/*! 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 60 An industrial flywheel with a mo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An industrial flywheel with a moment of inertia of \(4.25 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates with a speed of \(7500 \mathrm{rpm}\) (a) How much work is required to bring the flywheel to rest? (b) If this work is done uniformly in \(1.5 \mathrm{~min}\), how much power is required?

Short Answer

Expert verified
(a) 1.31 x 10^8 J, (b) 1.46 x 10^6 W

Step by step solution

01

Find Initial Angular Velocity

First, we need to convert the given rotational speed from revolutions per minute (rpm) to radian per second (rad/s). The conversion is given by the formula:\[\omega = \text{rpm} \times \frac{2\pi}{60}\]Substitute the value:\[\omega = 7500 \times \frac{2\pi}{60} = 785.4\, \text{rad/s}\]
02

Calculate Initial Kinetic Energy

The kinetic energy of a rotating object is given by the formula:\[KE = \frac{1}{2}I\omega^2\]Where \(I = 4.25 \times 10^2\, \text{kg} \cdot \text{m}^2\) is the moment of inertia. Substitute the values:\[KE = \frac{1}{2} \times 4.25 \times 10^2 \times (785.4)^2\]Calculate to find:\[KE = 1.31 \times 10^8\, \text{Joules}\]
03

Determine Work Required

To bring the flywheel to rest, the work done is equal to the initial kinetic energy since the final kinetic energy is zero. Thus, the required work is:\[W = 1.31 \times 10^8\, \text{Joules}\]
04

Calculate the Time Interval in Seconds

Convert the given time from minutes to seconds. Since 1.5 minutes is given, multiply by 60 to convert to seconds:\[1.5 \times 60 = 90\, \text{seconds}\]
05

Find the Required Power

Power is the rate at which work is done and is calculated with the formula:\[P = \frac{W}{t}\]Where \(W\) is the work (\(1.31 \times 10^8\, \text{Joules}\)) and \(t\) is the time (90 seconds). Substitute these values:\[P = \frac{1.31 \times 10^8}{90} \approx 1.46 \times 10^6\, \text{Watts}\]

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 91Ó°ÊÓ!

Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in rotational motion. It measures how hard it is to change the rotational motion of an object. This property is like mass in linear motion but applies to rotating bodies. It's influenced by the mass of the object and how that mass is distributed relative to the rotation axis. The formula for the moment of inertia (\(I\)) depends on the shape and mass distribution of the object. For instance, for a solid cylinder rotating about its central axis, the moment of inertia is given as \(I = \frac{1}{2}mr^2\), where \(m\) is the mass and \(r\) is the radius.
  • Unit: kilogram meter squared (\(\text{kg} \cdot \text{m}^2\))
  • Depends on axis: Changing the rotation axis can change the moment of inertia.
  • Crucial for calculations: It is essential in determining kinetic energy in rotation and analyzing angular velocity.
Understanding it helps to predict how much effort is required to make an object start or stop rotating.
Kinetic Energy
In the world of rotational motion, kinetic energy corresponds to how much energy a rotating object possesses due to its motion. While linear kinetic energy is calculated with the familiar \(KE = \frac{1}{2}mv^2\), in rotational motion, we use:\[ KE = \frac{1}{2}I\omega^2 \]Here, \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
  • Depends on rotation: Larger moment of inertia or angular velocity increases kinetic energy.
  • Unit: Joules, the same as linear kinetic energy.
  • Practical insights: Knowing the kinetic energy tells us how much work is needed to change the rotational motion—like bringing a wheel to a halt.
The kinetic energy essentially shows how much rotational energy is contained and how it can be transferred or modified.
Angular Velocity
Angular velocity is the rate at which an object rotates or revolves around an axis. It tells us how fast the rotation is happening and is a key player in understanding rotational dynamics. Measured in radians per second (\(\text{rad/s}\)), it can be derived from rotational speeds given in revolutions per minute (rpm) using:\[ \omega = \text{rpm} \times \frac{2\pi}{60} \]Conversion:- Radians allow us to use the rotational analogs of linear motion equations with ease.
  • Radial measurement: Instead of meters, we deal in radians, which relate to the arc length traveled by a point on the rotating object.
  • Units: \(\text{rad/s}\) provide simplicity in calculations with constants like \(\pi\).
Understanding angular velocity gives insight into how quickly a system can perform work or how it responds to forces applied in rotation.
Power Calculation
Power is an important concept because it represents the rate at which work is being done or energy is being transferred or converted. In rotational dynamics, power (\(P\)) is crucial for systems like engines and turbines, where energy conversion efficiency matters a lot. It is calculated with:\[ P = \frac{W}{t} \]where \(W\) is the work done, and \(t\) is the time in which this work is performed.
  • Power units: Watts (\(\text{W}\)), where 1 Watt equals 1 Joule per second.
  • Energy flow rate: Gives an understanding of how fast energy gets transferred.
In scenarios like stopping a rotating flywheel, calculating power can indicate how much energy must be managed to safely and efficiently stop the motion in a given time frame.

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

A planetary space probe is in the shape of a cylinder. To protect it from heat on one side (from the Sun's rays), operators on the Earth put it into a "barbecue mode," that is, they set it rotating about its long axis. To do this, they fire four small rockets mounted tangentially as shown in \(v\) Fig. 8.51 (the probe is shown coming toward you). The object is to get the probe to rotate completely once every \(30 \mathrm{~s}\), starting from no rotation at all. They wish to do this by firing all four rockets for a certain length of time. Each rocket can exert a thrust of \(50.0 \mathrm{~N}\). Assume the probe is a uniform solid cylinder with a radius of \(2.50 \mathrm{~m}\) and a mass of \(1000 \mathrm{~kg}\) and neglect the mass of each rocket engine. Determine the amount of time the rockets need to be fired.

A flat cylindrical grinding wheel is spinning at 2000 rpm (clockwise when viewed head-on) when its power is suddenly turned off. Normally, if left alone, it takes 45.0 s to coast to rest. Assume the grinder has a moment of inertia of \(2.43 \mathrm{~kg} \cdot \mathrm{m}^{2}\). (a) Determine its angular acceleration during this process. (b) Determine the tangential acceleration of a point on the grinding wheel if the wheel is \(7.5 \mathrm{~cm}\) in diameter. (c) The slowing down is caused by a frictional torque on the axle of the wheel. The axle is \(1.00 \mathrm{~cm}\) in diameter. Determine the frictional force on the axle. (d) How much work was done by friction on the system?

Two objects of different masses are joined by a light rod. (a) Is the moment of inertia about the center of mass the minimum or the maximum? Why? (b) If the two masses are \(3.0 \mathrm{~kg}\) and \(5.0 \mathrm{~kg}\) and the length of the rod is \(2.0 \mathrm{~m},\) find the moments of inertia of the system about an axis perpendicular to the rod, through the center of the rod and the center of mass.

For the following objects, which all roll without slipping, determine the rotational kinetic energy about the center of mass as a percentage of the total kinetic energy: (a) a solid sphere, (b) a thin spherical shell, and (c) a thin cylindrical shell.

(a) How many uniform, identical textbooks of width \(25.0 \mathrm{~cm}\) can be stacked on top of each other on a level surface without the stack falling over if each successive book is displaced \(3.00 \mathrm{~cm}\) in width relative to the book below it? (b) If the books are \(5.00 \mathrm{~cm}\) thick, what will be the height of the center of mass of the stack above the level surface?

See all solutions

Recommended explanations on Physics Textbooks

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.