Question

A sample of blood is placed in a centrifuge of radius 12.0 cm. The mass of a red blood cell is 3.0 ✕ 10^−16 kg, and the magnitude of the force acting on it as it settles out of the plasma is 4.0 ✕ 10^−11 N. At how many revolutions per second should the centrifuge be operated?

Answer 1

Answer:

167.85 rev / s

Explanation:

r = 12 cm = 0.12 m, m = 3 x 10^-16 kg, F = 4 x 10^-11 N

F = m r w^2

where, w is the angular velocity.

4 x 10^-11 = 3 x 10^-16 x 0.12 x w^2

w = 1054.1 rad / s

w = 2 π f

f = w / 2 π = 1054.1 / (2 x 3.14) = 167.85 rev / s

Answer 2

The number of revolutions given by the calculated frequency value in which the centrifuge would be operated is 167.8 Hz.

Recall :

• Frequency, f = ω/2π

• Force, F = mω²r

• Mass, m = $3 \times 10^{-16}$

• Force, F = $4 \times 10^{-11} N$

• Radius, r = 12 cm = 12/100 = 0.12 m

We calculate the angular velocity, ω thus :

ω² = F/mr

ω² = $\frac{4 \times 10^{-11}}{3 \times 10^{-16} \times 0.12 = 11.11 \times 10^{5}$

ω = $\sqrt{1.11 \times 10^{6}} = 1053.56 rad/s$

Frequency = 1053.56 ÷ (2π)

Frequency = 167.68 Hz

Therefore, the Number of revolutions per seconds would be about 167.8 Hz

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