Solution Found!
Verify that the linear speed of an ultracentrifuge is
Chapter 6, Problem 18(choose chapter or problem)
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
\(\begin{aligned} \omega= & \frac{50,000 \mathrm{rev}}{\mathrm{min}} \\ & =\frac{50,000 \mathrm{rev}}{1 \mathrm{~min}} \times \frac{1 \mathrm{~min}}{60 \mathrm{~s}} \times \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \\ & =\mathbf{5 2 3 5 . 9 8 \mathrm { rad } / \mathrm { s }} \end{aligned}\)