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(II) How fast (in rpm) must a centrifuge rotate if a particle 7.0 \(\mathrm{cm}\) from the axis of rotation is to experience an acceleration of \(100,000 g^{\prime} s ?\)

Short Answer

Expert verified
The centrifuge must rotate at approximately 357,900 RPM.

Step by step solution

01

Understand the Given Values

We are given that the radius of the rotation, \( r \), is 7.0 cm, which can be converted to meters as \( 0.07 \;\text{m} \). The required acceleration is \( 100,000 \; g\). Since \( g = 9.8 \; m/s^2 \), the acceleration becomes \( 100,000 \times 9.8 \; m/s^2 \).
02

Recall the Relation Between Centripetal Acceleration and Angular Velocity

The formula for centripetal acceleration \( a \) is given by \( a = r\omega^2 \), where \( \omega \) is the angular velocity in radians per second and \( r \) is the radius of rotation. We need to find \( \omega \) in terms of \( a \) and \( r \).
03

Solve for Angular Velocity (\( \omega \))

Rearrange the centripetal acceleration formula to solve for \( \omega \): \[ \omega = \sqrt{\frac{a}{r}} \]Substitute the given values: \[ \omega = \sqrt{\frac{100,000 \times 9.8}{0.07}} \]Calculate to find \( \omega \).
04

Convert Angular Velocity to Revolutions Per Minute (RPM)

Angular velocity \( \omega \) in radians per second can be converted to revolutions per minute by using the conversion factor:\[ \text{RPM} = \frac{\omega}{2\pi} \times 60 \]Substitute the value of \( \omega \) from Step 3 and calculate RPM.
05

Calculate Final Value

Using the steps above, perform the calculations:1. \( \omega = \sqrt{\frac{100,000 \times 9.8}{0.07}} \approx 37,412 \; \text{rad/s} \)2. Convert \( \omega \) to RPM: \[ \text{RPM} = \frac{37,412}{2\pi} \times 60 \approx 357,900 \text{ RPM} \]

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is a key concept when dealing with objects in circular motion. It refers to the acceleration that keeps an object moving along a circular path. This acceleration points towards the center of the circle, a direction known as "radial" or "inward." It is responsible for changing the direction of the object's velocity continuously, resulting in circular motion.
To calculate centripetal acceleration, we use the formula:
  • \( a = r\omega^2 \)
where \( a \) is the centripetal acceleration, \( r \) is the radius of the circle (distance from the axis), and \( \omega \) is the angular velocity. This formula shows that centripetal acceleration depends on how fast the object is rotating (angular velocity) and how far it is from the center of rotation (radius).
In our exercise, the acceleration was given in terms of \( g \), where \( g = 9.8 \, m/s^2 \). So, if a centrifuge needs to provide an acceleration of \( 100,000 \, g \), we multiply \( 100,000 \) by \( 9.8 \, m/s^2 \) to find the actual acceleration in \( m/s^2 \). Understanding this process is crucial for solving problems involving rotational motion.
Angular Velocity
Angular velocity is the measure of how quickly something is spinning or rotating around a central point. When dealing with circular motion, angular velocity indicates the angle through which an object rotates in a certain amount of time. It is usually measured in radians per second (\( rad/s \)).
The connection between angular velocity and centripetal acceleration is pivotal because it helps us find how fast an object must spin to achieve a certain acceleration. Using the formula:
  • \( \omega = \sqrt{\frac{a}{r}} \)
we can determine the angular velocity \( \omega \) needed to produce a given centripetal acceleration \( a \) at a radius \( r \).
Given the problem conditions, substituting \( a = 100,000 \times 9.8 \) and \( r = 0.07 \) meters into the formula will provide us with the angular velocity required to give a particle the specified acceleration. This is a crucial step before we can switch from radians per second to revolutions per minute (RPM).
Revolutions per Minute (RPM)
Revolutions per minute (RPM) is a common unit of rotational speed or rate of spin, indicating how many complete turns an object makes in one minute. It is widely used to describe the rotation speed of mechanical components like centrifuges, motors, and wheels.
To convert angular velocity from radians per second to RPM, we use the conversion formula:
  • \( \text{RPM} = \frac{\omega}{2\pi} \times 60 \)
where \( \omega \) is the angular velocity in \( rad/s \). This formula considers that one complete revolution is \( 2\pi \) radians and adjusts the time frame from seconds to one minute.
For example, in our solution, we calculated that \( \omega \) was approximately \( 37,412 \; rad/s \). Substituting into the formula, we found the RPM to be about \( 357,900 \; \text{RPM} \). This conversion is essential in practical applications, as RPM is a more intuitive way to express rotational speed, especially in engineering and everyday life contexts.

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Most popular questions from this chapter

A narrow but solid spool of thread has radius \(R\) and mass \(M\). If you pull up on the thread so that the \(\mathrm{CM}\) of the spool remains suspended in the air at the same place as it unwinds, (a) what force must you exert on the thread? (b) How much work have you done by the time the spool turns with angular velocity \(\omega ?\)

When bicycle and motorcycle riders "pop a wheelie," a large acceleration causes the bike's front wheel to leave the ground. Let \(M\) be the total mass of the bike-plus-rider system; let \(x\) and \(y\) be the horizontal and vertical distance of this system's cM from the rear wheel's point of contact with the ground (Fig. \(72 ) .(a)\) Determine the horizontal acceleration \(a\) required to barely lift the bike's front wheel off of the ground. \((b)\) To minimize the acceleration necessary to pop a wheelie, should \(x\) be made as small or as large as possible? How about \(y ?\) How should a rider position his or her body on the bike in order to achieve these optimal values for \(x\) and \(y ?\) (c) If \(x=35 \mathrm{cm}\) and \(y=95 \mathrm{cm},\) find \(a\) .

A thin \(7.0-\mathrm{kg}\) wheel of radius \(32 \mathrm{~cm}\) is weighted to one side by a \(1.50-\mathrm{kg}\) weight, small in size, placed \(22 \mathrm{~cm}\) from the center of the wheel. Calculate \((a)\) the position of the center of mass of the weighted wheel and (b) the moment of inertia about an axis through its \(\mathrm{CM},\) perpendicular to its face.

An automobile engine slows down from 3500 rpm to 1200 rpm in \(2.5 \mathrm{~s} .\) Calculate \((a)\) its angular acceleration, assumed constant, and \((b)\) the total number of revolutions the engine makes in this time.

A crucial part of a piece of machinery starts as a flat uniform cylindrical disk of radius \(R_{0}\) and mass \(M .\) It then has a circular hole of radius \(R_{1}\) drilled into it (Fig. \(10-73\) ). The hole's center is a distance \(h\) from the center of the disk. Find the moment of inertia of this disk (with off-center hole) when rotated about its center, C. [Hint: Consider a solid disk and "subtract" the hole; use the parallel-axis theorem.]

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