Life scientists use ultracentrifuges to separate biological components or to remove molecules from suspension. Samples in a symmetric array of containers are spun rapidly about a central axis. The centrifugal acceleration they experience in their moving reference frame acts as "artificial gravity" to effect a rapid separation. If the sample containers are \(10.0 \mathrm{~cm}\) from the rotation axis, what rotation frequency is required to produce an acceleration of \(1.00 \cdot 10^{5} g ?\)

Short Answer

Expert verified
Answer: The required rotation frequency is approximately \(25,320.35\) rpm.

Step by step solution

01

Convert acceleration to m/s²

First, we need to convert the given acceleration from \(1.00\times 10^5 g\) to m/s². The gravitational acceleration on Earth's surface is approximately \(9.81 \mathrm{m/s²}\). Thus, the desired acceleration in m/s² can be found by multiplying the given acceleration by the gravitational constant: \(a = 1.00\times 10^5g \cdot 9.81 \mathrm{m/s^2}\).
02

Calculate the angular velocity

Now, we can use the formula for the centrifugal acceleration to find the angular velocity: \(a_c = \omega^2r\) We can rearrange the formula to solve for \(\omega\): \(\omega = \sqrt{\frac{a_c}{r}}\) Plug in the values for \(a_c\) and \(r\): \(\omega = \sqrt{\frac{1.00\times 10^5 \cdot 9.81 \mathrm{m/s^2}}{0.1 \mathrm{m}}}\)
03

Calculate the rotation frequency

Now that we have the angular velocity in radians per second, we can convert it to the rotation frequency in revolutions per minute (rpm). We do this using the following conversions: 1 revolution = \(2\pi\) radians 1 minute = 60 seconds Therefore, the rotation frequency in rpm can be calculated as: \(f = \frac{\omega}{2\pi} \cdot 60\; \mathrm{rpm}\) Plug in the value for \(\omega\) derived in Step 2: \(f = \frac{\sqrt{\frac{1.00\times 10^5 \cdot 9.81 \mathrm{m/s^2}}{0.1 \mathrm{m}}}}{2\pi} \cdot 60\; \mathrm{rpm}\)
04

Simplify and find the final answer

Simplify the expression to get the final answer: \(f = \frac{\sqrt{\frac{1.00\times 10^5 \cdot 9.81 \mathrm{m/s^2}}{0.1 \mathrm{m}}}}{2\pi} \cdot 60\; \mathrm{rpm} \approx 25320.35 \; \mathrm{rpm}\) The required rotation frequency to produce an acceleration of \(1.00 \times 10^5 g\) is approximately \(25,320.35\) rpm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ultracentrifuges in Biology
Ultracentrifuges are pivotal instruments in the field of biology, particularly for separating cellular components such as proteins, nucleic acids, and organelles. By spinning samples at extremely high speeds, centrifugal forces are generated, analogous to gravity but much more powerful, called centrifugal acceleration. This 'artificial gravity' causes denser substances to migrate outwards and settle more quickly than they would using Earth’s gravity alone.

For instance, in the isolation of DNA, the ultracentrifuge can rapidly sediment DNA into a pellet at the bottom of a tube, while other components stay in the solution. This efficiency in separation is crucial for many diagnostic tests and research experiments that require high purity samples. The ability to control the rotational speed allows for precise manipulation of the centrifugal force, enabling biologists to target specific components with varying densities and sizes.
Angular Velocity
Angular velocity is a measure of the speed of rotation. It tells us how fast an object is rotating and is usually denoted by the Greek letter \(\omega\). When it comes to ultracentrifuges, angular velocity is key to understanding the forces at play. It's the angular velocity that relates directly to the centrifugal acceleration experienced by the objects being centrifuged.

The calculation of angular velocity involves the rotational speed of the object and the distance from the center of rotation. If you picture a point on the spinning sample tube in the ultracentrifuge, that point's angular velocity tells you how quickly it is moving in its circular path. In the context of our exercise, we calculate angular velocity using the formula \(\omega = \sqrt{\frac{a_c}{r}}\), where \(a_c\) is the centrifugal acceleration, and \(r\) is the radial distance from the axis of rotation.
Rotation Frequency
Rotation frequency, often measured in revolutions per minute (rpm), is a more intuitive metric for many people compared to angular velocity, because it corresponds directly to the number of complete revolutions a spinning object makes in one minute. When we speak of rotation frequency in biology labs, we're usually referring to the setting on the ultracentrifuge that determines how fast the rotor spins.

The relationship between angular velocity and rotation frequency is given by the formula \(f = \frac{\omega}{2\pi} \cdot 60\), turning the angular velocity in radians per second into the number of revolutions per minute. This makes it easier for biologists and lab technicians to set their equipment to achieve the desired centrifugal forces. Understanding how to control rotation frequency helps in tailoring protocols for separating biological molecules based on size and density.

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Most popular questions from this chapter

Two masses hang from two strings of equal length that are attached to the ceiling of a car. One mass is over the driver's seat; the other is over the passenger's seat. As the car makes a sharp turn, both masses swing away from the center of the turn. In their resulting positions, will they be farther apart, closer together, or the same distance apart as they were when the car wasn't turning?

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