Chapter 7, Problem 7.4CP

To determine

(a)

Determine a differential equation for the oscillation $\theta \left(t\right)$ ?

Answer to Problem 7.4CP

$\frac{d^2\theta }{d{t}^2}+\frac{C_D\rho \pi L_{eff}{D}^2}{8m}{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$

Explanation of Solution

Given information:

Cup diameter is $D$

Density of the air is equal to $\rho$

To find the differential equation, we should apply Newton’s second law of motion to the direction that the cup is moving,

The Newton’s second lay of motion is defined as,

${\displaystyle \sum F=ma}$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A$

Where,

$C_D$ - Drag co-efficient

$\rho$ - Density of the fluid

$V$ - Velocity

$A$ - Characteristic area

Calculation:

Find the effective length of the pendulum,

$L_{eff}=L+\frac{D}{2}$

Apply Newton’s second law of motion in the direction of the motion of the cup,

${\displaystyle \sum F=ma}$

The sum of tangential forces will be,

$-mg\sin \theta -F_{drag}=ma$

Substitute for $F_{drag}$

$-mg\sin \theta -C_D\frac{\rho }{2}{V}^2\frac{\pi }{4}{D}^2=m\frac{dV}{dt}$

Where, $V=L_{eff}\frac{d\theta }{dt}$

$-mg\sin \theta -C_D\frac{\rho }{2}\frac{\pi }{4}{D}^2{\left(L_{eff}\right)}^2{\left(\frac{d\theta }{dt}\right)}^2=m{L}_{eff}\frac{d^2\theta }{d{t}^2}$

Rearrange,

$\frac{d^2\theta }{d{t}^2}+\frac{C_D\rho \pi L_{eff}{D}^2}{8m}{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$

Conclusion:

The differential equation for the oscillation $\theta \left(t\right)$ is,

$\frac{d^2\theta }{d{t}^2}+\frac{C_D\rho \pi L_{eff}{D}^2}{8m}{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$.

To determine

(b)

Non-dimensionalize the obtained equation in sub-part a?

Answer to Problem 7.4CP

$\frac{d^2\theta }{d{t}^2}+K{\left(\frac{d\theta }{dt}\right)}^2+\theta =0$

Explanation of Solution

Given information:

Cup diameter is $D$

Density of the air is equal to $\rho$

According to the sub-part a,

The oscillation $\theta \left(t\right)$ can be given as a differential equation as below,

$\frac{d^2\theta }{d{t}^2}+\frac{C_D\rho \pi L_{eff}{D}^2}{8m}{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$

Calculation:

In below equation,

$\frac{d^2\theta }{d{t}^2}+\frac{C_D\rho \pi L_{eff}{D}^2}{8m}{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$

Assume,

$K=\frac{C_D\rho \pi L_{eff}{D}^2}{8m}$

Therefore,

$\frac{d^2\theta }{d{t}^2}+K{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$

In above equation $\theta$ and $K$ are already dimensionless

Therefore, take

$t=t{\left(\frac{g}{L_{eff}}\right)}^{1/2}$

Therefore, the dimensionless equation will be,

For small angle $\theta$, $\sin \theta =\theta$

$\frac{d^2\theta }{d{t}^2}+K{\left(\frac{d\theta }{dt}\right)}^2+\theta =0$

Conclusion:

The dimensionless equation is equal to,

$\frac{d^2\theta }{d{t}^2}+K{\left(\frac{d\theta }{dt}\right)}^2+\theta =0$.

To determine

(c)

Determine the natural frequency for $\theta <1rad$ ?

Answer to Problem 7.4CP

$\omega =\sqrt{\frac{g}{L_{eff}}}$

Explanation of Solution

Given information:

Cup diameter is $D$

Density of the air is equal to $\rho$

According to the sub-part a,

The oscillation $\theta \left(t\right)$ can be given as a differential equation as below,

$\frac{d^2\theta }{d{t}^2}+\frac{C_D\rho \pi L_{eff}{D}^2}{8m}{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$

Calculation:

In below equation,

$\frac{d^2\theta }{d{t}^2}+\frac{C_D\rho \pi L_{eff}{D}^2}{8m}{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$

Assume,

$K=\frac{C_D\rho \pi L_{eff}{D}^2}{8m}$

Therefore,

$\frac{d^2\theta }{d{t}^2}+K{\left(\frac{d\theta }{dt}\right)}^2+\frac{g}{L_{eff}}\sin \theta =0$

Neglect $K$

For small angle $\theta$, $\sin \theta =\theta$

$\frac{d^2\theta }{d{t}^2}+\frac{g}{L_{eff}}\theta =0\to \left(1\right)$

Rearrange,

$\frac{d^2\theta }{d{t}^2}+{\omega }^2\theta =0\to \left(2\right)$

Therefore,

$\omega =\sqrt{\frac{g}{L_{eff}}}$

Conclusion:

The natural frequency for small angles is equal to $\omega =\sqrt{\frac{g}{L_{eff}}}$.

To determine

(d)

Find the time required?

Answer to Problem 7.4CP

Pendulum is very lightly damped; therefore it takes almost $10-15$ cycles to get down from $30°$ to $22°$

Therefore, the number of cycles to get down to $\theta =1°$ is tough to calculate numerically.

Explanation of Solution

Given information:

$\begin{array}{l}L=1m\\ D=10cm\\ m=50g\\ \theta (0)=30°\end{array}$

According to sub-part b,

$K=\frac{C_D\rho \pi L_{eff}{D}^2}{8m}$

Assume, the air at $20°C$ will have

$\rho =1.2kg/m^3$

According to the definitions,

Take the drag co-efficient as $C_D=0.4$ when moving to right

Take the drag co-efficient as $C_D=1.4$ when moving to left

Calculate the effective length,

$L_{eff}=L+\frac{D}{2}=1.05m$

Calculate the value of $K$ when moving to right,

$K=\frac{C_D\rho \pi L_{eff}{D}^2}{8m}=\frac{\left(0.4\right)\left(1.2kg/m^3\right)\left(1.05m\right)\pi {\left(0.1m\right)}^2}{8\left(0.05kg\right)}=0.0395$

Calculate the value of $K$ when moving to left,

$K=\frac{C_D\rho \pi L_{eff}{D}^2}{8m}=\frac{\left(1.4\right)\left(1.2kg/m^3\right)\left(1.05m\right)\pi {\left(0.1m\right)}^2}{8\left(0.05kg\right)}=0.1385$

Calculate the natural frequency,

$\omega =\sqrt{\frac{g}{L_{eff}}}=\sqrt{\frac{9.81m/s^2}{1.05m}}=3.056rad/s$

To find the required time integrate numerically until,

$\theta =1°$

Pendulum is very lightly damped; therefore it takes almost $10-15$ cycles to get down from $30°$ to $22°$

Therefore, the number of cycles to get down to $\theta =1°$ is tough to calculate numerically.

Conclusion:

Pendulum is very lightly damped; therefore it takes almost $10-15$ cycles to get down from $30°$ to $22°$

Therefore, the number of cycles to get down to $\theta =1°$ is tough to calculate numerically.