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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

Small blocks, each with mass \(m\), are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point one-fourth of the length from one end.

At \(t\) \(=\) 3.00 s a point on the rim of a 0.200-m-radius wheel has a tangential speed of 50.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.0 m/s\(^2\). (a) Calculate the wheel’s constant angular acceleration. (b) Calculate the angular velocities at \(t\) \(=\) 3.00 s and \(t\) \(=\) 0. (c) Through what angle did the wheel turn between \(t\) \(=\) 0 and \(t\) \(=\) 3.00 s? (d) At what time will the radial acceleration equal g?

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 safety device brings the blade of a power mower from an initial angular speed of \(\omega_1\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_3\) that was three times as great, \(\omega_3 = 3\omega_1\)?

You are to design a rotating cylindrical axle to lift 800-N buckets of cement from the ground to a rooftop 78.0 m above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 cm/s when it is turning at 7.5 rpm? (b) If instead the axle must give the buckets an upward acceleration of 0.400 m/s\(^2\), what should the angular acceleration of the axle be?

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