Question

A light, cubical container of volume ${\mathit{a}}^{\mathbf{3}}$is initially filled with a liquid of mass density$\mathit{\rho }$as shown in Figure P15.89a. The cube is initially supported by a light string to form a simple pendulum of length${\mathit{L}}_{\mathbf{i}}$, measured from the center of mass of the filled container, where ${\mathit{L}}_{\mathbf{i}}\mathbf{>}\mathbf{>}\mathit{a}$. The liquid is allowed to flow from the bottom of the container at a constant rate $\left(\mathbf{dM}/\mathbf{dt}\right)$. At any time$\mathit{t}$, the level of the liquid in the container is$\mathit{h}$and the length of the pendulum is$\mathit{L}$(measured relative to the instantaneous center of mass) as shown in Figure P15.89b. (a) Find the period of the pendulum as a function of time. (b) What is the period of the pendulum after the liquid completely runs out of the container?

Short answer

(a) $\frac{2\pi }{\sqrt{g}}\sqrt{L_i+\left(\frac{\text{1}}{\text{2}\rho a^2}\right)\frac{dM}{dt}t}$

(b)$2\pi \sqrt{\frac{L_i}{g}}$

Step-by-step solution

Identification of given data

Volume of cubical container is$a^3$

Liquid of mass density is$\rho$

Initial length of string is$L_i$

level of the liquid in the container is$h$

Significance of time period of simple pendulum

The letter "$\mathit{T}$" stands for the period of time needed for the pendulum to complete one complete oscillation.

$T=2\pi \sqrt{\frac{L}{g}}$ ...(i)

Where $L$is the length of the pendulum and $g$is the acceleration due to gravity

(a) Determining the period of the pendulum as a function of time

When the cube is initially supported by a light string to form a simple pendulum then its period for length $L$is,

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

Differentiate the above equation with respect to time $t$

$\frac{dT}{dt}=\frac{\pi }{\sqrt{g}}\left(\frac{1}{\sqrt{L}}\right)\frac{dL}{dt}$ ...(ii)

Where,

$L=L_i+\frac{a}{2}-\frac{h}{2}$

Differentiate the above equation with respect to time $t$

$\frac{dL}{dt}=-\left(\frac{1}{2}\right)\frac{dh}{dt}$ ...(iii)

The liquid is allowed to flow from the bottom of the container at a constant rate role="math" localid="1660201440977" $\left(\mathrm{dM}/\mathrm{dt}\right)$

$\begin{array}{rcl}\frac{dM}{dt}& =& \rho \frac{dV}{dt}\\ \frac{dM}{dt}& =& -\rho \mathrm{A}\frac{dh}{dt}\\ \frac{dh}{dt}& =& -\frac{\text{1}}{\rho \mathrm{A}}\frac{dM}{dt}\end{array}$

By using equation (iii)

$\frac{dL}{dt}=-\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt}$ ...(iv)

$$\begin{gathered} {\int }_{L_i}^{L}dL=\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt} \\ L-L_i=\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt} \end{gathered}$$ ...(v)

Substitute equation (iv) and (v) in equation (ii)

$\begin{array}{rcl}\frac{dT}{dt}& =& \frac{\pi }{\sqrt{g}}\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt}\frac{1}{\sqrt{L}}\\ & =& \frac{\pi }{\sqrt{g}}\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt}\frac{1}{\sqrt{L_i+\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt}t}}\end{array}$

Where $A$is the area containing the cubical container and it is $a^2$

Substitute$A$to the above equation

$\begin{array}{rcl}\frac{dT}{dt}& =& \frac{\pi }{\sqrt{g}}\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt}\frac{1}{\sqrt{L_i+\left(\frac{\text{1}}{\text{2}\rho a^2}\right)\frac{dM}{dt}t}}\\ {\int }_{T_i}^{T}dT& =& \frac{\pi }{\sqrt{g}}\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt}\frac{1}{\sqrt{L_i+\left(\frac{\text{1}}{\text{2}\rho a^2}\right)\frac{dM}{dt}t}}\\ T-T_i& =& \frac{\pi }{\sqrt{g}}\left(\frac{\text{1}}{\text{2}\rho \mathrm{A}}\right)\frac{dM}{dt}\left(\frac{2}{\frac{\text{1}}{\text{2}\rho a^2}\frac{dM}{dt}}\right)\left[\sqrt{L_i+\left(\frac{\text{1}}{\text{2}\rho a^2}\right)\frac{dM}{dt}t}-\sqrt{L_i}\right]\\ T-T_i& =& \frac{2\pi }{\sqrt{g}}\left[\sqrt{L_i+\left(\frac{\text{1}}{\text{2}\rho a^2}\right)\frac{dM}{dt}t}-\sqrt{L_i}\right]\end{array}$

Where

$T_i=2\pi \sqrt{\frac{L_i}{g}}$

So from equation

$T=\frac{2\pi }{\sqrt{g}}\sqrt{L_i+\left(\frac{\text{1}}{\text{2}\rho a^2}\right)\frac{dM}{dt}t}$

Hence the period of the pendulum as a function of time is$T=\frac{2\pi }{\sqrt{g}}\sqrt{L_i+\left(\frac{\text{1}}{\text{2}\rho a^2}\right)\frac{dM}{dt}t}$

(b) Determining the period of the pendulum after the liquid completely runs out of the container

Period of the pendulum after the liquid completely runs out of the container is

$T_i=2\pi \sqrt{\frac{L_i}{g}}$