Chapter 6, Problem 6.49AP

(a)

To determine

The slope of the straight line, including units.

Answer to Problem 6.49AP

The slope of the straight line is $0.0162\text{kg}/\text{m}$.

Explanation of Solution

From the Figure, the terminal speed of the filters is $5{\left(\text{m}/\text{s}\right)}^2$ and the resistive force towards the speed is $0.081\text{kg-}\text{m}/{\text{s}}^2$.

Formula to calculate the slope is,

$\tan \theta =\frac{R}{{\left(v\right)}^2}$

Substitute $0.081\text{kg-}\text{m}/{\text{s}}^2$ for $R$ and $5{\left(\text{m}/\text{s}\right)}^2$ for $v$ in the above equation.

$\begin{array}{c}\tan \theta =\frac{0.081\text{kg-}\text{m}/{\text{s}}^2}{5{\left(\text{m}/\text{s}\right)}^2}\\ =0.0162\text{kg}/\text{m}\end{array}$

Conclusion:

Therefore, the slope of the straight line is $0.0162\text{kg}/\text{m}$.

(b)

To determine

The theoretical slope of a graph of resistive force versus squared speed.

Answer to Problem 6.49AP

The theoretical slope of a graph of resistive force versus squared speed is $\frac{1}{2}D\rho A$.

Explanation of Solution

Given info:

The expression for the resistive force is,

$R=\frac{1}{2}D\rho A{v}^2$

Here,

$\rho$ is the density of the coffee.

$A$ is the cross sectional area of the coffee filters.

$D$ is the drag coefficient.

$v$ is the terminal velocity.

Formula to calculate the slope is,

$\tan \theta =\frac{R}{{\left(v\right)}^2}$

Substitute $\frac{1}{2}D\rho A{v}^2$ for $R$ in the above equation.

$\begin{array}{c}\tan \theta =\frac{\frac{1}{2}D\rho A{v}^2}{{\left(v\right)}^2}\\ =\frac{1}{2}D\rho A\end{array}$

Conclusion:

Therefore, the theoretical slope of a graph of resistive force versus squared speed is $\frac{1}{2}D\rho A$.

(c)

To determine

The drag coefficient of the filters.

Answer to Problem 6.49AP

The drag coefficient of the filters is $0.778$.

Explanation of Solution

Given info: Radius of the circle is $10.5\text{cm}$. The density of the air is $1.20\text{kg}/{\text{m}}^{\text{3}}$.

Formula to calculate the area of the circle is,

$A=\pi r^2$

Here,

$r$ radius of the circle.

$A$ is the area of the circle.

Substitute $10.5\text{cm}$ for $r$ in the above equation.

$\begin{array}{c}A=\pi {\left(10.5\text{cm}\right)}^2\\ =346.36\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =3.463\text{m}\end{array}$

Thus, the area of the circle is $3.463\text{m}$.

Formula to calculate the drag coefficient is,

$D=\frac{2R}{\rho A}$

Here,

$\rho$ is the density of the coffee.

$A$ is the cross sectional area of the coffee filters.

$D$ is the drag coefficient.

$v$ is the terminal velocity.

Substitute $0.0162\text{kg}/\text{m}$ for $R$ and $1.20\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $3.463\text{m}$ for $A$ in the above equation.

$\begin{array}{c}D=\frac{2\left(0.0162\text{kg}/\text{m}\right)}{\left(1.20\text{kg}/{\text{m}}^{\text{3}}\right)3.463\text{m}}\\ =0.7796\\ \approx 0.778\end{array}$

Conclusion:

Therefore, the drag coefficient of the filters is $0.778$.

(d)

To determine

The vertical separation from the line best fit for the eight data point.

Answer to Problem 6.49AP

The vertical separation from the line best fit for the eight data point is $1.5\%$.

Explanation of Solution

Given info:

Form the Figure (1), the force at point 8 in the graph, the mass off the coffee is $0.164\text{kg}$.

Formula to calculate the force at point 8 is,

$F=mg$

Substitute $0.164\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation.

$\begin{array}{c}F=8\left(1.64\text{kg}\right)9.81\text{m}/{\text{s}}^{\text{2}}\\ =128.70\text{kg-}\text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1000\text{kg-}\text{m}/{\text{s}}^{\text{2}}}\right)\\ =0.1287\text{N}\\ =\text{0}\text{.129}\text{N}\end{array}$

The terminal speed of the filters is $2.8{\left(\text{m}/\text{s}\right)}^2$ and the resistive force towards the speed is $0.162\text{kg}/\text{m}$.

The vertical separation from the line best fit for the eight data point is,

$\begin{array}{c}d=\frac{0.129\text{N}-0.127\text{N}}{0.127\text{N}}\\ =1.57\%\\ \approx 1.5\%\end{array}$

Conclusion:

Therefore, the vertical separation from the line best fit for the eight data point is $1.5\%$.

(e)

To determine

The explanation for what graph explains and compare it with the theoretical prediction.

Answer to Problem 6.49AP

The graph for the coffee filter falling in air at terminal speed shows the resistance force is a function of the terminal speed squared which gives the air resistance.

Explanation of Solution

Given info:

The drag coefficient of the filters is $0.778$ and the vertical separation from the line best fit for the eight data point is $1.5\%$.

Thus, the constant slope of the graph is,

$D=0.778\pm 2\%$

The graph for the coffee filter falling in air at terminal speed shows the resistance force is a function of the terminal speed squared which gives the air resistance.

The expression for the resistive force is,

$R=\frac{1}{2}D\rho A{v}^2$

From this given expression,

$R\propto v^2$

Thus, the graph of the resistive force is directly proportional to the terminal speed squared.

Conclusion:

Therefore, the graph for the coffee filter falling in air at terminal speed shows the resistance force is a function of the terminal speed squared which gives the air resistance.