Centrifuges are commonly used in biological laboratories for the isolation and maintenance of cell preparations. For cell separation, the centrifugation conditions are typically \(1.0 \times 10^{3}\) rpm using an 8.0 -cm-radius rotor. (a) What is the radial acceleration of material in the centrifuge under these conditions? Express your answer as a multiple of \(g .\) (b) At $1.0 \times 10^{3}$ rpm (and with a 8.0-cm rotor), what is the net force on a red blood cell whose mass is \(9.0 \times 10^{-14} \mathrm{kg} ?\) (c) What is the net force on a virus particle of mass \(5.0 \times 10^{-21} \mathrm{kg}\) under the same conditions? (d) To pellet out virus particles and even to separate large molecules such as proteins, superhigh-speed centrifuges called ultracentrifuges are used in which the rotor spins in a vacuum to reduce heating due to friction. What is the radial acceleration inside an ultracentrifuge at 75000 rpm with an 8.0 -cm rotor? Express your answer as a multiple of \(g\).

Step by step solution

01

(a) Calculate the radial acceleration

Firstly, we have to convert rpm to radians per second. 1.0 x 10^3 rpm = (1.0 × 10^3 * 2π) / 60 rad/s ≈ 104.7 rad/s Now, we can find the radial acceleration using the formula \(a_r = \omega^2 R\): \(a_r = (104.7 \,\text{rad/s})^2 \times 0.08\, \text{m} \approx 914\, \text{m/s}^2\) To express the answer as a multiple of \(g\), divide the radial acceleration by \(g\) (9.8 m/s²): \(a_r / g = 914\, \text{m/s}^2 / 9.8\, \text{m/s}^2 \approx 93.3 g\)

02

(b) Calculate the net force on a red blood cell

We'll use the equation \(F = ma_r\) to find the net force on a red blood cell with mass \(9.0 \times 10^{-14}\, \mathrm{kg}\): \(F_{\text{blood cell}} = (9.0 \times 10^{-14}\, \mathrm{kg}) \times (914\, \text{m/s}^2) \approx 8.23 \times 10^{-11}\, \text{N}\)

03

(c) Calculate the net force on a virus particle

Using the same equation, we'll find the net force on a virus particle with mass \(5.0 \times 10^{-21}\, \mathrm{kg}\): \(F_{\text{virus particle}} = (5.0 \times 10^{-21}\, \mathrm{kg}) \times (914\, \text{m/s}^2) \approx 4.57 \times 10^{-18}\, \text{N}\)

04

(d) Calculate the radial acceleration inside an ultracentrifuge

Since we have an ultracentrifuge spinning at 75000 rpm, we first convert that to radians per second: 75000 rpm = (75000 * 2π) / 60 rad/s ≈ 7854 rad/s Next, we find the radial acceleration using the formula \(a_r = \omega^2 R\) with the same radius as before: \(a_r = (7854 \,\text{rad/s})^2 \times 0.08\, \text{m} \approx 4.89 \times 10^6\, \text{m/s}^2\) To express the answer as a multiple of \(g\), divide the radial acceleration by \(g\) (9.8 m/s²): \(a_r / g = 4.89 \times 10^6 \, \text{m/s}^2 / 9.8\, \text{m/s}^2 \approx 4.99 \times 10^5 g\)

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