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Verify that the linear speed of an ultracentrifuge is

Chapter 6, Problem 18

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QUESTION:

Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

Questions & Answers

QUESTION:

Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

ANSWER:

Step 1 of  4

(a)

The linear speed of  a point on an ultracentrifuge can be calculated using the expression for angular velocity (\(\omega\))

\(\omega=\frac{v}{r}, where v = linear speed in m/s , r = radius of the ultracentrifuge = 0.1 m

\(v=\omega r\)

Now , the ultracentrifuge is rotating at 50,000 rev/min . So the angular velocity \(\omega\) is

\(\begin{aligned} \omega= & \frac{50,000 \mathrm{rev}}{\mathrm{min}} \\ & =\frac{50,000 \mathrm{rev}}{1 \mathrm{~min}} \times \frac{1 \mathrm{~min}}{60 \mathrm{~s}} \times \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \\ & =\mathbf{5 2 3 5 . 9 8 \mathrm { rad } / \mathrm { s }} \end{aligned}\)

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