Chapter 5, Problem 84P

(a)

To determine

The radial acceleration of the material in the centrifugal conditions.

Answer to Problem 84P

The radial acceleration of the material in the centrifugal conditions is $\underset{\_}{89.36g}$.

Explanation of Solution

Write the expression for the speed of the material,

    $v=2\pi rf$        (I)

Here, $v$ is the speed of the material, $r$ is the radius of the rotation and $f$ is the frequency.

Write the expression for the radial acceleration of the material in the centrifugal conditions,

    $a=\frac{v^2}{r}$        (II)

Here, $a$ is the radial acceleration of the material in the centrifugal conditions.

Conclusion:

Substitute $0.080\text{m}$ for $r$ and $1000\text{rev/min}$ for $f$ in (I) to find $v$,

    $\begin{array}{c}v=2\pi \left(0.080\text{m}\right)\left(\frac{1000\text{rev}}{\min }\right)\times \left(\frac{1\min }{60\text{s}}\right)\\ =8.37\text{m/s}\end{array}$

Substitute $8.37\text{m/s}$ for $v$ and $0.080\text{m}$ for $r$ in (II) to find $a$ ,

    $\begin{array}{c}a=\frac{{\left(8.37\text{m/s}\right)}^2}{0.080\text{m}}\\ =875.71{\text{m/s}}^{\text{2}}\end{array}$

In terms of $g$,

    $\begin{array}{c}a=\frac{875.71{\text{m/s}}^{\text{2}}}{9.8{\text{m/s}}^{\text{2}}}g\\ =89.36g\end{array}$

Therefore, the radial acceleration of the material in the centrifugal conditions is $\underset{\_}{89.36g}$.

(b)

To determine

The net force on the red blood cell.

Answer to Problem 84P

The net force on the red blood cell is $\underset{\_}{7.88\times {10}^{-11}\text{N}}$.

Explanation of Solution

Write the expression for the net force on the red blood cell,

    $F=ma$        (III)

Here, $F$ is the net force on the red blood cell, $a$ is the acceleration and $m$ is the mass of the red blood cell.

Conclusion:

Substitute $9.0\times {10}^{-14}\text{kg}$ for $m$ and $875.71{\text{m/s}}^{\text{2}}$ for $a$ in (III) to find $F$,

    $\begin{array}{c}F=\left(9.0\times {10}^{-14}\text{kg}\right)\left(875.71{\text{m/s}}^{\text{2}}\right)\\ =7.88\times {10}^{-11}\text{N}\end{array}$

Therefore, the net force on the red blood cell is $\underset{\_}{7.88\times {10}^{-11}\text{N}}$.

(c)

To determine

The net force on the virus particle.

Answer to Problem 84P

The net force on the virus particle is $\underset{\_}{4.38\times {10}^{-18}\text{N}}$.

Explanation of Solution

Write the expression for the net force on the virus particle,

    $F=ma$        (IV)

Here, $F$ is the net force on the virus particle, $a$ is the acceleration and $m$  is the mass of the virus particle.

Conclusion:

Substitute $5.0\times {10}^{-21}\text{kg}$ for $m$ and $875.71{\text{m/s}}^{\text{2}}$ for $a$ in (IV) to find $F$,

    $\begin{array}{c}F=\left(5.0\times {10}^{-21}\text{kg}\right)\left(875.71{\text{m/s}}^{\text{2}}\right)\\ =4.38\times {10}^{-18}\text{N}\end{array}$

Therefore, the net force on the virus particle is $\underset{\_}{4.38\times {10}^{-18}\text{N}}$.

(d)

To determine

The radial acceleration of the material in the ultra-centrifuge conditions.

Answer to Problem 84P

The radial acceleration of the material in the ultra-centrifuge conditions is $\underset{\_}{503040.8g}$.

Explanation of Solution

Write the expression for the speed of the material,

    $v=2\pi rf$        (V)

Here, $v$ is the speed of the material, $r$ is the radius of the rotation and $f$ is the frequency.

Write the expression for the radial acceleration of the material in the ultra-centrifuge conditions,

    $a=\frac{v^2}{r}$        (VI)

Here, $a$ is the radial acceleration of the material in the ultra-centrifuge conditions.

Conclusion:

Substitute $0.080\text{m}$ for $r$ and $75000\text{rev/min}$ for $f$ in (V) to find $v$,

    $\begin{array}{c}v=2\pi \left(0.080\text{m}\right)\left(\frac{75000\text{rev}}{\min }\right)\times \left(\frac{1\min }{60\text{s}}\right)\\ =628\text{m/s}\end{array}$

Substitute $628\text{m/s}$ for $v$ and $0.080\text{m}$ for $r$ in (VI) to find $a$ ,

    $\begin{array}{c}a=\frac{{\left(628\text{m/s}\right)}^2}{0.080\text{m}}\\ =4929800{\text{m/s}}^{\text{2}}\\ =4.93\times {10}^6{\text{m/s}}^{\text{2}}\end{array}$

In terms of $g$ ,

    $\begin{array}{c}a=\frac{4929800{\text{m/s}}^{\text{2}}}{9.8{\text{m/s}}^{\text{2}}}g\\ =503040.8g\end{array}$

Therefore, the radial acceleration of the material in the ultracentrifuge conditions is $\underset{\_}{503040.8g}$.