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

Short Answer

Expert verified
The required moment of inertia is approximately 0.600 kg·m².

Step by step solution

01

Convert Angular Velocities from Rev/Min to Rad/Sec

First, we need to convert the angular velocities from revolutions per minute (rev/min) to radians per second (rad/s).Knowing that 1 revolution is equal to \(2\pi\) radians and 1 minute is 60 seconds, the conversion can be computed as:For \(\omega_1 = 650\) rev/min:\[ \omega_1 = 650 \times \frac{2\pi}{60} \approx 68.0668 \text{ rad/s} \]For \(\omega_2 = 520\) rev/min:\[ \omega_2 = 520 \times \frac{2\pi}{60} \approx 54.4531 \text{ rad/s} \]
02

Use the Change in Kinetic Energy Formula

The change in kinetic energy for a rotating object is given by:\[ \Delta KE = \frac{1}{2} I (\omega_1^2 - \omega_2^2) \]Given that the change in kinetic energy \( \Delta KE = 500 \text{ J} \), we can rearrange the formula to solve for the moment of inertia \( I \):\[ I = \frac{2 \times \Delta KE}{\omega_1^2 - \omega_2^2} \]
03

Calculate Angular Velocities Squared

Now calculate the squares of the angular velocities:\[ \omega_1^2 = (68.0668)^2 \approx 4633.1 \]\[ \omega_2^2 = (54.4531)^2 \approx 2966.9 \]
04

Substitute Values and Solve for Moment of Inertia

Substitute the known values into the equation for \( I \):\[ I = \frac{2 \times 500}{4633.1 - 2966.9} \]Calculate the denominator:\[ 4633.1 - 2966.9 = 1666.2 \]Thus:\[ I = \frac{1000}{1666.2} \approx 0.600 \text{ kg} \cdot \text{m}^2 \]

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Key Concepts

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

Angular Velocity Conversion
When dealing with rotational dynamics, it's common to convert angular velocities from revolutions per minute (rev/min) to radians per second (rad/s). This conversion is essential for calculating rotational kinetic energy and other physics formulas. Remember that one full revolution is equal to the angle of the circumference in radians, which is precisely \(2\pi\) radians. Therefore, to convert revolutions to radians per second, you multiply by \(\frac{2\pi}{60}\) (since there are 60 seconds in a minute).
For example, to convert an angular velocity of 650 rev/min:
  • Use the relation: \(\omega = 650 \times \frac{2\pi}{60} \approx 68.07 \text{ rad/s}\).
This conversion allows you to work directly within the International System of Units (SI), which is crucial for consistency in physics equations.
Kinetic Energy Formula
The kinetic energy of a rotating object is a crucial concept in understanding rotational dynamics. It can be described using the kinetic energy formula specific to rotational motion:
  • \( KE = \frac{1}{2} I \omega^2 \)
Where \( KE \) stands for kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
This formula indicates that the kinetic energy of rotation depends on how the mass is distributed relative to the axis (moment of inertia) and the speed of rotation (angular velocity). By squaring the angular velocity, we emphasize that changes in speed have a significant impact on the energy involved.
Change in Kinetic Energy
Understanding how to calculate the change in kinetic energy is vital, especially when dealing with systems that undergo changes in motion. For rotational systems, this becomes essential when analyzing energy aspects of spinning objects:
  • \( \Delta KE = \frac{1}{2} I (\omega_1^2 - \omega_2^2) \)
In this formula, \( \Delta KE \) is the change in kinetic energy, \( I \) is the moment of inertia, and \( \omega_1 \) and \( \omega_2 \) are the initial and final angular velocities, respectively. This calculation reflects how the energy stored in rotational motion is affected by the change in speed. Recognizing this difference helps in solving problems related to energy transformations in mechanical systems.
Rotational Dynamics
Rotational dynamics is a branch of physics focusing on the motion of rotating bodies and is closely related to linear dynamics. It involves concepts such as torque, angular momentum, and moment of inertia, as well as angular velocity and acceleration. In this context, the moment of inertia plays a similar role to mass in linear dynamics, determining a body's resistance to change in rotational motion.
Understanding these relationships is vital for analyzing how forces and torques affect an object's rotation, thus impacting its energy and motion. In practical applications, like the one given in the problem, rotational dynamics provide insights into energy efficiency and mechanical performance of systems such as engines or turbines.

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Most popular questions from this chapter

A compound disk of outside diameter 140.0 \(\mathrm{cm}\) is made up of a uniform solid disk of radius 50.0 \(\mathrm{cm}\) and area density 3.00 \(\mathrm{g} / \mathrm{cm}^{2}\) surrounded by a concentric ring of inner radius \(50.0 \mathrm{cm},\) outer radius \(70.0 \mathrm{cm},\) and area density 2.00 \(\mathrm{g} / \mathrm{cm}^{2} .\) Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b .\) Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

Energy is to be stored in a 70.0 -kg flywheel in the shape of a uniform solid disk with radius \(R=1.20 \mathrm{m.}\) To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 \(\mathrm{m} / \mathrm{s}^{2} .\) What is the maximum kinetic energy that can be stored in the flywheel?

A sphere with radius \(R=0.200 \mathrm{m}\) has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(\rho=3.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}-\left(9.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{4}\right) r .\) (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

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?

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