Chapter 5, Problem 84P

(a)

To determine

The radial acceleration of the material in terms of $g$.

Answer to Problem 84P

The radial acceleration is $90g$.

Explanation of Solution

The angular speed of centrifuge is $1.0\times {10}^3\text{rev}/\text{min}$ and the radius of rotor is $8.0\text{cm}$.

Write the relation between the radial acceleration and the angular speed.

$a_r={\omega }^{2}r$

Here, the radial acceleration is $a_r$, angular speed is $\omega$, and the radius of rotor is $r$.

Conclusion:

Substitute $1.0\times {10}^3\text{rev}/\text{min}$ for $\omega$ and $8.0\text{cm}$ for $r$ in the above equation to find $a_r$.

$\begin{array}{c}{a}_r={\left(1.0\times {10}^3\text{rev}/\text{min}\left(\frac{\text{2π}\text{rad}}{\text{1}\text{rev}}\right)\left(\frac{\text{1}\text{min}}{\text{60}\text{s}}\right)\right)}^2\left(8.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)\left(\frac{g}{\text{9}\text{.80}\text{m}/{\text{s}}^{\text{2}}}\right)\\ =\left(1.09\times {10}^4\right)\left(0.0082\right)g\\ =90g\end{array}$

Therefore, the radial acceleration is $90g$.

(b)

To determine

The net force on the red blood cell.

Answer to Problem 84P

The net force is $7.9\times {10}^{-11}\text{N}$.

Explanation of Solution

The angular speed of centrifuge is $1.0\times {10}^3\text{rev}/\text{min}$, radius of rotor is $8.0\text{cm}$, and the mass of blood cell is $9.0\times {10}^{-14}\text{kg}$.

Write the equation for net force.

$F_{net}=m{a}_r$

Here, the net force is $F_{net}$, mass of red blood cell is $m$.

Conclusion:

Substitute $9.0\times {10}^{-14}\text{kg}$ for $m$ and $90g$ for $a_r$ in the above equation to find $F_{net}$.

$\begin{array}{c}{F}_{net}=\left(9.0\times {10}^{-14}\text{kg}\right)\left(90g\left(\frac{9.8\text{m}/{\text{s}}^{\text{2}}}{g}\right)\right)\\ =7.9\times {10}^{-11}\text{N}\end{array}$

Therefore, the net force is $7.9\times {10}^{-11}\text{N}$.

(b)

To determine

The net force on the red blood cell.

Answer to Problem 84P

The net force is $7.9\times {10}^{-11}\text{N}$.

Explanation of Solution

The angular speed of centrifuge is $1.0\times {10}^3\text{rev}/\text{min}$, radius of rotor is $8.0\text{cm}$, and the mass of blood cell is $9.0\times {10}^{-14}\text{kg}$.

Write the equation for net force.

$F_{net}=m{a}_r$

Here, the net force is $F_{net}$, mass of red blood cell is $m$.

Conclusion:

Substitute $9.0\times {10}^{-14}\text{kg}$ for $m$ and $90g$ for $a_r$ in the above equation to find $F_{net}$.

$\begin{array}{c}{F}_{net}=\left(9.0\times {10}^{-14}\text{kg}\right)\left(90g\left(\frac{9.8\text{m}/{\text{s}}^{\text{2}}}{g}\right)\right)\\ =7.9\times {10}^{-11}\text{N}\end{array}$

Therefore, the net force is $7.9\times {10}^{-11}\text{N}$.

(c)

To determine

The net force on the virus particle.

Answer to Problem 84P

The net force is $4.4\times {10}^{-18}\text{N}$.

Explanation of Solution

The angular speed of centrifuge is $1.0\times {10}^3\text{rev}/\text{min}$, radius of rotor is $8.0\text{cm}$, and the mass of virus is $5.0\times {10}^{-21}\text{kg}$.

Write the equation for net force.

$F_{net}=m_{virus}{a}_r$

Here, the mass of virus is $m_{virus}$.

Conclusion:

Substitute $5.0\times {10}^{-21}\text{kg}$ for $m_{virus}$ and $90g$ for $a_r$ in the above equation to find $F_{net}$.

$\begin{array}{c}{F}_{net}=\left(5.0\times {10}^{-21}\text{kg}\right)\left(90g\left(\frac{9.8\text{m}/{\text{s}}^{\text{2}}}{g}\right)\right)\\ =4.4\times {10}^{-18}\text{N}\end{array}$

Therefore, the net force is $4.4\times {10}^{-18}\text{N}$.

(d)

To determine

The radial acceleration inside the ultracentrifuge in terms of $g$.

Answer to Problem 84P

The radial acceleration is $5.0\times {10}^{5}g$.

Explanation of Solution

The angular speed of centrifuge is $75,000\text{rev}/\text{min}$ and the radius of rotor is $8.0\text{cm}$.

Write the equation for net force.

Write the relation between the radial acceleration and the angular speed.

$a_r={\omega }^{2}r$

Conclusion:

Substitute $75,000\text{rev}/\text{min}$ for $\omega$ and $8.0\text{cm}$ for $r$ in the above equation to find $a_r$.

$\begin{array}{c}{a}_r={\left(75,000\text{rev}/\text{min}\left(\frac{\text{2π}\text{rad}}{\text{1}\text{rev}}\right)\left(\frac{\text{1}\text{min}}{\text{60}\text{s}}\right)\right)}^2\left(8.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)\left(\frac{g}{\text{9}\text{.80}\text{m}/{\text{s}}^{\text{2}}}\right)\\ =\left(61.62\right)\left(0.0082\right)g\\ =5.0\times {10}^{5}g\end{array}$

Therefore, the net force is $5.0\times {10}^{5}g$.