/*! 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 75 9.75\(\cdots .\) It has been arg... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

9.75\(\cdots .\) It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. Consider a flywheel made of iron (density 7800 \(\mathrm{kg} / \mathrm{m}^{3}\) ) in the shape of a 10.0 -cm-thick uniform disk. (a) What would the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 rpm about an axis perpendicular to the disk at its center? (b) What would be the centripetal acceleration of a point on its rim when spinning at this rate?

Short Answer

Expert verified
(a) Diameter \( \approx 2.74 \, \text{m} \). (b) Centripetal acceleration \( \approx 121.45 \, \text{m/s}^{2} \).

Step by step solution

01

Convert Angular Speed to Radians per Second

First, convert 90.0 revolutions per minute (rpm) to radians per second (rad/s) by using the conversion: \[ 1 \text{ rpm} = \frac{2\pi}{60} \, \text{rad/s} \].So, for 90 rpm, the angular speed \( \omega = 90 \times \frac{2\pi}{60} \approx 9.42 \, \text{rad/s} \).
02

Express the Moment of Inertia and Kinetic Energy

For a disk, the moment of inertia \( I \) about an axis through its center is \( I = \frac{1}{2} m r^{2} \), where \( m \) is the mass and \( r \) is the radius. The kinetic energy \( K \) of a rotating object is given by \( K = \frac{1}{2} I \omega^{2} \).
03

Calculate Mass of the Disk

The mass \( m \) of the disk can be expressed using volume and density: \[ m = \text{Volume} \times \text{Density} = \pi r^{2} t \times \rho \].Given thickness \( t = 0.1 \, \text{m} \) and density \( \rho = 7800 \, \text{kg/m}^{3} \),\[ m = \pi r^{2} \times 0.1 \times 7800 \].
04

Set Up the Kinetic Energy Equation

Substitute \( m \) from Step 3 into the kinetic energy formula:\[ 10 \times 10^{6} = \frac{1}{2} \left( \frac{1}{2} \pi r^{2} \times 0.1 \times 7800 \right) r^{2} \times (9.42)^{2} \].Simplify and solve for \( r \).
05

Solve for the Radius and Calculate Diameter

Rearrange the equation:\[ r^{4} = \frac{10 \times 10^{6}}{0.5 \times 0.5 \times \pi \times 0.1 \times 7800 \times 9.42^{2}} \].Calculate \( r \) and then multiply by 2 to find the diameter. The diameter is \( 2r \approx 2.74 \, \text{m} \).
06

Calculate Centripetal Acceleration on the Rim

The centripetal acceleration \( a_c \) is given by the formula \( a_c = r \times \omega^{2} \).Substitute the values: \( r \approx 1.37 \, \text{m} \) and \( \omega = 9.42 \, \text{rad/s} \):\[ a_c = 1.37 \times (9.42)^{2} \approx 121.45 \, \text{m/s}^{2} \].

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.

Flywheel Mechanics
Flywheels are fascinating mechanical devices used to store energy. At their core, they are spinning disks that can hold significant amounts of kinetic energy. This makes them ideal for applications where energy needs to be stored during times of low demand and released during peak demand. Flywheels are often used in power generation and other industries requiring energy efficiency.

The primary advantage of using flywheels for energy storage is their ability to deliver quick bursts of energy without degradation over time. When designing a flywheel, crucial factors include its shape, size, and the materials used in its construction. In this exercise, the focus is on a disk-shaped flywheel made of iron, known for its density and mechanical robustness. Its function is simple: it stores energy when it rotates and releases it when the rotation slows down. By minimizing friction with advanced bearings, these flywheels can efficiently store energy for extended periods.
Angular Speed Conversion
Angular speed, often measured in revolutions per minute (rpm), needs conversion to radians per second (rad/s) for most calculations in physics. This conversion is essential because radians are the standard unit for angular displacement, making calculations consistent and manageable.

To convert angular speed from rpm to rad/s, we use the fact that one complete revolution equals two pi radians. Thus, the conversion factor is given by:
  • 1 rpm = \( \frac{2\pi}{60} \) rad/s
  • To convert 90 rpm to rad/s: \( 90 \times \frac{2\pi}{60} \approx 9.42 \) rad/s
This conversion allows for precise calculations when determining the kinetic energy or centripetal force in rotating objects, as seen in this exercise.
Centripetal Acceleration
Centripetal acceleration is a crucial concept when analyzing anything moving in a circular path. It describes the acceleration that keeps an object moving along this path, always directed towards the center of the circle. In the context of a rotating flywheel, centripetal acceleration can affect both the safety and performance of the flywheel.

Centripetal acceleration can be calculated using the formula:
  • \( a_c = r \times \omega^2 \)
where:
  • \( r \) is the radius of the circle
  • \( \omega \) is the angular velocity in rad/s
In this exercise, for a flywheel with a radius of approximately 1.37 meters and an angular speed of 9.42 rad/s, the centripetal acceleration at the rim would be:
  • \( a_c = 1.37 \times (9.42)^2 \approx 121.45 \) m/s²
This indicates how fast the outer edge of the wheel must accelerate to maintain its circular path.
Moment of Inertia Calculation
The moment of inertia is a measure of how resistant an object is to changes in its rotation. For a disk-shaped flywheel, it determines how effectively the flywheel can store and release energy.

For a uniform disk, the moment of inertia \( I \) around its center axis is given by:
  • \( I = \frac{1}{2} m r^2 \)
Here, \( m \) is the mass of the disk, and \( r \) is its radius. Mass can be calculated from the volume and density:
  • \( m = \text{Volume} \times \text{Density} = \pi r^2 t \times \rho \)
where \( t \) is the thickness and \( \rho \) is the density.In this exercise, substituting the mass of the disk and other known values into the kinetic energy formula allows for calculating the necessary radius to achieve the desired energy storage:
  • \( 10 \times 10^6 = \frac{1}{2} \left( \frac{1}{2} \pi r^2 \times 0.1 \times 7800 \right) r^2 \times (9.42)^2 \)
After solving, we find the diameter of the disk, crucial for the design of practical energy storage systems.

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 thin, uniform rod is bent into a square of side length \(a\) . If the total mass is \(M\) , find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (Hint: Use the parallel-axis theorem.)

A passenger bus in Zurich, Switzerland, derived its motive power from the energy stored in a large flywheel. The wheel was brought up to speed periodically, when the bus stopped at a station, by an electric motor, which could then be attached to the electric power lines. The flywheel was a solid cylinder with mass 1000 \(\mathrm{kg}\) and diameter 1.80 \(\mathrm{m}\) ; its top angular speed was 3000 rev/min. (a) At this angular speed, what is the kinetic energy of the flywheel? (b) If the average power required to operate the bus is \(1.86 \times 10^{4} \mathrm{W},\) how long could it operate between stops?

A uniform sphere with mass 28.0 \(\mathrm{kg}\) and radius 0.380 \(\mathrm{m}\) is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is \(176 \mathrm{J},\) what is the tangential velocity of a point on the rim of the sphere?

A wheel of diameter 40.0 \(\mathrm{cm}\) starts from rest and rotates with a constant angular acceleration of 3.00 \(\mathrm{rad} / \mathrm{s}^{2} .\) At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship \(a_{\text { rad }}=\omega^{2} r\) and \((b)\) from the relationship \(a_{\text { rad }}=v^{2} / r\) .

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a \(\mathrm{CD}\) player, the track is scanned at a constant linear speed of \(v=1.25 \mathrm{m} / \mathrm{s}\) Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the CD is played. (See Exercise \(9.20 .\) ) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta)=r_{0}+\beta \theta\) , where \(r_{0}\) is the radius of the spiral at \(\theta=0\) and \(\beta\) is a constant. On a \(\mathrm{CD}, r_{0}\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d \theta\) , the distance scanned along the track is \(d s=r d \theta .\) Using the above expression for \(r(\theta),\) integrate \(d s\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) since the track is scanned at a constant linear speed \(v,\) the distance \(s\) found in part (a) is equal to \(v t\) . Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta\) ; choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_{z}\) and the angular acceleration \(\alpha_{z}\) as functions of time. Is \(\alpha_{z}\) constant? (d) On a CD, the inner radius of the track is 25.0 \(\mathrm{mm}\) , the track radius increases by 1.55\(\mu \mathrm{m}\) per revolution, and the playing time is 74.0 min. Find the values of \(r_{0}\) and \(\beta,\) and find the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_{z}\) \((\) in \(\operatorname{rad} / \mathrm{s})\) versus \(t\) and \(\alpha_{z}\) \((\) in \(\operatorname{rad} / \mathrm{s}^2)\) versus \(t\) between \(t=0\) and \(t=74.0\) min.

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.