Chapter 6, Problem 35PQ

(a)

To determine

The terminal speed of the sphere.

Answer to Problem 35PQ

The terminal speed of the sphere is $118\text{m/s}$.

Explanation of Solution

Write the given expression for resistive force on the sphere.

  ${\overrightarrow{F}}_D=-b\overrightarrow{v}$ (I)

Here, ${\overrightarrow{F}}_D$ is the drag force, $b$ is the constant, and $v$ is the speed of the object.

The negative sign indicates that the direction of the drag force is opposite to the velocity’s direction.

Write the velocity of the sphere.

  $v=\frac{mg}{b}\left[1-\exp \left(\frac{-bt}{m}\right)\right]$ (II)

Here, $m$ is the mass of the object, $g$ is the gravitational acceleration, and $t$ is the time period.

If the time period $t\to \infty$ velocity changes to $v\to v_T$.

Rearrange the equation (II) for $v\to v_T$.

  $v_T=\frac{mg}{b}$ (III)

Here, $v_T$ is the terminal velocity.

If the object moves in its speed reaches one fourth of its terminal speed.

Substitute the equation (III) in equation (II).

  $0.250v_T=v_T\left[1-\exp \left(\frac{-bt}{m}\right)\right]$ (IV)

Conclusion:

Substitute $0.500\text{kg}$ for $m$ and $3.45\text{s}$ for $t$ in equation (IV).

  $\begin{array}{c}0.250v_T=v_T\left[1-\exp \left(\frac{-b\left(3.45\text{s}\right)}{0.500\text{kg}}\right)\right]\\ 0.250=1-\exp \left(\frac{-b\left(3.45\text{s}\right)}{0.500\text{kg}}\right)\\ 0.750=\exp \left(\frac{-b\left(3.45\text{s}\right)}{0.500\text{kg}}\right)\end{array}$

Solve the above equation for $b$.

  $\begin{array}{c}\ln \left(0.750\right)=\left(\frac{-b\left(3.45\text{s}\right)}{0.500\text{kg}}\right)\\ -0.288=\left(\frac{-b\left(3.45\text{s}\right)}{0.500\text{kg}}\right)\\ b=\frac{\left(0.288\right)\left(0.500\text{kg}\right)}{\left(3.45\text{s}\right)}\\ =0.0417\text{kg/s}\end{array}$

Substitute $0.500\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, and $0.0417\text{kg/s}$ for $b$ in the equation (III) to find $v_T$.

  $\begin{array}{c}{v}_T=\frac{\left(0.500\text{kg}\right)\left(9.80{\text{m/s}}^2\right)}{\left(0.0417\text{kg/s}\right)}\\ =117.5\text{m/s}\\ \simeq \text{118}\text{m/s}\end{array}$

Therefore, the terminal speed of the sphere is $118\text{m/s}$.

(b)

To determine

The distance traveled by the sphere.

Answer to Problem 35PQ

The distance traveled by the sphere is $52.7\text{m}$.

Explanation of Solution

Write the derivative form of the equation of velocity.

  $v=\frac{dx}{dt}$ (V)

Here, $dx/dt$ is the rate of change of distance and $v$ is the velocity.

Compare the equation (II) and (V).

  $\begin{array}{c}\frac{dx}{dt}=\frac{mg}{b}\left[1-\exp \left(\frac{-bt}{m}\right)\right]\\ {\displaystyle \underset{x_0}{\overset{x}{\int }}dx}={\displaystyle \underset{0}{\overset{t}{\int }}\frac{mg}{b}\left[1-\exp \left(\frac{-bt}{m}\right)\right]dt}\\ x-x_0=\frac{mgt}{b}+{\left[\left(\frac{m^{2}g}{b^2}\right)\exp \left(\frac{-bt}{m}\right)\right]}_0^t\end{array}$

Rearrange the above equation for $x$.

  $x-x_0=\frac{mgt}{b}+\left(\frac{m^{2}g}{b^2}\right)\left[\exp \left(\frac{-bt}{m}\right)-1\right]$ (VI)

Conclusion:

Substitute $0.500\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $0.0417\text{kg/s}$ for $b$, $0$ for $x_0$ and $3.45\text{s}$ for $t$ in equation (VI) to find $x$.

  $\begin{array}{c}x-0=\frac{\left(0.500\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\left(3.45\text{s}\right)}{\left(0.0417\text{kg/s}\right)}+\left(\frac{{\left(0.500\text{kg}\right)}^2\left(9.80{\text{m/s}}^2\right)}{{\left(0.0417\text{kg/s}\right)}^2}\right)\left[\exp \left(\frac{-\left(0.0417\text{kg/s}\right)\left(3.45\text{s}\right)}{\left(0.500\text{kg}\right)}\right)-1\right]\\ x=405\text{m}+1409\text{m}\left[\exp \left(-0.288\right)-1\right]\\ =405\text{m}+1409\text{m}\left[-0.250\right]\\ =52.7\text{m}\end{array}$

Therefore, the distance traveled by the sphere is $52.7\text{m}$.