/*! 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 96 The operation of the Princeton T... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 96

The operation of the Princeton Tokomak Fusion Test Reactor requires large bursts of energy. The power needed exceeds the amount that can be supplied by the utility company. Prior to pulsing the reactor, energy is stored in a giant flywheel of mass \(7.27 \times 10^{5} \mathrm{kg}\) and rotational inertia $4.55 \times 10^{6} \mathrm{kg} \cdot \mathrm{m}^{2} .$ The flywheel rotates at a maximum angular speed of 386 rpm. When the stored energy is needed to operate the reactor, the flywheel is connected to an electrical generator, which converts some of the rotational kinetic energy into electric energy. (a) If the flywheel is a uniform disk, what is its radius? (b) If the flywheel is a hollow cylinder with its mass concentrated at the rim, what is its radius? (c) If the flywheel slows to 252 rpm in 5.00 s, what is the average power supplied by the flywheel during that time?

Short Answer

Expert verified
Step 1: Calculate the radius of the uniform disk flywheel Using the given values, we can find the radius of the uniform disk flywheel: $$ r_d = \sqrt{\frac{2 \times 4.55 \times 10^{6}}{7.27 \times 10^{5}}} $$ $$ r_d 鈮� 2.53\, \mathrm{m} $$ Step 2: Calculate the radius of the hollow cylinder flywheel Similarly, we can calculate the radius of the hollow cylinder flywheel: $$ r_c = \sqrt{\frac{4.55 \times 10^{6}}{7.27 \times 10^{5}}} $$ $$ r_c 鈮� 2.00\, \mathrm{m} $$ Step 3: Convert angular velocities to rad/s Converting the angular velocities to rad/s: $$ \omega_i = 386 \times \frac{2 \pi}{60} 鈮� 40.37\, \mathrm{rad/s} $$ $$ \omega_f = 252 \times \frac{2 \pi}{60} 鈮� 26.39\, \mathrm{rad/s} $$ Step 4: Calculate the average power supplied by the flywheel Using the given rotational inertia and angular velocities, we can find the total energy lost by the flywheel: $$ \Delta E = \frac{1}{2} \times 4.55 \times 10^6 \times (26.39^2 - 40.37^2) 鈮� - 2.67 \times 10^9\, \mathrm{J} $$ Now, we can find the average power supplied during the time interval: $$ P = \frac{-2.67 \times 10^9}{5.00} 鈮� -5.34 \times 10^8\, \mathrm{W} $$ In conclusion, the radius of the flywheel as a uniform disk is approximately 2.53 m, and as a hollow cylinder is approximately 2.00 m. The average power supplied by the flywheel during the given time interval is approximately -5.34 x 10^8 W.

Step by step solution

01

Calculate the radius of the uniform disk flywheel

Using the formula for the rotational inertia of a uniform disk-$$ I_d = \frac{1}{2} m_d r_d^2 $$, where \(I_d\) is the rotational inertia of the uniform disk, \(m_d\) is the mass, and \(r_d\) is the radius. Given the values: \(I_d = 4.55 \times 10^{6} \mathrm{kg} \cdot \mathrm{m}^{2}\) and \(m_d = 7.27 \times 10^{5} \mathrm{kg}\), We can solve for \(r_d\):$$ r_d = \sqrt{\frac{2 \times I_d}{m_d}} $$
02

Calculate the radius of the hollow cylinder flywheel

Using the formula for the rotational inertia of a hollow cylinder-$$ I_c = m_c r_c^2 $$, where \(I_c\) is the rotational inertia of the hollow cylinder, \(m_c\) is the mass, and \(r_c\) is the radius. Given the values: \(I_c = 4.55 \times 10^{6} \mathrm{kg} \cdot \mathrm{m}^{2}\) and \(m_c = 7.27 \times 10^{5} \mathrm{kg}\), We can solve for \(r_c\):$$ r_c = \sqrt{\frac{I_c}{m_c}} $$
03

Convert angular velocities to rad/s

Given the initial angular velocity, \(\omega_i\) = 386 rpm, and final angular velocity, \(\omega_f\) = 252 rpm, We need to convert these values to radians per second using the conversion factor: \(1 \mathrm{rpm} = \frac{2 \pi \mathrm{rad}}{60s}\)
04

Calculate the average power supplied by the flywheel

The total energy lost by the flywheel during the time interval can be calculated using the rotational kinetic energy formula:$$ \Delta E = \frac{1}{2}I(\omega_f^2 - \omega_i^2) $$ The average power supplied during this time interval can then be calculated as:$$ P = \frac{\Delta E}{\Delta t} $$, where \(\Delta t\) is the time interval. Now let's substitute the numbers into the formulas and calculate the answers.

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!

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 diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.00 complete rotations in \(1.33 \mathrm{s}\). If his rotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board?
The pull cord of a lawnmower engine is wound around a drum of radius $6.00 \mathrm{cm} .\( While the cord is pulled with a force of \)75 \mathrm{N}$ to start the engine, what magnitude torque does the cord apply to the drum?
A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber's waist \(15 \mathrm{cm}\) to the right of his center of gravity. If the climber weighs \(770 \mathrm{N},\) find (a) the tension in the rope and (b) the magnitude and direction of the contact force exerted by the wall on the climber's feet.
A person is trying to lift a ladder of mass \(15 \mathrm{kg}\) and length $8.0 \mathrm{m} .$ The person is exerting a vertical force on the ladder at a point of contact \(2.0 \mathrm{m}\) from the center of gravity. The opposite end of the ladder rests on the floor. (a) When the ladder makes an angle of \(60.0^{\circ}\) with the floor, what is this vertical force? (b) A person tries to help by lifting the ladder at the point of contact with the floor. Does this help the person trying to lift the ladder? Explain.
A tower outside the Houses of Parliament in London has a famous clock commonly referred to as Big Ben, the name of its 13 -ton chiming bell. The hour hand of each clock face is \(2.7 \mathrm{m}\) long and has a mass of \(60.0 \mathrm{kg}\) Assume the hour hand to be a uniform rod attached at one end. (a) What is the torque on the clock mechanism due to the weight of one of the four hour hands when the clock strikes noon? The axis of rotation is perpendicular to a clock face and through the center of the clock. (b) What is the torque due to the weight of one hour hand about the same axis when the clock tolls 9: 00 A.M.?
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Electric Charge Field and Potential

Read Explanation

Space Physics

Read Explanation

Rotational Dynamics

Read Explanation

Force

Read Explanation

Circular Motion and Gravitation

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.