/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 A centrifuge rotor has a moment ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 63

A centrifuge rotor has a moment of inertia of \(4.25 \times 10^{-2} \mathrm{~kg} \cdot \mathrm{m}^{2} .\) How much energy is required to bring it from rest to 9750 rpm?

Short Answer

Expert verified
The energy required is approximately 22145.94 Joules.

Step by step solution

01

Understand the Problem

We need to find the kinetic energy required to bring a centrifuge from rest to a certain speed. For rotating objects, this is determined using the rotational kinetic energy formula.
02

Know the Formula

The rotational kinetic energy \( K \) of a rotating object is given by the formula \( K = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity in radians per second.
03

Convert RPM to Radians per Second

The given speed is 9750 rpm. To use it in our formula, we need to convert it to radians per second. The conversion factor is \( 2\pi/60 \) radians per second for each rpm.\[ \omega = 9750 \times \frac{2\pi}{60} \approx 1021.017 \text{ radians/second} \]
04

Calculate Rotational Kinetic Energy

Substitute the values of \( I = 4.25 \times 10^{-2} \text{ kg} \cdot \text{m}^2 \) and \( \omega \approx 1021.017 \text{ rad/s} \) into the kinetic energy formula.\[ K = \frac{1}{2} \times 4.25 \times 10^{-2} \times (1021.017)^2 \]After calculation:\[ K \approx 22145.94 \text{ Joules} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Moment of Inertia
The moment of inertia, often abbreviated as "I," is a fundamental concept when dealing with rotational motion. It is essentially a measure of how difficult it is to change an object's rotation. Just like mass is a measure of an object's resistance to changes in its linear motion, the moment of inertia serves the same purpose for rotational motion. Key points about moment of inertia:
  • The moment of inertia depends on both the mass of an object and how that mass is distributed relative to the axis of rotation.
  • For example, if you have two cylinders with the same mass, but one is hollow and one is solid, the hollow one will have a larger moment of inertia when rotating around its central axis.
  • The moment of inertia is measured in units of \( ext{kg} \cdot \text{m}^2\).
In the context of our exercise with the centrifuge, its moment of inertia is given as \(4.25 \text{ kg} \cdot \text{m}^2\), indicating it should require a certain amount of energy to change its rotational speed from zero to the desired speed.
Angular Velocity
Angular velocity, denoted by the symbol \(\omega\), describes how fast an object is rotating. It's similar to linear velocity, but for circular motion.
  • Measured in radians per second ( ext{rad/s}), angular velocity gives us the rate at which an object rotates around a particular axis.
  • To find angular velocity when you're given revolutions per minute (rpm), like in our exercise, you use the conversion formula: \(\omega = \frac{2\pi}{60} \times \text{rpm}\).
In the exercise, we converted 9750 rpm to approximately 1021.017 rad/s. This conversion is essential because the equation for rotational kinetic energy uses angular velocity in radians per second. Understanding how quickly an object rotates helps determine the energy needed to reach that speed from a standstill.
Energy Conversion
Energy conversion in rotational dynamics involves transforming energy into rotational kinetic form. It's the same principle as accelerating a car, but instead of linear motion, we're dealing with motion around an axis.
  • Rotational kinetic energy is given by the formula \( K = \frac{1}{2} I \omega^2\), where \( I\) is moment of inertia and \( \omega\) is angular velocity.
  • This formula calculates the energy required to accelerate an object from rest to a particular angular speed.
  • In the exercise, using the given moment of inertia and calculated angular velocity, we find that bringing the centrifuge rotor from rest to 9750 rpm requires approximately 22145.94 Joules of energy.
This energy conversion is specific to rotational scenarios and plays a critical role in designing mechanical systems involving spinning parts, like engines and turbines. Understanding these concepts can help you solve complex rotational problems by breaking them down into more manageable parts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The angle through which a rotating wheel has turned in time \(t\) is given by \(\theta=8.5 t-15.0 t^{2}+1.6 t^{4},\) where \(\theta\) is in radians and \(t\) in seconds. Determine an expression \((a)\) for the instantaneous angular velocity \(\omega\) and \((b)\) for the instantaneous angular acceleration \(\alpha .(c)\) Evaluate \(\omega\) and \(\alpha\) at \(t=3.0 \mathrm{~s} . \quad(d)\) What is the average angular velocity, and (e) the average angular acceleration between \(t=2.0 \mathrm{~s}\) and \(t=3.0 \mathrm{~s} ?\)

A wheel with rotational inertia \(I=\frac{1}{2} M R^{2}\) about its central axle is set spinning with initial angular speed \(\omega_{0}\) and is then lowered onto the ground so that it touches the ground with no horizontal speed. Initially it slips, but then begins to move forward and eventually rolls without slipping. \((a)\) In what direction does friction act on the slipping wheel? (b) How long does the wheel slip before it begins to roll without slipping? \((c)\) What is the wheel's final translational speed? [Hint: Use \(\Sigma \overrightarrow{\mathbf{F}}=m \overrightarrow{\mathbf{a}}, \Sigma \tau_{\mathrm{CM}}=I_{\mathrm{CM}} \alpha_{\mathrm{CM}},\) and recall that only when there is rolling without slipping is \(v_{\mathrm{CM}}=\omega R .\)

(I) Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.20 \(\mathrm{m}\) high. Assume it starts from rest and rolls without slipping.

A hollow cylinder (hoop) is rolling on a horizontal surface at speed \(v=3.3 \mathrm{~m} / \mathrm{s}\) when it reaches a \(15^{\circ}\) incline (a) How far up the incline will it go? (b) How long will it be on the incline before it arrives back at the bottom?

A uniform thin rod of length \(\ell\) and mass \(M\) is suspended freely from one end. It is pulled to the side an angle \(\theta\) and released. If friction can be ignored, what is its angular velocity, and the speed of its free end, at the lowest point?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Electricity

Read Explanation

Waves Physics

Read Explanation

Kinematics Physics

Read Explanation

Fields in Physics

Read Explanation

Wave Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.