/*! 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 39 The flywheel of a gasoline engin... [FREE SOLUTION] | 91Ó°ÊÓ

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The flywheel of a gasoline engine is required to give up 500 J of kinetic energy while its angular velocity decreases from 650 rev/min to 520 rev/min. What moment of inertia is required?

Short Answer

Expert verified
The required moment of inertia is approximately 0.599 kg·m².

Step by step solution

01

Convert Angular Velocities to Radians per Second

First, we need to convert the initial and final angular velocities from revolutions per minute (rev/min) to radians per second (rad/s).Using the conversion factor, \( 1 ext{ rev/min} = \frac{2\pi ext{ rad}}{60 ext{ s}} \).So, the initial angular velocity \( \omega_i \) is:\[ \omega_i = 650 \times \frac{2\pi}{60} = \frac{650 \times 3.1416}{60} = 68.068 \text{ rad/s} \]And the final angular velocity \( \omega_f \) is:\[ \omega_f = 520 \times \frac{2\pi}{60} = \frac{520 \times 3.1416}{60} = 54.453 \text{ rad/s} \]
02

Set up the Kinetic Energy Change Formula

The change in kinetic energy \( \Delta KE \) is given by:\[ \Delta KE = \frac{1}{2} I (\omega_i^2 - \omega_f^2) \]where \( I \) is the moment of inertia, \( \omega_i \) is the initial angular velocity, and \( \omega_f \) is the final angular velocity.In this problem, \( \Delta KE = 500 \, \text{Joules} \).
03

Calculate Moment of Inertia

Rearrange the kinetic energy formula to solve for \( I \):\[ I = \frac{2 \cdot \Delta KE}{\omega_i^2 - \omega_f^2} \]Substitute \( \Delta KE = 500 \, \text{J} \), \( \omega_i = 68.068 \, \text{rad/s} \), and \( \omega_f = 54.453 \, \text{rad/s} \):\[I = \frac{2 \times 500}{68.068^2 - 54.453^2} = \frac{1000}{4633.2246 - 2964.537} \= \frac{1000}{1668.6876} = 0.599 \text{ kg·m}^2\]
04

Verify and State the Required Moment of Inertia

The calculations show that the required moment of inertia is approximately \( 0.599 \, \text{kg·m}^2 \). It is verified based on the known values and formula used.

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins. It tells us the angle through which a point or a line has been rotated in a specified time.
Imagine the hands of a clock moving around its center; that motion can be described using angular velocity. It is usually represented by the symbol \( \omega \).
Angular velocity can be expressed in different units such as revolutions per minute (rev/min) or radians per second (rad/s). When solving physics problems, rad/s is the preferred unit because it is part of the International System (SI) of units.- **Example Usage:** In circular motion, a car moving around a racetrack uses angular velocity to describe its motion.- **Equation:** Angular velocity can be calculated using the formula \( \omega = \frac{\theta}{t} \), where \( \theta \) is the angle moved through, and \( t \) is time.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. Anything that moves - from a car rolling down a hill to a flywheel spinning in an engine - has kinetic energy.
In rotational motion, the kinetic energy has a specific formula that involves the moment of inertia and angular velocity:- **Formula:** \( KE = \frac{1}{2} I \omega^2 \)- \( KE \) stands for kinetic energy.- \( I \) is the moment of inertia, which measures how much an object resists rotational motion.- \( \omega \) is the angular velocity.
Key takeaway: The faster an object spins, or the bigger its moment of inertia, the more kinetic energy it has.
Conversion to Radians per Second
Many angular velocity problems begin with values given in revolutions per minute (rev/min), which are practical for everyday contexts but need conversion into radians per second (rad/s) for use in scientific calculations.
The conversion factor is crucial here:- **Conversion Factor:** \( 1 \text{ rev/min} = \frac{2\pi \text{ rad}}{60 \text{ s}} \)- **Conversion Steps:** 1. Multiply the number of revolutions by \( 2\pi \) to change it to radians. 2. Divide by 60 to convert from per minute to per second.
**Example:** To convert 650 rev/min: - Multiply by \( \frac{2\pi}{60} \): \[ 650 \times \frac{2\pi}{60} = 68.068 \text{ rad/s} \]
This conversion simplifies using angular velocity in other formulas.
Change in Kinetic Energy Formula
The change in kinetic energy formula is a key concept in rotational dynamics. It helps calculate how much energy is gained or lost as an object speeds up or slows down.
The formula is:- \( \Delta KE = \frac{1}{2} I (\omega_i^2 - \omega_f^2) \)- \( \Delta KE \) represents the change in kinetic energy.- \( I \) is the moment of inertia, indicating object resistance to changes in its rotation.- \( \omega_i \) and \( \omega_f \) are the initial and final angular velocities, respectively.
In the given exercise, this formula helps determine how much energy is lost when the angular velocity of the flywheel decreases. It's essential for solving many physics problems related to rotational motion.

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Most popular questions from this chapter

The Crab Nebula is a cloud of glowing gas about 10 lightyears across, located about 6500 light-years from the earth (\(\textbf{Fig. P9.86}\)). It is the remnant of a star that underwent a \(supernova\) \(explosion\), seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about 5 \(\times\) 10\(^3$$^1\) W, about 10\(^5\) times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning \(neutron\) \(star\) at its center. This object rotates once every 0.0331 s, and this period is increasing by 4.22 \(\times 10^{-13}\) s for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock (3000 kg/m\(^3\)) and to the density of an atomic nucleus (about 10\(^{17}\) kg/m\(^3\)). Justify the statement that a neutron star is essentially a large atomic nucleus.

At \(t =\) 0 the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t) =\) (250 rad/s)\(t -\) (20.0 rad/s\(^2\))\(t^2 -\) (1.50 rad/s\(^3\))\(t^3\). (a) At what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t =\) 0, when the current was reversed? (e) Calculate the average angular velocity for the time period from \(t =\) 0 to the time calculated in part (a).

You must design a device for shooting a small marble vertically upward. The marble is in a small cup that is attached to the rim of a wheel of radius 0.260 m; the cup is covered by a lid. The wheel starts from rest and rotates about a horizontal axis that is perpendicular to the wheel at its center. After the wheel has turned through 20.0 rev, the cup is the same height as the center of the wheel. At this point in the motion, the lid opens and the marble travels vertically upward to a maximum height \(h\) above the center of the wheel. If the wheel rotates with a constant angular acceleration \(\alpha\), what value of a is required for the marble to reach a height of \(h =\) 12.0 m?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

At \(t =\) 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0 rad/s\(^2\) until a circuit breaker trips at \(t =\) 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t =\) 0 and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

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