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Chapter 5: Problem 51

A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to settle through the less dense blood serum and collect at the bottom of the container. Suppose the centripetal acceleration of the sample is \(6.25 \times 10^{3}\) times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of \(5.00 \mathrm{~cm}\) from the axis of rotation?

Short Answer

Expert verified
The sample makes approximately 4750 RPM.

Step by step solution

01

Understand the Problem

We are asked to find the revolutions per minute (RPM) of a centrifuge sample given its centripetal acceleration and the radius of rotation. The centripetal acceleration is expressed as being a multiple of gravitational acceleration.
02

Convert Physical Quantities

The acceleration due to gravity is approximately 9.81 m/s虏. The problem states that the centripetal acceleration is \(6.25 \times 10^{3}\) times the acceleration due to gravity. Therefore, the centripetal acceleration \(a_c\) is:\[a_c = 6.25 \times 10^{3} \times 9.81 \ \text{m/s}^2.\]
03

Relate Centripetal Acceleration to RPM

Centripetal acceleration \(a_c\) is given by the formula:\[a_c = r \omega^2,\]where \(r\) is the radius in meters, and \(\omega\) is the angular velocity in radians per second (rad/s). Convert the radius from centimeters to meters: \(r = 0.05 \text{ m}.\)
04

Solve for Angular Velocity

From the equation \(a_c = r \omega^2\), solve for \(\omega\):\[\omega = \sqrt{\frac{a_c}{r}}.\]
05

Calculate Angular Velocity

Substitute the values into the equation:\[\omega = \sqrt{\frac{(6.25 \times 10^{3} \times 9.81)}{0.05}} \ \text{rad/s}.\]
06

Convert Angular Velocity to RPM

First, convert \(\omega\) from rad/s to revolutions per second (RPS) using the conversion factor \(1 \text{ revolution} = 2\pi \text{ radians}\):\[\text{RPS} = \frac{\omega}{2\pi}.\]Then convert RPS to RPM by multiplying by 60.
07

Calculate Final RPM

Substitute the calculated \(\omega\) into the conversion formulas:1. Calculate \(\text{RPS}\).2. Multiply by 60 to find RPM.

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is a key concept in the motion of objects along a circular path. It is the acceleration that is directed towards the center of the circle, causing the object to maintain its circular path. Without this acceleration, the object would move off in a straight line due to inertia.

  • This acceleration depends on the speed of the object and the radius of the circular path.
  • In the case of the centrifuge, it is expressed as a multiple of the gravitational acceleration, which is approximately 9.81 m/s虏.
In our specific exercise, we learned that the centrifuge's centripetal acceleration is 6,250 times the acceleration due to gravity. Thus, the formula for centripetal acceleration is:\[a_c = 6.25 imes 10^3 imes 9.81 ext{ m/s}^2\]This shows the immense force being applied to keep the sample rotating at high speeds, which helps to separate substances like blood cells in the centrifuge.
Angular Velocity
Angular velocity measures how fast an object rotates or spins. It is crucial in understanding how quickly an object moves around a circular path and is expressed in radians per second (rad/s).
  • It captures the change in angular position of the object per unit time.
  • The formula connecting angular velocity \(\omega\), centripetal acceleration \(a_c\), and radius \(r\) is: \[a_c = r \omega^2\]
In our example, the formula is rearranged to find \(\omega\):\[\omega = \sqrt{\frac{a_c}{r}}\]By substituting known values into the formula, one can determine the angular velocity of the centrifuge, steering us closer to finding the revolutions per minute (RPM). This velocity reveals the swiftness of the unyielding circular motion.
Revolutions Per Minute (RPM)
Revolutions per minute (RPM) is a common unit used to measure rotational speed. It states how many complete turns an object makes in one minute. This is invaluable in the context of centrifuges, which often operate at rapid speeds.
  • To find RPM, we can convert from angular velocity, initially calculated in radians per second.
  • First, convert angular velocity to revolutions per second (RPS) using the formula: \[ \text{RPS} = \frac{\omega}{2\pi} \]
  • Next, multiply the RPS by 60 to get RPM, as there are 60 seconds in a minute.
This two-step process ensures that we correctly translate the unit of rotation into RPM, making it easier to comprehend how fast the centrifuge is spinning in real-world terms.

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Most popular questions from this chapter

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of \(3.2 \mathrm{~m} / \mathrm{s},\) and an \(83-\mathrm{kg}\) person feels a \(560-\mathrm{N}\) force pressing against his back. What is the radius of a chamber?

Concept Questions The following table lists data for the speed and the radius in three examples of uniform circular motion. $$ \begin{array}{lcc} & \text { Radius } & \text { Speed } \\ \hline \text { Example 1 } & 0.50 \mathrm{~m} & 12 \mathrm{~m} / \mathrm{s} \\\ \text { Example } 2 & \text { Infinitely large } & 35 \mathrm{~m} / \mathrm{s} \\\ \text { Example 3 } & 1.8 \mathrm{~m} & 2.3 \mathrm{~m} / \mathrm{s} \end{array} $$ (a) Without doing any calculations, identify the example with the smallest centripetal acceleration. (b) Similarly identify the example with the greatest centripetal acceleration. In each case, justify your answer. Problem Find the value for the centripetal acceleration for each example. Verify that your answers are consistent with your answers to the Concept Questions.

The moon orbits the earth at a distance of \(3.85 \times 10^{8} \mathrm{~m}\). Assume that this distance is between the centers of the earth and the moon and that the mass of the earth is \(5.98 \times 10^{24} \mathrm{~kg}\). Find the period for the moon's motion around the earth. Express the answer in days and compare it to the length of a month.

Concept Simulation 5.1 at reviews the concepts that are involved in this problem. A child is twirling a \(0.0120-\mathrm{kg}\) ball on a string in a horizontal circle whose radius is 0.100 \(\mathrm{m}\). The ball travels once around the circle in \(0.500 \mathrm{~s}\). (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?

In an automatic clothes dryer, a hollow cylinder moves the clothes on a vertical circle (radius \(r=0.32 \mathrm{~m}\) ), as the drawing shows. The appliance is designed so that the clothes tumble gently as they dry. This means that when a piece of clothing reaches an angle of \(\theta\) above the horizontal, it loses contact with the wall of the cylinder and falls onto the clothes below. How many revolutions per second should the cylinder make in order that the clothes lose contact with the wall when \(\theta=70.0^{\circ} ?\)
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