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(II) How fast (in rpm) must a centrifuge rotate if a particle 8.0 cm from the axis of rotation is to experience an acceleration of 100,000 g's?

Short Answer

Expert verified
The centrifuge must rotate at approximately 33,414 rpm.

Step by step solution

01

Understanding the Problem

We need to find the rotational speed of a centrifuge such that a particle located 8.0 cm from its axis experiences an acceleration of 100,000 times the acceleration due to gravity (g). Here, the acceleration due to gravity is approximately 9.8 m/s².
02

Convert Units

First, we convert the given acceleration to meters per second squared. Since 1 g is 9.8 m/s², 100,000 g's is equivalent to 100,000 x 9.8 m/s² = 980,000 m/s².
03

Use the Centripetal Acceleration Formula

The formula for centripetal acceleration is given by: \[ a = \frac{v^2}{r} \]where \( a \) is the centripetal acceleration, \( v \) is the linear velocity, and \( r \) is the radius from the axis of rotation (in this case, 8.0 cm or 0.08 m).
04

Solve for Linear Velocity

Rearrange the formula to solve for \( v \): \[ v = \sqrt{a \cdot r} \] Substitute the known values: \[ v = \sqrt{980,000 \times 0.08} \] Calculate \( v \).
05

Calculate Linear Velocity

After substituting and calculating: \[ v = \sqrt{78,400} \approx 280 \text{ m/s} \]
06

Convert Linear Velocity to Angular Velocity

Use the relationship between linear velocity and angular velocity: \[ v = \omega \cdot r \] where \( \omega \) is angular velocity in radians per second. Rearrange the formula to solve for \( \omega \):\[ \omega = \frac{v}{r} = \frac{280}{0.08} \approx 3500 \text{ rad/s} \]
07

Convert Angular Velocity to Revolutions Per Minute (RPM)

Convert radians per second to revolutions per minute:1 revolution = 2\( \pi \) radians, and 1 minute = 60 seconds:\[ \frac{3500}{2\pi} \times 60 \approx 33,414 \text{ rpm} \]

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration plays a critical role in the rotation of objects. It is the acceleration that keeps an object moving in a circular path, always pointing towards the center of the circle. This happens because the direction of the object's velocity changes as it follows the circular path.

For objects moving in a circle with a constant speed, like in our centrifuge problem, centripetal acceleration (\(a\)) can be calculated using the formula: \[ a = \frac{v^2}{r} \]where \(v\) is the linear velocity, and \(r\) is the radius of the circle.

In practical applications, such as in a centrifuge, this acceleration allows particles to be separated based on their density. Adjusting it ensures that particles experience a certain force that will drive their separation efficiently.
Angular Velocity
Angular velocity (\(\omega\)) describes how fast something spins. It's a commonly used term in rotational motion, indicating the rate of rotation around an axis, and it is typically measured in radians per second (rad/s).

In the context of a centrifuge, converting linear velocity to angular velocity involves using the relationship: \[ v = \omega \cdot r \]where \(v\) is the linear velocity and \(r\) is the radius.

By rearranging this formula, students can find angular velocity: \[ \omega = \frac{v}{r} \]This relationship also illustrates that a larger radius leads to a smaller angular velocity for the same linear speed, explaining why adjusting distances in practical applications affects the rotational speed required.
Unit Conversion
Understanding unit conversion is essential in physics, especially when dealing with different units of measurement for things like acceleration, velocity, and distance.

In rotational motion, such as with a centrifuge, acceleration is often given in g-forces rather than \(\text{m/s}^2\). Thus, to calculate physical quantities correctly, you need to translate these into a usable form. Here, the acceleration of 100,000 g's was converted to \(980,000\text{ m/s}^2\) by recognizing that \(1 \text{ g} = 9.8 \text{ m/s}^2\).

Furthermore, when expressing speed in revolutions per minute (rpm), converting from radians per second involves using the conversion factors: 1 revolution = \(2\pi\) radians and 1 minute = 60 seconds. Thorough comprehension of these conversions is necessary for successfully solving physics problems.
Centrifuge Calculations
Calculating the required speed of a centrifuge involves a step-by-step understanding of rotational motion concepts. The aim is to achieve a certain centripetal acceleration, necessary for effective particle separation.

First, the acceleration must be expressed in appropriate units (e.g., \(\text{m/s}^2\)). Next, using the centripetal acceleration formula, you can calculate the linear velocity required, given the radius.

Once the linear velocity is known, it is converted to angular velocity, often needed in the laboratory settings.Finally, this angular velocity is converted to more practical terms like rpm for monitoring or setting the centrifuge.
  • Begin with acceleration conversion and computation.
  • Determine linear velocity using radius data.
  • Convert to useful rotation speeds for application (\(\omega\) to rpm).
These steps illustrate how physics principles are applied in calculating speeds and settings in real-world applications like centrifuges.

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Most popular questions from this chapter

(I) Express the following angles in radians: \((a)\) 45.0\(^{\circ}\), \((b)\) 60.0\(^{\circ}\), \((c)\) 90.0\(^{\circ}\), \((d)\) 360.0\(^{\circ}\), and \((e)\) 445\(^{\circ}\). Give as numerical values and as fractions of \(\pi\).

(II) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is \(1360 kg\cdot m^2\). Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. \((a)\) What is the angular velocity of the merry- go-round now? \((b)\) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

(II) Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms, \((a)\) that due to its daily rotation about its axis, and \((b)\) that due to its yearly revolution about the Sun. [Assume the Earth is a uniform sphere with \(mass = 6.0 \times 10^{24} kg\), \(radius = 6.4 \times 10^6 m\), and is \(1.5 \times 10^8 km\) from the Sun.]

A cyclist accelerates from rest at a rate of \(1.00 m/s^2\). How fast will a point at the top of the rim of the tire \((diameter = 68.0 cm)\) be moving after 2.25 s? [\(Hint\): At any moment, the lowest point on the tire is in contact with the ground and is at rest-see Fig. 8-57.

(II) In traveling to the Moon, astronauts aboard the \(Apollo\) spacecraft put the spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis. Determine \((a)\) the angular acceleration, and \((b)\) the radial and tangential components of the linear acceleration of a point on the skin of the ship 6.0 min after it started this acceleration.

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