Chapter 5, Problem 84QAP

To determine

(a)

The value the drag constant c.

Answer to Problem 84QAP

The value the drag constant c is $4.54\times {10}^{-6}\text{kg/m}$.

Explanation of Solution

Given:

Terminal velocity of the raindrop

  $v=8.50\text{ m/s}$

Diameter of the drop

  $2r=4.00\text{ mm}$

Density of water

  $\rho =1.00\times {10}^3{\text{kg/m}}^3$

Equation for the drag force

  $F_{\text{drag}}=c{v}^2$

Formula used:

If the drop moves with terminal velocity, then the drag force acting upwards is equal to the drop's weight w.

  $F_{\text{drag}}=w$

If m is the mass of the drop, V its volume and g the acceleration of free fall, then,

  $w=mg$

And,

  $m=\rho V$

Since the volume is given by,

  $V=\frac{4}{3}\pi r^3$, therefore,

  $\begin{array}{c}m=\rho V\\ =\frac{4}{3}\pi r^3\rho \end{array}$

This follows that,

  $F_{\text{drag}}=\frac{4}{3}\pi r^3\rho g$

Since $F_{\text{drag}}=c{v}^2$,

  $c{v}^2=\frac{4}{3}\pi r^3\rho g$

Write an expression for c.

  $c=\frac{4\pi r^3\rho g}{3v^2}$......(1)

Calculation:

Determine the radius of the drop and express it in m.

  $\begin{array}{c}r=\frac{2r}{2}=\frac{4.00\text{ mm}}{2}\\ =2.00\text{ mm}\times \frac{{10}^{-3}\text{m}}{1\text{ mm}}\\ =2.00\times {10}^{-3}\text{m}\end{array}$

Substitute the given values in the expression and calculate the value of c.

  $\begin{array}{c}c=\frac{4\pi r^3\rho g}{3v^2}\\ =\frac{4\left(3.14\right){\left(2.00\times {10}^{-3}\text{m}\right)}^3\left(1.00\times {10}^3{\text{kg/m}}^3\right)\left(9.80{\text{ m/s}}^2\right)}{3{\left(8.50\text{ m/s}\right)}^2}\\ =4.54\times {10}^{-6}\text{kg/m}\end{array}$

Conclusion:

The value the drag constant c is $4.54\times {10}^{-6}\text{kg/m}$.

To determine

(b)

The value of the terminal velocity of a drop of diameter 8.00 mm under the same conditions.

Answer to Problem 84QAP

The value of the terminal velocity of a drop of diameter 8.00 mm

is 24.04 m/s.

Explanation of Solution

Given:

The terminal velocity of the drop of diameter 4.00 mm

  $v=8.50\text{ m/s}$

Diameter of the drop 1

  $2r=4.00\text{ mm}$

Diameter of the drop 2

  $2r\text{'}=8.00\text{ mm}$

Formula used:

From the equation

  $c=\frac{4\pi r^3\rho g}{3v^2}$,

it can be seen that, since

  $v=\sqrt{\frac{4}{3c}\pi r^3\rho g}$

Therefore,

  $v\propto \sqrt{r^3}$

Therefore,

  $\frac{v\text{'}}{v}=\sqrt{\frac{r{\text{'}}^3}{r^3}}$......(2)

Calculation:

The diameter of the drop 2 is 8.00 mm, which is 2 times the diameter of drop 1.

Therefore,

  $r\text{'}=2r$

The equation (2) reduces to,

  $\begin{array}{c}\frac{v\text{'}}{v}=\sqrt{\frac{r{\text{'}}^3}{r^3}}\\ =\sqrt{{\left(\frac{2r}{r}\right)}^3}\\ =\sqrt{8}\end{array}$

Substitute the value of the terminal velocity of drop1 and calculate the value of the terminal velocity of drop 2.

  $\begin{array}{c}v\text{'}=\sqrt{8}v\\ =2.82\left(8.50\text{ m/s}\right)\\ =24.04\text{ m/s}\end{array}$

Conclusion:

The value of the terminal velocity of a drop of diameter 8.00 mm

is 24.04 m/s.