Chapter 11, Problem 111P

To determine

The variation of velocity with time.

The graph between the variations of velocity with time.

Time taken by ball to reach $99\%$ of its terminal velocity.

Answer to Problem 111P

The time taken to reach $99\%$ terminal velocity is $0.0126\text{s}$.

The graph between the variations of velocity with time is.

Explanation of Solution

Given information:

The diameter of the ball is $2\text{mm}$, the density of the ball is $2700\text{kg}/{\text{m}}^{\text{3}}$, the temperature of the oil is $40°C$, the density of the fluid is $876\text{kg}/{\text{m}}^{\text{3}}$ and the dynamic viscosity of the fluid is $0.2177\text{kg}/\text{m}\cdot \text{s}$.

The below figure represent the free body diagram of the ball.

Figure-(1)

Write the expression of net force acting on the ball.

  $F_{\text{net}}=W-F_D-F_B$...... (I)

Here, the weight of the ball is $W$, the drag force is $F_D$ and the bouncy force is $F_B$.

Write the expression of weight of the ball.

  $W=mg$

Here, the mass of the ball is $m$ and the acceleration due to gravity is $g$.

Write the expression of bouncy force.

  $F_D=m_{f}g$

Here, the mass of the fluid is $m_f$.

Substitute $mg$ for $W$ and $m_{f}g$ for $F_D$ in Equation (I).

  $F_{\text{net}}=mg-F_D-m_{f}g$....... (II)

Write the expression of the mass of the ball.

  $m={\rho }_s{V}_1$

Here, the density of the ball is ${\rho }_s$ and the volume is $V_1$.

Write the expression of the mass of the fluid.

  $m_f={\rho }_f{V}_1$

Here, the density of the fluid is ${\rho }_f$ and the volume is $V_1$.

Substitute ${\rho }_f{V}_1$ for $m_f$ and ${\rho }_s{V}_1$ for $m$ in Equation (II).

  $\begin{array}{c}{F}_{\text{net}}=\left({\rho }_s{V}_1\right)g-F_D-\left({\rho }_f{V}_1\right)g\\ ={\rho }_s{V}_{1}g-F_D-{\rho }_f{V}_{1}g\end{array}$...... (III)

Write the expression of net force by Newton's law of motion.

  $F_{\text{net}}=ma$...... (IV)

Here, the acceleration is $a$.

Write the expression of acceleration.

  $a=\frac{dV}{dt}$

Here, the change of velocity with respect to time is $\frac{dV}{dt}$.

Substitute ${\rho }_s{V}_1$ for $m$ and $\frac{dV}{dt}$ for $a$ in Equation (IV).

  $F_{\text{net}}=\left({\rho }_s{V}_1\right)\left(\frac{dV}{dt}\right)$

Substitute $\left({\rho }_s{V}_1\right)\left(\frac{dV}{dt}\right)$ for $F_{\text{net}}$ in Equation (III).

  $\left({\rho }_s{V}_1\right)\left(\frac{dV}{dt}\right)={\rho }_s{V}_{1}g-F_D-{\rho }_f{V}_{1}g$...... (V)

Write the expression of volume of the ball.

  $V_1=\frac{\pi D^3}{6}$

Here, the diameter of the ball is $D$.

Write the expression of drag force by using strokes equation.

  $F_D=3\pi \mu DV$

Here, the dynamic viscosity is $\mu$.

Substitute $3\pi \mu DV$ for $F_D$ and $\frac{\pi D^3}{6}$ for $V_1$ in Equation (V).

  $\begin{array}{l}\left({\rho }_s\frac{\pi D^3}{6}\right)\left(\frac{dV}{dt}\right)={\rho }_s\frac{\pi D^3}{6}g-3\pi \mu DV-{\rho }_f\frac{\pi D^3}{6}g\\ \left(\frac{dV}{dt}\right)=\frac{\left({\rho }_s\frac{\pi D^3}{6}g-3\pi \mu DV-{\rho }_f\frac{\pi D^3}{6}g\right)}{\left({\rho }_s\frac{\pi D^3}{6}\right)}\\ \left(\frac{dV}{dt}\right)=g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V\\ \frac{dV}{g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V}=dt\end{array}$...... (VI)

Integrate Equation (VI) on both sides as initial velocity is $0$ and the final velocity is $V$ with respect to time as initial time is $0$ and the final time is $t$.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{V}{\int }}\frac{dV}{g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V}}={\displaystyle \underset{0}{\overset{t}{\int }}dt}\\ {\left[\frac{\ln \left(g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V\right)}{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)}\right]}_0^V={\left[t\right]}_0^t\\ \left[\frac{\ln \left(g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V\right)}{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)}-\frac{\ln \left(g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)0\right)}{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)}\right]=t-0\\ \left(\frac{18\mu }{{\rho }_s{D}^2}\right)t=\ln \left(g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V\right)-\ln \left(g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]\right)\end{array}$

  $\begin{array}{l}\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t=\ln \left[\frac{\left(g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V\right)}{g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]}\right]\\ \frac{\left(g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V\right)}{g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]}=e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\\ \left(g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V\right)=g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\\ \left(\frac{18\mu }{{\rho }_s{D}^2}\right)V=g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]-g\left[1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right]e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\\ \end{array}$

  $\begin{array}{l}\left(\frac{18\mu }{{\rho }_s{D}^2}\right)V=g\left\{1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right\}\left[1-e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\right]\\ V=\frac{g\left\{1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right\}\left[1-e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\right]}{\left(\frac{18\mu }{{\rho }_s{D}^2}\right)}\\ V=g\left\{1-\left(\frac{{\rho }_f}{{\rho }_s}\right)\right\}\left[1-e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\right]\times \left(\frac{{\rho }_s{D}^2}{18\mu }\right)\\ V=g{D}^2\left[\frac{{\rho }_s-{\rho }_f}{18\mu }\right]\left[1-e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\right]\end{array}$...... (VII)

Substitute $V_{\text{terminal}}$ for $V$ and $\infty$ for $t$ in Equation (VII).

  $\begin{array}{c}{V}_{\text{terminal}}=g{D}^2\left[\frac{{\rho }_s-{\rho }_f}{18\mu }\right]\left[1-e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)\infty }\right]\\ =g{D}^2\left[\frac{{\rho }_s-{\rho }_f}{18\mu }\right]\left[1-0\right]\\ =g{D}^2\left[\frac{{\rho }_s-{\rho }_f}{18\mu }\right]\end{array}$....... (VIII)

Calculation:

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$, $2\text{mm}$ for $D$, $2700\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_s$, $876\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_f$ and $0.2177\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (VIII).

  $\begin{array}{c}{V}_{\text{terminal}}=\left(9.81\text{m}/{\text{s}}^2\right){\left(2\text{mm}\right)}^2\left[\frac{\left(2700\text{kg}/{\text{m}}^{\text{3}}\right)-\left(876\text{kg}/{\text{m}}^{\text{3}}\right)}{18\times \left(0.2177\text{kg}/\text{m}\cdot \text{s}\right)}\right]\\ =\left(9.81\text{m}/{\text{s}}^2\right){\left\{\left(2\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2\left[\frac{1824\text{kg}/{\text{m}}^{\text{3}}}{18\times \left(0.2177\text{kg}/\text{m}\cdot \text{s}\right)}\right]\\ =\left(9.81\text{m}/{\text{s}}^2\right)\left(0.000004{\text{m}}^{\text{2}}\right)\left(465.47\text{s}/{\text{m}}^2\right)\\ =0.01826\text{m}/\text{s}\end{array}$

Substitute $99\%V_{\text{terminal}}$ for $V$ and $V_{\text{terminal}}$ for $g{D}^2\left[\frac{{\rho }_s-{\rho }_f}{18\mu }\right]$ in Equation (VII).

  $\begin{array}{l}99\%V_{\text{terminal}}=V_{\text{terminal}}\left[1-e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\right]\\ \frac{99}{100}=1-e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}\\ e^{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t}=1-\frac{99}{100}\\ -\left(\frac{18\mu }{{\rho }_s{D}^2}\right)t=\ln \left(0.01\right)\end{array}$

  $t=\frac{-4.605}{-\left(\frac{18\mu }{{\rho }_s{D}^2}\right)}$...... (IX)

Substitute $2\text{mm}$ for $D$, $2700\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_s$ and $0.2177\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (IX).

  $\begin{array}{c}t=\frac{-4.605}{-\left(\frac{18\left(0.2177\text{kg}/\text{m}\cdot \text{s}\right)}{\left(2700\text{kg}/{\text{m}}^{\text{3}}\right){\left(2\text{mm}\right)}^2}\right)}\\ =\frac{4.605}{\left(\frac{18\left(0.2177\text{kg}/\text{m}\cdot \text{s}\right)}{\left(2700\text{kg}/{\text{m}}^{\text{3}}\right){\left\{\left(2\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2}\right)}\\ =\frac{4.605\text{s}}{362.83}\\ =0.0126\text{s}\end{array}$

The time taken to reach $99\%$ terminal velocity is $0.0126\text{s}$.

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$, $2\text{mm}$ for $D$, $2700\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_s$, $876\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_f$ and $0.2177\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (VII).

  $\begin{array}{l}V=\left(9.81\text{m}/{\text{s}}^2\right){\left(2\text{mm}\right)}^2\left[\frac{\left(2700\text{kg}/{\text{m}}^{\text{3}}\right)-\left(876\text{kg}/{\text{m}}^{\text{3}}\right)}{18\left(0.2177\text{kg}/\text{m}\cdot \text{s}\right)}\right]\left[1-e^{-\left(\frac{18\left(0.2177\text{kg}/\text{m}\cdot \text{s}\right)}{\left(2700\text{kg}/{\text{m}}^{\text{3}}\right){\left(2\text{mm}\right)}^2}\right)t}\right]\\ V=\left[\begin{array}{l}\left(9.81\text{m}/{\text{s}}^2\right){\left\{\left(2\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2\left[\frac{\left(2700\text{kg}/{\text{m}}^{\text{3}}\right)-\left(876\text{kg}/{\text{m}}^{\text{3}}\right)}{18\left(0.2177\text{kg}/\text{m}\cdot \text{s}\right)}\right]\times \\ \left[1-e^{-\left(\frac{18\left(0.2177\text{kg}/\text{m}\cdot \text{s}\right)}{\left(2700\text{kg}/{\text{m}}^{\text{3}}\right){\left\{\left(2\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2}\right)t}\right]\end{array}\right]\\ V=\left(0.018\text{m}/\text{s}\right)\left[1-e^{-362.83/\text{s}t}\right]\end{array}$

The variation of velocity with time.

Time $\left(t\right)$	Velocity $V=\left(0.018\text{m}/\text{s}\right)\left[1-e^{-362.83/\text{s}t}\right]$
$0\text{s}$	$0$
$0.01\text{s}$	$0.0175\text{m}/\text{s}$
$0.02\text{s}$	$0.0179\text{m}/\text{s}$
$0.03\text{s}$	$0.017996\text{m}/\text{s}$
$0.04\text{s}$	$0.17999\text{m}/\text{s}$
$0.05\text{s}$	$0.0179999\text{m}/\text{s}$
$0.06\text{s}$	$0.01799999\text{m}/\text{s}$
$0.07\text{s}$	$0.01799999\text{m}/\text{s}$
$0.08\text{s}$	$0.017999999\text{m}/\text{s}$
$0.09\text{s}$	$0.0179999999\text{m}/\text{s}$
$0.10\text{s}$	$0.018\text{m}/\text{s}$

The graph between the variation of velocity with time is.

Figure-(2)

Conclusion:

The time taken to reach $99\%$ terminal velocity is $0.0126\text{s}$.