Chapter 2, Problem 2.1P

To determine

The approximate speed at which the two drag forces are equally important, the approximate range of speeds at which it is safe to treat the drag force as purely quadratic, the conditions under which it is a good approximation to ignore the linear term for both baseball and beach ball.

Answer to Problem 2.1P

The approximate speed at which the two drag forces are equally important for the baseball is $1\text{ cm/s}$ , the drag force will be purely quadratic if $v>>1\text{ cm/s}$ and it is the best approximation to neglect the linear force, the approximate speed at which the two drag forces are equally important for the beachball is $1\text{ mm/s}$ , the drag force will be purely quadratic if $v>>1\text{ mm/s}$ and it is the best approximation to neglect the linear force.

Explanation of Solution

Write the equation for the ratio of the quadratic drag force to the linear drag force.

  $\frac{f_{\text{quard}}}{f_{\text{linear}}}=\left(1.6\times {10}^3{\text{ m/s}}^2\right)Dv$        (I)

Here, $f_{\text{quard}}$ is the quadratic drag force, $f_{\text{linear}}$ is the linear drag force, $D$ is the diameter of the ball and $v$ is the speed of the ball.

It is given that the two forces have same strength.

Write the expression for the ratio of two drag forces in the given case.

  $\frac{f_{\text{quard}}}{f_{\text{linear}}}=1$        (II)

Put equation (I) in equation (II).

  $\left(1.6\times {10}^3{\text{ m/s}}^2\right)Dv=1$

Rewrite the above equation for $v$ .

  $v=\frac{1}{\left(1.6\times {10}^3{\text{ m/s}}^2\right)D}$        (III)

Conclusion:

Substitute $7\text{ cm}$ for $D$ in equation (III) to find $v$ for the baseball.

  $\begin{array}{c}v=\frac{1}{\left(1.6\times {10}^3{\text{ m/s}}^2\right)\left(7\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}\\ =0.89\times {10}^{-2}\text{ m/s}\left(\frac{1\text{ cm}}{{10}^{-2}\text{ m}}\right)\\ =0.89\text{ cm/s}\\ \approx \text{1 cm/s}\end{array}$

If $v>>1\text{ cm/s}$ , the drag force will be purely quadratic and it is the best approximation to neglect the linear force.

Substitute $70\text{ cm}$ for $D$ in equation (III) to find $v$ for the beachball.

  $\begin{array}{c}v=\frac{1}{\left(1.6\times {10}^3{\text{ m/s}}^2\right)\left(70\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}\\ =0.89\times {10}^{-3}\text{ m/s}\left(\frac{1\text{ mm}}{{10}^{-3}\text{ m}}\right)\\ =0.89\text{ mm/s}\\ \approx \text{1 mm/s}\end{array}$

If $v>>1\text{ mm/s}$ , the drag force will be purely quadratic and it is the best approximation to neglect the linear force.

Therefore, approximate speed at which the two drag forces are equally important for the baseball is $1\text{ cm/s}$ , the drag force will be purely quadratic if $v>>1\text{ cm/s}$ and it is the best approximation to neglect the linear force, the approximate speed at which the two drag forces are equally important for the beachball is $1\text{ mm/s}$ , the drag force will be purely quadratic if $v>>1\text{ mm/s}$ and it is the best approximation to neglect the linear force.