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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 a single differential equation: $$ \frac{\mathrm{d}}{\mathrm{d}t}\left[\ell(t)^2\frac{\mathrm{d}\theta}{\mathrm{d}t}\right]+g\ell(t)\theta=0,\tag{1} $$ which 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, $$ r_\text{CoM}=\frac{\int r\rho\mathop{}\!\mathrm{d}V}{\int \rho\mathop{}\!\mathrm{d}V}, $$ 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.}

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.}

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).

Since you are trying to analyze the motion of a varying mass system, you can't use Newton's $\mathbf F=m\mathbf a$, you have to use the variable mass system of equations (sans vectors since we're really working in 1D): $$ F_{ext}+v_{rel}\frac{\mathrm{d}m}{\mathrm{d}t}=m\frac{\mathrm{d}v}{\mathrm{d}t}, $$ which leads to an equation of motion of the form,¹ $$ \frac{\mathrm{d}}{\mathrm{d}t}\left[m(t)\ell^2\frac{\mathrm{d}\theta}{\mathrm{d}t}\right]+m(t)g\ell\sin\theta=0. $$ Then assuming the small angle approximation, reduces to, $$ \frac{\mathrm{d}}{\mathrm{d}t}\left[m(t)\frac{\mathrm{d}\theta}{\mathrm{d}t}\right]+m(t)\frac{g}{\ell}\theta=0,\tag{1} $$ which is a Sturm-Liouville equation.

Since $\mathrm dm/\mathrm dt=\text{const}$, then $m(t)\triangleq m_b+m_w-\alpha t$ (with $m_b$ the mass of the bob shell, $m_w$ the total mass of the water at $t=0$ and $\alpha>0$ the rate of flow). So as stated at the onset, at both $t=0$ and $t\geq m_w/\alpha$, the solution should be identical to the standard pendulum system (i.e., $T^2\sim\ell/g$). In between, however, the mass-loss clearly plays a role in the dynamics & cannot be "divided out" as suggested by other answers.

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.²
 

<sup>1. Alternatively, if you are familiar with the Lagrangian approach, then simple analysis of the energy of the system shows $$ L=T-V=\frac{1}{2}m(t)\left(\ell\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2-m(t)g\ell\left(1-\cos\theta\right). $$ Applying this Lagrangian to the Euler-Lagrange equation gives the exact same solution that the Newtonian formulation above gives--which it must, since the two formalisms are equivalent.
2. 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.</sup>

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 a single differential equation: $$ \frac{\mathrm{d}}{\mathrm{d}t}\left[\ell(t)^2\frac{\mathrm{d}\theta}{\mathrm{d}t}\right]+g\ell(t)\theta=0,\tag{1} $$ which 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, $$ r_\text{CoM}=\frac{\int r\rho\mathop{}\!\mathrm{d}V}{\int \rho\mathop{}\!\mathrm{d}V}, $$ 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.}

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).

Since you are trying to analyze the motion of a varying mass system, you can't use Newton's $\mathbf F=m\mathbf a$, you have to use the variable mass system of equations (sans vectors since we're really working in 1D): $$ F_{ext}+v_{rel}\frac{dm}{dt}=m\frac{d v}{dt} $$ which leads to the not-so-nice equation of motion,¹ $$ \frac{dm}{dt}\ell\frac{d\theta}{dt}+m\ell\frac{d^2\theta}{dt^2}=mg\sin\theta $$ If you can assume the small angle approximation, then, since $dm/dt$ is a constant, you have a second order differential equation with constant coefficients, with a solution that goes $$ \theta(t)=A\exp\left[-\frac{t}{2m}\lambda_-\right]+B\exp\left[-\frac{t}{2m}\lambda_+\right] $$ where $$ \lambda_\pm=\frac{dm}{dt}\pm\sqrt{\frac{4 g m^2+\ell \left(\frac{dm}{dt}\right)^2}{\ell}} $$ and appropriately determined constants $A$ and $B$. If you cannot make that approximation, you probably would have to solve this numerically using, for example, Runge-Kutta integration methods.

^{1 Alternatively, if you are familiar with the Lagrangian approach, then simple analysis of the energy of the system shows $$ L=T-V=\frac{1}{2}m(t)\left(\ell\frac{d\theta}{dt}\right)^2-mg\ell\left(1-\cos\theta\right) $$ Applying this Lagrangian to the Euler-Lagrange equation gives the exact same solution that the Newtonian formulation above gives--which it must, since the two formalisms are equivalent.}

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).

Since you are trying to analyze the motion of a varying mass system, you can't use Newton's $\mathbf F=m\mathbf a$, you have to use the variable mass system of equations (sans vectors since we're really working in 1D): $$ F_{ext}+v_{rel}\frac{\mathrm{d}m}{\mathrm{d}t}=m\frac{\mathrm{d}v}{\mathrm{d}t}, $$ which leads to an equation of motion of the form,¹ $$ \frac{\mathrm{d}}{\mathrm{d}t}\left[m(t)\ell^2\frac{\mathrm{d}\theta}{\mathrm{d}t}\right]+m(t)g\ell\sin\theta=0. $$ Then assuming the small angle approximation, reduces to, $$ \frac{\mathrm{d}}{\mathrm{d}t}\left[m(t)\frac{\mathrm{d}\theta}{\mathrm{d}t}\right]+m(t)\frac{g}{\ell}\theta=0,\tag{1} $$ which is a Sturm-Liouville equation.

Since $\mathrm dm/\mathrm dt=\text{const}$, then $m(t)\triangleq m_b+m_w-\alpha t$ (with $m_b$ the mass of the bob shell, $m_w$ the total mass of the water at $t=0$ and $\alpha>0$ the rate of flow). So as stated at the onset, at both $t=0$ and $t\geq m_w/\alpha$, the solution should be identical to the standard pendulum system (i.e., $T^2\sim\ell/g$). In between, however, the mass-loss clearly plays a role in the dynamics & cannot be "divided out" as suggested by other answers.

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.²

<sup>1. Alternatively, if you are familiar with the Lagrangian approach, then simple analysis of the energy of the system shows $$ L=T-V=\frac{1}{2}m(t)\left(\ell\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2-m(t)g\ell\left(1-\cos\theta\right). $$ Applying this Lagrangian to the Euler-Lagrange equation gives the exact same solution that the Newtonian formulation above gives--which it must, since the two formalisms are equivalent.
2. 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.</sup>