frac {(1+x)L_w}2 \end{aligned}$

Then when it is full, $L(0) \approx L_r + \dfrac {L_w}2$ . And when $x=0.1$ , $L(0.1) \approx L_r + \dfrac {1.1L_w}2 > L(0)$ .

But when the cylinder is empty or $x=1$ , then $L(1) = \dfrac {(1-1)m_w \left( 2L_r + (1+1)L_w\right) + m_rL_r}{2\left((1-1)m_w+m_r\right)} = \dfrac {L_r}2 < L(0)$ .

Therefore, $L(x)$ and hence the period $T$ first increases then decreases .

My intuition says:

If the cylinder is NOT leaking: (potential energy at position 1) = (potential energy at position 2) - (kinetic energy lost to the system). Therefore, each swing results in a steady loss of potential energy, ultimately slowing the swing to a stop.

If the cylinder IS leaking: (potential energy at position1) > (potential energy at position 1) - (kinetic energy lost to the system). Therefore, each swing gets a bit of a "boost" and swings faster because there is MORE energy at the start of the swing than at the end...that is, untill the leaking cylinder is empty. Then, the system reverts to a simpler "lossy" system, ultimately slowing the swing to a stop.

Make sense?

As far as I concern, it's simple to determine the answer. By common sense, when it starts swinging the water is eventually emptying, leading to less energy used to swing, and it'll swing further. Once the cylinder gets empty, it has to stop moving. Just saying, ignore me if you disagree, I'm not a scientific.