Chapter 11, Problem 113P

To determine

The error involved for each case and asses the accuracy of strokes law.

Answer to Problem 113P

In first ball $9.375\%$ error is involved.

In second ball $10.156\%$ error is involved.

In third ball $-4.3\%$ error is involved.

Explanation of Solution

Given information:

The diameter of the first aluminum ball is $2\text{mm}$, the diameter of the second aluminum ball is $4\text{mm}$, the diameter of the third aluminum ball is $10\text{mm}$, the density of aluminum ball is $2600\text{kg}/{\text{m}}^{\text{3}}$, the temperature of glycerin is $22°C$, the density of the glycerin is $1274\text{kg}/{\text{m}}^{\text{3}}$, the dynamic viscosity of glycerin is $1\text{kg}/\text{m}\cdot \text{s}$, terminal setting velocity of the first ball is $3.2\text{mm}/\text{s}$, the terminal velocity of second ball is $12.8\text{mm}/\text{s}$ and the terminal velocity of third ball is $60.4\text{mm}/\text{s}$.

Write the expression of drag force acting on the freely falling body.

  $F_D=W-F_B$...... (I)

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

Write the expression of drag force by strokes law.

  $F_D=3\pi VD+\left(\frac{9\pi }{16}\right)\rho V^2{D}^2$...... (II)

Here, the terminal velocity is $V$, the diameter of the ball is $D$ and the density of the fluid is $\rho$.

Write the expression of the weight of the body.

  $W={\rho }_{s}g{V}_1$...... (III)

Here, the density of the ball is ${\rho }_s$, the acceleration due to gravity is $g$ and the volume of the ball is $V_1$.

Write the expression of bouncy force applied by a body.

  $F_B=\rho g{V}_1$...... (IV)

Here, the density of the fluid is $\rho$.

Substitute $3\pi VD+\left(\frac{9\pi }{16}\right)\rho V^2{D}^2$ for $F_D$, $\rho g{V}_1$ for $F_B$ and ${\rho }_{s}g{V}_1$ for $W$ in Equation (I).

  $\begin{array}{l}3\pi VD+\left(\frac{9\pi }{16}\right)\rho V^2{D}^2={\rho }_{s}g{V}_1-\rho g{V}_1\\ \left(3\pi D\right)V+\left(\frac{9\pi \rho D^2}{16}\right)V^2=g{V}_1\left({\rho }_s-\rho \right)\\ \left(\frac{9\pi \rho D^2}{16}\right)V^2+\left(3\pi D\right)V-g{V}_1\left({\rho }_s-\rho \right)=0\end{array}$...... (V)

Write the expression of volume of the ball.

  $V_1=\frac{\pi D^3}{6}$...... (VI)

Write the expression of percentage error.

  $\text{Error=}\left(\frac{V_g-V}{V_g}\right)\times 100$...... (VII)

Here, the given terminal velocity is $V_g$.

Calculation:

For First ball,

Substitute $2\text{mm}$ for $D$ in Equation (VI).

  $\begin{array}{c}{V}_1=\frac{\pi {\left(2\text{mm}\right)}^3}{6}\\ =\frac{\pi {\left\{\left(2\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^3}{6}\\ =\frac{\pi \times 8\times {10}^{-9}}{6}{\text{m}}^{\text{3}}\\ =4.189\times {10}^{-9}{\text{m}}^{\text{3}}\end{array}$

Substitute $1274\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_s$, $9.81{\text{m}/\text{s}}^2$ for $g$ $2\text{mm}$ for $D$ and $4.189\times {10}^{-9}{\text{m}}^{\text{3}}$ for $V_1$ in Equation (V).

$\left[\begin{array}{l}\left\{\frac{9\pi \times \left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(2\text{mm}\right)}^2}{16}\right\}{V}^2+\left\{3\pi \left(2\text{mm}\right)\right\}V\\ -\left(9.81{\text{m}/\text{s}}^2\right)\times \left(4.189\times {10}^{-9}{\text{m}}^{\text{3}}\right)\times \left\{\left(2600\text{kg}/{\text{m}}^{\text{3}}\right)-\left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\right\}\end{array}\right]=0$

$\left[\begin{array}{l}\left\{\frac{9\pi \times \left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left\{\left(2\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2}{16}\right\}{V}^2\\ +\left\{3\pi \left(2\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}V-\left[\begin{array}{l}\left(9.81{\text{m}/\text{s}}^2\right)\times \left(4.189\times {10}^{-9}{\text{m}}^{\text{3}}\right)\\ \times \left\{\left(2600\text{kg}/{\text{m}}^{\text{3}}\right)-\left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\right\}\end{array}\right]\end{array}\right]=0$

$\left[\left(9.005\times {10}^{-3}\text{kg}/\text{m}\right)V^2+\left(0.0188\text{m}\right)V-\left(5.3455\times {10}^{-4}{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right]=0$

$V=\frac{\left[-\left(0.0188\text{m}\right)\pm \sqrt{\begin{array}{l}{\left(0.0188\text{m}\right)}^2-\\ \left\{4\times \left(9.005\times {10}^{-3}\text{kg}/\text{m}\right)\times \left\{-\left(5.3455\times {10}^{-4}{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(9.005\times {10}^{-3}\right)}$

Taking positive sign.

  $\begin{array}{c}V=\frac{\left[-\left(0.0188\text{m}\right)+\sqrt{\begin{array}{l}{\left(0.0188\text{m}\right)}^2-\\ \left\{4\times \left(9.005\times {10}^{-3}\text{kg}/\text{m}\right)\times \left\{-\left(5.3455\times {10}^{-4}{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(9.005\times {10}^{-3}\text{kg}/\text{m}\right)}\\ =0.0029\text{m}/\text{s}\end{array}$

Taking Negative sign.

  $\begin{array}{c}V=\frac{\left[-\left(0.0188\text{m}\right)-\sqrt{\begin{array}{l}{\left(0.0188\text{m}\right)}^2-\\ \left\{4\times \left(9.005\times {10}^{-3}\text{kg}/\text{m}\right)\times \left\{-\left(5.3455\times {10}^{-4}{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(9.005\times {10}^{-3}\text{kg}/\text{m}\right)}\\ =-2.8967\text{m}/\text{s}\end{array}$

Here, the negative value of velocity is not desirable so the velocity is $0.0029\text{m}/\text{s}$.

Substitute $0.0029\text{m}/\text{s}$ for $V$ and $3.2\text{mm}/\text{s}$ for $V_g$ in Equation (VII).

  $\begin{array}{c}\text{Error=}\left\{\frac{\left(3.2\text{mm}/\text{s}\right)-\left(0.0029\text{m}/\text{s}\right)}{\left(3.2\text{mm}/\text{s}\right)}\right\}\times 100\\ =\left\{\frac{\left(3.2\text{mm}/\text{s}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)-\left(0.0029\text{m}/\text{s}\right)}{\left(3.2\text{mm}/\text{s}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)}\right\}\times 100\\ =\left\{\frac{\left(0.0032\text{m}/\text{s}\right)-\left(0.0029\text{m}/\text{s}\right)}{\left(0.0032\text{m}/\text{s}\right)}\right\}\times 100\\ =9.375\%\end{array}$

In first ball $9.375\%$ error is involved.

For Second ball,

Substitute $4\text{mm}$ for $D$ in Equation (VI).

  $\begin{array}{c}{V}_1=\frac{\pi {\left(4\text{mm}\right)}^3}{6}\\ =\frac{\pi {\left\{\left(4\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^3}{6}\\ =\frac{\pi \times 6.4\times {10}^{-8}}{6}{\text{m}}^{\text{3}}\\ =3.35\times {10}^{-8}{\text{m}}^{\text{3}}\end{array}$

Substitute $1274\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_s$, $9.81{\text{m}/\text{s}}^2$ for $g$ $4\text{mm}$ for $D$ and $3.35\times {10}^{-8}{\text{m}}^{\text{3}}$ for $V_1$ in Equation (V).

$\left[\begin{array}{l}\left\{\frac{9\pi \times \left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(4\text{mm}\right)}^2}{16}\right\}{V}^2+\left\{3\pi \left(4\text{mm}\right)\right\}V\\ -\left(9.81{\text{m}/\text{s}}^2\right)\times \left(3.35\times {10}^{-8}{\text{m}}^{\text{3}}\right)\times \left\{\left(2600\text{kg}/{\text{m}}^{\text{3}}\right)-\left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\right\}\end{array}\right]=0$

$\left[\begin{array}{l}\left\{\frac{9\pi \times \left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left\{\left(4\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2}{16}\right\}{V}^2\\ +\left\{3\pi \left(4\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}V-\left[\begin{array}{l}\left(9.81{\text{m}/\text{s}}^2\right)\times \left(3.35\times {10}^{-8}{\text{m}}^{\text{3}}\right)\\ \times \left\{\left(2600\text{kg}/{\text{m}}^{\text{3}}\right)-\left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\right\}\end{array}\right]\end{array}\right]=0$

$\left[\left(0.036\text{kg}/\text{m}\right)V^2+\left(0.03768\text{m}\right)V-\left(4.4\times {10}^{-4}{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right]=0$

$V=\frac{\left[-\left(0.0188\text{m}\right)\pm \sqrt{\begin{array}{l}{\left(0.03768\text{m}\right)}^2-\\ \left\{4\times \left(0.036\text{kg}/\text{m}\right)\times \left\{-\left(5.3455\times {10}^{-4}{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(0.036\text{kg}/\text{m}\right)}$

Taking positive sign.

  $\begin{array}{c}V=\frac{\left[-\left(0.0188\text{m}\right)+\sqrt{\begin{array}{l}{\left(0.03768\text{m}\right)}^2-\\ \left\{4\times \left(0.036\text{kg}/\text{m}\right)\times \left\{-\left(5.3455\times {10}^{-4}{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(0.036\text{kg}/\text{m}\right)}\\ =0.0115\text{m}/\text{s}\end{array}$

Taking Negative sign.

  $\begin{array}{c}V=\frac{\left[-\left(0.0188\text{m}\right)-\sqrt{\begin{array}{l}{\left(0.03768\text{m}\right)}^2-\\ \left\{4\times \left(0.036\text{kg}/\text{m}\right)\times \left\{-\left(5.3455\times {10}^{-4}{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(0.036\text{kg}/\text{m}\right)}\\ =-1.058\text{m}/\text{s}\end{array}$

Here, the negative value of velocity is not desirable so the velocity is $0.0115\text{m}/\text{s}$.

Substitute $0.0115\text{m}/\text{s}$ for $V$ and $12.8\text{mm}/\text{s}$ for $V_g$ in Equation (VII).

  $\begin{array}{c}\text{Error=}\left\{\frac{\left(12.8\text{mm}/\text{s}\right)-\left(0.0115\text{m}/\text{s}\right)}{\left(12.8\text{mm}/\text{s}\right)}\right\}\times 100\\ =\left\{\frac{\left(12.8\text{mm}/\text{s}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)-\left(0.0115\text{m}/\text{s}\right)}{\left(12.8\text{mm}/\text{s}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)}\right\}\times 100\\ =\left\{\frac{\left(0.0128\text{m}/\text{s}\right)-\left(0.0115\text{m}/\text{s}\right)}{\left(0.0128\text{m}/\text{s}\right)}\right\}\times 100\\ =10.156\%\end{array}$

In second ball $10.156\%$ error is involved.

For Third ball.

Substitute $10\text{mm}$ for $D$ in Equation (VI).

  $\begin{array}{c}{V}_1=\frac{\pi {\left(10\text{mm}\right)}^3}{6}\\ =\frac{\pi {\left\{\left(10\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^3}{6}\\ =\frac{\pi \times 1\times {10}^{-6}}{6}{\text{m}}^{\text{3}}\\ =5.235\times {10}^{-7}{\text{m}}^{\text{3}}\end{array}$

Substitute $1274\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_s$, $9.81{\text{m}/\text{s}}^2$ for $g$ $10\text{mm}$ for $D$ and $5.235\times {10}^{-7}{\text{m}}^{\text{3}}$ for $V_1$ in Equation (V).

$\left[\begin{array}{l}\left\{\frac{9\pi \times \left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(10\text{mm}\right)}^2}{16}\right\}{V}^2+\left\{3\pi \left(10\text{mm}\right)\right\}V\\ -\left(9.81{\text{m}/\text{s}}^2\right)\times \left(5.235\times {10}^{-7}{\text{m}}^{\text{3}}\right)\times \left\{\left(2600\text{kg}/{\text{m}}^{\text{3}}\right)-\left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\right\}\end{array}\right]=0$

$\left[\begin{array}{l}\left\{\frac{9\pi \times \left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left\{\left(10\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2}{16}\right\}{V}^2\\ +\left\{3\pi \left(10\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}V-\left[\begin{array}{l}\left(9.81{\text{m}/\text{s}}^2\right)\times \left(5.235\times {10}^{-7}{\text{m}}^{\text{3}}\right)\\ \times \left\{\left(2600\text{kg}/{\text{m}}^{\text{3}}\right)-\left(1274\text{kg}/{\text{m}}^{\text{3}}\right)\right\}\end{array}\right]\end{array}\right]=0$

$\left[\left(0.225\text{kg}/\text{m}\right)V^2+\left(0.0942\text{m}\right)V-\left(0.0068{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right]=0$

$V=\frac{\left[-\left(0.0942\text{m}\right)\pm \sqrt{\begin{array}{l}{\left(0.0942\text{m}\right)}^2-\\ \left\{4\times \left(0.225\text{kg}/\text{m}\right)\times \left\{-\left(0.0068{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(0.225\text{kg}/\text{m}\right)}$

Taking positive sign.

  $\begin{array}{c}V=\frac{\left[-\left(0.0942\text{m}\right)+\sqrt{\begin{array}{l}{\left(0.0942\text{m}\right)}^2-\\ \left\{4\times \left(0.225\text{kg}/\text{m}\right)\times \left\{-\left(0.0068{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(0.225\text{kg}/\text{m}\right)}\\ =0.063\text{m}/\text{s}\end{array}$

Taking Negative sign.

  $\begin{array}{c}V=\frac{\left[-\left(0.0942\text{m}\right)+\sqrt{\begin{array}{l}{\left(0.0942\text{m}\right)}^2-\\ \left\{4\times \left(0.225\text{kg}/\text{m}\right)\times \left\{-\left(0.0068{\text{kg}\cdot \text{m}/\text{s}}^2\right)\right\}\right\}\end{array}}\right]}{2\times \left(0.225\text{kg}/\text{m}\right)}\\ =-0.48\text{m}/\text{s}\end{array}$

Here, the negative value of velocity is not desirable so the velocity is $0.063\text{m}/\text{s}$.

Substitute $0.063\text{m}/\text{s}$ for $V$ and $60.4\text{mm}/\text{s}$ for $V_g$ in Equation (VII).

  $\begin{array}{c}\text{Error}=\left\{\frac{\left(60.4\text{mm}/\text{s}\right)-\left(0.063\text{m}/\text{s}\right)}{\left(12.8\text{mm}/\text{s}\right)}\right\}\times 100\\ =\left\{\frac{\left(60.4\text{mm}/\text{s}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)-\left(0.063\text{m}/\text{s}\right)}{\left(60.4\text{mm}/\text{s}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)}\right\}\times 100\\ =\left\{\frac{\left(0.0604\text{m}/\text{s}\right)-\left(0.063\text{m}/\text{s}\right)}{\left(0.0604\text{m}/\text{s}\right)}\right\}\times 100\\ =-4.3\%\end{array}$

In third ball $-4.3\%$ error is involved.

Conclusion:

In first ball $9.375\%$ error is involved.

In second ball $10.156\%$ error is involved.

In third ball $-4.3\%$ error is involved.