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\(\bullet\) Ultracentrifuge. Find the required angular speed (in rpm) of an ultracentrifuge for the radial acceleration of a point 2.50 \(\mathrm{cm}\) from the axis to equal \(400,000 g .\)

Short Answer

Expert verified
The required angular speed is approximately 1,882,604 rpm.

Step by step solution

01

Understand the Relationship

The problem asks for the angular speed of the ultracentrifuge so that the radial acceleration equals \(400,000g\), where \(g\) refers to the acceleration due to gravity. The given distance from the axis is 2.50 cm.
02

Convert Units

Convert the distance from centimeters to meters since SI units will be used. Thus, 2.50 cm is equal to 0.025 m.
03

Express Radial Acceleration

The radial acceleration \(a_r\) at a distance \(r\) is given by \(a_r = \omega^2 r\), where \(\omega\) is the angular speed in rad/s and \(r\) is the radius in meters. We set \(a_r = 400,000g = 400,000 \, 9.81 \, \text{m/s}^2\).
04

Substitute and Solve for \(\omega\)

Substituting the values into the radial acceleration formula: \[ 400,000 \, 9.81 = \omega^2 \, 0.025 \]Solve for \(\omega^2\):\[ \omega^2 = \frac{400,000 \, 9.81}{0.025} \]Calculate \(\omega\) by taking the square root.
05

Convert to RPM

The angular speed \(\omega\) is in rad/s. Convert this speed into revolutions per minute (rpm) using the conversion:\[ \text{rpm} = \frac{\omega \times 60}{2\pi} \]Substitute the value obtained from the square root calculation into this conversion to find the rpm.

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Key Concepts

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

Angular Speed Calculation
Calculating angular speed is essential in various physics problems, especially in scenarios involving circular motion like those in ultracentrifuges. Angular speed, often denoted by the Greek letter \(\omega\), refers to how fast an object rotates or revolves relative to another point, typically the center of the circle. It is measured in radians per second (rad/s).
To calculate angular speed when given radial acceleration and radius, we use the formula for radial acceleration:
  • \(a_r = \omega^2 r\) - where \(a_r\) is the radial acceleration, \(\omega\) is the angular speed, and \(r\) is the radius.
  • Rearrange this formula to solve for \(\omega\): \(\omega = \sqrt{\frac{a_r}{r}}\).
By substituting the given radial acceleration and the radius into this formula, we can find the angular speed necessary for a point on the ultracentrifuge to achieve the required acceleration.
Radial Acceleration
Radial acceleration is a key concept in understanding circular motion. It refers to the acceleration experienced by an object moving along a circular path, directed towards the center of the circle.
In this context, we talk about the ultracentrifuge, where points on the device need to achieve extremely high radial accelerations, sometimes even to the tune of hundreds of thousands of times the acceleration due to gravity (denoted as \(g\)).
  • Radial acceleration formula: \(a_r = \omega^2 r\).
  • \(g\) is the standard gravitational acceleration on Earth, approximately \(9.81 \text{ m/s}^2\).
When given a radial acceleration value expressed in terms of \(g\), such as \(400,000g\), it is essential to convert this value into \(\text{m/s}^2\) by multiplying it by \(9.81\). This converted value is then used in calculations to determine angular speed.
Unit Conversion
Proper unit conversion is crucial in physics to ensure consistent calculations and accurate results. In this exercise, converting the units of distance and acceleration is necessary for determining the angular speed of an ultracentrifuge.
Firstly, convert the distance from centimeters to meters to align with the International System of Units (SI):
  • 2.50 cm = 0.025 m
Next, consider the conversion of radial acceleration often given in terms of \(g\):
  • Convert \(400,000g\) to \(\text{m/s}^2\) by multiplying by \(9.81\), the acceleration due to gravity.
Finally, when obtaining \(\omega\) in rad/s, convert to revolutions per minute (rpm) for practical understanding in the context of centrifuges:
  • Use the formula: \( \text{rpm} = \frac{\omega \times 60}{2\pi} \)
These conversions ensure results are meaningful and applicable in real-world situations.
Centrifuge Physics
Centrifuge physics revolves around the principles of rotational motion, allowing for the separation of substances based on density. An ultracentrifuge is a high-speed centrifuge that achieves incredibly high speeds, generating massive radial accelerations that force denser particles to the bottom.
The ultracentrifuge operates by spinning samples at high angular velocities. The key elements of centrifuge physics include:
  • Rotational motion and angular speed: Characterized by \(\omega\).
  • Radial acceleration: Centripetal force acting on the particles, moving them outward.
  • Radial distance: Distance from the axis of rotation to the point of interest in the sample tube.
The principles of centrifugal force are employed to efficiently separate microscopic substances, such as proteins or nucleic acids, based on their mass. Understanding these concepts is vital for the precise operation and application of ultracentrifuges in scientific research and medical labs.

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Most popular questions from this chapter

\(\cdot\) A car is traveling at a speed of 63 \(\mathrm{mi} / \mathrm{h}\) on a freeway. If its tires have diameter 24.0 in and are rolling without sliding or slipping, what is their angular velocity?

A A wall clock on Planet \(X\) has two hands that are aligned at midnight and turn in the same direction at uniform rates, one at 0.0425 \(\mathrm{rad} / \mathrm{s}\) and the other at 0.0163 \(\mathrm{rad} / \mathrm{s}\) . At how many seconds after midnight are these hands (a) first aligned and (b) next aligned?

\(\bullet\) (a) A cylinder 0.150 \(\mathrm{m}\) in diameter rotates in a lathe at 620 \(\mathrm{rpm} .\) What is the tangential speed of the surface of the cylinder? (b) The proper tangential speed for machining cast iron is about 0.600 \(\mathrm{m} / \mathrm{s} .\) At how many revolutions per minute should a piece of stock 0.0800 \(\mathrm{m}\) in diameter be rotated in a lathe to produce this tangential speed?

A \(\mathrm{A} 392\) -N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2} .\) Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 J. Calculate \(h .\)

A solid uniform marble and a block of ice, each with the same mass, start from rest at the same height \(H\) above the bottom of a hill and move down it. The marble rolls without slipping, but the ice slides without friction. (a) Find the speed of each of these objects when it reaches the bottom of the hill. (b) Which object is moving faster at the bottom, the ice or the marble? (c) Which object has more kinetic energy at the bottom, the ice or the marble?

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